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ECE 320 Energy Conversion and Power Electronics
Dr. Tim Hogan
Chapter 1: Introduction and Three Phase Power
1.1 Review of Basic Circuit Analysis Definitions: Node - Electrical junction between two or more devices. Loop - Closed path formed by tracing through an ordered sequence of nodes without passing through any node more than once. Element Constraints:
Ohm’s Law iRv =
Capacitor EquationdtdvCi =
Inductor Equation dtdiLv =
Connection Constraints: Kirchhoff’s Current Law - The algebraic sum of currents entering a node is zero at every instant in time. 0=∑
Nodeki (1.1)
Kirchhoff’s Voltage Law - The algebraic Sum of all voltages around a loop is zero at every instant in time. 0=∑
Loopkv (1.2)
Passive Sign Convention:
Whenever the reference direction of current into a two terminal device is in the direction of the reference voltage drop across the device, then the power absorbed (or dissipated) is positive.
+ _ vi
Figure 1. Circuit element and passive sign convention. ( ) ( ) ( )tvtitp ⋅= (1.3)
1- 1
When the above convention is used, p(t) > 0 for absorbed power, and p(t) < 0 for delivered power. Time Varying Signals
Although a number of exceptions can be found throughout the world, the predominance of electric power follows 60Hz or 50Hz frequencies. North America, part of Japan, and ships at sea use 60Hz while most of the rest of the world uses 50Hz. The historical reasons for these two frequencies stem from the differences in lighting (filaments in vacuum or filaments in a gas atmosphere). The lower frequencies caused an annoying flicker for lights having filaments in a gas atmosphere, and thus a higher frequency was adopted in part of the world that initially used such lighting.
While not strictly adhered to within your textbook and these notes, an attempt to use the following conventions has been made. Scalar time varying signals (examples):
( )tvv ωsinmax= (V) ( ) ( )tvtv ωcosmax= (V)
( tv ⋅= 377sin185 ) (kV) ( ) ( )titi ⋅= π100cosmax (A)
Spatial vectors (bold or arrow overhead or line overhead):
F or FG
or B If necessary, unit vectors will be used, (for example): zByBxB zyx ˆˆˆ ++=B These unit vectors should not be confused with phasors below
Phasors (phasor representation of a time varying signal):
A phasor can be represented as a complex number with real and imaginary components such that a phasor of magnitude (or length) R that is at a phase angle of θ with respect to the x-axis (the Real axis) can be written as ( ) ( )θθθ sincosˆ jRReRjyxR j +=⋅=+= where Euler’s formula ( ) ( )θθθ sincos je j += was used for the last representation. Note
that R is the magnitude and θ the phase of the phasor, and 22 yxR += and
⎟⎠⎞
⎜⎝⎛=
xyarctanθ . The phasor can be shown graphically in Figure 2.
1- 2
θωt
Real
Imaginary
R
x
y
Figure 2. Phasor of magnitude R and phase θ.
For a sinusoidal function ( ) ( )θω += tVtv cos , the phasor representation is
VVeV j == θˆ /θ
If the phasor is rotating counterclockwise about the origin at a rate of ω radians per second, then we multiply the phasor by such that tje ω ( )θωωθω +== tjtjjtj AeeVeeV and using Euler’s formula ( ) ( ) ( )θωθωθωω +++== + tjVtVVeeV tjtj sincosˆ . Then we see that
( ) ( )θω += tVtv cos is the real part of , or tjeV ωˆ
( ) ( )θωω +== tVeVtv tj cosˆRe .
In the above description of the phasor, the peak value is used, however we will use the RMS value of V as described in (1.6) instead of the peak value. This simplifies many of the calculations, particularly those associated with power as shown below.
Phasors will be represented with a hat (or caret) above the variable. Impedance is understood to be a complex quantity in general, and the hat (or caret ^) is left off the impedance variable
where R is the resistance, and X is the reactance component respectively. jXRZ +=
For a circuit element such as the one shown in Figure 1 representing a load in the circuit with i(t) as the instantaneous value of current through the load and v(t) is the instantaneous value of the voltage across the load.
In quasi-steady state conditions, the current and voltage are both sinusoidal, with corresponding
amplitudes of Vmax, Imax, and initial phases, φv and φi, and the same frequency T
f ππω 22 ==
( ) ( )vtVtv φω += cosmax (1.4) ( ) ( )itIti φω += cosmax (1.5)
The root-mean-squared (RMS) value of the voltage and current are then:
( )[ ]2
cos1 max2max
VdttVT
VT
o v =+= ∫ φω (1.6)
1- 3
( )[ ]2
cos1 max2max
IdttIT
IT
o i =+= ∫ φω (1.7)
Phasor representations for the above signals use the RMS values as VV =ˆ /φv and II =ˆ /φi. The instantaneous power is the product of voltage times current or:
( ) ( ) ( ) ( ) ( ) ( ) ( iviv ttIVtItVtitvtp )φωφωφωφω ++=+⋅+=⋅= coscoscoscos maxmaxmaxmax ( ) ( )iv ttVI φωφω ++= coscos2
( ) ([ iviv tVI φφωφφ +++−= 2coscos212 )] (1.8)
The average value is found by integrating over a period of time and then dividing the result by
that same time interval. The first term in (1.8) is independent of time, while the second term varies from -1 to +1 symmetrically about zero. Because it is symmetric about zero, integration over an integer number of cycles (or periods) gives a value of zero for the second term in (1.8), and the average power which we define as the real power with units of watts (W) is:
( )ivVIP φφ −= cos (1.9)
This shows the power is not only proportional to the RMS values of voltage and current, but also proportional to ( iv )φφ −cos . The cosine of this angle is defined as the displacement factor, DF. In more general terms for periodic, but not necessarily sinusoidal signals, the power factor is defined as:
VIPpf ≡ (1.10)
For sinusoidal signals, the power factor equals the displacement factor, or
( )ivpf φφ −= cos (1.11)
For comparison, the voltage, current, and power for various angles between voltage and current are shown below:
For leading power factors:
-10
-5
0
5
10
0 0.5 1 1.5 2 2.5 3Vol
tage
(V),
Cur
rent
(A),
Pow
er (W
)
# of Periods
( ) ( )ttv ωcos5 ⋅= ( ) ( )tti ωcos2 ⋅=
( ) ( ) 5cos22
cos maxmax =−=−= ivivIVVIP φφφφ (W)
Real
Imaginary
V
I
1- 4
1- 5
-10
-5
0
5
10
0 0.5 1 1.5 2 2.5 3Vol
tage
(V),
Cur
rent
(A),
Pow
er (W
)
# of Periods
( ) ( )ttv ωcos5 ⋅= ( ) ( )º30cos2 +⋅= tti ω
( ) ( ) 33.4cos22
cos maxmax =−=−= ivivIVVIP φφφφ (W)
Real
Imaginary
V
I
-10
-5
0
5
10
0 0.5 1 1.5 2 2.5 3Vol
tage
(V),
Cur
rent
(A),
Pow
er (W
)
# of Periods
( ) ( )ttv ωcos5 ⋅= ( ) ( )º60cos2 +⋅= tti ω
( ) ( ) 5.2cos22
cos maxmax =−=−= ivivIVVIP φφφφ (W)
Real
Imaginary
V
I
-10
-5
0
5
10
0 0.5 1 1.5 2 2.5 3Vol
tage
(V),
Cur
rent
(A),
Pow
er (W
)
# of Periods
( ) ( )ttv ωcos5 ⋅= ( ) ( )º90cos2 +⋅= tti ω
( ) ( ) 0cos22
cos maxmax =−=−= ivivIVVIP φφφφ (W)
Real
Imaginary
V
I
-10
-5
0
5
10
0 0.5 1 1.5 2 2.5 3Vol
tage
(V),
Cur
rent
(A),
Pow
er (W
)
# of Periods
( ) ( )ttv ωcos5 ⋅= ( ) ( )º120cos2 +⋅= tti ω
( ) ( ) 5.2cos22
cos maxmax −=−=−= ivivIVVIP φφφφ (W)
Real
Imaginary
V
I
-10
-5
0
5
10
0 0.5 1 1.5 2 2.5 3Vol
tage
(V),
Cur
rent
(A),
Pow
er (W
)
# of Periods
( ) ( )ttv ωcos5 ⋅= ( ) ( )º150cos2 +⋅= tti ω
( ) ( ) 33.4cos22
cos maxmax −=−=−= ivivIVVIP φφφφ (W)
Real
Imaginary
V
I
-10
-5
0
5
10
0 0.5 1 1.5 2 2.5 3Vol
tage
(V),
Cur
rent
(A),
Pow
er (W
)
# of Periods
( ) ( )ttv ωcos5 ⋅= ( ) ( )º180cos2 +⋅= tti ω
( ) ( ) 5cos22
cos maxmax −=−=−= ivivIVVIP φφφφ (W)
Real
Imaginary
V
I
For lagging power factors:
-10
-5
0
5
10
0 0.5 1 1.5 2 2.5 3Vol
tage
(V),
Cur
rent
(A),
Pow
er (W
)
# of Periods
( ) ( )ttv ωcos5 ⋅= ( ) ( )tti ωcos2 ⋅=
( ) ( ) 5cos22
cos maxmax =−=−= ivivIVVIP φφφφ (W)
Real
Imaginary
V
I
1- 6
1- 7
-10
-5
0
5
10
0 0.5 1 1.5 2 2.5 3Vol
tage
(V),
Cur
rent
(A),
Pow
er (W
)
# of Periods
( ) ( )ttv ωcos5 ⋅= ( ) ( )º30cos2 −⋅= tti ω
( ) ( ) 33.4cos22
cos maxmax =−=−= ivivIVVIP φφφφ (W)
Real
Imaginary
VI
-10
-5
0
5
10
0 0.5 1 1.5 2 2.5 3Vol
tage
(V),
Cur
rent
(A),
Pow
er (W
)
# of Periods
( ) ( )ttv ωcos5 ⋅= ( ) ( )º60cos2 −⋅= tti ω
( ) ( ) 5.2cos22
cos maxmax =−=−= ivivIVVIP φφφφ (W)
Real
Imaginary
V
I
-10
-5
0
5
10
0 0.5 1 1.5 2 2.5 3Vol
tage
(V),
Cur
rent
(A),
Pow
er (W
)
# of Periods
( ) ( )ttv ωcos5 ⋅= ( ) ( )º90cos2 −⋅= tti ω
( ) ( ) 0cos22
cos maxmax =−=−= ivivIVVIP φφφφ (W)
Real
Imaginary
VI
-10
-5
0
5
10
0 0.5 1 1.5 2 2.5 3Vol
tage
(V),
Cur
rent
(A),
Pow
er (W
)
# of Periods
( ) ( )ttv ωcos5 ⋅= ( ) ( )º120cos2 −⋅= tti ω
( ) ( ) 5.2cos22
cos maxmax −=−=−= ivivIVVIP φφφφ (W)
Real
Imaginary
VI
-10
-5
0
5
10
0 0.5 1 1.5 2 2.5 3Vol
tage
(V),
Cur
rent
(A),
Pow
er (W
)
# of Periods
( ) ( )ttv ωcos5 ⋅= ( ) ( )º150cos2 −⋅= tti ω
( ) ( ) 33.4cos22
cos maxmax −=−=−= ivivIVVIP φφφφ (W)
Real
Imaginary
VI
-10
-5
0
5
10
0 0.5 1 1.5 2 2.5 3Vol
tage
(V),
Cur
rent
(A),
Pow
er (W
)
# of Periods
( ) ( )ttv ωcos5 ⋅= ( ) ( )º180cos2 −⋅= tti ω
( ) ( ) 5cos22
cos maxmax −=−=−= ivivIVVIP φφφφ (W)
Real
Imaginary
VI
We call the power factor leading or lagging, depending on whether the current of the load leads or
lags the voltage across it. We know that current does not change instantly through an inductor, so it is clear that for a resistive - inductive (RL) load, the power factor is lagging. Likewise the voltage can not change instantly across a capacitor, therefore for a resistive – capacitive (RC) load, the power factor is leading (the current changes before the voltage can). Also, for a purely inductive or capacitive load the power factor is 0, while for a purely resistive load it is 1.
The product of the RMS values of voltage and current at a load is the apparent power, S having units of volt-amperes (VA):
VIS = (1.12)
The reactive power is Q with units of volt-amperes reactive (VA reactive, or VAr):
( )ivVIQ φφ −= sin (1.13)
The reactive power represents the energy oscillating in and out of an inductor or capacitor. Since the energy oscillation in an inductor is 180º out of phase with the energy oscillating in an capacitor, the reactive power of these two have opposite signs with the convention that it is positive for the inductor and negative for the a capacitor.
Using phasors, the complex apparent power, is: S
V== *ˆˆˆ IVS /φv I/-φi (1.14) or
1- 8
(1.15) jQPS +=ˆ
As an example, consider the following voltage and current for a given load:
( ) ⎟⎠⎞
⎜⎝⎛ +=
6377sin2120 πttv (V) (1.16)
( ) ⎟⎠⎞
⎜⎝⎛ +=
4377sin25 πtti (A) (1.17)
then (W), while the power factor is 6005120 =⋅== VIS 966.046
cos =⎟⎠⎞
⎜⎝⎛ −=
ππpf leading. Also,
the complex apparent power is: 120== *ˆˆˆ IVS /π/6 · 5/-π/4 = 600/- π/12 = 579.6 (W) – j155.3 (VAr) (1.18)
It should also be noted that when the angles are represented in radian, care must be taken to
assure your calculator is in radians mode. Often we represent the argument of the sine or cosine term in mixed units. For example we might write cos(377t + 30º). The first term (377t) has units of radians, while the second (30º) has units of degrees, and one of these must be converted before making calculations with your calculator or computer either in radians mode or in degrees mode.
As a phasor the complex apparent power can be shown on a complex plane for the cases of leading and lagging power factors as shown in Figure 3.
Real
Imaginary
V
I
Real
Imaginary
PQ
S
Real
Imaginary
V
I
Real
Imaginary
P
SQ
(a) Leading power factor (b) Lagging power factor
Figure 3. Phasor of magnitude R and phase θ.
1- 9
Recall that the leading power factor corresponds to a resistive-capacitive load. We see from Figure 3(a), this also corresponds to a negative value for Q. Similarly, lagging power factor has the current lagging the voltage corresponding to an inductive-resistive load, and Figure 3(b) shows this corresponds to a positive value for Q.
To summarize, we can use the following tables:
Equation Units Real Power ( )
pfSpfVI
VIP iv
⋅=⋅=
−= φφcos
Watts
Reactive Power ( )
leading)(for 1
lagging)(for 1
sin
2
2
pfS
pfS
VIQ iv
−⋅−=
−⋅=
−= φφ Volt-Amperes-Reactive
Apparent Power VIS = == *ˆˆˆ IVS V/φv I/-φi
jQPS +=ˆ 222 QPS +=
Volt-Amperes
Type of Load Reactive Power Power Factor Inductive Q > 0 lagging Capacitive Q < 0 leading Resistive Q = 0 1
From the interdependence of the four quantities, S, P, Q, pf, if we know any two of these
quantities, the other two can be determined. For example, if S = 100 (kVA), and pf = 0.8 leading, then:
( ) ( )[ ]( )iv
iv
SQ
pfSQ
pfSP
φφφφ
−==−
−=−−=
=⋅=
sin then8.0arccossinsin
or (kVAR), 601
(kW) 8002
It is important to notice that Q < 0, such that ( )iv φφ −sin is a negative quantity. This can be seen
when it is understood that there are two possible answers for the arccos(0.8), that is cos(36.87º) = 0.8, and cos(-36.87º) = 0.8 so to obtain a Q < 0 we use (φv – φi) = −36.87º.
Generally, in systems that contain more than one load (or source), the real and reactive power can be found by adding individual contributions, but this is not the case with the apparent power. That is
∑=
i itotal PP
∑= (1.19) i itotalQ Q
∑≠i itotal SS
1- 10
For the above example, if the load voltage is VL = 2000 (V), then the load current would be IL = S/VL = [100×103 (VA)]/[2×103 (V)] = 50 (A). If we use the load voltage as the reference, then:
2000ˆ =V /0º (V) 50ˆ =I /φi = 50 /+36.87º (A)
== *ˆˆˆ IVS [2000 /0º][50 /−36.87º] = P + jQ = 80×103 (W) – j60×103 (VAr)
1.2 Three Phase Balanced Systems
Three-phase systems offer significant advantages over single phase systems: for the same power and voltage there is less copper in the windings, and the total power absorbed remains constant rather than oscillating about an average value.
For a three phase system consisting of three current sources having the same amplitude and frequency, but with phases differing by 120º as:
( ) ( )
( )
( ) ⎟⎠⎞
⎜⎝⎛ ++=
⎟⎠⎞
⎜⎝⎛ −+=
+=
32sin2
32sin2
sin2
3
2
1
πφω
πφω
φω
tIti
tIti
tIti
(1.20)
If these are connected as shown in Figure 4, then at node n or n’, the current adds to zero, and the
neutral line n-n’ (dashed) is not needed.
( ) ( ) ( ) ( ) 03
2sin3
2sinsin2321 =⎭⎬⎫
⎩⎨⎧
⎟⎠⎞
⎜⎝⎛ +++⎟
⎠⎞
⎜⎝⎛ −+++=++
πφωπφωφω tttItititi
i1
nn'
i2i3
Figure 4. Balanced three phase Y-connected system with zero neutral current.
If instead we had three voltage sources Y-connected as in Figure 5 with the following values
1- 11
( ) ( )
( )
( ) ⎟⎠⎞
⎜⎝⎛ ++=
⎟⎠⎞
⎜⎝⎛ −+=
+=
32sin2
32sin2
sin2
3
2
1
πφω
πφω
φω
tVtv
tVtv
tVtv
(1.21)
then, the current through each of the three loads (assuming the loads are equal), would have equal magnitudes, but each current would have a phase that is shifted by an equal amount with respect to the voltages, v1(t), v2(t), v3(t).
v1
nn'
v2v3
+
+ +
ia
ib
ic
Figure 5. Balanced voltage fed three phase Y-connected system with zero neutral current.
With equal impedances for the loads, then
( ) ( )
( )
( ) ⎟⎠⎞
⎜⎝⎛ +++=
⎟⎠⎞
⎜⎝⎛ +−+=
++=
θπφω
θπφω
θφω
32sin2
32sin2
sin2
tZVti
tZVti
tZVti
c
b
a
(1.22)
and again the currents sum to zero at nodes n or n’, and for this balanced three phase system, the neutral wire (dashed) is not required.
In comparison to a single phase system, where two wires are required per phase, the three phase system delivers three times the power, and requires only three transmission wires total. This is a significant advantage considering the hundreds of miles of wire needed for power transmission.
Y and Δ Connections
The loads in the previous two figures, as well as in Figure 6 are connected in a Y or star configuration. If the load of Figure 6 is for a balanced Y system, then the voltages between each phase and the neutral are: VV n =1 /φ , VV n =2 /φ − 2π/3 , and VV n =3 /φ + 2π/3 .
Kirchhoff’s voltage law (KVL) states that the sum of voltages around a closed loop equals zero. This is also the case here however the voltages are complex numbers or phasors, and as such must be 1- 12
added as vectors. The phase φ can be any value, but the relative position of the phase to neutral phasors must be 120º with respect to each other as shown in Figure 7.
nV12
1
3
2
+ +
+
+
V1n
V3n
V2n
Figure 6. Y-connected loads with voltages relative to neutral identified. By KVL: , or as shown in 0ˆˆˆ
n2n112 =−+− VVV n2n112ˆˆˆ VVV −= Figure 7.
V3n
V1n
V2n
2n-V-V3n
1n-V
V12
V31
V23
Figure 7. Voltage phasors of the Y-connected loads shown in Figure 6.
We could also use the phasor representation VV n =1 /φ , VV n =2 /φ − 2π/3 ,
and VV n =3 /φ + 2π/3 to determine the line-to-line voltages as
VVVV nn 3ˆˆˆ2112 =−= /φ + π/6 (1.23)
This shows the RMS value of the line-to-line voltage, Vl-l , at a Y load is 3 times the line-to-
neutral or phase voltage, Vln. In the Y connection, the phase current is equal to the line current, and the power supplied to the system is three times the power supplied to each phase, since the voltage and current amplitudes and phase differences between them are the same in all three phases. If the power factor in one phase is ( )ivpf φφ −= cos , then the total power to the system is:
1- 13
( ) ( )ivlllivlll
nn
IVjIV
IV
jQPS
φφφφ
φφφ
−+−=
=
+=
−− sin3cos3
ˆˆ3
ˆ
*11
333
(1.24)
Similarly, for a connection of the loads in the Δ configuration (as in Figure 8), the phase voltage
is equal to the line voltage however; the phase currents are not equal to the line currents for the Δ configuration. If the phase currents are II =12
ˆ /φ , II =23ˆ /φ − 2π/3 , and II =31
ˆ /φ + 2π/3 then using Kirchhoff’s current law (KCL) the current of line 1, as shown in Figure 8 is:
IIII 3ˆˆˆ
31121 =−= /φ - π/6 (1.25) Thus for the Δ configuration, the line current is 3 times the IΔ current.
I1
2
1
3
I2
I3
I12
I31
I23
I12
I31
I23
-I23
I1
I3
I2-I12
-I31
Figure 8. Δ connected load, line, and phase currents. To calculate the power in the three-phase Δ connected load:
( ) ( )ivlllivlll IVjIV
IV
jQPS
φφφφ
φφφ
−+−=
=
+=
−− sin3cos3
ˆˆ3
ˆ
*1212
333
(1.26)
which is the same value as for the Y connected load.
For a balanced system, the loads of the three phases are equal. Also, a Δ configured load, can be replaced with a Y configured load (and visa versa) if:
Δ= ZZY 31 (1.27)
1- 14
Under these conditions, the two loads are indistinguishable by the power transmission lines. You might recall the Δ to Y transformation for resistor circuits can be remembered by overlaying
the Δ and Y configurations such as in Figure 9.
R
C B
A
B
RA
RCR 1
R 3
R 2
Figure 9. Δ to Y or Y to Δ resistor network transformation.
1
133221
1
RRRRRRRR
RRRRRR
A
CBA
CB
++=
++=
(1.28)
Equation (1.28) gives a similar result to that of (1.27) when RA = RB = RB C and R1 = R2 = R3.
1.3 Calculations in Three-Phase Systems Calculations of quantities like currents, voltages, and power in three-phase systems can be
simplified by the following procedure:
1. transform the Δ circuits to Y, 2. connect a neutral conductor, 3. solve one of the three 1-phase systems 4. convert the results back to the Δ systems
1.3.1 Example For the 3-phase system in Figure 10 calculate the line-to-line voltage, real power, and power
factor at the load. To solve this by the procedure outlined above, first consider only one phase as shown in Figure
11.
1- 15
v1
nn'
v2v3
+
+ +
= 120 (V)
j1 (Ω)
7 + j5 (Ω)
Figure 10. Three phase system with Y connected load, and line impedance.
n
v1
n'
+ = 120 (V)
j1 (Ω)
7 + j5 (Ω)
I
Figure 11. One phase of the three phase system shown in Figure 10.
For the one-phase in Figure 11,
( ) 02.13571
120ˆ =++
=jj
I /-40.6º (A)
02.13ˆ1 == Ln ZIV /-40.6º (7+j5) = 112/-5.1º (V)
== *1,
ˆˆ IVS LL φ (112/-5.1º)(13.02/40.6º) = 1458.3/35.5º = 1.187×103 + j0.848×103
=φ1,LP 1.187 (kW), 0.848 (kVAr) =φ1,LQpf = cos(-5.1º - (-40.6º)) = 0.814 lagging
For the three-phase system of Figure 10 the load voltage (line-to-line), the real, and reactive
power are:
1941123, =⋅=−llLV (V) =φ3,LP 3.56 (kW) =φ3,LQ 2.544 (kVAr)
1- 16
1.3.2 Example For the Y to Δ three-phase system in Figure 12, calculate the power factor and the real power at
the load, as well as the phase voltage and current. The source voltage is 400 (V) line-to-line.
v1
n'
v2v3
++ +
j1 (Ω)
18 + j6 (Ω)
Figure 12. Δ connected load. First convert the load to an equivalent Y connected load, then work with one phase of the system.
The line to neutral voltage of the source is 2313
400ln ==V (V).
231 (V)
nn'
+
+ +
j1 (Ω)
6 + j2 (Ω)
Figure 13. Equivalent Y connected load.
n
v1
n'
+ = 231 (V)
j1 (Ω)
6 + j2 (Ω)
IL
Figure 14. One phase of the Y connected load.
1- 17
( ) 44.34261
231ˆ =++
=jj
IL /-26.6º (A)
( ) 8.21726ˆˆ =+= jIV LL /-8.1º (V)
The power factor at the load is:
( ) ( ) 948.0º6.26º1.8coscos =+−=−= ivpf φφ lagging Converting back to a Δ connected load gives:
88.19344.34
3=== LIIφ (A)
22.37738.217 =⋅=−llV (V)
At the load the power is:
,3 3 3 377.22 34.44 0.948 21.34L l l LP V I pfφ −= = ⋅ ⋅ ⋅ = (kW)
1.3.3 Example Two three-phase loads are connected as shown in . Load 1 draws from the system PL1 = 500
(kW) at 0.8 pf lagging, while the total load is ST = 1000 (kVA) at 0.95 pf lagging. What is the power factor of load 2?
Power System
Load 2
Load 1
Figure 15. Two three-phase loads connected to the same power source.
1- 18
For the total load we can add the real and reactive power for each of the two loads (we can not add the apparent power).
21
21
21
LLT
LLT
LLT
SSSQQQPPP
+≠+=+=
From the information we have for the total load we can write the following:
=⋅= TTT pfSP 950 (kW)
( )[ ] 25.31295.0cossin 1 =⋅= −TT SQ (kVAr)
The reactive power, QT, is positive since the power factor is lagging.
For the load L1, PL1 = 500 (kW), pf1 = 0.8 lagging, thus:
=×
=8.010500 3
1LS 625 (kVA)
=−= 21
211 LLL PSQ 375 (kVA)
Again, QL1 is positive since the power factor is lagging. This leads to:
=−= 12 LTL PPP 450 (kW)
=−= 12 LTL QQQ -62.75 (kVAr)
and
99.075.62450
45022
2
22 =
+==
L
LL S
Ppf leading.
Chapter Notes • A sinusoidal signal can be described uniquely by:
1. Time dependent form as for example: ( ) ( )vfttv φπ += 2sin5 (V) 2. by a time dependent graph of the signal 3. as a phasor along with the associated frequency of the phasor
one of these descriptions is enough to produce the other two.
• It is the phase difference that is important in power calculations, not the phase. The phase is arbitrary depending on the defined time (t = 0). We need the phase to solve circuit problems after we take one quantity (some voltage or current) as a reference. For that reference quantity we assign an arbitrary phase (often zero).
• In both three-phase and one-phase systems the total real power is the sum of the real power from the individual loads. Likewise the total reactive power is the sum of the reactive power of the individual loads. This is not the case for the apparent power or the power factor.
1- 19
1- 20
• Of the four quantities: real power, reactive power, apparent power, and power factor, any two describe a load adequately. The other two quantities can be calculated from the two given.
• To calculate real, reactive, and apparent power when using equations (1.9), (1.12), and (1.13) we must use absolute values, not complex values for the currents and voltages. To calculate the complex power using equation (1.14) we do use complex currents and voltages and find directly both the real and reactive power (as the real and imaginary components respectively).
• When solving a circuit to calculate currents and voltages, use complex impedances, currents and voltages.
ECE 320 Energy Conversion and Power Electronics
Dr. Tim Hogan
Chapter 2: Magnetic Circuits and Materials Chapter Objectives In this chapter you will learn the following: • How Maxwell’s equations can be simplified to solve simple practical magnetic problems • The concepts of saturation and hysteresis of magnetic materials • The characteristics of permanent magnets and how they can be used to solve simple problems • How Faraday’s law can be used in simple windings and magnetic circuits • Power loss mechanisms in magnetic materials • How force and torque is developed in magnetic fields
2.1 Ampere’s Law and Magnetic Quantities Ampere’s experiment is illustrated in Figure 1 where there is a force on a small current element
I2l when it is placed a distance, r, from a very long conductor carrying current I1 and that force is quantified as:
lIr
IF 21
2πμ
= (N) (2.1)
I1
rI2
F
Conductor1
Current Elementof Length l
Figure 1. Ampere’s experiment of forces between current carrying wires.
The magnetic flux density, B, is defined as the first portion of equation (2.1) such that:
lBIF 2= (N) (2.2)
1- 1
From (2.1) and (2.2) we see the magnetic flux density around conductor 1 is proportional to the current through conductor 1, I1, and inversely proportional to the distance from conductor 1. Looking
at the units and constants handout given in class, or from (2.2) the units of, B, are seen as ⎟⎠⎞
⎜⎝⎛
⋅mAN ,
thus µ, called permeability, has units of ⎟⎠⎞
⎜⎝⎛
2AN . More commonly, the relative permeability of a given
material is given where 0μμμ r= and 90 2
N400 10 A
μ π − ⎛= × ⎜⎝ ⎠
⎞⎟ . Since a Newton-meter is a Joule, and
a Joule is a Watt-second: ⎟⎠⎞
⎜⎝⎛ ⋅
=⎟⎠⎞
⎜⎝⎛
⋅⋅⋅
=⎟⎠⎞
⎜⎝⎛
⋅=⎟
⎠⎞
⎜⎝⎛
⋅⋅
=⎟⎠⎞
⎜⎝⎛
⋅ 2222 msV
mAsAV
mAJ
mAmN
mAN . This shows B is a
per meter squared quantity, and the (V·s) units represents the magnetic flux and is given units of Webers (Wb). This flux can be found by integrating the normal component of B over the area of a given surface:
∫ ⋅=S
dsnB ˆφ (2.3)
The magnetic field intensity is related to the magnetic flux density by the permeability of the
media in which the magnetic flux exists.
μBH ≡ (2.4)
For the system in Figure 1, r
IBHπμ 21== and have units of ⎟
⎠⎞
⎜⎝⎛
mA . If there were multiple
conductors in place of conductor 1, for example in a coil, then the units would be ampere-turns per meter. A line integration of H over a closed circular path gives the current enclosed by that path, or for the system in Figure 1:
1
2
0
1
2Idl
rIdlHH
r
C==⋅= ∫∫
π
π (2.5)
again, if the system contained multiple conductors within the enclosed path, the result would give ampere-turns. Equation (2.5) is Ampere’s circuital law.
An alternative approach as described in your textbook is to begin with Maxwell’s equations
which include Ampere’s circuital law in a more general form as shown in Table I below:
Table I. Maxwell’s equations. Name Point Form Integral Form
Faraday’s Law tB∂∂
−=×∇ E ∫ ∫ ⋅−=⋅C S
dsnBdtddl ˆE
Ampere’s Law Modified by MaxwelltDJH∂∂
+=×∇ ∫∫ ⋅⎥⎦
⎤⎢⎣
⎡∂∂
+=⋅SC
dsntDJdlH ˆ
Gauss’s Law ρ=⋅∇ D ∫∫ =⋅VS
ˆ dvdsnD ρ
Gauss’s Law for Magnetism 0=⋅∇ B 0 ˆS
=⋅∫ dsnB
1- 2
where ∫C represents the integral over a closed path, ∫S represents the integral over the surface of a
closed volume of space, is a surface normal vector, n E is the spatial vector of electric field, B is the spatial vector of magnetic flux density, J is the spatial vector of the electric current density, D is
the displacement charge vector, ρ is the electric charge density, ∫V is a volume integral ×∇ is the
curl and the divergence of the vector being acted upon. ⋅∇
Some assumptions commonly used in electromechanical energy conversion include a low enough
frequency, that the displacement current, tD∂∂
, can be neglected, and the assumption of homogeneous
and isotropic media used in the magnetic circuit. Under these assumptions, Ampere’s circuital law is modified to remove the displacement current component such that
∫∫ ⋅=⋅SC
dsnJdlH ˆ (2.6)
which for the system of Figure 1 reduces to equation (2.5).
2.2 Magnetic Circuits
From Ampere’s circuital law, ∫∫ ⋅=⋅SC
dsnJdlH ˆ , we see the magnetic field intensity around a
closed contour is a result of the total electric current density passing through any surface linking that
contour. Gauss’s Law for magnetism, 0 ˆS
=⋅∫ dsnB states that there are no magnetic monopoles –
that is to say there for a closed surface there is as much magnetic field density leaving that closed surface as there is entering the closed surface. If the integration is for an area, but not a closed
surface area, then we obtain the flux or ∫ ⋅=S
dsnB ˆφ .
The permeability of free space is 1, while the permeability of magnetic steel is a few hundred thousand. Magnetic flux can be confined to the structures or paths formed by high permeability materials. In this way, magnetic circuits can be formed such as the one shown in Figure 2.
Figure 2. Simple magnetic circuit [1].
1- 3
The driving force for the magnetic field is the magnetomotive force (mmf), F, which equals the
ampere-turn product iN ⋅=F (2.7)
The analysis of a magnetic circuit is similar to the analysis of an electric circuit, and an analogy
can be made for the individual variables as shown in Table II.
Table II. Comparison of electric and magnetic circuits. Electric Circuit Magnetic Circuit
V r
I
I
a
b
c
Electrically ConductiveMaterial
V r
φ
I
a
b
c
High PermeabilityMaterial
Driving Force applied battery voltage = V applied ampere-turns = F
Response
resistance electricforce driving current =
or
RVI =
reluctance magneticforce driving flux =
or
RF
=φ
Impedance Impedance is used to indicate the impediment to the driving force in establishing a response.
AlRσ
==resistance
where l = 2πr, σ = electrical conductivity, A = cross-sectional area
Alμ
== Rreluctance
where l = 2πr, μ = permeability, A = cross-sectional area
Equivalent Circuit
V
I
R
V = IR
F
φ
R
F = φR
1- 4
1- 5
Electric Circuit Magnetic Circuit
Fields Electric Field Intensity
rV
lV
π2=≡E (V/m)
or
∫ =⋅ VdlE
Magnetic Field Intensity
rlH
π2FF
=≡ (A-t/m)
or
∫ =⋅ FdlH
Potential Electric Potential Difference
abababb
a
b
aab IRl
Al
lIl
lIRdl
lVdlV ====⋅= ∫∫ σ
E
Magnetic Potential Difference
ababababb
aab l
Al
ll
ll
ldlH RRFF φ
μφφ
====⋅= ∫Flow Densities
Current Density
( ) EE σσ
===≡A
lAl
ARV
AIJ
Flux Density
H
AlA
HlAA
B μ
μ
φ=
⎟⎠⎞⎜
⎝⎛
==≡R
F
An example magnetic circuit is shown in Figure 3, below.
Figure 3. Simple magnetic circuit with an air gap [1].
For this circuit, we will assume: • the magnetic flux density is uniform throughout the magnetic core’s cross-sectional area and is
perpendicular to the cross-sectional area • the magnetic flux remains within the core and the air gap defined by the cross-sectional area of the
core and the length of the gap (no leaking of field, no fringe fields at the gap). Then ggcc
AABABAdB
c
==⋅= ∫φ (2.8)
with Ac = Ag
HH
BB
gc
gc
μμ =
= (2.9)
Since, F = Hl
g
g
gc
c
c
gcc
lB
lB
gHlH
μμ+=
+=F (2.10)
or
( )gc
ggcc
c
Ag
Al
RR
F
+=
⎟⎟⎠
⎞⎜⎜⎝
⎛+=
φ
μμφ
(2.11)
Thus the magnetic circuit shown in Figure 3 can be represented as
F
φ
R
R
c
g
gcgc
iNRRRR
F+⋅
=+
=φ
Figure 4. Equivalent circuit for the magnetic circuit in Figure 3.
This concept is helpful for more complex configurations such as shown in Figure 5.
1- 6
AreaA1
AreaA1
AreaA1
Area A2
AreaA3
g2
I
Area A2
g1
I21
AreaA1
Figure 5. Magnetic circuit with various cross-sectional areas and two coils.
Using the lengths defined in Figure 6, and paying attention to the direction of the magnetomotive
force from each of the coils by use of the right hand rule, the equivalent circuit can be drawn as seen in Figure 7.
Area A
Area A2
AreaA3
g2
I
g1
I21
1
Area A1
Area A1
l 1
·l221
·l321
Figure 6. Lengths for each section of the magnetic circuit of Figure 5.
Then,
( ) 121111 11φφφ glIN RRF ++=⋅= (2.12)
and ( )
2231 21222 lglgIN RRRRF ++−=⋅= φφ (2.13) For a system with the dimensions, number of turns, current, and permeability known, then
equations (2.12) and (2.13) give two equations with two unknowns such that φ1 and φ2 can be found.
1- 7
F
Rg
Rg
F2
11
2
R l1
Rl2
Rl3
φ 1
φ2
Figure 7. Equivalent circuit for the configuration of Figure 5.
Assuming the gaps are air gaps, the value of each reluctance can be found as:
1
11 A
ll μ=R
10
11 A
gg μ=R
2
22 A
ll μ=R
3
33 A
ll μ=R
20
22 A
gg μ=R
Then we can find the magnetic flux density for each gap as:
2
2
1
1
2
1
AB
AB
g
g
φ
φ
=
= (2.14)
Note that the permeability of the core material can be much larger than the permeability of the gaps such that the total reluctance is dominated by the reluctances of the gaps.
2.3 Inductance From circuit theory we recall that the voltage across an inductor is proportional to the time rate of
change of the current through the inductor.
( ) ( )dt
tdiLtv LL = (2.15)
and while the power of an inductor can be positive or negative, the energy is always positive as
( ) 2
21
LL iLtw ⋅= (2.16)
In Table I, Faraday’s law is ∫ ∫ ⋅−=⋅C S
dsnBdtddl ˆE or the electric field intensity around a
closed contour C is equal to the time rate of change of the magnetic flux linking that contour. Integrating over the closed contour of the coil itself gives us the negative of the voltage at the terminals of the coil. On the right side of Faraday’s law we then must integrate over the surface of the full coil, thus including the N turns of the coil. Then Faraday’s law gives
1- 8
( ) ( )dt
tdNtvLφ
= (2.17)
Comparing equation (2.15) and (2.17) gives
didNL φ
= (2.18)
For linear inductors, the flux φ is directly proportional to current, i, for all values such that
i
NL φ= (2.19)
with units of (Weber-turns per ampere), or Henry’s (H). The flux linkage, λ, is defined as φλ N=
and combining this with the relationship between flux and total circuit reluctance tottot
NiRR
F==φ
along with (2.19) gives
tot
NLR
2
= (2.20)
Thus inductance can be increased by increasing the number of turns, using a metal core with a higher permeability, reducing the length of the metal core, and by increasing the cross-sectional area of the metal core. This is information that is not readily seen by either the circuit laws for inductors, or through Faraday’s equation alone.
Mutual Inductance
For a magnetic circuit containing two coils and an air gap such as in Figure 8, with each coil wound such that the flux is additive, then the total magnetomotive force is given by the sum of contributions from the two coils as 2211 iNiN +=F (2.21)
i i21
Area Ac lc
φ
g
λ 1 λ 2
Figure 8. Mutual inductance magnetic circuit.
The equivalent circuit is shown in Figure 9.
1- 9
F2
Rg
R lc
F1
φ
Figure 9. Equivalent circuit for Figure 8.
If the permeability of the core is large such that glc RR << and the cross-sectional area of the gap
is assumed equal to the cross-sectional area of the core (Ag = Ac), then
( )gAiNiN c0
2211μφ +≈ (2.22)
The flux linkage, λ1, in Figure 8 is φλ 11 N= , or
212111
20
21102
111
iLiL
igANNi
gANN cc
+=
⎟⎟⎠
⎞⎜⎜⎝
⎛+⎟⎟
⎠
⎞⎜⎜⎝
⎛==
μμφλ (2.23)
where
⎟⎟⎠
⎞⎜⎜⎝
⎛=
gANL c02
111μ (2.24)
is the self-inductance of coil 1 and L11i1 is the flux linkage of coil 1 due to its own current i1. The mutual inductance between coils 1 and 2 is
⎟⎟⎠
⎞⎜⎜⎝
⎛=
gANNL c0
2112μ (2.25)
and L12i2 is the flux linkage of coil 1 due to current i2 in the other coil. Similarly, for coil 2
222121
202
210
2122
iLiL
igANi
gANNN cc
+=
⎟⎟⎠
⎞⎜⎜⎝
⎛+⎟⎟
⎠
⎞⎜⎜⎝
⎛==
μμφλ (2.26)
where L21 = L12 is the mutual inductance and L22 is the self-inductance of coil 2.
⎟⎟⎠
⎞⎜⎜⎝
⎛=
gANL c02
222μ (2.27)
1- 10
2.4 Magnetic Material Properties A simplification we have used is that the permeability of a given material is constant for different
applied magnetic fields. This is true for air, but not for magnetic materials. Materials that have a relatively large permeability are ferromagnetic materials in which the magnetic moments of the atoms can align in the same direction within domains of the material when and external field is applied. As more of these domains align, saturation is reached when there is no further increase in flux density of that of free space for further increases in the magnetizing force. This leads to a changing permeability of the material and a nonlinear B vs. H relationship as shown in Figure 10.
H
B
Figure 10. Nonlinear B vs H normal magnetization curve.
When the field intensity is increased to some value and is then decreased, it does not follow the curve shown in Figure 10, but exhibits hysteresis as shown by the abcdea loop in Figure 11. The deviation from the normal magnetization curve is caused by some of the domains remaining oriented in the direction of the originally applied field. The value of B that remains after the field intensity H is removed is called residual flux density. Its value varies with the extent to which the material is magnetized. The maximum possible value of the residual flux density is called retentivity and results whenever values of H are used that cause complete saturation. When the applied magnetic field is cyclically applied so as to form the hysteresis loop such as abcdea in Figure 11, the field intensity required to reduce the residual flux density to zero is called the coercive force. The maximum value of the coercive force is called the coercivity.
The delayed reorientation of the domains leads to the hysteresis loops. The units of BH is
32 mJ
mN
meter-amperenewtons
meteramperes of units ==×=HB (2.28)
1- 11
or an energy density.
H (A·t/m)
B (Wb/m2)
O
a
b
c
d
e
f
Residual flux
Retentivity
Coercive force
Coercivity
Figure 11. Hysteresis loops. The normal magnetization curve is in bold.
H
B
O
a
b
c
d
e
f
Figure 12. Energy relationship for hysteresis loop per half-cycle.
1- 12
The full shaded area in Figure 12 outlined by eafe represents the energy stored in the magnetic field during the positive half cycle of H. The hatched area eabe represents the hysteresis loss per half cycle. This energy is what is required to move around the magnetic dipoles and is dissipated as heat. The energy released by the magnetic field during the positive half cycle of they hysteresis loop is given by the cross-hatched area outlined by bafb and is energy that is returned to the source.
The power loss due to hysteresis is given by the area of the hysteresis loop times the volume of the ferromagnetic material times the frequency of variation of H. This power loss is empirically given as (2.29) n
hh fBkP maxν=
where n lies in the range 1.5 ≤ n ≤ 2.5 depending on the material used, ν is the volume of the ferromagnetic material, and the value of the constant, kh, also depends on the material used. Some typical values for kh are: cast steel 0.025, silicon sheet steel 0.001, and permalloy 0.0001.
In addition to the hysteresis power loss, eddy current losses also exist for time-varying magnetic fluxes. Circulating currents within the ferromagnetic material follow from the induced voltages described by Faraday’s law. To reduce these eddy current losses, thin laminations (typically 14-25 mils thick) are commonly used where the magnetic material is composed of stacked layers with an insulating varnish or oxide between the thin layers. An empirical equation for the eddy-current loss is (2.30) 2
max22 BfkP ee ντ=
where ke = constant dependent on the material f = frequency of the variation of flux BBmax = maximum flux density τ = lamination thickness ν = total volume of the material
The total magnetic core loss is the sum of the hysteresis and eddy current losses.
If the value of H, when increasing towards some maximum, Hmax, does not increase continuously, but at some point, H1, decreases to H = 0 then increases again to it maximum value of Hmax, then a minor hysteresis loop is created as shown in Figure 13. The energy loss in one cycle includes these additional minor loop surfaces.
1- 13
H
B
O H1
Hmax
Figure 13. Minor hysteresis loops.
2.5 Permanent Magnets For a ring of iron with a uniform cross-section and hysteresis curve shown in Figure 15, the
magnetic field is zero when there is a nonzero flux density, BBr called the remnant flux density. To achieve a zero flux density, we could wind a coil around a section of the iron, and send current through the coil to reach a field intensity of –Hc (the coercive field). In practice a permanent magnet operates on a minor loop as shown in that can be approximated as a straight line, recoil line, such that
Figure 13
rm m
c
BrB H B
H= + (2.31)
Magnetization curves for some important permanent-magnet materials is shown in figure 1.19 from your textbook and shown below in Figure 14.
Example In the magnetic circuit shown with the length of the magnet, lm = 1cm, the length of the air gap is g = 1mm and the length of the iron is li = 20cm. For the magnet B
½·li
½·li
glm
Br = 1.1 (T), Hc = 750 (kA/m), what is the flux density in the air gap if the iron is assumed to have an infinite permeability and the cross-section is uniform? Since the cross-section is uniform, and there is no current: Hi[0.2(m)] + Hg[g] + Hm[lm]=0 With the iron assumed to have infinite permeability, Hi = 0 and
1- 14
( ) 01.1
00
0
0
=⎟⎟⎠
⎞⎜⎜⎝
⎛−+
=⎥⎦
⎤⎢⎣
⎡−+==+
mr
cmg
mcr
mcgmmg
lBHBgB
lHB
BHgB
lHgH
μ
μ
or B·795.77 + (B-1.1)·6818=0
B=0.985 (T)
Figure 14. Magnetization curves for common permanent-magnet materials [1].
1- 15
H (A·t/m)
B (Wb/m2)
O-Hc
Br
(Hm, Bm)
Figure 15. Remnant flux and coercive field for a piece of iron.
2.6 Torque and Force In Figure 12 we found the energy density in the field for the positive half-cycle is the total area or
(2.32) ∫=a
e
B
Bf dBHw
If this is simply approximated as a triangular area, then wf = ½ BH (J/m3). The total energy, Wf, would be found by multiplying this by the volume of the core
( )( ) ( )
R
F
221
21
21
21
φ
φ
=
=== HlBAAlBHW f (2.33)
Mechanical energy is done by reducing the reluctance (as the armatures of a relay are brought
together, for example), thus
RddWm2
21φ−= (2.34)
Since force is given as , the magnetic force is given by mdWFdx =
1- 16
dxdF R2
21φ−= (2.35)
and has units of newtons.
In a mechanical system with a force F acting on a body and moving it at velocity v, the power Pmech is vFPmech ·= (2.36) For a rotating system with torque T, rotating a body with angular velocity ωmech: mechmech TP ω·= (2.37) On the other hand, an electrical source e, supplying current, i, to a load provides electrical power Pelec iePelec ·= (2.38) Since power has to balance, if there is no change in the field energy, mechmechelec TiePP ω·· === (2.39) Notes • It is more reasonable to solve magnetic circuits starting from the integral form of Maxwell’s
equations than finding equivalent resistance, voltage and current. This also makes it easier to use saturation curves and permanent magnets.
• Permanent magnets do not have flux density equal to BBr. Equation ( ) defines the relation between the variables, flux density Bm
2.31B
and field intensity Hm in a permanent magnet. • There are two types of iron losses: eddy current losses that are proportional to the square of the
frequency and the square of the flux density, and hysteresis losses that are proportional to the frequency and to some power n of the flux density.
1 A. E. Fitzgerald, C. Kingsley, Jr., S. D. Umans, Electric Machinery, 6th edition, McGraw-Hill, New York, 2003.
1- 17
ECE 320 Energy Conversion and Power Electronics
Dr. Tim Hogan
Chapter 3: Transformers (Textbook Chapter 2) Chapter Objectives In this chapter you will be able to: • Choose the correct rating and characteristics of a transformer for a specific application • Calculate the losses, efficiency, and voltage regulation of a transformer under specific operating
conditions. • Experimentally determine the transformer parameters given its ratings. • Utilize the per unit system.
3.1 Introduction Transformers do not have moving parts, nor are they energy conversion devices, however their
ability to modify the current-voltage characteristics of a given load or source, make them invaluable components in energy conversion systems. They are utilized for power applications and in low power signal processing systems. One application in power transmission is the use of transformers on a transmission line utility pole commonly seen as a cylinder with a few wires sticking out. These wires enter the transformer through bushings that provide isolation between the wires and the tank. Inside the tank there is an iron core commonly made of silicon-steel laminations that are 14 mils (0.014”) thick. The insulation often used is paper with the whole coil system immersed in insulating oil. The oil increases the dielectric strength of the paper and helps to transfer heat from the core/coil assembly. An drawing of one such distribution transformer is shown in Figure 2.2 in your textbook. Connection of the transformer to the transmission lines can take several electrical configurations. A relatively simple connection to a 2.4 kV three phase transmission line is shown in Figure 1.
2400 voltthree-phasethree-wire Δprimary system
A
B
C
2400
24002400
H1 H2
X1X2X3
120 voltone-phasetwo-wire service
Figure 1. Example configuration of a distribution pole transformer connection to
three phase power lines to provide 120 (V) service to your home.
1- 1
2400 voltthree-phasethree-wire Δprimary system
A
B
C
2400
24002400
H1 H2
X1X2X3
120 (V)
120 (V)240 (V)
Figure 2. Example configuration of a distribution pole transformer connection to
three phase power lines and provides 120 (V) and 240 (V) service. If a neutral line is also part of the three phase transmission line (perhaps between the substation
and your home), then the connection could be made as shown in Figure 3.
4160Y/2400 voltthree-phasefour-wire Ywith neutral
B
C
4160
41604160
H1 H2
X1X3
120 (V)
120 (V)240 (V)
A
NN2400
2400
2400
X2
Figure 3. Distribution transformer connection to provide 120 (V) and 240 (V)
service from a 4160Y/2400 (V) four-wire transmission line.
1- 2
For a three phase line at the service end, a system could be connected to a four wire three phase transmission line source as shown in Figure 4 below.
4160Y/2400 voltthree-phasefour-wire Ywith neutralprimary service
B
C
4160
41604160
H1 H2
X1X3
208 (V)
208 (V)208 (V)
A
NN2400
2400
2400
X2
H1 H2
X1X3 X2
H1 H2
X1X3 X2
B
C
A
Nthree-phasefour-wire Ygrounded neutralsecondary service
120 (V)
120 (V)
120 (V)
Figure 4. Three phase to three phase distribution transformer connection providing a four-wire three phase distribution of 120 (V) and 208 (V) service from a 4160Y/2400 (V) four-wire transmission line.
Transformers are very common place in society, and have had significant impact at many power
levels. They also continue to be improved on today, with such studies as amorphous metals to further decrease core losses. For more understanding, we begin with the ideal transformer.
3.2 Ideal Transformer Under ideal conditions, the iron core has infinite permeability and the coils have zero electrical
resistance. Coil 1 has N1 turns, and coil 2 has N2 turns, and all of the magnetic flux is maintained in the iron (no flux leakage).
i i21
φ
e1 e2
φ
F2F1
Figure 5. Transformer and equivalent magnetic circuit. 1- 3
The electromotive force, or emf, is represented with the symbol e. For an ideal transformer, this
is the voltage at the terminals of a given coil. The flux linkage in each coil is φλ 11 N= and φλ 22 N= . The electromotive force induced in each coil is then the time derivative of the flux
linkage or
dtdN
dtde φλ
11
1 == (V) (3.1)
dtdN
dtde φλ
22
2 == (V) (3.2)
The ratio of the voltage at the terminals of coil 1 to the voltage at coil 2 is then
2
1
2
1NN
ee
= (3.3)
Using an equivalent circuit shown in Figure 5 with the magnetomotive force F1 for coil 1 and F2 for coil 2 we would write: 0221121 =−=− iNiNFF (3.4) or
1
2
2
1NN
ii= (3.5)
Transformers are often used with a voltage source connected to one coil and a load connected to
the other coil. The dots above the coils help to indicate the voltage reference marks for the coils such
that for an ideal transformer a positive voltage, dtdNev φ
111 == in coil 1 (primary coil) results in a
positive voltage, dtdNev φ
222 == in the coil 2 (secondary coil) as shown in Figure 6.
i i21
φ
v1 v2
φ
LoadN1
N2
Figure 6. Circuit utilizing a transformer.
1- 4
The circuit symbol for the transformer is shown in Figure 7. Since the voltages across the coils
transform in the direct ratio of the turns and the current transforms in the inverse ratio of the turns, then the impedance can also be transformed through the transformer.
i 1 i2
N1 N2
Figure 7. Transformer circuit symbol. For a voltage applied to the primary coil, and a load impedance, ZL, connected to the secondary as
shown in Figure 8.
ZL
+
V1 V2
+
N1 N2
I 1 I 2
Figure 8. Transformer with a load connected to the secondary.
The voltages and currents from secondary to primary are related by (3.3) and (3.5), or
22
11 ˆˆ V
NNV = (3.6)
21
21 ˆˆ I
NNI = (3.7)
Then the impedance measured at the terminals of the primary is
2
22
1
21
1
1ˆˆ
ˆˆ
IV
NNZ
IV
⎟⎟⎠
⎞⎜⎜⎝
⎛== (3.8)
Thus the following circuits are equivalent
1- 5
ZL
+
V1 V2
+
N1 N2
I 1 I 2
ZL
+
V1
N1 N2
I 1
I 2
N1( )N2
2
ZLN1( )N2
2+
V1
I 1
(a) (b) (c)
Figure 9. Three equivalent circuits assuming an ideal transformer.
3.3 Non-Ideal Transformer There are several non-ideal properties that we can take into account by modifying the equivalent
circuit shown in Figure 7. These include a finite core permeability (µc < ∞), leakage flux, core losses, and coil resistance. The contribution from each of these is discussed below.
Finite Core Permeability
For the core of the transformer to have a finite permeability, then the circuit in Figure 5 is modified to include the reluctance of the core.
F2F1
R φ
Figure 10. Transformer shown in Figure 5, but with a core of finite permeability.
Then we can write φRFF =−=− 221121 iNiN (3.9) Defining a magnetomotive force that is equal to the drop across the reluctance of the core as: φRF == exc iN ,11 (3.10) Solving for the flux gives
R
exiN ,11=φ (3.11)
The rate of change in the flux is proportional to the induced voltage as
1- 6
dtdiN
dt
iNd
NdtdNe
ex
ex
,121
,11
111
⎟⎟⎠
⎞⎜⎜⎝
⎛=
⎟⎟⎠
⎞⎜⎜⎝
⎛
==
R
Rφ (3.12)
This induced voltage is proportional to the time derivative of the current i1,ex which can be
represented by an inductor in our equivalent circuit with a value of R
21NLm = . The equivalent circuit
for the ideal transformer is then modified to account for the finite permeability of the core by placing an additional inductor across the primary coil as shown in Figure 11.
+
e 1 e2
+
N1 N2
i'1
L
i2
m
i1
i1,ex
Ideal transformer Figure 11. Equivalent circuit for a transformer with finite core permeability.
We can also see from Figure 11 that 1,11 iii ex ′=− which is in agreement with equations (3.9) and (3.10).
Leakage Flux
With finite core permeability, not all of the flux will be confined to the metal core, but some will “leak” outside the core in the surrounding air. The influence of this leakage flux can also be included in the equivalent circuit, by considering an additional reluctance associated with the leakage flux, φl1, for coil 1, and φl2 for coil 2 such that
2
222
1
111
ll
ll
iN
iN
R
R
=
=
φ
φ (3.13)
These reduce the magnetizing flux, φm, of the core, and modify the induced voltages for the primary (coil 1) and secondary (coil 2) to be
dtdiNe
dtdN
dtdN
dtdv
dtdiNe
dtdN
dtdN
dtdv
l
lm
l
lm
2
2
22
22
222
2
1
1
21
11
111
1
⎟⎟⎠
⎞⎜⎜⎝
⎛+=⎟
⎠⎞
⎜⎝⎛+⎟
⎠⎞
⎜⎝⎛==
⎟⎟⎠
⎞⎜⎜⎝
⎛+=⎟
⎠⎞
⎜⎝⎛+⎟
⎠⎞
⎜⎝⎛==
R
R
φφλ
φφλ
(3.14)
1- 7
The last term for v1 can be represented by an inductor, ⎟⎟⎠
⎞⎜⎜⎝
⎛=
1
21
1l
lNLR
and the last term for v2 can be
represented by ⎟⎟⎠
⎞⎜⎜⎝
⎛=
2
22
2l
lNLR
. These can be incorporated into the equivalent circuit as shown in
Figure 12.
+
e 1e2
+
N1 N2
i'1
L
i2
m
i1
i1,ex
Ideal transformer
Ll1 Ll2
Figure 12. Equivalent circuit for a transformer with finite core permeability and
leakage flux.
Core Losses The dissipation of heat in the core due to the flux through it can be represented by a resistor in the
equivalent circuit. It is placed in parallel to Lm since the power loss in the core is proportional to the flux through the core squared, and thus is proportional to . 2
1e
Resistance of the Coils The resistance of copper is approximately 16.8×10-9 (Ω·m), however with hundreds to thousands
of turns for the primary and secondary coils it can lead to an appreciable resistance. Losses to these resistances are proportional to the current through the coils squared.
Adding the core losses and resistance of the coils to the equivalent circuit we obtain our
completed model of the transformer as show in Figure 13.
+
e 1e2
+
N1 N2i'1
L
i2
m
i1
i1,ex
Ideal transformer
Ll1 Ll2
Rc
R1 R2
V2
++
V1
Figure 13. Equivalent circuit for a non-ideal transformer.
1- 8
Example Consider a transformer with a turns ratio of 4000/120, with a primary coil resistance R1 = 1.6 (Ω),
a secondary coil resistance of R2 = 1.44 (mΩ), leakage flux that corresponds to Ll1 = 21 (mH), and Ll2 = 19 (µH), and a realistic core characterized by Rc = 160 (kΩ) and Lm = 450 (H). The low voltage side of the transformer is at 60 (Hz), and V2 = 120 (V), and the power there is P2 = 20 (kW) at pf = 0.85 lagging. Calculate the voltage at the high voltage side and the efficiency of the transformer.
7.169450602 =⋅== πω mm LX (kΩ)
92.7021.060211 =⋅== πω lLX (Ω)
16.71019602 622 =×⋅== −πω lLX (mΩ)
( )pfVPI⋅
=2
22ˆ /-31.8º = 196.08/-31.8º (A)
( ) 045.198.120ˆˆˆ 22222 jjXRIVE +=++= (V)
83.347.4032ˆˆ 22
11 jE
NNE +=⎟⎟
⎠
⎞⎜⎜⎝
⎛= = 4032.8/0.49º (V)
1017.3001.5ˆˆ 21
21 jI
NNI −=⎟⎟
⎠
⎞⎜⎜⎝
⎛=′ (A)
0236.00254.011ˆˆ 1,1 jjXR
EImc
ex −=⎟⎟⎠
⎞⎜⎜⎝
⎛+= (A)
=−=′+= 125.30255.5ˆˆˆ 1,11 jIII ex 5.918/-31.87º (A)
( ) 2.695.4065ˆˆˆ 11111 jjXRIEV +=++= (V) = 4066/0.9º (V)
The power losses in the windings and core are:
39.550144.008.196 22
222 =⋅== RIPR (W)
04.566.1918.5 21
211 =⋅== RIPR (W)
64.101101608.4032
3
221 =
×==
cc R
EP (W)
08.21321 =++= cRRloss PPPP (W)
( ) 9895.008.2131020
10203
3
2
2 =+×
×=
+==
lossin
outPP
PPPη
3.4 Losses and Ratings The impedances shown in Figure 13 can be reflected into either the primary side, or the secondary
side of the transformer as shown in Figure 14.
1- 9
N1 N2
Lm
Ll1
Rc
R1 Ll2N1( )N2
2
R2N1( )N2
2
+
V1 V2
+
V'2
+
N1 N2
V'1
+
V1
+
Lm
Ll2
Rc
R2Ll1
N2( )N1
2
R1N2( )N1
2
V2
+
N2( )N1
2 N2( )N1
2
Figure 14. Reflection of the impedances to the primary or secondary side of the transformer. For a given frequency, the power losses in the core (iron losses) increase with the voltage e1 (or
e2). If these losses exceed a particular limit, a hot spot in the transformer will reach a temperature that dramatically reduces the life of the insulation. Limits are therefore put on E1 and E2 which are the voltage limits for the transformer. The current limits for the transformer limit I1 and I2 to avoid excessive Joule heating due to winding resistances.
Typically a transformer is described by its rated voltages, E1N and E2N, that give both the limits
and turns ratio. The ratio of the rated currents N
NII2
1 , is the inverse of the ratio of the voltages when
the magnetizing current is neglected. Instead of listing these rated current limits, a transformer is described by its rated apparent power as:
NNNNN IEIES 2211 == (3.15)
When a transformer is operated at its rated conditions, that is at its maximum current and maximum voltage, the magnetizing current, I1,ex, typically does not exceed 1% of the current in the transformer. Its effect on the voltage drop on the leakage inductance and on the winding resistance is therefore negligible.
Under maximum (rated) current, the total voltage drops on the winding resistances, and leakage inductances typically do not exceed 6% of the rated voltage. Their effect on E1 and E2 is small and their effect on the magnetizing current can be neglected.
Because these effects are small, we can modify the equivalent circuits shown in Figure 14 to a slightly inaccurate, but much more useful on shown in Figure 15.
1- 10
N1 N2
Lm
Ll1
Rc
R1
Ll2N1( )N2
2R2
N1( )N2
2+
V1 V2
+
V'2
+
N1 N2
V'1
+
V1
+
Lm
Ll2
Rc
R2
Ll1N2( )N1
2R1
N2( )N1
2
V2
+
N2( )N1
2 N2( )N1
2
Figure 15. Slightly inaccurate, but highly simplified equivalent circuits for the transformer.
When we utilize this simplification and work with the reflected voltages, the transformer
equivalent circuits can be shown as:
cR mL
Req + R'2
+
V1 V'2
+
= R1
Leq + L'l2= Ll1i'1
i1
i1,ex
V'1
+
L'mR'c V2
+Req + R2= R'1
Leq + Ll2= L'l1
Figure 16. Working with V’2 or V’1 the above approximate circuits for the transformers can simplify the analysis.
1- 11
Example (repeat with simplified equivalent circuits) Consider a transformer with a turns ratio of 4000/120, with a primary coil resistance R1 = 1.6 (Ω),
a secondary coil resistance of R2 = 1.44 (mΩ), leakage flux that corresponds to Ll1 = 21 (mH), and Ll2 = 19 (µH), and a realistic core characterized by Rc = 160 (kΩ) and Lm = 450 (H). The low voltage side of the transformer is at 60 (Hz), and V2 = 120 (V), and the power there is P2 = 20 (kW) at pf = 0.85 lagging. Calculate the voltage at the high voltage side and the efficiency of the transformer.
Using the simplified equivalent circuits of Figure 16, we can first find Req, and Xeq
2.32
2
2
11 =⎟⎟
⎠
⎞⎜⎜⎝
⎛+= R
NNRReq (Ω)
876.15602 2
2
2
11 =⎟
⎟
⎠
⎞
⎜⎜
⎝
⎛⎟⎟⎠
⎞⎜⎜⎝
⎛+⋅= lleq L
NNLX π (Ω)
then
( )pfVPI⋅
=2
22ˆ /-31.8º = 196.08/-31.8º (A)
22 ˆˆ VE =
102.35ˆˆ1
221 j
NNII −=⎟⎟
⎠
⎞⎜⎜⎝
⎛=′ = 5.884/-31.8º (A)
4000ˆˆ2
121 =⎟⎟
⎠
⎞⎜⎜⎝
⎛=
NNEE (V)
0235.00258.011ˆˆ 1,1 jjXR
EImc
ex −=⎟⎟⎠
⎞⎜⎜⎝
⎛+= (A)
=−=′+= 125.30259.5ˆˆˆ 1,11 jIII ex 5.918/-31.87º (A)
( ) 79.694065ˆˆˆ 111 jjXRIEV eqeq +=++= (V) = 4066/0.98º (V)
The power losses in the windings and core are:
79.1102.3884.5 221 =⋅=′= eqR RIP
eq(W)
28.10310160
40653
221 =
×==
cc R
VP (W)
07.214Re =+= cqloss PPP (W)
( ) 9894.007.2141020
10203
3
2
2 =+×
×=
+==
lossin
outPP
PPPη
These values agree well with the previous analysis using the more accurate model.
1- 12
3.4 Per Unit System A simplification in the analysis can come from expressing each of the values as a fraction of a
defined base system of quantities. When this is done, simple problems can be made more complex, however more complex problems can be made easier to solve. As an example consider a simple problem of a load impedance of 10 + j5 (Ω) that has a voltage of 100 (V) connected to it. Calculate the power at the load.
The traditional solution is found as:
94.8510
100ˆˆ =+
==jZ
VIL
LL /-26.57º (A)
and the power is
( ) 800º57.26cos94.8100 =⋅⋅=⋅= pfIVP LLL (W) Using the per unit system to find the solution: 1. First define a consistent system of values for the base. If we choose Vb = 50 (V), Ib = 10 (A),
then Zb = Vb/Ib = 5 (Ω), and Pb = Vb·Ib = 500 (W), Qb = 500 (VAr), and Sb = 500 (VA). 2. Convert all units to pu: VL,pu = VL/Vb = 2pu, ZL,pu = (10 + j5)/5 = 2 + j1 (pu) 3. Solve the problem using the pu system:
894.012
2ˆˆ
,
,, =
+==
jZV
IpuL
puLpuL /-26.57º (pu)
( ) 6.1º57.26cos894.02,,, =⋅⋅=⋅= pfIVP puLpuLpuL (pu) 4. Convert back to the SI system
94.810894.0, =⋅=⋅= bpuLL III (A)
8005006.1, =⋅=⋅= bpuLL PPP (W) For the more complicated example of a transformer we choose the bases for each side of the
transformer such that
2
1
2
1NN
VV
b
b = (3.16)
1
2
2
1NN
II
b
b = (3.17)
This leads to the two base apparent powers being equal
bbbbbb SIVIVS 222111 === (3.18)
The bases are often chosen to be the rated quantities of the transformer on each side. This is
convenient since most of the time transformers operate at rated voltage (making the pu voltage unity), and the currents and power are seldom above rated (seldom above 1 pu).
1- 13The base impedances are related by:
b
bb I
VZ1
11 = (3.19)
1b
b
b
bb I
VNN
IVZ 1
2
1
2
2
22 ⎟⎟
⎠
⎞⎜⎜⎝
⎛== (3.20)
or
bb ZNNZ 1
2
1
22 ⎟⎟
⎠
⎞⎜⎜⎝
⎛= (3.21)
To move impedances from one side of the transformer to the other, they get multiplied or divided
by the square of the turns ratio,2
1
2⎟⎟⎠
⎞⎜⎜⎝
⎛NN , but so does the base impedance, hence the pu value of an
impedance stays the same regardless of which side of the transformer it is on. Through our choice of the bases in (3.16) and (3.17), we also see that for an ideal transformer
pupu
pupu
II
EE
,2,1
,2,1
=
=
Thus when using the per unit system, an ideal transformer has voltages and currents on one side that are identical to the voltages and currents on the other side, and the ideal transformer can be eliminated.
Example (repeat and solved using pu system) Consider a 30 (kVA) rated transformer with a turns ratio of 4000/120, with a primary coil
resistance R1 = 1.6 (Ω), a secondary coil resistance of R2 = 1.44 (mΩ), leakage flux that corresponds to Ll1 = 21 (mH), and Ll2 = 19 (µH), and a realistic core characterized by Rc = 160 (kΩ) and Lm = 450 (H). The low voltage side of the transformer is at 60 (Hz), and V2 = 120 (V), and the power there is P2 = 20 (kW) at pf = 0.85 lagging. Calculate the voltage at the high voltage side and the efficiency of the transformer.
1. First calculate the impedances of the equivalent circuit.
40001 =bV (V) 301 =bS (kVA)
5.71041030
3
31 =
×
×=bI (A)
5331
21
1 ==b
bb S
VZ (Ω)
1202 =bV (V)
3012 == bb SS (kVA)
2502
22 ==
b
bb V
SI (A)
1- 14
48.02
22 ==
b
bb I
VZ (Ω)
2. Convert everything to per unit.
003.01
1,1 ==
bpu Z
RR (pu)
003.02
2,2 ==
bpu Z
RR (pu)
3001
, ==b
cpuc Z
RR (pu)
0149.0602
1
1,1 =
⋅⋅=
b
lpul Z
LX π (pu)
0149.0602
2
2,2 =
⋅⋅=
b
lpul Z
LX π (pu)
318602
1, =
⋅⋅=
b
mpum Z
LX π (pu)
12
2,2 ==
bpu V
VV (pu)
6667.02
2,2 ==
bpu S
PP (pu)
3. Solve in the per unit system.
)(ˆ
,2
,2,2 pfV
PI
pu
pupu ⋅
= /arccos(pf) = 0.666 – j0.413(pu)
( ) 0.017361.0163ˆˆˆ ,2,2,1 jjXRIVV eqeqpupupu +=++= (pu)
0031.00034.0ˆˆ
ˆ,
,1
,
,1, j
jXV
RV
Ipum
pu
puc
pupum −=+= (pu)
4163.06701.0ˆˆˆ ,2,,1 jIII pupumpu −=+= (pu)
00369.0,2,2 == pueqpuR RIP
eq(pu)
00344.0,
2,1
, ==puc
pupuc R
VP (pu)
( ) 9894.00034.00037.06667.0
6667.0)( ,,2
,2 =++
=+
=pulosspu
puPP
Pη
1- 15
4. Convert back to SI units. The efficiency is unitless and thus stays the same. Converting gives
1V
( ) =+=+=⋅= 46.698.40650.017361.01634000ˆˆ ,111 jjVVV pub 4065.8/0.979º (V)
3.5 Testing Transformers Purchased transformers often give information on the frequency, winding ratio, power, and
voltage ratings, but not the impedances. These impedances are important in calculating the voltage regulation, efficiency, etc. Use of open circuit and short circuit tests we will determine Req, Leq, Rc, and Lm.
Open Circuit Test Leaving one side of the transformer open circuited, while the other has the rated voltage
Vin-oc = 1(pu) applied to it, we measure the current and power. The current that flows into the transformer is mostly determined by the impedances Xm and Rc, and it is much lower than the rated current for the transformer. It is often the case that the rated voltage for the low voltage side (low tension side) of the transform since for the above example it would be easier applying 120 (V) instead of 4000 (V). Since the units we use indicate if we are using the per unit system, the following calculations will drop the subscript pu. Using the following equivalent circuit with the primary open circuited, and V2 = Voc = 120 (V) applied to the secondary.
V'1
+
L'mR'c V2
+Req + R2= R'1
Leq + Ll2= L'l1
Figure 17. Open circuit testing of a 4000/120 rated transformer. With 120 (V) applied to the low voltage side V2 = 120 (V), the primary voltage for this open circuit test is 4000 (V).
For this open circuit test, the following can be compared to the measured values of current and
power as:
cc
ococ RR
VP 12== (pu) (3.22)
m
oc
c
ococ jX
VR
VI +=ˆ (3.23)
1- 16
22111mc
ocXR
I += (pu) (3.24)
Then from the open circuit test with measurements of the current and the power, we determine Rc and Xm.
Short Circuit Test For the above open circuit test, the low voltage side of the transformer was chosen since it is
easier to apply the lower voltage during that test. Similarly, the rated current is lower on the high voltage side of the transformer. With the short circuit test, the rated current is commonly applied to the high voltage side of the transformer (since with a short circuited secondary, the applied voltage required to reach the rated current is relatively low). With the rated current applied to the high voltage side, we measure the voltage, Vsc which is V1 for this example, and the power, Psc.
(pu) (3.25) eqeqscsc RRIP ⋅== 12
( )eqeqscsc jXRIV += ˆˆ (3.26)
221 eqeqsc XRV += (pu) (3.27)
cR mL
Req + R'2
+
V1 V'2
+
= R1
Leq + L'l2= Ll1i'1
i1
i1,ex
= 0
Figure 18. Short circuit testing of a 4000/120, 30 (kVA) rated transformer. This corresponds to the rated current on the high voltage side of 7.5 (A).
Then from the short circuit test with measurements of the current and the power, we determine and 21 RRReq ′+= ( )21602 llm LLX ′+⋅= π .
Example A 60Hz transformer is rated 30 (kVA), 4000(V)/120(V). The open circuit test, with the high
voltage side open, gives Poc = 100 (W), Ioc = 1.1455 (A). The short circuit test, measured with the low voltage side shorted, gives Psc = 180 (W), Vsc = 129.79 (V). Determine the equivalent circuit for this transformer by using the per unit system.
1- 17
1. Bases: 40001 =bV (V)
301 =bS (kVA)
5.71041030
3
31 =
×
×=bI (A)
5331
21
1 ==b
bb S
VZ (Ω)
1202 =bV (V)
3012 == bb SS (kVA)
250120
1030 3
2
12 =
×==
b
bb V
SI (A)
48.0250120
2
22 ===
b
bb I
VZ (Ω)
2. Convert to (pu):
006.01030
1803, =
×=puscP (pu)
0324.010479.129
3, =×
=puscV (pu)
00333.01030
1003, =
×=puocP (pu)
00458.02501455.1
, ==puocI (pu)
3. Calculate the components of the equivalent circuit dropping the (pu) subscripts
eqscsc RIP 2= or 006.012 =⋅== scsc
sceq P
IPR (pu)
221ˆˆ eqeqeqeqscscsc XRjXRIVV +⋅=+⋅== or 0318.022 =−= eqsceq RVX (pu)
c
ococ R
VP2
= or 30012===
ococ
occ PP
VR (pu)
1- 18
2211ˆˆˆmcm
oc
c
ocococ
XRjXV
RVII +=+== or 318
11
22
=−
=
coc
m
RI
X (pu)
With this knowledge, we can address the more common example of: A 60Hz transformer is rated 30 (kVA), 4000(V)/120(V). The short circuit impedance is
0.0324 (pu) and the open circuit current is 0.0046 (pu). The rated core losses are 100 (W) and the rated winding losses are 180 (W). Calculate the efficiency and the necessary primary voltage when the load at the secondary is at rated voltage, 20 (kW), and at 0.8 pf lagging.
Using (pu) system:
0324.0=scZ (pu)
006.01030
1801 32 =
×=⋅== eqeqscsc RRIP (pu)
017.022 =−= eqsceq RZX (pu)
coc R
P 1= or 300
1030100
11
3
=⎟⎠
⎞⎜⎝
⎛
×
==oc
c PR (pu)
2211
mcoc
XRI += or 318
11
22
=−
=
coc
m
RI
X (pu)
Now we have the equivalent circuit components of the transformer and can work on the full
circuit which includes the load that has a power of 20 (kW) or:
6667.010301020
3
32 =
×
×=P (pu)
but the power at the load is
pfIVP ⋅= 222 or 8.016667.0 2 ⋅⋅= I thus (pu) 8333.02 =I
as a phasor: 8333.0ˆ2 =I /-36.87º = 0.6667 – j0.5 (pu)
( ) =+=++= 008334.00125.1ˆˆˆ 221 jjXRIVV eqeq 1.0125/0.472º (pu) V1 = 1.0125 (pu)
0062.022 =⋅= eqR RIP
eq(pu)
1- 19
034.02==
c
cc R
VP (pu)
986.02
2 =++
=cR PPP
P
eq
η
converting V1 to SI units gives
405040000125.11,11 =⋅=⋅= bpu VVV (V)
3.6 Three-Phase Transformers If we consider three-phase transformers as consisting of three identical one-phase transformers,
then we have an accurate representation as far as equivalent circuits and two-port models are concerned, but it does not give insight into the magnetic circuit of the three-phase transformers.
The primaries and the secondaries of the one-phase transformers can be connected either in Δ or in Y configurations. In either case, the rated power of the three-phase transformer is three times that of the one-phase transformers.
For the Δ connection: φ1VV ll =− (3.28)
φ13IIl = (3.29) For the Y connection:
φ13VV ll =− (3.30)
φ1IIl = (3.31)
Connections to the three-phase transformer that are Y connected in the primary are shown in Figure 19.
1- 20
I1
I1
I1
I2
V1 V2
V1 V2
I2
V1 V2
I2
I1
I1
I1
I2
V1 V2
V1 V2
I2
V1 V2
I2
Figure 19. Y – Y and Y – Δ connections of three-phase transformers.
Connections to the three-phase transformer that are Δ connected in the primary are shown in Figure 20.
I1
I1
I1
I2
V1 V2
V1 V2
I2
V1 V2
I2
I1
I1
I1
I2
V1 V2
V1 V2
I2
V1 V2
I2
Figure 20. Δ – Y and Δ – Δ connections of three-phase transformers.
1- 21
3.7 Autotransformers An autotransformer is a transformer where the two windings (of turns N1 and N2) are not isolated
from each other, but are connected as shown in Figure 21.
I1
V1 I2
V2N2
N1
I1 I2
Figure 21. An autotransformer.
From this figure that the voltage ratio in an autotransformer is:
2
21
2
1N
NNVV +
= (3.32)
and the current ratio is:
2
21
1
2N
NNII +
= (3.33)
An interesting note on autotransformers is that the coil of turns N1 carries current I1, while the coil of turns N2 carries the (vectorial) sum of the two currents, . So if the voltage ratio were 1, no current would flow through the N
21 ˆˆ II −2 coil. This characteristic leads to a significant reduction in the size
of the autotransformer compared to a similarly rated transformer, especially if the primary and secondary voltages are of the same order of magnitude. These savings come at a serious disadvantage of the loss of isolation between the two sides.
1- 22
1- 23
Chapter Notes: • To understand the operation of transformers we have to use both the Gauss’s law for magnetism (or
Biot-Savart law) and Faraday’s law.
• Most transformers operate under or near rated voltage. The voltage drop in the winding resistance and leakage reactance are usually small.
• In both transformers and in rotating machines, the net mmf of all the currents must accordingly adjust itself to create the resultant flux required by this voltage balance.
• Leakage fluxes induce voltage in the windings that are accounted for in the equivalent circuit as leakage reactance (elements Ll1 and Ll2). The leakage-flux paths are dominated by paths through air, and are thus almost linearly proportional to the currents producing them. The leakage reactances therefore are often assumed to be constant (independent of the degree of saturation of the core material).
• The open- and short-circuit tests provide the parameters for the equivalent circuit of the transformer.
• Three-phase transformers can be considered to be made of three single-phase transformers for the purpose of this course. The main issue is then to calculate the ratings, voltages, and currents of each.
• Autotransformers are used mostly to vary the voltage a little. It is seldom that an autotransformer will have a voltage ratio greater than two.
ECE 320 Energy Conversion and Power Electronics
Dr. Tim Hogan
Chapter 4: Concepts of Electrical Machines: DC Motors (Textbook Sections 3.1-3.4, and 4.1-4.2)
Chapter Objectives
DC machines have faded from use due to their relatively high cost and increased maintenance requirements. Nevertheless, they remain good examples for electromechanical systems used for control. We’ll study DC machines here, at a conceptual level, for two reasons:
1. DC machines although complex in construction, can be useful in establishing the concepts of
emf and torque development, and are described by simple equations.
2. The magnetic fields in them, along with the voltage and torque equations can be used easily to develop the ideas of field orientation.
In doing so we will develop basic steady state equations, again starting from fundamentals of the electromagnetic field. We are going to see the same equations in ‘Brushless DC’ motors, when we discuss synchronous AC machines.
4.1 Geometry, Fields, Voltages, and Currents The geometry shown in Figure 1 describes an outer iron frame (stator), through which (i.e. its
center part) a uniform magnetic flux, , is established. The flux could be established by a current in a coil or by a permanent magnet for example.
φ
In the center part of the frame there is a solid iron cylinder (called rotor), free to rotate around its axis. A coil of one turn is wound diametrically around the cylinder, parallel to its axis, and as the stator and its coil rotate, the flux through the coil changes. Figure 2 shows consecutive locations of the rotor and we can see that the flux through the coil changes both in value and direction. The top graph of Figure 3 shows how the flux linkages of the coil through the coil would change, if the rotor were to rotate at a constant angular velocity, ω.
( )tωφλ cosˆ= (4.1)
Figure 1. Geometry of an elementary DC motor.
1- 1
1 2 3 4 5 Figure 2. Flux through the stator coil of the simple dc motor shown in Figure 1.
Since the flux linking the coil changes with time, than a voltage will be induced in this coil, vcoil, as shown below:
( tdtdvcoil ωφ )λ sinˆ−== (V) (4.2)
-1.5
-1
-0.5
0
0.5
1
1.5
0 100 200 300 400 500 600 700
λ coil
Angle (º)
1
2
3
4
5
-1.5
-1
-0.5
0
0.5
1
1.5
0 100 200 300 400 500 600 700
v dcoi
l
Angle (º)
1
2
3
4
5
Figure 3. Flux and voltage in a coil of the motor in Figure 2 with coil positions 1-5.
1- 2
The points marked on the cosine and sine waveforms of Figure 3 correspond to the positions of the rotor as shown in Figure 2.
Torque on the coil follows the Lorentz Force Law: which gives the force, F, in newtons on a charge, q, in coloumbs that is exposed to a vectoral electric field, , in volts per meter and magnetic flux density, B, as
E
( )BvF ×+= Eq (4.3)
where v is the velocity of a charged particle q. The first term gives the force on the electron that moves it through the wire or the voltage applied to the wire. The second term gives the force on the charge due to the magnetic field in which the wire is placed. With many electrons contributing to the electrical current, I, (as a vector) in the wire, the force on the wire is
BIF ×= (4.4) The force on the full coil is then twice this value to account for each of the wires shown in Figure 2.
To maintain this torque in a direction that continues the rotary motion of the rotor, the ends of the rotor coil are connected to metal electrode ring segments called a commutator as shown in Figure 4. These ring segments are attached to the rotor so as to rotate with it, and are electrically contacted using two stationary brushes typically made of carbon and copper. The brushes are spring loaded and pushed against the commutator. As the rotor spins, the brushes make contact with the opposite segments of the commutator and they switch between segments of the commutator just as the induced voltage goes through zero and switches sign.
B
ω
commutat
or
brush
brush
Figure 4. A coil of a DC motor and brushes that are pushed up against the rotating
commutator (wire of the coil is highlighted red and soldered to the commutator segments).
Because the current switches direction through the coil each time the brushes rotate to the
opposite segment of the commutator (and that this occurs as the voltage goes through zero), the 1- 3
voltage at the terminals of the brushes is a rectified version of the voltage induced in the coil as shown in Figure 5.
-1.5
-1
-0.5
0
0.5
1
1.5
v coil
Time (arb. units) -1.5
-1
-0.5
0
0.5
1
1.5
v term
inal
Time (arb. units)
Figure 5. Induced voltage in a coil and terminal voltage in an elementary DC machine. If a number of coils are placed on the rotor, as shown in Figure 6, with each coil connected to a
different segment of the commutator, then the total induced voltage to the coils, E, will be:
(4.5) ωφkE =
where k is proportional to the number of coils.
Figure 6. Multiple coils on the rotor of a DC machine.
We stated in equation (2.39) that ieTPP mechmechelec ⋅=⋅⇒= ω thus with the multiple coil DC
machine: ω⋅=⋅ TiE (4.6) (4.7) ωωφ ⋅=⋅ Tik ˆ
(4.8) ikT φ=
1- 4
If the DC machine is connected to a load or a source as in Figure 7, then the induced voltage and terminal voltage will be related by: wdgg RiEV −=terminals for a generator (4.9)
wdgmRiEV +=terminals for a motor (4.10)
Load
or
Source
Rwdg
E
gimi
Figure 7. Circuit with a DC machine with the current reference directions defined as
indicated for DC machine used as a generator, ig, or as a motor, im. Example 4.1.1
A DC motor, when connected to a 100 (V) source and with no load connected to the motor runs at 1200 (rpm). Its stator resistance is 2 (Ω). What should be the torque and current if it is fed from a 220 (V) supply and its speed is 1500 (rpm)? Assume the field is constant.
With no load connected to the motor, we will assume the torque is zero (assuming no friction in bearings). With the torque equal to zero, the current is zero since: KiikT == φ . This means for this operation:
ωφω KkEV ===
and with ( ) (rad/s) 66.125)s( 60
(min) 1revolution
(rad) 2rpm 1200 =⋅⋅=πω , then
s)(V 796.0
66.125(V) 100
⋅=
⋅=
K
K
Then at 1500 (rpm)
( ) (rad/s) 08.157)s( 60
(min) 1revolution
(rad) 2rpm 1500 =⋅⋅=πωo
(V) 125== ωKE
1- 5
For a motor:
(N·m) 81.37
(A) 5.47
(V) 125(V) 220
==
=
+=
+=
KIT
I
IR
IREV
wdg
wdg
4.2 Energy Considerations Conservation of energy governs that the total energy supplied to an electromechanical system
equals the energy output. Some of the energy output could be mechanical, some could be lost as heat, and some could be stored. This is summarized in equation (3.10) of your textbook as:
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛+
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛+
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛=
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
heat intoconvertedEnergy
field magneticin the stored
energyin Increase
outputenergyMechanical
sourceselectric from
inputEnergy
In a lossless system, the last term is zero. When the losses are negligible, a differential change in the electrical energy input corresponds to the sum of a differential mechanical energy output and a differential change in the energy stored in the magnetic field, or
fldmechelec dWdWdW += (4.11)
Mechanical energy is force times distance, and the time rate of change of electrical energy is power or and (dteidW =elec 4.11) can be written as:
fldfld dWdxFdtei += (4.12)
where Ffld is the force produced by the magnetic field. From equation (3.1) in the handout notes, we
saw dtde λ
= . Using the linear inductor assumption of (2.19) which was ii
NL λφ== , then (4.12)
becomes: dxFdidW fldfld −= λ (4.13)
The energy in the field is given as the energy stored in an inductor
L
LiW2
2fld 2
121 λ
== (4.14)
The current and force produced by the magnetic field can then be found from (4.13) as:
( )x
xWiλλ
∂∂
=,fld (4.15)
( ) ( )dx
xdLix
xWF2
, 2fld
fld =∂
∂−=
λ
λ (4.16)
If the magnetic force results in a torque as in a rotating mechanical terminal, the mechanical energy from the field is then replaced with torque and angular displacement, θdTdxF fldfld ⇒ , as:
1- 6
θλ dTdidW fldfld −= (4.17) thus leading to
( )θθ
ddLiT
2
2fld = (4.18)
In a rotary machine with multiple windings, the inductance in equation (4.18) would contain self and mutual inductances of all coils involved. Chapter Notes: • The field of a DC motor can be created either by a DC current or a permanent magnet.
• The two fields, the one coming from the stator and the one coming from the moving rotor, are both stationary (despite rotation) and they are perpendicular to each other.
• If the directions of current in the stator and in the rotor reverse together, torque will remain in the same direction. Hence if the same current flows in both windings, it could be AC and the motor will not reverse.
1- 7
ECE 320 Energy Conversion and Power Electronics
Dr. Tim Hogan
Chapter 5: Three Phase Windings (Textbook Sections 4.3-4.7)
Chapter Objectives
Flux linkage plays a crucial role in the operation of both DC and AC machines. In this chapter, the geometry and the operation of windings in AC machines is discussed. The flux varies in time, and can also vary in position, or be stationary. To understand how these machines operate, the concept of space vectors (or space phasors) is introduced.
5.1 Introduction Electric machines often have defined an armature winding which is the winding that is power
producing, and a field winding that generates the magnetic field. Either could be on the stator or rotor depending on the specific motor or generator; however it is more common with AC machines such as synchronous or induction machines that the armature winding is on the stator (the stationary portion of the motor). Synchronous machines have field windings on the rotor that is excited by direct current delivered to the rotor windings by slip rings or collector rings by carbon brushes. The field winding produces the north and south poles, thus the image shown in Figure 1 is for a two-pole, single phase (one armature winding) synchronous generator. The magnetic axis for the armature winding is perpendicular to the area defined by the armature winding (armature winding is the perimeter of this area).
Figure 1. Figures 4.4 and 4.5 from your textbook showing a simple two-pole, single phase synchronous generator, the spatial distribution of the magnetic field relative to the magnetic axis of the armature winding, and the time dependent induced voltage in the armature winding [1].
1- 1
In a three phase device, the armature has three coils each with a magnetic axis that is rotated spatially by 120º as shown in Figure 2.
Figure 2. A three-phase, two-pole synchronous generator as shown in Figure 4.12(a) in your textbook [1].
More poles for the field winding and more armature windings are also possible as shown in
Figure 3. Such a configuration can deliver three phase power by interconnecting the armature windings in a Y connection configuration as an example.
Figure 3. A four-pole synchronous generator with multiple armature windings that can be wired together in a three-phase Y connection is shown (from Figure 4.12(b) and 4.12(c) in your textbook [1]).
5.2 Control of the Magnetomotive Force Distribution As generators, the synchronous machines typically use the stator windings as a source of
electrical power. As motors, the stator (or armature) windings are commonly supplied electrical power to generate a spatially varying field (we will consider the time variation of these fields later). Insight to the distribution (in space) of the field from the armature windings can be seen by considering a single N-turn coil on the stator that spans 180 electrical degrees (in 180º it has gone
1- 2
from a +F to a –F). Such a coil is known as a full pitch coil. The mmf distribution for such a full pitch coil is depicted in Figure 4.19 from your textbook as shown below in Figure 4.
Figure 4. A full pitch coil on the stator. Windings on the rotor are left off for clarity. This is Figure 4.19 in your textbook. (a) Schematic view of flux produced by a concentrated, full-pitch winding in a machine with a uniform air gap. (b) The air-gap mmf produced by current in this winding [1]. Assuming the reluctances of the stator and rotor negligible compared to the gaps, then the full
mmf of N·i would drop across two gaps as the flux traverses a full loop. This gives rise to a +(N·i/2) maximum, and a –(N·i/2) minimum for the mmf as one spatially maps F for the stator winding as a function of the angle relative to its magnetic axis. Plotting F as a function of this angle, θa, shows abrupt changes at the wires of the coil. The Fourier transform of this square wave gives a fundamental sinusoidal distribution of Fag1 as shown in Figure 4.
Harmonics exist since it is non-sinusoidal. In an attempt to make the spatial distribution more sinusoidal in form, multiple coils can be placed within specific groves of the stator as shown for one of the phases in a three-phase winding in Figure 5. Here there are equal numbers of wires in each of the groves, and equal current in each of the wires.
1- 3
1- 4
Figure 5. The mmf of one phase of a distributed two-pole, three phase winding with full-pitch
coils. This is image 4.20 from your textbook [1].
By adjusting the number of turns in each slot a more sinusoidal mmf distribution can be also be
obtained - as shown for windings of a rotor in the configuration shown in Figure 6.
Figure 6. The air-gap mmf of a distributed winding on the rotor of a round-rotor generator. This
is image 4.21 from your textbook [1]. Some simple methods for controlling the shape of the magnetomotive force for a set of coils have
been outlined. Now we turn our attention to how we use this information for analyzing the electric machine through the concept of space vectors.
5.3 Current Space Vectors Consider three identical windings placed in a space of uniform permeability as shown in Figure 7.
Each winding carries a time dependent current, i1(t), i2(t), and i3(t), We also require that
( ) ( ) ( ) 0321 ≡++ tititi (5.1)
Each current produces a flux in the direction of the coil axis, and if we assume the magnetic medium to be linear, we can find the total flux by adding the individual fluxes. This means that we could produce the same flux by having only one coil, identical to the three, but placed in the direction of the total flux, carrying an appropriate current. If the coils in Figure 7 carry the following currents i1 = 5 (A), i2 = -8 (A), i3 = 3 (A), then the vectoral sum of these currents, oriented in the same direction as the corresponding coil can be shown in Figure 8.
I1
I2
I3
φ1
φ2
φ3
Figure 7. Three phase windings spatially oriented 120º apart.
The direction of the resultant coil and current it should carry, we create three vectors, each in the direction of one coil, and equal in amplitude to the current of the coil it represents. If, for example, the coils are placed at angles of 0º, 120º, 240º. Then their vectoral sum will be:
i=i /φ = (5.2) 2403
12021
jj eieii ++ This represents the vector of a current oriented in space, and thus we define i as a space vector. If
i1, i2, and i3 are functions of time, so will be the amplitude and angle of i. If we consider a horizontal axis in space as a real axis, and a vertical axis as an imaginary axis, then we can find the real (id = Rei) and imaginary (iq = Imi) components of the space vector. With this representation, we can determine the three currents from the space vector as:
1- 5
( ) ( )
( ) ( ) ( ) ( ) γ
γ
23
2
1
323232
j
j
etti
etti
tti
−
−
=
=
=
i
i
i
Re
Re
Re
(5.3)
rad3
2120 πγ == (5.4)
I
II = 5 (A)1
I = - 8 (A)2
I = 3 (A)3
I1
I3
I2
Figure 8. (a) Currents in the three windings of Figure 7. (b) Resultant space
vector and (c) corresponding winding position and current of an equivalent single coil.
Consider if the three coils in Figure 7 were to represent the three stator windings of an AC
machine as shown in Figure 9, with currents for each phase also shown in Figure 9.
(a) (b)
Figure 9. Simplified two pole 3-phase stator winding and the instantaneous currents for each phase. These are Figures 4.29 and 4.30 from your textbook [1].
1- 6
Then for:
( ) ( )( ) ( )( ) ( )º120cos
º120coscos
+=−=
=
tItitItitIti
emc
emb
ema
ωωω
(5.5)
the space vector is:
( ) ( ) ( ) ( )[ ]
( ) ( ) ( ) ( ) ( ) ( ) ([ ]
( ) ( ) ( )
)
( ) ( ⎥⎦
⎤⎢⎣
⎡+−−++−−−=
+++−++=++−+=
º120cosº120cos23º120cos5.0º120cos5.0cos
º120cosº240sinº240cosº120cosº120sinº120coscosº120cosº120coscos º240º120
ttjtttI
tjtjtItetetIt
eeeeem
eeem
ej
ej
em
ωωωωω
ωωωωωωi
)
(5.6)
Using: ( ) ( ) ( ) ( ) ( )yxyxyx sinsincoscoscos ∓=±
( ) ( ) ( ) ( ) ( ) ( )
( ) ( ) ( ) ( )
( ) ( )[ ]tjtI
ttttjI
tttttIt
eem
eeeem
eeeeem
ωω
ωωωω
ωωωωω
sincos23
sin23cos5.0sin
23cos5.0
23
sin23cos5.05.0sin
23cos5.05.0cos
−=
⎥⎦
⎤⎢⎣
⎡+++−+
⎥⎥⎦
⎤
⎢⎢⎣
⎡
⎭⎬⎫
⎩⎨⎧
−−−⎭⎬⎫
⎩⎨⎧
+−−=i
(5.7)
This is a space vector that rotates in space as a function time at an angular frequency of ωe. Graphically, this can be seen from summation of ia, ib, and ic in Figure 9 (b) at different points in time (ωet = 0, π/3, and 2π/3), the resultant space vector at each point in time is represented by the corresponding magnetomotive force, F, associated with that space vector as shown in Figure 10.
Figure 10. The production of a rotating magnetic field by means of three-phase currents.
1- 7
Section 5.2 in these notes described ways of forming magnetomotive force spatial distributions
that were more sinusoidal in profile. Representing such windings as sinusoidally concentrated windings in the stator as depicted in Figure 11, then the density of turns of a the coil would vary sinusoidally in space as function of the angle θ.
Stator Winding
Stator
Rotor
Air Gap
Figure 11. The production of a rotating magnetic field by means of three-phase currents. Thus the number of turns, dNs, covering an angle dθ at a position θ over dθ is a sinusoidal
function of the angle θ. The turns density, ns1(θ ) is then:
( ) θθθ
sinˆ1 sss nn
ddN
== (5.8)
For the total number of turns, Ns, in the winding:
( )∫=π
θθ0
1 dnN ss (5.9)
This leads to
( ) θθ sin21s
sNn = (5.10)
With i1 current flowing through this winding, the flux density in the air gap between the rotor and the stator can be found for the integration path shown in Figure 12. The path of integration is defined by the angle q and we can notice that because of symmetry of the flux density at the two air gap segments in the path is the same. 1- 8
Stator Winding
Rotor
Stator
Air Gap
Current in positive direction
Current in negative direction
θ
Figure 12. Integration path to calculate flux density in the air gap. Assuming the permeability of the stator and rotor is infinite, then Hiron = 0 and:
( ) ( )
( )
( ) θμθ
θμ
θ
φφθπθ
θ
cos2
cos2
2
011
10
1
111
gNiB
NigB
dnigH
sg
sg
sg
=
=
= ∫+
(5.11)
For a given current, i1, in the coil the flux density in the air gap varies sinusoidally with angle, but
as shown in Figure 13 it reaches a maximum when the angle θ is zero. For the same machine and conditions as in Figure 13, Figure 14 shows the plot of turns density, ns(θ) and flux density, BBg(θ) in cartesian coordinates with θ as the horizontal axis. For this one coil, if the current, i1, were to vary sinusoidally in time, then the flux density would also change in time. The direction of the space vector would be maintained however the amplitude would change in time. The nodes of the flux density where it is equal to zero will remain at 90º and 270º, while the extrema of the flux will be maintained at 0º and 180º.
1- 9
Figure 13. Sketch of the flux (red) in the air gap.
-1
-0.5
0
0.5
1
0 60 120 180 240 300 360
n s(θ)
θ
-1
-0.5
0
0.5
1
0 60 120 180 240 300 360
Bg(θ
)
θ Figure 14. Turns density on the stator and the air gap flux density vs. θ.
Consider now an additional winding, identical to the first, but rotated with respect to it by 120º.
For a current in this winding we will get a similar air gap flux density as before, but with nodes at 210º = 90º + 120º and at 30º = 270º + 120º. If a current, i2, is flowing in this winding, then the air gap flux density due to it will follow a form similar to equation (5.11) but rotated by 120º = (2π/3).
( ) ⎟⎠⎞
⎜⎝⎛ −=
32cos
20
22πθμθ
gNiB s
g (5.12)
Similarly, a third winding, rotated 240º relative to the first winding and carrying current i3, will produce an air gap flux density of:
( ) ⎟⎠⎞
⎜⎝⎛ −=
34cos
20
33πθμθ
gNiB s
g (5.13)
1- 10
Combining these three flux densities, we obtain a sinusoidally distributed air gap flux density, that could equivalently come from a winding placed at an angle φ and carrying current i as:
( ) ( ) ( ) ( ) ( )φθμθθθθ +=++= cos2
0321 g
NiBBBB sgggg (5.14)
This means that as the currents change, the flux could be due instead to only one sinusoidally
distributed winding with the same number of turns. The location, φ (t), and current, i(t), of this winding can be determined from the current space vector:
( ) ( )tit =i /φ = (5.15) ( ) ( ) ( ) º240
3º120
21jj etietiti ++
5.3.1 Balanced, Symmetric Three-phase Currents If the currents i1, i2, i3 form a balanced three-phase system of frequency fs = ωs/2π, then we can
write:
( ) [ ]( ) ( )[ ]( ) ( )[ ]3434
13
323212
11
22
34cos2
22
32cos2
22cos2
πωπω
πωπω
ωω
πφω
πφω
φω
−−−
−−−
−
+=⎟⎠⎞
⎜⎝⎛ −+=
+=⎟⎠⎞
⎜⎝⎛ −+=
+=+=
tjs
tjss
tjs
tjss
tjs
tjss
ss
ss
ss
eetIi
eetIi
eetIi
II
II
II
(5.16)
where I is the phasor corresponding to the current in phase 1. The resultant space vector is:
( ) ( )122
23
22
23 φωω +== tjtj
s ss Ieet Ii (5.17)
The resultant flux density wave is then:
( ) ( )θφωμθ −+= 10 cos
22
23, t
gNItB s
s (5.18)
which shows a travelling wave, with a maximum value of 0
max 223
μsNIB = . This wave travels
around the stator at a constant speed ωs, as shown Figure 15.
1- 11
-1
-0.5
0
0.5
1
0 60 120 180 240 300 360
Bg(θ
)
Angle (θ)
t1
t2
t3
Figure 15. Air gap flux density profile vs. θ for three times t3>t2>t1.
5.4 Phasors and Space Vectors This is a good point to reflect on the differences between phasors and space vectors. A current
phasor, , describes one sinusoidally varying current of frequency ω, amplitude 0ˆ φjIeI = I2 , and initial phase φ0. The sinusoid can be reconstructed from the phasor as:
( ) [ ] ( ) ( )tjtjtj etIeeti ωωω φω III Re=+=+= −0
* cos222 (5.19)
Although rotation is implicit in the definition of the phasor, no rotation is described by it.
On the other hand, the definition of current space vector requires three currents that sum to zero. These currents are implicitly in windings that are symmetrically placed, but the currents are not necessarily sinusoidal. Generally the amplitude and angle of the space vector changes with time, but no specific pattern is a priori defined. We can reconstruct the three currents that constitute the space vector from equation (5.3). When these constituent currents form a balanced, symmetric system, of frequency ωs, then the resultant space vector is of constant amplitude, rotating at a constant speed. In that case, the relationship between the phasor of one current and the space vector is shown in equation (5.17).
5.5 Magnetizing Current, Flux, and Voltage To see how this rotating magnetic flux influences the windings we use Faraday’s law. From here
on we will use sinusoidal symmetric three-phase quantities. The three stationary windings are linked by a rotating flux as shown in Figure 16. When the
current is maximum in phase 1, the flux is as shown in Figure 16(a) and is linking all of the turns in phase 1. Later, the flux has rotated as show in Figure 16(b) then the flux linkages with coil 1 have decreased. When the flux has rotated 90º as in Figure 16(c), the flux linkages with the phase 1 windings are zero. 1- 12
(a) (b) (c)
Figure 16. Rotating flux and flux linkages. Sinusoidal windings containing many turns in the stator are represented by single wires to show current flow direction.
To calculate the flux linkages, l, we determine the flux through a differential number of turns at
an angle θ as shown in Figure 17.
θdθ
Figure 17. Flux linkage through a differential number of turns (such as one turn) for coil 1.
The flux through this differential section of coil is then:
( ) ( ) ( ) θθθθφθ
πθ
θ
πθdtBlrdAtBt gg ∫∫ −−
== ,,, (5.20)
1- 13
where l is the axial length of the coil (into the page), and r is the radius to the coil. The number of turns linked by this flux is ( ) ( ) θθθ dndn ss = , so the flux linkages for these few (or one) turns is: ( ) ( )θφθθλ ⋅= dnd s (5.21) To find the flux linkages, λ1, for all of coil 1, then we must integrate the flux linkages over all turns of coil 1 or:
( ) ( )∫∫ ⋅==π
θφθθλλ0
1 dnd s (5.22)
When these two integrals are taken, then we find:
( ) ( ) ( )1102
1 cos2cos238
φωφωπμλ +=+= tILtIglrNt sMs
s (5.23)
which means the flux linkages in coil 1 are in phase with the current in this coil and proportional to it. The flux linkages of the other two coils, 2 and 3, are identical to that of coil 1, but lagging in time by 120º and 240º respectively. With these three quantities we can create a flux-linkage space vector, λ as: (5.24) iλ M
jj Lee =++= º2403
º12021 λλλ
Since the flux linkages of each coil vary, and in our case sinusoidally, a voltage is induced in each of these coils. The induced voltage in each coil is 90º ahead of the current in it, bringing to mind the relationship of current and voltage of the inductor. Notice though, that it is not just the current in the winding that causes the flux linkages and the induced voltages, but rather the current in all three windings. Although this is the case, we sill call the constant LM the magnetizing inductance, and the induced voltages in each coil can be found as:
( )
( )
( ) ⎟⎠⎞
⎜⎝⎛ −++==
⎟⎠⎞
⎜⎝⎛ −++==
⎟⎠⎞
⎜⎝⎛ ++==
34
2cos2
32
2cos2
2cos2
33
3
22
2
11
1
ππφωωλ
ππφωωλ
πφωωλ
tIdt
dte
tIdt
dte
tIdt
dte
(5.25)
and the voltage space vector, e can be defined as:
(5.26) ie Mjj Ljeeeee ω=++= º240
3º120
21
The flux linkage space vector is aligned with the current space vector, while the voltage space vector, e, is ahead of both by 90º. This agrees with the fact that the individual phase voltages lead the current by 90º, as shown in
1- 14
ωt
i
λe
Figure 18. Magnetizing current, flux-linkage, and induced voltage space vectors.
1 A. E. Fitzgerald, C. Kingsley, Jr., S. D. Umans, Electric Machinery, 6th edition, McGraw-Hill, New York, 2003.
1- 15
ECE 320 Energy Conversion and Power Electronics
Dr. Tim Hogan
Chapter 6: Induction Machines (Textbook Sections 6.1-6.5)
Chapter Objectives
The popularity of induction machines has helped to label them as the ‘workhorse of industry’. They are relatively easy to fabricate, rugged and reliable, and find their way into most applications. For variable speed applications, inexpensive power electronics can be used along with computer hardware and this has allowed induction machines to become more versatile. In particular, vector or field-oriented control allows induction motors to replace DC motors in many applications.
6.1 Description The stator of an induction machine is a typical three-phase one, as described in the previous
chapter. The rotor can be one of two major types – either (a) it is wound in a fashion similar to that of the stator with the terminals connected to slip rings on the shaft, as shown in Figure 1, or (b) it is made with shorted bars.
Shaft
SlipRings
Rotor
Figure 1. Wound rotor, slip rings, and connections. Figure 2 shows the rotor of such a machine, while the images in Figure 3 show the shorted bars
and the laminations. The bars in Figure 3 are formed by casting aluminum in the openings of the rotor laminations. In
this case the iron laminations were chemically removed.
1- 1
Figure 2. (a) Cutaway view of a three-phase induction motor with a wound rotor and slip rings connected to the three-phase rotor winding shown in Figure 6.1 in your textbook [1]. (b) Cutaway view of a three-phase squirrel-cage motor as shown in Figure 6.3 in your textbook [1].
(a) (b)
Figure 3. (a) The rotor of a small squirrel-cage motor. (b) The squirrel-cage structure after the rotor laminations have been chemically etched away as shown in Figure 6.3 in your textbook [1].
6.2 Concept of Operation As these rotor windings or bars rotate within the magnetic field created by the stator magnetizing
currents, voltages are induced in them. If the rotor were to stand still, then the induced voltages would be very similar to those induced in the stator windings. In the case of a squirrel cage rotor, the voltage induced in a bar will be slightly out of phase with the voltage in the next bar, since the flux linkages will change in it after a short delay. This is depicted in Figure 4.
If the rotor is moving at synchronous speed, together with the field, no voltage will be induced in the bars or the windings.
1- 2
Bg12
34
567
13
19
-1
-0.5
0
0.5
1
0 50 100 150 200 250 300 350
e(t)
ωt
bar 3bar 1
bar 2
bar 7
(a) (b)
Figure 4. (a) Rotor bars in the stator field and (b) voltages in the rotor bars.
Generally when the synchronous speed is ss fπω 2= , and the rotor speed ω 0, the frequency of the
induced voltages will be fr, where osrf ωωπ −=2 . Maxwell’s equation becomes here:
gBv ×=GG
E (6.1)
where EG
is the electric field and vG is the relative velocity of the rotor with respect to the field:
( )rv os ωω −= (6.2) Since a voltage is induced in the bars, and these are short-circuited, currents will flow in them. The current density will be: ( )θJ
G
( ) EGG
ρθ 1
=J (6.3)
where ρ is the resistivity of the bars.
These currents are out of phase in different bars, just like the induced voltages. To simplify the analysis we can consider the rotor as one winding carrying currents sinusoidally distributed in space. This will be clearly the case for a wound rotor. It will also be the case for uniformly distributed rotor bars, but now each bar, located at an angle θ will carry different current, as shown in Figure 5(a).
1- 3
Bg
Bg
(a) (b)
Figure 5. (a) Currents in rotor bars and (b) equivalent current sheet in the rotor.
( ) θωρ gosω BJ ⋅−=1 ( ) (6.4)
( ) ( ) θωρ
θ sinˆ1gos BωJ −= ( ) (6.5)
The bars can also be replaced with a conductive cylinder as shown in Figure 5(b) with a
distributed current. Slip, s, is defined as the ratio:
s
ossω
ωω −= (6.6)
Thus at starting, the speed is zero and s = 1, and at synchronous speed, os ωω = and s = 0. Above synchronous speed s < 0, and when the rotor rotates in a direction opposite of the magnetic field, then s > 1. Example 6.2.1
The rotor of a two-pole 3-phase induction machine rotates at 3300 (rpm), while the stator is fed by a three-phase system of voltages at 60 (Hz). What are the possible frequencies of the rotor voltages?
At 3300 (rpm):
6.3456023300 ==πωo (rad/s) while 377=sω (rad/s)
These two speeds can be in the opposite or the same direction such that:
6.345377 ±=−= osr ωωω (rad/s) = 722.58 (rad/s) or 31.43 (rad/s) or f = 115 (Hz) or f = 5 (Hz)r r
1- 4
To better understand the currents induced in the squirrel cage as caused by the magnetic flux density from the stator coils consider the simplified view of the squirrel cage in Figure 6. Adjacent bars of the squirrel cage form a coil since the ends of the cage are electrically shorted and the largest coils are formed by pairs of opposite side bars of the cage. The bars of the squirrel cage work collectively to respond to any changing magnetic field. If the squirrel cage is rotating at the same angular velocity as the stator magnetic flux density, then the loops of the squirrel cage experience a constant magnetic field, and no current flows through the bars of the cage. Under this condition, no magnetomotive force is generated by the squirrel cage and no torque on the rotor exists. The rotor then begins to slow down relative to the stator magnetic field. As is does so, the magnetic flux density for a given loop of the squirrel cage changes (some loops experiencing an increase in magnetic flux density, and some loops experiencing a decrease in magnetic flux density). The response to this change in magnetic field is a current that flows in the bars of the squirrel cage so as to oppose this change in magnetic field. Since the stator magnetic field is sinusoidally distributed in space around the rotor, and the most rapid change in this field is determined by its spatial derivative, then as the squirrel cage slips with respect to ωs, the field from the rotor caused by this slip is spatially oriented 90º with respect to the stator magnetic field (derivative of the cosine field distribution).
For example, if the stator magnetic flux density is oriented from left to right as was shown in Figure 5 and is rotating clockwise, then as a motor with s near zero (near synchronous speed s > 0) the rotor is also rotating clockwise, but not quite as fast as the stator magnetic flux density. As this slip occurs, the largest change in the field occurs from the top to the bottom of the squirrel cage or at the zero magnetic field points of the stator field where the spatial slope of the stator field is largest. As the slip occurs, the field through the squirrel cage, BBg in , begins to rotate in a clockwise fashion relative to the rotor. The largest change in field occurs in a direction from top to bottom of the rotor and the coils of the squirrel cage collectively generate a counter magnetomotive force so as to oppose the change in magnetic field. Thus current flows in the bars of the squirrel cage to generate a counter field directed from bottom to top, or into the page on the right side, and out of the page on the left as indicated in .
Figure 5
Figure 5
B
ω − ωs o
ωs
ωo Figure 6. Squirrel cage isometric view.
1- 5
6.3 Torque Development To calculate the torque on the rotor we use the following two equations:
BliF = and rFT ⋅= (6.7)
since the flux density is perpendicular to the current producing it. For the length of the conductor, l, we will use the depth of the rotor. The thickness, de, of the equivalent conducting sheet shown in Figure 5 is set equal to the total area of the bars of the squirrel cage.
rdnd
rddnA
e
e
8
24
2barsrotor
2barsrotor
⋅=
== ππ
(6.8)
where d is the diameter of a single bar of the squirrel cage.
For a small angle dθ centered at a given angle θ, we find the contribution to the total force and torque as:
( )( )
( )osge
ge
g
BldrdTT
rdFdT
BrdJddF
lBJdAdF
ωωρ
π
θ
πθ
θ−⎟
⎟⎠
⎞⎜⎜⎝
⎛==
=
=
⋅⋅=
∫=
=
222
0ˆ2
(6.9)
where and ( )θsinˆgg BB = ( ) ( ) θωρ
θ sinˆ1gos BωJ −= ( ) from equation (6.5) with ρ equal to the
resistivity of the bars of the squirrel cage. Using the relationship between flux density (or flux linkages), Λs, and the rotor voltage, Es, the
torque can be expressed as:
( osss
elN
dT ωωρπ
−Λ⎟⎟⎠
⎞⎜⎜⎝
⎛= 2
28 ) where
s
ss
Eω
=Λ (6.10)
where the stator voltage is related to the flux density as gsss lrBNe2πω= . Focusing on the variables
of this equation we see the torque is proportional to the frequency of the rotor currents, ( )osω ω− and to the square of the flux density. This is so since the torque comes from the interaction of the flux density, BBg, and the rotor currents, but the rotor currents are induced due to the flux, BgB
), and the
relative speed ( osω ω− . Equation (6.10) gives torque as a function of more accessible quantities of stator induced voltage, Es, and frequency ωs. This comes as a result of the simple and direct relationship between stator induced voltage, flux (or flux linkages), and frequency.
1- 6
6.4 Operation of the Induction Machine near Synchronous Speed We already determined that the voltages induced in the rotor bars are of slip frequency,
fr = (ωs - ωo)/2π. At rotor speeds near synchronous, fr is small. The rotor bars in a squirrel cage machine possess resistance and leakage inductance, but at very low frequencies (near synchronous speed) we can neglect this leakage inductance. The rotor currents are therefore limited near synchronous speed by the rotor resistance only.
The induced rotor-bar voltages and currents form space vectors. These are perpendicular to the stator magnetizing current and in phase with the space vectors of the voltages induced in the stator as shown in Figure 7.
Bg
ir
is,m
Figure 7. Stator magnetizing current, airgap flux, and rotor currents.
These rotor currents, ir, produce additional airgap flux, which is 90º out of phase of the
magnetizing flux. The stator voltage, es, is applied externally and is proportional to and 90º out of phase with the airgap flux. Thus additional currents, isr, will flow in the stator windings in order to cancel the flux due to the rotor currents. These additional currents are shown in Figure 8(a), and the resulting stator current and space vectors are depicted in Figure 8(b).
Bg
ir
is,m
is,r
Bg ir
is
i s
(a) (b)
Figure 8. Rotor and stator currents in an induction motor (a) Rotor and stator current components (b) total stator current and space vector.
1- 7
is,mBg
ir
i sis,r
es
Figure 9. Space vectors of the stator and rotor current and induced voltages.
A few items to note from the above analysis: • isr is 90º ahead of the stator magnetizing current, is,m. This means that it corresponds to
currents in the windings i1r, i2r, i3r, leading by 90º the magnetizing currents i1m, i2m, i3m. • The amplitude of the magnetizing component of the stator current is proportional to the stator
frequency, fs, and induced voltage. On the other hand, the amplitude of this component of the stator currents, isr, is proportional to the current in the rotor, ir, which is proportional to the flux and slip speed, ωr = ωs – ωo, or proportional to the developed torque.
• The stator current of one phase, is1, can be split into two components. One in phase with the voltage, isr1, and one 90º behind it, ism1. The first reflects the rotor current, while the second depends on the voltage and frequency. In an equivalent circuit, this means that isr1 will flow through a resistor, and ism1 will flow through an inductor.
• Since isr1 is equal to the rotor current (through a factor), it will be inversely proportional to ωs – ωr, or better stated as proportional to ωs/(ωs – ωr). The equivalent circuit for the stator shown in Figure 10 reflects these considerations.
es
+
Xm RR
is,m
is,r
ωsω − ωs o
is,1
Figure 10. Equivalent circuit of one stator phase.
If the inductor motor is supplied with a three-phase, balanced sinusoidal voltage, then it is
expected that the rotor will develop a torque according to equation (6.10). The relationship between speed, ωo, and torque near synchronous speed is shown in Figure 11. This curve is accurate as long as the speed does not vary more than ±5% around the rated synchronous speed, ωs.
1- 8
ωsωo
T
Figure 11. Torque-speed characteristics near synchronous speed.
As the speed exceeds synchronous, the torque produced by the machine is in the opposite direction to the speed (i.e. the machine operates as a generator), developing a torque opposite to the rotation (or counter torque) and transferring power from the shaft to the electrical system.
We already know the relationship of the magnetizing current, Ism, to the induced voltage, Esm, through our analysis of the three-phase windings. Now we relate the currents, ir and isr to the same induced voltage.
The current density, , on the rotor conducting sheet is related to the air gap flux density as: JG
( ) gos BωJGG
⋅−= ωρ1 (6.11)
This current density corresponds to a space vector ir that is opposite to the isr in the stator. This current space vector will correspond to the same current density:
rd
NiJ ssr1
= (6.12)
while the stator voltage es is also related to the flux density BBg. The amplitude of es is:
gsss lrBNe2πω= (6.13)
Substitution of the relating phasors instead of space vectors in equation (6.11) we obtain:
sros
sRs IRE
ωωω−
= (6.14)
The torque is then three times (three phases) the power Es·Isr over the stator speed or
s
gr
R
s
s
os
Rs
s PRR
ETω
ωω
ωωω
331322
=Λ
=−
= (6.15)
1- 9
where Λ = (Es/ωs). Here Pg is the power transferred to the resistance os
sRR
ωωω−
, through the
airgap. Of this power a portion is converted to mechanical power represented by losses on the
resistance os
oRR
ωωω−
, and the remaining is losses in the rotor resistance, represented by the losses
on resistance RR. In Figure 12 this split in the equivalent circuit is shown; note the resistance
os
oRR
ωωω−
can be negative, indicating that mechanical power is absorbed in the induction machine.
es
+
Xm
RR
is,m
is,r
ωoω − ωs o
is,1
RR
RRωs
ω − ωs o
Figure 12. Equivalent circuit of one stator phase separating the loss and torque rotor
components. 6.4.1 Example
A 2-pole three-phase induction motor is connected in a Y configuration and is fed from a 60 (Hz), 208 (V) (l-l) system. Its equivalent one-phase rotor resistance is RR = 0.1125 (Ω). At what speed and slip is the developed torque 28 (N·m)?
(rad/s) 6.366
0275.0377
364.10(rad/s) 364.10
1125.01
377120328
)V( 120 with 13
s
r
2
2
=−=
===
=
⎟⎠⎞
⎜⎝⎛=
=⎟⎟⎠
⎞⎜⎜⎝
⎛=
rso
r
r
srRs
s
s
VR
VT
ωωωωω
ω
ω
ωω
1- 10
6.5 Leakage Inductances and Their Effects In the previous discussion we assumed that all the flux crosses the airgap and links both the stator
and the rotor windings. In addition to this flux there are flux components which link only the stator or the rotor windings and are proportional to the currents there, producing voltages in these windings 90º ahead of the stator and rotor currents and proportional to the amplitude of these currents and their frequency.
This is a simple model for the stator windings, since the equivalent circuit we are using is for the stator, and we can model the effects of this flux with an inductor added to the circuit. The rotor leakage flux can be modeled in the rotor circuit with an inductance L1s, as well, but corresponding to
frequency of πωω
2os
rf−
= , the frequency of the rotor currents. Its effects on the stator can be
modeled with an inductance L1r at frequency fs, as shown in the complete 1-phase equivalent circuit in Figure 13.
Es
+
Xm
RR
Is,mIs,r
ωoω − ωs o
Is,1 RR
RRωs
ω − ωs o
Xls XlrRs
Vs
+
Figure 13. Complete equivalent circuit of one stator phase including leakage flux contributions. Here is the phasor of the voltage induced into the rotor windings from the airgap flux, while
is the phasor of the applied 1-phase stator voltage. The torque equation (sE
sV 6.15) still holds here, but give us slightly different results. We can develop torque-speed curves, by selecting speeds, solving the equivalent circuit, calculating power Pg, and using equation (6.15) for the torque. Figure 14 shows the stator current per phase, torque for the three phase induction machine, and the power factor as a function of ωo for ωo given as a percentage of ωs and ranging from negative values to greater than ωs.
1- 11
-300
-200
-100
0
100
200
0
0.2
0.4
0.6
0.8
1
-50 0 50 100 150
I s1 (A
), an
d T
(N·m
)
pf
ωo (percent of synchronous speed)
Is1
(A)
T (N·m)
pf
Figure 14. Stator current (single phase), torque and power factor of an induction machine vs. speed.
6.6 Operating Characteristics Figure 14 shows the developed torque, current, and power factor of an induction machine over a
speed range from below zero (slip > 1 or braking region) to above synchronous (slip < 0 or generator region). There are three regions of interest:
1. For speeds in the range 0 ≤ ωo ≤ ωs the torque is of the same sign as the speed, and the machine operates as a motor. There are a few interesting points on this curve and on the corresponding current and power factor curves.
2. For speeds in the range ωo ≤ 0, torque and speed have opposite signs, and the machine is in breaking mode. Notice the current is very high, resulting in high winding losses.
3. For speeds in the range ωo ≥ ωs the speed and torque are of opposite signs and the machine is in the generating mode.
These regions are identified on an extended plot of torque in Figure 15. If we consider the motor
operation region 0 ≤ ωo ≤ ωs the operating point is often designed to be near, or at, the point where the power factor is maximized. It is for this point that the motor characteristics are given on the nameplate, rated speed, current, power factor, and torque. When designing an application it is this point that we have to consider primarily. Will the torque suffice? Will the efficiency and power factor be acceptable?
Starting the motor is also of interest (where slip s = 1) where the torque is not necessarily high, but the current often is. When selecting a motor for an application, we have to make sure this starting torque is adequate to overcome the load torque which may also include a static component. In
1- 12
addition, the starting current is often 3-5 times the rated current of the machine. If the developed torque at starting is not adequately higher than the load starting torque, their difference, called the accelerating torque, will be small and it may take too long to reach the operating point. This means that the current will remain high for a long time, and fuses or circuit breakers could have their limits exceeded.
0
-100 -50 0 50 100 150 200 250
Torq
ue
ωo (percent of synchronous speed)
Mot
orG
ener
ator
Breakingregion
Motorregion
Generator region
Figure 15. Stator torque vs. speed identifying three regions of operation. A third point of interest is the maximum torque, Tmax, corresponding to speed ωTmax. We can find
it by analytically calculating torque as a function of slip, and equating the derivative to zero. This point is interesting, since speeds higher than this generally correspond to stable operating conditions, while lower speeds generally correspond to unstable operating conditions. To study this point of operation we use the Thevenin equivalent circuit for the left side of the circuit as seen looking into terminal A-B in Figure 17.
Es
+
Xm
RR
Is,mIs,r
ωoω − ωs o
Is,1 RR
RRωs
ω − ωs o
Xls XlrRs
Vs
+A
B
Thevenin
Figure 16. Equivalent circuit for induction machine indicating terminals for Thevenin equivalent circuit.
1- 13
Es
+
RR
Is,r
ωoω − ωs o
RR
RRωs
ω − ωs o
XTH XlrRTH
VTH
+A
B
Figure 17. Equivalent circuit for induction machine with the stator circuit replaced with a Thevenin equivalent circuit.
We find
( ) ( ) ( )( )mlss
msmlsmlssTHTH XXjR
XjRXXjXjXRjXR++++−
=+=+ (6.16)
and
( )mlss
msTH XXjR
jXVV++
= ˆˆ (6.17)
Then from the circuit in Figure 17, we can find as: rsI ,ˆ
( )lrTH
RTH
THrs
XXjs
RR
VI++⎟
⎠⎞
⎜⎝⎛ +
=ˆˆ , (6.18)
where s is in the per unit system, that is ( ) soss ωωω /−= . From equation (6.15)
( )2
2
22,
333
lrTHR
TH
RTH
s
Rrs
sg
s XXs
RR
sRV
sRIPT
++⎟⎠⎞
⎜⎝⎛ +
⎟⎠⎞
⎜⎝⎛
=⎟⎠⎞
⎜⎝⎛==
ωωω (6.19)
This torque reaches a maximum as a slip, smaxT , that can be found by taking the derivative of
equation (6.19) and setting it equal to zero. When this is done we find:
( ) ( )
( ) ( )22max
22
maxor
lrTHTH
RT
lrTHTHT
R
XXR
Rs
XXRs
R
++=
++=
(6.20)
1- 14
This gives the maximum torque of:
( )22
2max 2
3
lrTHTHTH
TH
s XXRR
VT+++
=ω
(6.21)
If the stator resistance is negligible, then RTH ≈ 0 and
ss
ssTT
T
T
max
max
max2
+≈ (6.22)
If we neglect both the stator resistance and the magnetizing inductance, we can find simple equations for Tmax and ωTmax. To do so, we must assume operation near synchronous speed, where the value of
os
sRR
ωωω−
is much larger than ωsLlr.
lslr
RsT LL
R+
−≈ ωω max (6.23)
( ) ⎟⎟⎠
⎞⎜⎜⎝
⎛+⎟⎟
⎠
⎞⎜⎜⎝
⎛=−⎟⎟
⎠
⎞⎜⎜⎝
⎛≈
lrlss
sTs
Rs
sLL
VR
VT 1231
23
2
max
2
max ωωω
ω (6.24)
The slip frequency at this torque, maxTsr ωωω −= , for a constant flux s
ss
Eω
=Λ is independent
of frequency, and is proportional to the resistance RR. We already know that RR is proportional to the rotor resistance, so if the rotor resistance is increased, the torque-speed characteristic is shifted to the left, as shown in Figure 18. If we have convenient ways to increase the rotor resistance, we can increase the starting torque, while decreasing the starting current. Increasing the rotor resistance can be easily accomplished in a wound-rotor induction machine. For the squirrel cage motor, more complex structures such as double or deep rotor bars can be used to increase the rotor resistance.
In the formula developed we notice the maximum torque is a function of the flux. This means that we can change the frequency of the stator voltage, but as long as the voltage amplitude changes so that the flux stays the same, the maximum torque will also stay the same as shown in Figure 19. This is called Constant Volts per Hertz Operation and is a first approach to controlling the speed of the motor through its supply.
Near synchronous speed the effect of the rotor leakage inductance can be neglected, and led to the torque-speed equation (6.15) repeated below:
s
gr
R
s
s
os
Rs
s PRR
ETω
ωω
ωωω
331322
=Λ
=−
=
Figure 20 shows both the exact and the approximate torque-speed characteristics. It is important to notice that the torque calculated from the approximate equation is grossly incorrect away from synchronous speed.
1- 15
0 20 40 60 80 100
I s
ωo (percent of synchronous speed)
RR = r
1RR = 2·r
1
0 20 40 60 80 100
Torq
ueω
o (percent of synchronous speed)
RR = r
1RR = 2·r
1
Figure 18. Effect of changing the rotor resistance on the torque-speed and
current-speed characteristics of an induction motor.
-50 0 50 100 150
I s
ωo (rad/s)
60 (Hz)
45 (Hz)20 (Hz)
-50 0 50 100 150
Torq
ue
ωo (rad/s)
60 (Hz)
45 (Hz)
20 (Hz)
Figure 19. Effect on the torque-speed characteristic of changing frequency
while keeping the flux constant.
1- 16
40 50 60 70 80 90 100 110
Torq
ueω
o (percent of synchronous speed)
LinearApprox.
Figure 20. Exact and approximate torque-speed characteristics.
6.7 Starting Characteristics of Induction Motors From the above analysis we see that a particular challenge with induction motors is the high
current and low torque during starting. A simple way to decrease the starting current is to decrease the stator terminal voltage during
startup. From equation (6.15), we see the torque is proportional to the stator voltage squared, while the current is directly proportional to the stator voltage. If a transformer is used to decrease the stator voltage, then both the developed torque and the line current will decease by the square of the turns ratio of the transformer.
A commonly used method is to change the connection configuration during starting of a motor. For example a motor designed to operate with the stator windings connected in Δ are changed to a Y connection during starting. The voltage ratio of these configurations is:
Δ= ,, 31
sYs VV (6.25)
then
Δ= ,, 31
sYs II (6.26)
Δ= ,, 31
sYs TT (6.27)
+
Vl-l
I line Y
Is,Y+
Vs,Y
I line Δ
Is,Δ +
Vs,Δ
+
Vl-l
Figure 21. Y – Delta starting of an induction motor.
1- 17
In the Δ connection, phline II 3= , leading to:
Δ= ,, 31
lineYline II (6.28)
Once the machine has approached the desired operating point, we can reconfigure the connection
to Δ, and provide better efficiency. This decrease in current is often adequate to allow a motor to start low load starting torque.
Using a variable frequency and voltage supply we can comfortably increase the starting torque, as shown in Figure 19, while decreasing the starting current.
6.8 Multiple Poles If we consider that an induction machine will operate close to synchronous speed 3000 (rpm) for
50 (Hz) and 3600 (rpm) for 60 (Hz) we may find that the speed of the machine is too high for an application. If we recall the pictures of the flux in AC machines we have shown before, we notice the flux has a relatively long path to travel in the stator. This makes the stator heavy and lossy.
In machines with more than one pair of poles, the sinusoidal distribution for the windings covers a smaller angle. For example, in a 4 pole machine each side of a sinusoidally distributed winding of one phase covers only 90º instead of 180º (as was the case for a 2 pole machine).
Figure 22 shows at one instant the equivalent windings resulting from rotor windings of a 2-pole, 4-pole, and 6-pole machines.
Figure 22. Multipole induction machines 2-pole, 4-pole, 6-pole rotor windings
(stator windings not shown). The effects of a large number of poles on the operation of the machine are not difficult to predict.
If the machine has p poles, or p/2 pole pairs, then in one period of the voltage, the flux will travel 2ωs/p (rad/s). This leads to the rotor speed corresponding to synchronous of ωsm:
ssm pωω 2
= (6.29)
We now introduce the actual mechanical speed of the rotor as ωm, while we keep the term ωo as
the rotor speed of a two pole motor. We generally measure ωm in (rad/s), while we measure ωo in (electrical rad/s). We retain the same definition for slip based on the electrical speed ωo.
1- 18
om pωω 2
= (6.30)
s
ms
s
osp
sω
ωω
ωωω 2
−=
−= (6.31)
This shows that for a 4-pole machine, supplied by a 60 (Hz) source, and operating close to rated
conditions, the speed will be near 1800 (rpm), while for a 6-pole machine, the speed will be near 1200 (rpm). While increasing the number of poles results in a decrease of the synchronous and operating speeds of the machine, it also results in an increase of the developed torque of the machine by the same ratio. Hence, the corrected torque formula will be:
o
m
s
g PpPpTωω 2
32
3 == (6.32)
Similarly, the torque near the synchronous speed is:
s
g
R
rs
s
os
Rs
s PpR
pR
EpTω
ωω
ωωω 2
32
312
322
=Λ
=−
= (6.33)
while the previously developed formulas for maximum torque will become:
( )22
2max 2
12
3lrTHTHTH
TH
s XXRR
VpT+++
=ω
(6.34)
and
( )lrlss
sTs
Rs
sLL
VpR
VpT+⎟⎟
⎠
⎞⎜⎜⎝
⎛=−⎟⎟
⎠
⎞⎜⎜⎝
⎛≈
122
3122
32
max
2
max ωωω
ω (6.35)
Example 6.8.1
A 3-phase, 2-pole induction motor is rated at 190 (V), 60 (Hz), is connected in the Y configuration, and has RR = 6.6 (Ω), Rs = 3.1 (Ω), XM = 190 (Ω), Xlr = 10 (Ω), Xls = 3 (Ω). Calculate the motor starting torque, starting current and starting power factor under rated voltage. What will be the current and power factor if no load is connected to the shaft?
1. At starting, s = 1
[ ] ( ) 06.7106.619031.3
3190
ˆ =+++
=jjj
Is /-54.5º (A)
7.6190106.6
190ˆˆ =++
=jj
jII sr /-52.6º (A)
( ) 36.222
3776.67.63
23
2===
pPT
s
gω
(N·m)
1- 19
2. Under no load, the speed is synchronous and s = 0
57.019031.3
3190
ˆ =++
=jj
Is /-89.1º (A)
57.0=sI (A)
016.0=pf lagging
Example 6.8.2
A 3-phase, 2-pole induction motor is rated at 190 (V), 60 (Hz), is connected in the Y configuration, and has RR = 6.6 (Ω), Rs = 3.1 (Ω), XM = 190 (Ω), Xlr = 10 (Ω), Xls = 3 (Ω). It is operating from a variable speed – variable frequency source at a speed of 1910 (rpm), under a constant (V/f ) policy and the developed torque is 0.8 (N·m). What is the voltage and frequency of the source? (Hint: First calculate the slip).
The ratio (Vs/ωs) stays at 377
1903
1
.
)(V 110or (V) 4.6437711066.220(Hz) 35
(rad/s) 66.22065.2001.2202
(rad/s) 65.206.6
1377110318.0
132
-
2
2
llss
rms
rr
rRs
s
Vf
p
RVpT
==⇒=
=+=+=
=⇒⎟⎠⎞
⎜⎝⎛⋅=
⎟⎟⎠
⎞⎜⎜⎝
⎛=
ωωω
ωω
ωω
Example 6.8.3
A 3-phase, 4-pole induction machine is rated 230 (V), 60 (Hz). It is connected in the Y configuration, and has RR = 0.191 (Ω), LM = 35 (mH), Lls = 1.2 (mH). It is operated as a generator connected to a variable frequency/variable voltage source. Its speed is 2036 (rpm), with a counter-torque of 59 (N·m). What is the efficiency of this generator? (Hint: Here power in is mechanical, power out is electrical; also first calculate the slip).
Although we do not know the voltage or the frequency, we know their ratio is (132.8/377).
(rad/s) 14.15191.01
3778.1322359
132
2
2
−=
⎟⎠⎞
⎜⎝⎛⋅=−
⎟⎟⎠
⎞⎜⎜⎝
⎛=
r
r
rRs
sR
VpT
ω
ω
ωω
1- 20
Now the synchronous speed can be found by adding slip and rotor speeds as:
(V) 14460
1325.65(Hz) 5.65
(rad/s) 3.41114.152602036·2
2
=⋅=⇒=
=−=+=
ss
rms
Vf
p πωωω
We have to calculate the impedances of the equivalent circuit for the frequency of 65.5 (Hz):
)( 617.0)( 49.0
)( 4.143.4111035 3
Ω=
Ω=Ω=⋅×= −
lr
ls
m
XXX
then
)( 38.52 Ω−=+
r
mrR
p
Rω
ωω
[ ] [ ] 30617.038.5191.04.1449.02.0
144ˆ =+−++
=jjj
Is /-148º (A)
2.27ˆ =rI /-166.9º (A) Notice that with generation operation RR < 0. We can calculate now the losses, etc.:
919.0
(kW) 98.10
(W) 5402.0303
(W) 423191.02.273
(kW) 941.1138.52.273
,,
2,
2,
2
==
=−−=
=⋅⋅=
=⋅⋅=
=⋅⋅=
m
out
lossstatorlossrotormout
lossstator
lossrotor
m
PP
PPPPP
P
P
η
1 A. E. Fitzgerald, C. Kingsley, Jr., S. D. Umans, Electric Machinery, 6th edition, McGraw-Hill, New York, 2003.
1- 21
1- 1
ECE 320 Energy Conversion and Power Electronics
Dr. Tim Hogan
Chapter 7: Synchronous Machines and Drives (Textbook Chapter 5)
Chapter Objectives
For induction machines, as the rotor approaches synchronous speed, the frequency of the currents in the rotor decreases, as does the amplitude of these currents. The reason an induction motor produces no torque at synchronous speed is not that the currents are DC, but that their amplitude is zero.
It is possible to operate a three-phase machine at synchronous speed if DC is externally applied to the rotor and the rotor is rotating at synchronous speed. In this case torque will be developed only at this speed, i.e. if the rotor is rotated at speeds other than synchronous, the average torque will be zero.
Machines operating on this principle are called synchronous machines, and cover a great variety. As generators they can be quite large, rated a few hundred MVA, and almost all power generation is done using synchronous machines. Large synchronous motors are not very common, but can be an attractive alternative to induction machines for some applications. Small synchronous motors with permanent magnets in the rotor, rather than coils with DC, are rapidly replacing induction motors in automotive, industrial and residential applications since they lighter and more efficient.
7.1 Design and Principle of Operation The stator of a synchronous machine is of the type that we have already discussed with three
windings carrying a three-phase system of currents. As we saw in Chapter 5, this results in a stator magnetic field that spatially rotates around the stator at a constant angular speed, ωs. Unlike induction machines, synchronous machines have zero slip, and the rotor maintains the same angular speed, ωs, as the stator generated field. In a synchronous machine, the rotor windings carry DC current, or are composed of permanent magnets as discussed next.
7.1.1 Wound Rotor Carrying DC
In this case the rotor steel structure can be either cylindrical, that that in Figure 1(a), or salient like the one in Figure 1(b). In either case, the rotor winding carries DC, delivered through slip rings, or through a rectified voltage of an inside-out synchronous generator mounted on the same shaft. In this handout, discussion is limited to cylindrical rotors.
7.1.2 Permanent Magnet Rotor Instead of supplying DC to the rotor, the rotor contains permanent magnets in configurations such
as those shown in Fig. 2. The effects of permanent magnet rotors include: • The rotor flux can no longer be controlled externally. It is defined by the magnets and the
geometry • The machine becomes simpler to construct, at least for small sizes.
Sinusoidallydistributedwinding
Concentratedwinding
(a) (b)
Figure 1. Wound rotor configurations of synchronous machines.
(a) Surface PM (SPM) synchronousmachine
(b) Surface inset PM (SIPM)synchronous machine
(c) Interior PM (IPM) synchronousmachine
(d) Interior PM with circumferentialorientation synchronous machine
S
N
S
N
NN
N
S
S
S N
N
N
SS
S
N
S
N
S
N
S
N
S
NS
N
S
NS
N
S
S
N
S
S
S
N
NN
N
N
N
N
S
SS
S N N S S
NN
NN
NN S
S
SS
SS
Figure 2. Possible magnet placements in PMAC motors.
1- 2
7.2 Equivalent Circuit The flux in the air gap can be considered to be due to two sources: the stator currents, and the
rotor currents or permanent magnet. Recall the stator currents produce a flux that rotates at synchronous speed, and that this flux could also be produced by one equivalent winding that is rotating at synchronous speed and carrying a current equal to the magnitude of the stator-current space vector.
The rotor is itself such a winding, a real one, sinusoidally distributed, carrying DC and rotating at synchronous speed. It produces an airgap flux, which could also be produced by an additional set of three phase stator currents, giving a space vector iF. The amplitude of this space vector would be:
fR
SF i
NN
=i (7.1)
where NS is the number of the stator turns of the one equivalent winding (when the three stator windings are represented by a single equivalent rotating winding as discussed above) and NR is the number of the turns in the rotor winding. Its angle φR would be the same as the angle of the rotor position:
0RsR t φωφ += (7.2)
The stator current space vector has amplitude:
ss I223
=i (7.3)
where Is is the rms current of one phase. The stator current space vector will have an instantaneous angle, φis, of
0issis t φωφ += (7.4)
The airgap flux is then produced by both these current space vectors (rotor and stator). This flux
induces in the stator windings a voltage, es. In quasi steady-state everything is sinusoidal and the voltage space vector corresponds to three phase voltages E1, E2, E3. In this case we can create an equivalent circuit for the stator as shown in Figure 3. Here IF is the stator AC current, that if it were to flow in the stator windings, it would have the same effects as the rotor current, if. In our analysis we can use as reference either the stator voltage, Vs, or the stator current, Is as shown in Figure 4. From the relationship of the space vectors in Figure 4, the angle θ is the power factor angle (the angle between Is and Vs). We call β, the angle between Vs and IF, and the power angle, δ, is between IF and Is.
I
+
Vs jX
s
M
IM
Es IF = IF β
Figure 3. Stator equivalent circuit for a synchronous machine.
1- 3
β
δ
α
θ
I M
Vs
Is
IF Is
Figure 4. Space vector diagram of a synchronous machine.
From Figure 3 and Figure 4 we can note the following two relationships: • The space vector of the voltages induced in the stator, es, is 90º ahead of the magnetizing
current space vector, IM. This can be understood by the fact that IM is what causes all the airgap flux that links the stator and induces es. For a given frequency, the amplitude of this voltage, es, is proportional to the current IM.
• A permanent magnet machine can be considered equivalent to that with a winding, carrying a direct current, if, that is constant (and cannot be controlled).
7.3 Operation of the Machine Connected to a Bus of Constant Voltage and Frequency This is usually the case for large synchronous generators or motors. We can consider any bus as
one of constant voltage, by making a few modifications to the equivalent circuit as shown in Figure 5.
1- 4
I
jX
s
M
IM
IF = IF β
+
Vs
jXS
Is
VF = jX IF β
+
Vs
jXS
I
j(X + X )
s
M
IM
IF = F β
+
Vs
+M
jXM
S j(X + X )M S
jXM I
Figure 5. Equivalent circuits that account for a non-zero bus impedance in the system.
Synchronous machines are very efficient, and most of the time we can neglect the stator resistance. All power then is converted to mechanical power and:
p
TIVP sss2cos3 ωθ == (7.5)
βcos3 FsIVP −= (7.6)
FsM III += (7.7)
Ms IV MjX= (7.8)
In this operation, Vs and ωs (and therefore speed) remain constant. The only input variables are
the torque, T, which affects output power, p
TP s2
out ω= , and the field current, if, which is
proportional to IF; the magnetizing current IM is constant, since it is tied to the voltage Vs. Assuming the machine is operated so the power to it varies while the frequency and field current
remain constant. Since this is a synchronous machine, the speed will not vary with the load. From equation (7.6) we can see the power, and therefore the torque, varies sinusoidally with the angle β. Remember that β is the angle between the axis of the rotor winding, and the stator voltage space vector. Since this voltage space is 90º ahead of the space vector of the magnetizing current (β – 90º)
1- 5
is the angle between the rotor axis and the magnetizing current space vector (same as the airgap flux). When there is no torque this angle is 0, i.e. the rotor rotates aligned with the flux, when external torque is applied to the rotor in the direction of rotation the rotor will accelerate. As it accelerates (with the flux rotating at constant speed) the flux falls behind the rotor, and negative torque is developed, making the rotor slow down and rotate again at synchronous speed, but now ahead of the flux.
Similarly, when load torque is applied to the rotor, the rotor decelerates; as it does so, the angle β decreases beyond -90º, i.e. the rotor falls behind the flux. Positive torque is developed that brings the rotor back to synchronous speed, but now rotating behind the stator flux.
In both cases when the load torque on a motor or the torque on a motor or the torque of the prime mover in a generator increases beyond a maximum, corresponding to cos β = ±1, the machine cannot develop adequate torque and it loses synchronization.
Let us now discuss the effect of varying the field current while keeping the power constant. From equation (7.6), when power and voltage are kept constant, the product βcosFI remains constant as well, but this product is the projection of IF on the horizontal axis. This means that as the field current changes while power stays the same, the tip of IF travels on a vertical line, as shown in Figure 7(a). Similarly equation (7.5) means that at the same time the tip of Is travels on another vertical line, also shown in Figure 7(a).
It is clear from Figure 7(a) that once the field current has exceeded a value specific to the power level, the power factor becomes leading and the machine produces reactive power. This is different from the operation of an induction machine, which always absorbs reactive power.
When the machine operates as a generator, the input power is negative. Figure 7(b) shows this operation for both leading and lagging load power factor. Here the angle between stator voltage and stator current defined in the direction shown in the equivalent circuit (Figure 5) is outside the range -90º < θ < 90º.
Torq
ue
β
− π − π/2
π/2
Motor Generator
0
Figure 6. Torque and angle β in a synchronous motor.
1- 6
I M
Vs
Is1
IF3
IF2
IF1
Is1
Is2
Is3
Is1
Is2
I s3
Vs
I M
IF1
IF2
IF3
(a) (b)
Figure 7. Equivalent circuits that account for a non-zero bus impedance in the system.
Example 7.3.1 A 3-phase Y-connected synchronous machine is fed from a 2300(V), 60(Hz) source. The ratio of the AC stator equivalent current to the rotor DC is IF/if = 1.8. The magnetizing inductance of the machine is 200(mH). The machine is operated as a motor and is absorbing 110 (kW) at 0.89 pf leading. Calculate the required field current and the load angle. Draw the corresponding phasor diagrams. Using Figure 8:
-131º
27.1ºVs
Is
IF
I M
Figure 8. Synchronous machine as a motor.
We find:
4.752.0602 =⋅= πMX (Ω)
1- 7
13283
2300==sV (V)
89.01328
310110 3
⋅
⎟⎟⎠
⎞⎜⎜⎝
⎛ ×
=sI / 27.1º = 31/ 27.1º (A)
using the stator voltage we can calculate IM and from it IF.
62.174.75
1328==MI / -90º (A)
42=−= sMF III / -131º (A)
4.238.1
== Ff
Ii (A)
Repeating this process for operation as a generator at 110 (kW), 0.82 pf leading gives Figure 9
-145º
IG
35º
Is
3.56°Vs
I M
IF
Figure 9. Synchronous machine as a generator.
7.33=GI / 35º (A) 7.33=⇒ sI / -145º (A)
66.27=−= sMF III / 3.56º (A)
37.158.1
== Ff
Ii (A)
What is the maximum power the above machine can produce (or absorb when operating as a
generator and at the field current just calculated? 1- 8
We know that absorbed power is:
βcos3 FsIVP −= for if = 15.37 (A) we have IF = 27.66 (A), and P generated is maximum for β = 0, thus:
2.11066.2713283 =⋅⋅=P (kW) If the terminal voltage remains at 2300 (V), 60 (Hz), what is the minimal field current required to
maintain operation as a motor with load 70 (kW)?
3107013283cos3 ×=⋅⋅=⋅= FFs IIVP β (W) or
57.17=FI (A)
7.4 Operation from a Source of Variable Frequency and Voltage This operation requires that our synchronous machine is supplied by an inverter. The operation
now is entirely different than before. We no longer have an infinite bus, but rather whatever stator voltage or current and frequency we desire. Moreover, with a space-vector controlled inverter, the phase of this voltage or current can be arbitrarily set at any instant, i.e. we can define the stator current or voltage space vector, and obtain it at will. The considerations for the motor operation are also different:
• There is no concern for absorbing or supplying reactive power. Instead, there is a limit on the
total stator current, determined by thermal considerations. • There is a limit to the maximum voltage the source can supply, which leads to modifications of
the machine mode of operation at high speeds.
Operation from a source of variable frequency and voltage is most common for permanent magnet machines, where the value of |IF| is constant.
In simple terms, when the machine is starting as a motor the frequency applied should be zero, but the voltage space vector should be of such angle with respect to the rotor that torque is developed as discussed in the previous section. As torque develops, the machine accelerates, and the applied stator currents have to create a rotating space vector leading the rotor flux. Voltage and frequency have to be increased, so that this torque is maintained. It is important therefore to monitor the position of the rotor in order to determine the location of the stator current or voltage space vector.
Two possible control techniques are implemented: either voltage control, where the stator voltage space vector is determined and applied, or current control, where the stator current space vector is applied.
For a fixed stator voltage and power (and torque) level, the stator losses are minimal when the stator voltage and current are in phase. Figure 10 shows this condition.
1- 9
β
θ IsVs
IM
IF
Figure 10. Operation of a synchronous PM drive at constant voltage and frequency.
Notice that as the power changes with the voltage constant two things happen:
1. The voltage space vector varies in amplitude and the magnetizing current changes with it. 2. The amplitude of IF stays constant, but its angle with respect to the voltage changes.
From the developed torque and speed we can calculate the frequency, the values of IM and Is, and
the angle between the stator voltage space vector and the rotor, since:
sMMsss
IILpIVpT2
323
==ω
(7.9)
(7.10) 222sMF III +=
where IF is a constant in permanent magnet machines. More common is the case when the stator voltage is not constant. Here we monitor the position
of the rotor and since the rotor flux and rotor space current are attached to it, we are actually monitoring the position of IF. To make matters simple we use this current rather than the stator voltage as reference, as shown in Figure 11.
Is
jIF
Vs
IM
IF
XMjIs XM
θ
γ
Figure 11. Operation of a synchronous PM drive below base speed.
Although the previous equations for power and torque are still true, they are not as useful. New
formulae can be created that have the stator current, Is, and magnetizing current, IM, as variables. We
1- 10
also make use of the angle γ, between IF and Is, since we can control it. Starting from what we already know:
[ ] ( )[ ]*FsF
*Fs IIIIV +== Mg jXP ReRe (7.11)
[ ] [ ]*Fs
*FF IIII MM jXjX ReRe += (7.12)
[ ] γsinIm FsMM IIXX == *FsII (7.13)
For a given torque, minimum losses require minimum value of the stator current. To minimize the value of Is with constant power and IF we choose γ = 90º to obtain:
[ ] FsMMg IIXXP == *FsIIIm (7.14)
[ ] FsMMs
g IILpLpPpT2
3Im2
32
3 === *FsII
ω (7.15)
which means that for constant power, the projection of the stator current on an axis perpendicular to IM is constant.
As the rotor speed increases, even if IM stays constant, the stator voltage Vs = ωsLMIM increases. At some base speed, ωsB, the required voltage exceeds the maximum that the power source can provide. To increase the speed beyond the base speed, we no longer keep γ = 90º. On the other hand, at that speed we know that the voltage has reached its upper limit, Vs = Vs,max, therefore the value of IM = Vs,max/XM is known. In this case equations (7.9) and (7.10) become:
θθω
cos2
3cos23
sMMsss
IILpIVpT == (7.16)
(7.17) θsin2222sMsMF IIIII ++=
Figure 12 shows such an operation with the variable having the subscript 1. Note that we
calculate the torque from power.
γsin3 FsM IIXP = (7.18)
[ ] γω
sin2
3Im2
32 FsMM
sIILpLppPT === *
FsII (7.19)
1- 11
Is1
jIF
Vs1
IM1
XMjIs1XM
IF
IM2
Vs2
Is2
jIs2XMγ
γθ1
21
2θ
Figure 12. Field weakening of a PM AC motor. The two diagrams are at the same frequency, but the second one has γ > 90º and a lower Vs.
Example 7.4.1 A permanent magnet, Y connected, three-phase, 2-pole motor has IF = 40 (A), and XM = 0.9 (Ω) at 100 (Hz). 1. If it is absorbing P = 1.5 (kW) at 100 (Hz) with minimum stator current Is, calculate this
current, the angle between Is and IF, the speed, the stator voltage (line-neutral) and the power factor.
The minimum current Is will exist when the angle between Is and IF, γ, is 90º. Under this condition:
FsM IIXP 3= or 89.13409.03
1500=
⋅⋅=sI (A)
89.1340 +=+= sFM III /90º = 42.34/19.15º (A) 12.38== Ms IV MsLjω /109.15º (V)
where 6281002 =⋅= πωs (rad/s) or 6000 (rpm)
The power factor is then calculated as:
( ) 946.0º90º14.109cos =−=pf lagging
2. It is desired to increase the motor speed to 6900 (rpm) while keeping the power the same, P = 1500 (W), but the supplied voltage has reached its upper limit of Vs = 38.12 (V). Now the motor absorbs the same power at the voltage calculated in the previous question, but at frequency of 115 (Hz). This can be accomplished by having the stator current no longer at a minimum value and g no longer at 90º. Calculate again the angle between Is and IF, the speed, and the power factor.
1- 12
( )βcos3 FsIVP −= or ( ) 32.04012.383
1500cos −=⋅⋅
=β
º14.109−=β and 12.38=sV /109.14º (V)
83.369.0
100115
109.15º12.38===
jjXV
M
s /MI /19.15º (A)
16.13=−= FMs III /113.18º (A) 57.7221152 =⋅= πωs (rad/s) or 6900 (rpm)
The power factor is then calculated as:
( ) 997.0º18.113º14.109cos =−=pf leading
Is
jIF
Vs
IM
IF
XMjIs XM
θ
γ
Case #1
Is
jIF
Vs
IM
IF
XMjIs XM
θ γ
Case #2
1- 13
Example 7.4.2 A permanent magnet machine has 4-poles, IF = 40 (A), and the value of the magnetizing reactance is XM = 2.9 (Ω) at 100 (Hz). The maximum stator current is Is,max = 50 (A) and the maximum voltage the inverter can provide is Vln,max = 200 (V). First calculate the maximum speed the machine can operate with minimal stator Joule losses.
6.41002
=⋅
=π
mm
XL (mH)
For minimal losses sin(γ) = 1, so:
( ) ( )[ ] ( )[ ] ( )[ ]( ) 551A40A50H106.4243sin
23 3 =×== −γFsm IILpT (N·m)
Then calculate the maximum speed the machine can reach for operation as above:
( )A64ˆˆ =+= Fsm III
( )
( )[ ] ( )[ ] 679H106.4A64
V20031max, =
×== −mm
ss IL
Vω (rad/s)
and the maximum mechanical speed under these conditions will be:
5.33921max,1max, == smech p
ωω (rad/s)
Next assume operation above this speed, ωmech,max1, for example at 400 (rad/s). Obviously this cannot be done with minimal stator Joule losses, so the field has to be weakened by increasing γ above 90º. Calculate the maximum torque that can be developed. We now have the values of Vs and of Is, since they correspond to the limits, Vs = 200 (V), Is = 50 (A). From the voltage and speed we can find
( )[ ] ( )V200rad/s4002
=⎟⎠⎞
⎜⎝⎛= mms ILpV
( )A35.54106.4800
2003 =
×⋅= −mI
From the triangle of Is, Im, IF (case 2 in the last example):
cos2222mFmFs IIIII −+= (/IFIm)
cos35.5440235.544050 222 ⋅⋅−+= (/IFIm) (/IFIm) = º8.61
b = 61.8 + 90 = 151.8º
1- 14
( ) ( )mN8.52cos2
3 ⋅=−= βFmIIpT
Note that the demagnetizing component of Is is
( ) ( )A 6.23sin −== βssd II which might be too high.
7.5 Controllers for Permanent Magnet AC Machines Figure 13 shows a typical controller for an AC Machine. It requires a DC power supply, usually a rectifier fed from an AC source, an inverter and a controller. Figure 14 shows the controller in slightly higher detail.
Rectifier DCLink Inverter
Motor
Controller
Variable voltage and frequency
Voltage or Current Command
3-phase AC
Speed or Torque Command
Figure 13. Generic Controller for a PMAC Machine.
PI Controller
Calculate Is
T* Calculate
*** , , scsbsa iii
sI
Inverter
PMAC Motor
Position of Is
Rotor position
Rotor speed
s1
γ
ω∗ +
_
+
+
ωο
Optical encoder
Figure 14. Field Oriented controller for a PMAC Machine. The calculations for Is are based on equation (7.19) and the calculation of are calculated from the space vector I*** , , scsbsa iii s from equations (5.3).
1- 15
7.6 Brushless DC Machines While it would be difficult to find the differences between a PM AC machine described above and a brushless DC machine by just looking at them, the concept of operation is quite different as is the analysis. The windings in the stator in a brushless DC machine are not sinusoidally distributed but instead they are concentrated, each occupying one third of the pole pitch. The flux density on the magnet surface and in the airgap is also not sinusoidally distributed over the magnet but almost uniform in the air gap.
As the stator currents interact with the flux coming from the magnet torque is developed. It should be clear that for the same direction of flux, currents in opposite directions result in opposite forces, and therefore in reduction of the total torque. This in turn makes it necessary that all the current in the stator above the rotor is in the same direction. To accomplish this, the following are needed:
• Sensors on the stator that sense the direction of the flux coming from the rotor, • A fast power supply that will provide currents to the appropriate stator windings as determined
by the flux direction • A way to control these currents, e.g. through Pulse Width Modulation • A controller with inputs of the desired speed, the flux direction in the stator, and the stator
currents; and outputs of the desired currents in the stator
Figs show the rotor positions, the stator currents and the switches of the supply inverter for two rotor positions.
b'
b
a
a'
c
c'
North
South
ba
c
Figure 15. Energizing the windings in a brushless DC motor.
Nor
th
Sout
h
b'
b
a
a'
c
c'
ba
c
Figure 16. Energizing the windings in a brushless DC motor, a little later.
1- 16
The operation of the system can be described simply since the developed torque is proportional to the stator currents as: sIkT ⋅= (7.20) At the same time, the rotating flux induces a voltage in the energized windings:
ω⋅= kE (7.21)
Finally the terminal voltage differs from the induced voltage by a resistive voltage drop: RIEV sterm += (7.22) The equations are similar to those for a DC motor (4.6), (4.7), (4.8). This is the reason that although this machine is entirely different from a DC motor, it is commonly referred to as a brushless DC motor.
1- 17
ECE 320 Energy Conversion and Power Electronics
Dr. Tim Hogan
Chapter 8: Power Electronics (Textbook Chapter 10, and Sections 11.2, 11.3, and reserve book: Power Electronics)
Chapter Objectives
As we saw in the last chapter, control over the torque and speed of the motor can be gained through voltage and frequency control to the motor. This can be accomplished by converting the input AC source power to a DC source (rectifying it), then filtering it to reduce harmonics, and finally converting it back to an AC source having the desired frequency and amplitude (inverter).
8.1 Line Controlled Rectifiers We start with a description of how to draw power from a 1-phase or 3-phase system to provide DC to
a load. The characteristics of the systems here include that the devices used will turn themselves off (commutate) and that the systems draw reactive power from the loads.
8.1.1 One-Phase and Three-Phase Circuits with Diodes If the source is 1-phase, a diode is used and the load is purely resistive, as shown in Figure 1 then it is
a relatively simple configuration. When the source voltage is positive, the current flows through the diode and the voltage of the source equals the voltage of the load.
Rvd
+
i
vs
+
vdiode+
i
vs, v
d
vs, v
diode
vdiode t
Figure 1. Simple circuit with diode and resistive load. If the load includes an inductance and a source (such as a battery we wish to charge), as in Figure 2,
then the diode will continue to conduct even when the load voltage becomes negative as long as the current is maintained. This comes from the characteristics of the inductor:
( ) ( ) 0013
0
3=−=∫ itidtv
L
tL (8.1)
Thus, the shaded area A in Figure 2 must equal the shaded area B.
vd
+
i
vs
+
vdiode+
L
+ vL
Ed
+
00
0
00
00
0
00
0s 5ms 10ms 15ms 20ms 25ms0V
0V
0V
0s 5ms 10ms 15ms 20ms 25ms0V
0V
0V
0V
0V
0V
0V
0V
0V
i
Ed
t
vdiode
t
t
vL
t1 t2 t3
vs
A
B
Figure 2. Simple circuit with diode and inductive load with voltage source.
8.1.2 One-Phase Full Wave Rectifier More common is a single phase diode bridge rectifier such as shown in Figure 3. The load can be
modeled with one of two extremes: either as a constant current source, representing the case of a large
inductance that keeps the current through it almost constant, or as a resistor, representing the case of minimum line inductance. We will study the first case with AC and DC side current and voltage waveforms shown in Figure 4 for the ideal case of Ls = 0.
vs
+ L
+
vdCd
i
s
s
id
Figure 3. One-phase full wave rectifier.
0 5 10 15 20 250
0
0
0 5 10 15 20 250
0
0
vs
is
Idt
0
0
0
0
0
0
vd
id = Id
t
Figure 4. Waveforms for a one-phase full wave rectifier with inductive load.
If we analyze these waveforms, the output voltage will have a DC component, Vdo (where the
subscript o represents that this is the ideal case with Ls = 0):
ssdo VVV 9.022≈=
π (8.2)
where Vs is the RMS value of the input AC voltage. On the other hand the RMS value of the output voltage will be
ds VV = (8.3)
containing components of higher frequency. Similarly on the AC side the current is not sinusoidal, rather it changes abruptly between Id and –Id.
dds III 9.0221 ≈=
π (8.4)
and again, the RMS values are the same
sd II = (8.5)
The total harmonic distortion, THD, can then be found as:
%43.481
21
2≈
−=
s
ssI
IITHD (8.6)
It is important to note here that if the source has some inductance (and it usually does), then commutation will be delayed after the voltage reaches zero, until the current has dropped to zero as shown in Figure 5. This will lead to a decrease of the output DC voltage below what is expected from (8.2)
vs
+ L
+
vd
i
s
s
id
I d
16 20 24 28 32 36 400V
0V
0V
16 20 24 28 32 36 400V
0V
0V
vd
t
0V
0V
0V
0V
0V
0V
AA
vs
vL
t
is
t
Figure 5. One-phase full wave rectifier with inductive load and source inductance.
8.1.3 Three-Phase Full Diode Rectifiers The circuit of Figure 3 can be modified to handle three phases, by using 6 diodes as shown in Figure 6. Figure 7 shows the AC side currents and DC side voltage for the case of high load inductance. Similar analysis as before shows that on the DC side, the voltage is:
lllldo VVV −− == 35.123π
(8.7)
From Figure 6, we see that on the AC side, the RMS current, Is is:
dds III 816.032
== (8.8)
while the fundamental current, i.e. the current at power frequency is:
dds III 78.0611 ==
π (8.9)
Again, inductance on the AC side will delay commutation, causing a voltage loss, i.e. the DC voltage will be less than that predicted by equation (8.7).
a+
L
i
s
a
id
R
+
vdCdb+
c+
n
Figure 6. Three-phase full-wave rectifier with diodes.
0s 10ms 20ms 30ms 40ms0V
0V
0V
0s 10ms 20ms 30ms 40ms0V
0V
0V
0s 10ms 20ms 30ms 40ms0V
0V
0V
vd
Vdo
van vbn vcn
t
0A0A
AA
ia
D1
D4
120º
120º t
60ºib
0A0A
D2
D5 t
ic
D3
D6 t
Figure 7. Waveforms of a three-phase full-wave rectifier with diodes and inductive load.
8.1.4 Controlled Rectifiers with Tyristors Thyristors give us the ability to vary the DC voltage. Remember that to make a thyristor start conducting, the thyristor must be forward biased and a gate pulse provided to its gate. Also, to turn off the thyristor the current through it must reverse direction for a short period of time, trr, and return to zero.
8.1.5 One-Phase Controlled Rectifiers Fig. shows the same 1-phase bridge we have already studied, now with thyristors instead of diodes, and fig shows the output voltage and input current waveforms. In this figure α is the delay angle, corresponding to the time we delay triggering the thyristors after they became forward biased.
vs
+
+
vd
is
I d
T1 T3
T4 T2
Figure 8. One-phase full wave converter with thyristors.
Thyristors 1 and 2 are triggered together and of course so are 3 and 4. Each pair of thyristors is turned off immediately (or shortly) after the other pair is turned on by gating. Analysis similar to that for diode circuits will give:
( ) ( )ααπ
cos9.0cos22ssdo VVV == (8.10)
and the relation for the currents is the same
( ) ( )ααπ
cos9.0cos221 dds III == (8.11)
It should also be pointed out that in Figure 9 the current waveform on the AC side is offset in time with respect to the corresponding voltage by the same angle α, hence so is the fundamental of the current, resulting in a lagging power factor.
vd
is
ωt
α = 0 vs
1 6 2 0 2 4 2 8 321 4 3 500 V
0 V
00 V
1 6 2 0 2 4 2 8 321 4 3 500 V
0 V
00 V
vd
ωt
-100
0
100
-100
0
100
α > 0 vs
α
is
ωt
vs
Figure 9. Waveforms of one-phase full wave converter with thyristors.
On the DC side, only the DC component of the voltage carries power, since there is no harmonic
content in the current. On the AC side the power is carried only by the fundamental, since there are no harmonics in the voltage.
( ) ddss IVIVP == αcos1 (8.12)
8.1.5.1 Inverter Mode If the current on the DC side is sustained even if the voltage reverses polarity, then power will be
transferred from the DC to the AC side. The voltage on the DC side can reverse polarity when the delay angle exceeds 90º, as long as the current is maintained. This can only happen when the load voltage is as shown in Figure 10, e.g. a battery.
vs
+
+
vd
is
T1 T3
T4 T2
L
Ed+
id
Figure 10. Operation of a one-phase controlled converter as an inverter.
8.1.6 Three-Phase Controlled Converters
As with diodes, only 6 thyristors are needed to accommodate three phases. Figure 11 shows the schematic of the system, and Figure 12 shows the output voltage waveforms.
a+ia
id
b+
c+
n
+
vd
I dia
ia
T1 T3 T5
T2T6T4
Figure 11. Schematic of a three-phase full-wave converter based on thyristors.
15ms 20ms 25ms 30ms 35ms00
0
00
15ms 20ms 25ms 30ms 35ms00
0
00
15ms 20ms 25ms 30ms 35ms00
0
00
vdVdo
van vbn vcnia
ωt
α = 0
16ms 20ms 24ms 28ms 32ms16ms 20ms 24ms 28ms 32mss 16ms 20ms 24ms 28ms 32ms16ms 20ms 24ms 28ms 32mss 16ms 20ms 24ms 28ms 32ms16ms 20ms 24ms 28ms 32ms16ms 20ms 24ms 28ms 32mss
vd
Vdαvan vbn vcnia
ωt
α > 0
Figure 12. Waveforms of a three-phase full-wave converter based on thyristors. The delay angle α is again measured from the point that a thyristor becomes forward biased, but in this case the point is at the intersection of the voltage waveforms of two different phases. The voltage on the DC side is then (the subscript o here again meaning Ls = 0):
( ) ( )ααπ
cos35.1cos23lllldo VVV −− == (8.13)
while the power for both the AC and DC side is
( ) ( )αα cos3cos35.1 1slldllddo IVIVIVP ⋅⋅=== −− (8.14)
which leads to:
ds II 78.01 = (8.15)
and the relationship between Vdo and Vdα in Figure 12 is:
( )αα cosdod VV = (8.16)
Again, if the delay angle α is extended beyond 90º, the converter transfers power from the DC side to the AC side, becoming an inverter. We should keep in mind, though that even in this case the converter is drawing reactive power from the AC side.
8.1.7 Notes • For both 1-phase and 3-phase controlled rectifiers, a delay in α creates a phase displacement of the
phase current with respect to the phase voltage, equal to α. The cosine of this angle is the power factor of the fundamental harmonic.
• For both motor and generator modes the controlled rectifier absorbs reactive power from the three-phase AC system, although it can either absorb or produce real power. It also needs the power line to commutate the thyristors. This means that inverter operation is possible only with the rectifier is connected to a power line.
• When a DC motor or a battery is connected to the terminals of a controlled rectifier and α becomes greater than 90º, the terminal DC voltage changes polarity, but the direction of the current stays the same. This means that in order for the rectifier to draw power from the battery or a motor that operates as a generator turning in the same direction, the terminals have to be switched.
8.2 Inverters Here we study systems that can convert DC to AC through the use of devices that can be turned on and off such as GTOs, BJTs, IGBTs, and MOSFETs, which allows the transfer of power from the DC source to any AC load, and gives considerable control over the resulting AC signal. The general block diagram of the complete system is shown in Figure 13.
Rectifier
AC Motor
60 (Hz) AC
Filter Capacitor
Switch-mode inverter
+ Vd _
Figure 13. Typical variable voltage and variable frequency system. 8.2.1 One-Phase Inverter
Figure 14 shows the operation of one leg of an inverter regardless of the number of phases. To illustrate the point better, the input DC voltage is divided into two equal parts. When the upper switch,
S1, is closed while S2 is open, the output voltage VAo will be 2dV+ , and when the lower switch, S2, is
closed while S1 is open, the output voltage will be 2dV−
vAo +
vAN
+
S1
S2Vd
+Vd2
Vd2
Figure 14. One leg of an inverter.
To control the output waveform the switches can be controlled by pulsed width modulation, PWM,
where the time each switch is closed can be determined by the difference between a control waveform, and a carrier (or triangular) waveform as shown in Figure 15. When the control wave is greater than the triangular wave, S1 is closed, and S2 is open. When the control wave is less than the triangular wave, S1 is open, and S2 is closed. In this way, the width of the output is modulated (hence the name).
T im e
16 m s 1 8 ms 2 0 m s 2 2m s 2 4 m s 26 m s 2 8 ms 3 0 m s 3 2m s 34 m sV 1 (V t r i ) V 1( V c o nt )
- 1 . 0 V
- 0 . 5 V
0 . 0 V
0 . 5 V
1 . 0 V
Figure 15. Control (red sinewave) and carrier or triangular waveform (green) determine when the switches are closed. For the times when the control wave is greater than the triangular wave, S1 is closed, and S2 is open. When the control wave is less than the triangular wave, S1 is open, and S2 is closed.
Time
16ms 18ms 20ms 22ms 24ms 26ms 28ms 30ms 32ms 34msV(3)- V(2)
-60V
-40V
-20V
-0V
20V
40V
60V
Figure 16. Output voltage VAo corresponding to the control and carrier waves shown in Figure 15.
The frequency of the triangular wave is fc, and the frequency of the control wave is fo. We define the ratio of these as the frequency modulation index, mf.
o
cf f
fm = (8.17)
Likewise, the amplitude modulation index, ma, is defined as the ratio of the control voltage to the triangular wave voltage.
triangular
controlVVma = (8.18)
A one phase, bull wave inverter is shown in Figure 17. It has four controlled switches, each with an
antiparallel diode.
S1
S2
S4
S3AoVd
+Vd2
Vd2
B
v =AB v -Ao vBo
+
Figure 17. One-phase full wave inverter. The diagonal switches operate together such that S1 and S3 open and close together, and S2 and S4
open and close together. The output will oscillate between 2dV+ and 2
dV− . If the Fourier
transformation of the pulse width modulated square wave shown in Figure 18 is taken, the amplitude of the fundamental will be a linear function of the amplitude index Vo = maVd/2 as long as ma ≤ 1. Then the RMS value of the output voltage will be:
dada
o VmVmV 353.0221 == (8.19)
Time
16ms 18ms 20ms 22ms 24ms 26m s 28m s 30 ms 32 ms 34m sV(2)- V(3)
-12 0V
-8 0V
-4 0V
- 0V
4 0V
8 0V
12 0V
Figure 18. Output voltage VAB for the one-phase, full wave inverter of Figure 17.
When ma increases beyond 1, the output voltage increases also, but not linearly with ma. The output
amplitude can reach a peak value of dVπ4 when the reference signal becomes infinite and the output is a
square wave. Under this condition, the RMS value is:
dd
o VVV 45.02
221 ==
π (8.20)
Equating the power of the DC side with that of the AC side gives
pfIVIVP oodod ⋅== 11 (8.21)
Thus for normal operation:
pfImI oado ⋅= 1353.0 (8.22)
and in the limit for a square wave:
pfII odo ⋅= 145.0 (8.23)
8.2.2 Three-Phase Inverter For three-phase loads it makes more sense to use a three-phase inverter shown in Fig., rather than using three one-phase inverters.
dV+
S1
S4
S3
S6
A C
S2
S5
B
Figure 19. Three-phase, full wave inverter.
The basic PWM scheme for a three-phase inverter has one common carrier and three separate control
waveforms. If the waveforms we want to achieve are sinusoidal the frequency modulation index, mf, is low we use a synchronized carrier signal with mf an integer multiple of 3.
T i m e
1 6 m s 1 7 m s 1 8 m s 1 9 m s 2 0 m s 2 1 m s 2 2 m s 2 3 m s 2 4 m sV 1 ( V c o n t 1 ) V 1 ( V c o n t 2 ) V 1 ( V c o n t 3 ) V 1 ( V t r i )
- 1 . 0 V
- 0 . 5 V
0 . 0 V
0 . 5 V
1 . 0 V
(a)
Time
16m s 17 ms 1 8ms 19ms 20ms 21ms 22ms 23ms 24msV(2)
0V
40V
80V
110V
(b)
Time
16m s 17 ms 1 8ms 19ms 20ms 21ms 22ms 23ms 24msV(3)
0V
40V
80V
110V
(c)
Time
16ms 17ms 18ms 19ms 20ms 21ms 22ms 23ms 24msV(2)- V(3)
-120V
-80V
-40V
-0V
40V
80V
120V
(d) Figure 20. Three-phase, full wave inverter showing (a) control and triangular wave, (b) phase A line-to-
neutral, (c) phase B line-to-neutral, and (d) line-to-line signal from phase A to B.
Time
1 7.00ms 17 .04ms 17. 08ms 17. 12ms 17.1 6ms 17.20 ms 17.24m s 17.28ms 17.32ms 17.36ms 17.40m sV(2)
0V
40V
80V
110V
(a)
Time
1 7.00ms 17 .04ms 17. 08ms 17. 12ms 17.1 6ms 17.20 ms 17.24m s 17.28ms 17.32ms 17.36ms 17.40m sV(3)
0V
40V
80V
110V
(b)
Tim e
17.00ms 17.04ms 1 7.08ms 17.12ms 17. 16ms 17.20ms 17.24 ms 17.28ms 17.32ms 17 .36ms 17.40msV(2)- V( 3)
-120V
-80V
-40V
-0V
40V
80V
120V
(c)
Figure 21. Shorter time span for the (a) phase A line-to-neutral, (b) phase B line-to-neutral, and (c) line-to-
line signal from phase A to B of the outputs in Figure 20.
The PSPICE netlist for this three phase PWM circuit is given below: 3 phase FULL-BRIDGE INVERTER ************ OUTPUT IS V(2,3) **************** **********INPUT PARAMETERS **************** .PARAM Vsource = 100 ; DC input to inverter .PARAM Fo = 200 ; fundamental frequency .PARAM Mf = 21 ; carrier, multiple of Fo .PARAM Ma = .8 ; amplitude ratio .PARAM Fc = Mf*Fo ; carrier frequency VS 1 0 DC Vsource ; dc source ******* VOLTAGE-CONTROLLED SWITCHES ***** S1 1 2 40 30 SWITCH S2 1 3 50 30 SWITCH S3 1 4 60 30 SWITCH S4 2 0 30 40 SWITCH S5 3 0 30 50 SWITCH S6 4 0 30 60 SWITCH ************** FEEDBACK DIODES ************* D1 2 1 DMOD D2 3 1 DMOD D3 4 1 DMOD D4 0 2 DMOD D5 0 3 DMOD D6 0 4 DMOD ******************** LOAD ****************** R1 2 5 10 ; load between nodes 2 and 0 L1 5 0 60MH R2 3 6 10 ; load between nodes 3 and 0 L2 6 0 60MH R3 4 7 10 ; load between nodes 4 and 0 L3 7 0 60MH *************** TRIANGLE CARRIER ************** Vtri 30 0 PULSE (1 -1 0 1/(2*Fc) 1/(2*Fc) 1ns 1/Fc) ******************** REFERENCE ***************** Vcont1 40 0 SIN(0 Ma Fo 0 0 -90/Mf) Vcont2 50 0 SIN(0 Ma Fo 0 0 -90/Mf - 120) Vcont3 60 0 SIN(0 Ma Fo 0 0 -90/Mf - 240) ************ MODELS AND COMMANDS ************* .MODEL SWITCH VSWITCH(RON=0.001 VON=.007 VOFF=-.007) .MODEL DMOD D ; default diode .PROBE .TRAN 0.5MS 33.33MS 16MS 0.1MS .FOUR 200 25 I(R1) ; Fourier transform .OPTIONS NOPAGE ITL5=0 .END
As long as ma is less than 1, the RMS value of the fundamental of the output voltage is a linear function of it:
dadall VmVmV 612.022
31
≈=− (8.24)
In the limit, when the control voltage becomes infinite, the RMS value of the fundamental of the output is then:
dd
ll VVV 78.02
423
1≈=− π
(8.25)
Again, equating the power on the DC and AC sides we obtain:
pfIVIVP olldod ⋅== − 113 (8.26)
or for normal PWM operation:
pfImI oado ⋅= 106.1 (8.27)
and in the limit for the square wave:
pfII odo ⋅= 135.1 (8.28)
Finally, there are other ways to control the operation of an inverter. If it is not he output voltage
waveform we want to control, but rather the current, we can either impose a fast controller on the voltage waveform, driven by the error between the current signal and the reference, or we can apply a hysteresis band controller, shown for one leg of the inverter in Figure 22.
vAo +
vAN
+
S1
S2Vd
+Vd2
Vd2
Ti me
16 ms 18 ms 20 ms 22 ms 24 ms 26 ms 28 ms 30 ms 32 ms 34msI( R)
-4 .0A
-2 .0A
0A
2 .0A
4 .0A
iεΣ Switch-Mode
Inverter
iε
iA
*Ai
+ _
Comparator tolerance
band
Figure 22. Current control with hysteresis band.
8.2.3 Inverter Notes
• With a sine-triangle PWM the harmonics of the output voltages are of frequency around nfn, where n is an integer and fn is the frequency of the carrier (triangle) waveform. The higher this frequency is the easier to filter out these harmonics. On the other hand, increasing the switching frequency also increases proportionally the switching losses. For 6-step operation of a 3-phase inverter the harmonics are even, except the tripled ones, i.e. they are of order 5, 7, 11, 13, 17 etc.
• When the load of an inverter is inductive the current in each phase remains positive after the voltage in that phase became negative, i.e. after the top switch has been turned off. The current then flows through the antiparallel diode of the bottom switch, returning power to the DC link. The same
happens of course when the bottom switch is turned off and the current flows through the antiparallel diode of the top switch.
8.2.4 Example 1
A three-phase controlled rectifier is supplying a DC motor that has k = 1 (V·s) and R = 1 (Ω). The rectifier is fed from a 208 (Vl-l) source.
Filter 208 (V)
Figure 23. Figure for example problem 8.2.4.
a) Calculate the maximum no-load speed of the DC motor:
Without a load the current is zero, therefore:
ωω kIRkV =+= The maximum speed is then found by the maximum DC voltage
maxmax ωkV =
The maximum DC voltage is provided by the controlled rectifier for α = 0.
(V) 8.28135.1max == −llVV Therefore,
(rad/s) 8.280max =ω
b) The motor now is producing torque of 20 (N·m). What is the maximum speed the motor can achieve? Now that there is load torque, there is also current:
(A) 20=⇒= IkIT Then
(rad/s) 8.2601
1208.280=
⋅−=
−=
kIRVω
c) For the case in b) calculate the total RMS current of the fundamental and the power factor at the AC side
At the maximum, the fundamental of the AC current is:
(A) 6.1578.01 =≈ ds II The power factor is then 1.
d) If the motor is now connected as a generator with a counter torque of 20 (N·m) at 1500 (rpm). What should be the delay angle and AC current?
For a DC generator:
(V) 08.137112
60215001 =⋅−⋅⋅−=−=πωω R
kTkIRkV
Since this is a generator, the voltage is negative for the inverter.
( )αcos2081.35(V) 08.137 ⋅⋅=− º22.119=α
8.2.5 Example 2 For the system shown in Figure 24, the AC source is constant, and the load voltage is 150 (Vl-l),
20 (A), 52 (Hz), 0.85 pf lagging.
Filter 208 (V)
α
6-step inverte
Figure 24. Figure for example problem 8.2.5.
a) Calculate the voltage on the DC side and the DC component of the current. For the 6-step inverter
(V) 19278.01, =⇒=− ddll VVV
(A) 23192
85.02015035.13 00 =⋅⋅⋅
=⇒=⋅= − dddlll IIVpfIVP
b) Calculate the AC source side RMS and fundamental current and power factor.
For a 3-phase rectifier
( ) ( ) ( ) 685.0coscos20835.1192cos35.1 =⇒⋅⋅=⇒= − αααlld VV (A) 94.1778.01 == ds II
( ) lagging 685.0cos == αpf
8.3 DC-DC Conversion DC to DC conversion is often associated with stabilizing the output while the input varies, however
the converse is also required in some applications, which is to produce a variable DC from a fixed or variable source. The issues of selecting component parameters and calculating the performance of the system is the focus of this section. Since these converters are switched mode systems, they are often referred to as choppers.
8.3.1 Step-Down or Buck Converters The basic circuit of this converter is shown in Figure 25 connected to a purely resistive load. If we
remove the low pass filter shown and the diode, the output voltage vo(t) is equal to the input voltage Vd when the switch is closed, and zero when the switch is open. The average output, Vo, is then:
ds
onT
t
td
so V
TtdtdtV
TV
s
on
on=⎥
⎦
⎤⎢⎣
⎡+= ∫∫ 01
0 (8.29)
L
R(load)
vo
+
io
C
Low-pass filter
o= V
iL
+ vL
+
Vd
vo
oVdV
t0
sT = 1sf
ontofft
Figure 25. Topology of the buck chopper.
The ratio of ton/Ts = D, the duty ratio.
The low pass filter attenuates the high frequencies (multiples of the switching frequency) and leaves almost only the DC component. The energy stored in the filter inductor (or the load inductor) has to be absorbed somewhere other than the switch, hence the diode, which conducts when the switch is open. Here we study this converter in the continuous mode of operation such that the current through the inductor never becomes zero. As the switch opens and closes the circuit assumes one of the topologies of Figure 26.
L
R
+
C oV
iL
+ vL
+
Vd
L
R
+
C oV
iL
+ vL
t
vL
A
B
t
i L
0
ontofft
i L = Io
sT = 1sf
Vd - Vo
- Vo
Figure 26. Operation of the buck chopper.
We will use the fact that the average voltage across the inductor is zero, and assume a perfect filter
such that the voltage across the inductor is (Vd – Vo) during ton, and –Vo for the remainder of the cycle.
(8.30) ( ) ( ) 00
=−+−= ∫∫s
on
on T
to
todo dtVdtVVV
( ) ( ) 0=−−− onsoonod tTVtVV (8.31)
DTt
VV
s
on
d
o == (8.32)
Also using the fact that the input and output powers are the same gives:
oodd IVIV = (8.33)
DVV
II
d
o
o
d == (8.34)
We analyzed this under the assumption of continuous mode of operation. In the discontinuous mode, the output DC voltage is less than what is given here, and the chopper is less easy to control. At the boundary between continuous and discontinuous modes, the inductor current reaches zero for one instant every cycle, as shown in Figure 27.
0
sT = 1sf
ontofft
vo
t
i L
i L, avg= IL
Vd - Vo
- Vo
i L
sT= 8L
0 0.5 1.0D
IL,max
Vd
IL
Figure 27. Operation of the buck converter at the boundary of continuous conduction.
From this figure we can see that at this operating point, the average inductor current is IL = ½ iL, and:
( ) ( )ods
odonL VVL
DTVVtI −=−=22
1 (8.35)
Since the average inductor current is the average output current (the average capacitor current is zero), equation (8.35) defines the minimum load current that will sustain continuous condition.
Finally, a consideration is the output voltage ripple. We assume the ripple current is absorbed by the capacitor, i.e. the voltage ripple is small. The ripple voltage is then due to the deviation from the average of the inductor current as show in fig. Under these conditions:
222
11 sLo
TILC
QV Δ=
Δ=Δ (8.36)
where
( ) so
L TDL
VI −=Δ 1 (8.37)
( )DLCT
VV s
o
o −=Δ 1
81 2
(8.38)
t
vL
A
B
t
i L
0
ontofft
i L = Io
sT = 1sf
Vd - Vo
- Vo
oΔV
2
vo
t0
Vo
Figure 28. Analysis of the output voltage ripple of the buck converter.
Another way to view this is to define the switching frequency fs = 1/Ts and use the corner frequency of the filter ( )LCfcorn π21= :
( )22
12 ⎟⎟
⎠
⎞⎜⎜⎝
⎛−=
Δ
s
corn
o
of
fDVV π (8.39)
8.3.2 Step-Up or Boost Converter Here the output voltage is always higher than the input. The schematic is shown in Figure 29.
L
R
+
C oV
iL
+ vL
+
Vd
Figure 29. Schematic diagram of a boost converter. Based on the condition of the switch, there are two possible topologies as shown in Figure 30. Again,
the way to calculate the relationship between input and output voltage we take the average current of the inductor to be zero, and the output power equal to the input power giving:
( )( ) 0=−−+ onsodond tTVVtV (8.40)
DV
V
d
o−
=⇒1
1 (8.41)
DII
d
o −=⇒ 1 (8.42)
L
R
+
C oV
iL
+ vL+
Vd
L
R
+
C oV
iL
+ vL
t
vL
A
B
t
i L
0
ontofft
i L = Io
sT = 1sf
Vd - Vo
- Vo
+
Vd
Figure 30. Two circuit topologies of the boost converter.
To determine the values of inductance and capacitance we study the boundary of continuous
conduction like before and the output voltage ripple. At the boundary of the continuous conduction, the geometry of the current waveform gives:
( )212
DDLVTI os
o −= (8.43)
The output current must exceed this value for continuous conduction. Using the ripple analysis as shown in Figure 31 we find:
RCDT
VV s
o
o =Δ (8.44)
It is important to note that the operation of a boost converter depends on parasitic components, especially for duty cycle approaching unity. These components will limit the output voltage to levels well below those given by equation (8.41).
tΔQ
ΔQ
i D
i D= Io
oΔV
vo
t0
Vo
DTs (1-D)Ts
Figure 31. Calculating the output voltage ripple for a boost inverter.
8.3.3 Buck-Boost Converter This converter has a schematic shown in fig. and can provide output voltage that can be lower or higher than the input voltage.
Again the operation of the converter can be analyzed using the two topologies resulting from the operation of the switch as shown in fig.
By equating the integral of the inductor voltage to zero we get:
( )( ) 01 =−−+ sosd TDVDTV (8.45)
D
DVV
d
o−
=⇒1
(8.46)
At the boundary between continuous and discontinuous conduction we find
( )212
DLVTI os
o −= (8.47)
The output ripple, as calculated from Fig. is
RCDT
VV s
o
o =Δ (8.48)
L R
+
C oV
iD
+
vL
+
Vd
io
iL
Figure 32. Basic Buck-Boost converter.
ss
t
vL
A
B
t
i L
0
IL = (I + I )o
sT = 1sf
- Vo
Vd
L R
+
C oV
iD
+
vL
+
Vd
io
iL
L R
+
C oV
+
vL
+
Vd
io
iL
DT (1-D)T
d
Figure 33. Operation of a Buck-Boost chopper.
8.3 Example
The input of a step down converter varies from 30 (V) to 40 (V) and the output voltage is to be constant at 20 (V), with output power varying between 100 (W) and 200 (W). The switch is operating at 10 (kHz). What is the inductor needed to keep the inductor current continuous? What is then the filter capacitor needed to keep the output ripple below 2%?
The duty cycle will vary between D1 = 20/30 = 0.667 and D2 = 20/40 = 0.5. The load current will
range between Io1 = 100/20 = 5 (A), and Io2 = 200/20 = 10 (A). The minimum current needed to keep the inductor current continuous is:
( )ods
o VVL
DTI −=2min,
Since the constant is the output voltage, Vo, and the minimum load current must be greater than Io,min, we can express it as a function of Vo and make it less than or equal to 5 (A), or:
( ) ( ) ( )DLTVVV
LDTI so
ods
o −=−=≥ 122
A5 min,
Ts = 1/10(kHz), Vo = 20 (V), and the maximum value is achieved for D = 0.5, leading to Lmin = 50 (µH). For the ripple, the highest will occur at 1-D = 0.5, thus:
(Hz) 9001010
5.02
02.02
3
2=⇒⎟⎟
⎠
⎞⎜⎜⎝
⎛
×= corn
corn ffπ
(Hz) 90010502
16
=×
⇒− Cπ
C = 625 (µF)
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