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Transformer Model
where r = diag [r1 r2], a diagonal matrix, and
The resistances r1 and r2 and the flux linkages l1 and l2 are related to coils 1 and 2, respectively. Because it is assumed that 1 links the equivalent turns of coil 1 and 2 links the equivalent turns of coil 2, the flux linkages may be written as
Voltage Equation of a transformer in matrix form is:
Where
Linear Magnetic System
Reluctance is impossible to measure
accurately, could be determined using:
1 1 1 1 2 21
1
2 2 2 2 1 12
2
l m m
l m m
l
A
N i N i N i
N i N i N i
m
f
f
 =
= + +Â Â Â
= + +Â Â Â
Flux Linkages2 21 1 1 2
1 1 1 21
2 22 2 1 2
2 2 2 12
l m m
l m m
N N N Ni i i
N N N Ni i i
l
l
= + +Â Â Â
= + +Â Â Â
The coefficients of the 1st two terms on the right–hand side depend upon the turns of coil 1 and the reluctance of the magnetic system; i.e. independent of coil 2. Similar situation exist in equation for 2
Self Inductances
From the previous equations one can define Self-Inductances:
2 21 1
11 1 11
2 22 2
22 2 22
l ml m
l ml m
N NL L L
N NL L L
= + = +Â Â
= + = +Â Â
Where
Ll1 and Ll2 are leakage inductances of coil 1 and 2 respectively.
Lm1 and Lm2 are the magnetizing inductances of coils 1 and 2 respectively.
Magnetizing Inductances
The two magnetizing Inductances are related as:
2 12 22 1
m mL L
N N=
Where
m the Magnetizing Reluctance being common for both coils.
The mutual Inductances are defined: 1 2
12
2 121
m
m
N NL
N NL
=Â
=Â
Mutual Inductances
Mutual Reluctance being common for both
Circuits; Mutual Inductances are
related to Magnetizing Inductances too:
2 112 21 1 2
1 2m m
N NL L L L
N N= = =
Flux Linkages
Flux Linkages may be written as:
= Li
Where1 1 2
111 1211
221 222 12 2
l mm
mm
L L NLL L
NL NLL L
N L L
é ù+é ù ê úê ú ê ú= =ê ú ê úë û ê ú+ë û
Flux Linkages
The Flux Linkage may also be derived based on self and mutual inductances:
21 1 1 1 1 2
1
12 12 2 2 1 2
2
( )
( )
l m
m
NL i L i i
N
NL i L i i
N
l
l
= + +
= + +
Example 1A
It is instructive to illustrate the method of deriving an
equivalent T circuit from open- and short-circuit measurements. For
this purpose let us assume that when coil 2 of the two-winding
transformer shown in Fig. is open-circuited, the power input to coil 1
is 12 W with an applied voltage is 100 V (rms) at 60 Hz and the
current is 1 A (rms). When coil 2 is short-circuited, the current
flowing in coil 1 is 1 A when the applied voltage is 30 V at 60 Hz.
The power during this test is 22 W. If we assume Ll1 = L’l2, an approximate equivalent T circuit can be determined from these measurements with coil 1 selected as the reference coil.
Example 1…..
1 1 1 cosP V I f® ®
=
Vand Ir r
The Power may be expressed:
Where are phasor and is the phase angle between them.
Solving for during the open circuit test, we have:
1 1 01 12cos cos 83.7
110 1
P
xV If - -= = =r r
Example 1……
Vr
0
0
110 012 109.3
1 83.7
VZ j
I
Ð= = = + W
Ð-
r
r
as a reference phasor and in an inductive circuit of the transformer I phasor would lag behind by the angle of =83.70
Z, the impedance may therefore be determined by:
That suggests that Xl1+Xm1 = 109.3, while r1 =12
V
I
Example 1…….
For short circuit test i1= -I’2 because transformers are
designed so that Xm1>> |r’2+jX’12|. Hence using phase
angle equation: 1 022cos 42.8
30 1xf -= =
In this case input Impedance is (r1+r’2)+j(Xl1+X’l2) and that is determined by:
0
0
30 022 20.4
1 42.8Z j
Ð= = + W
Ð-
That means r’2 = 10 and Xl1=X’l2 both are 10.2
Example 1….
That leads to conclusion that:
Xm1= 109.3 -10.2 =99.1
Hence other parameters are:
r1 = 12 Lm1 = 262.9mH r’2 = 10
Ll1 = 27.1mH L’l2 = 27.1mH
V1
I1V’2
E1
Flux Linkage of a Coil
Fig. 1 shows a coil of N turns. All these N turns link flux lines of Weber resulting in the N flux linkages.In such a case:
Where
N is number of turns in a coil;
e is emf induced, and
is flux linking to each coil
N
de N
dt
y f
f
=
=
Design of Transformer
Let's try to proportion a transformer for 120 V, 60 Hz supply, with a full-load current of 10 A. The core material is to be silicon-steel laminations with a maximum operating flux density Bmax = 12,000 gauss. This is comfortably less than the saturation flux density, Bsat. The first requirement is to ensure that we have sufficient ampere-turns to magnetize the core to this level with a permissible magnetizing current I0A.
Design of Transformer….
Let's choose the magnetizing current to be 1% of the full-load current, or 0.1 A. The exact value is not sacred; this might be thought of as an upper limit. Here, we will assume a simple, uniform magnetic circuit for simplicity. If l is the length of the magnetic circuit, H is 0.4πN(√2I0)/l, and the magnetization curve for the core iron gives the H required for the chosen Bmax. From this, we can find the number of turns, N, required for the primary.
Design of Transformer
We could also estimate the ampere-turns required by using an assumed permeability μ. Experience will furnish a satisfactory value. It is not taken from the magnetization curve, but from the hysteresis loop. Let's take μ = 1000. Then, N = Bmaxl / 0.4π√2 μI0. If we estimate l = 20 cm, the number of primary turns required is N = 1350. The rms voltage induced per turn is determined from Faraday's Law:
√2 e = (2πf)BmaxA x 10-8. Now, e must be 120 / 1350 = 0.126 V/turn, f is 60, and Bmax = 12,000 gauss. We know everything but A, the cross-sectional area of the core. We find A = 2.8 cm2.
Design of Transformer
Powdered iron and ferrite cores have low Bsat and permeability values. A type 43 ferrite has Bsat = 2750 gauss, but a maximum permeability of 3000, and is recommended for frequencies from 10 kHz to 1 MHz. Silicon iron is much better magnetically, but cannot be used at these frequencies. The approximate dimensions of an FT-114 ferrite core (of any desired material) are OD 28 mm, ID 19 mm, thickness 7.5 mm. The magnetic dimensions are l = 74.17 mm, A = 37.49 mm2, and volume 2778 mm3. Similar information is available for a wide range of cores. There are tables showing how much wire can be wound on them, and even the inductance as a function of the number of turns.