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Phase Controlled Rectifiers ______ (AC/DC Converters)
Objectives
• Principle o f controlled rectification.
• Single phase and 3 phase converters.
• Half wave and full wave converters.
• Bridge converters — --------► semicoisemiconverter
* full bridge converter
• Resistive, inductive and motor (RLE) loads on converters.
• Continuous and discontinuous output current operation and its effects.
• Inverting operation (power flow from load to source) in case o f full converters.
• Effects o f feedback diode and freewheeling operation.
• Harmonic analysis o f converters.
3.1 Introduction
3.1.1 Principle of AC/DC Conversion (Controlled Rectifier)
• Controlled rectifiers arebasically AC to DC converters. The power transferred to the load is controlled by controlling triggering angle of the devices. Fig. 3.1.1 shows this operation.
Controlledrectifier Load
a
Controlcircuit
Fig. 3.1.1 Principle of operation of a controlled rectifier
( 3 - 1 )
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Power Devices and Machines 3 - 2 Phase Controlled Rectifiers (AC/DC Converters)
• The triggering angle ' a ’ of the devices is controlled by the control circuit.
• The input to the controlled rectifier is normally AC mains. The output of the controlled rectifier is adjustable DC voltage. Hence the power transferred across the load is regulated.
Applications :
The controlled rectifiers are used in battery chargers, DC drives, DC power supplies etc. The controlled rectifiers can be single phase or three phase depending upon the load power requirement.
3.1.2 Concept of Commutation
Answer following question after reading this topic
1. What do you mean by commutation o f SCR ? Give types o f commutations. Explain natural commutation in details.
Marks (6), May-2007 I \ Most likely andJ masked In previous
niversity Exam
D efinition : Commutation is the collective operation, which turns of the conducting SCR.
Commutation requires external conditions to be imposed in such a way that either current through SCR is reduced below holding current or voltage across it is reversed.
There are two types of commutation techniques.
Fig. 3.1.2
• Forced com m utation : It requires external components to store energy and it is used to apply reverse voltage across the SCR or reduce anode current below holding current of the SCR to turn it off.
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Power Devices and Machines 3 - 3 Phase Controlled Rectifiers (AC/DC Converters)
• Current com m utation : The SCR is turned off by reducing its anode current below holding current.
• V oltage com m utation : The SCR is turned off by applying large reverse voltage across it.
• Principle of line com m utation
The natural commutation does not need any external components. It uses supply (mains) voltage for turning off the SCR. Hence it is also called as line commutation.
• Explanation
Fig. 3.1.3 shows the circuit using natural commutation. It is basically half wave rectifier. The mains AC supply is applied to the input. The SCR is triggered in the positive half cycle at a. Since the SCR is forward biased, it starts conducting and load current i0 starts flowing. The waveforms of currents and voltages are shown in Fig. 3.1.4. Since the
Fig. 3.1.3 A half wave rectifier uses natural load is resistive, commutation to turn off SCR
Mains AC
j::::::::::::::::::: I::::::::::;,:::::::;:::::;:::::::::::: :::::::::::: .... . ....I..... . .................. *......... . • ..... .......... .31113S HIS sstS :!t H :3S !•t:S!?:!: SHi !H9 5SI! §8: HIE B i:! HHi IS
:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: I:::::::::::::::::;::::::::::::::::::::::::::::::::::::::::::::::::::::;:;::::-.;:
Fig. 3.1.4 Waveforms of half wave controlled rectifier to illustrate naturalcommutation
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Power Devices and Machines 3 - 4 Phase Controlled Rectifiers (AC/DC Converters)
Hence the shape of the output current is same as output voltage. Observe that the output current is basically SCR current. At 'rf the supply voltage is zero. Hence current through SCR becomes zero. Therefore the SCR turns off. The supply voltage is then negative. This voltage appears across the SCRs and it does not conduct. Thus natural commutation takes place without any external components. Here note that natural commutation takes place only when the supply voltage is AC. Thus the controlled rectifiers use natural commutation.
3.1.3 Forced Commutation
3.1.3.1 Principle of Forced Commutation
Forced commutation is used when the supply is D.C. A commutation circuit is connected across the SCR as shown in Fig. 3.1.5.
The commutation circuit is normally LC circuit.The LC circuit stores energy when the SCR is ON. This energy is used to turn-off the SCR. The LC circuit imposes reverse bias across the SCR due to stored energy. Hence forward current of SCR is dropped below holding current and the SCR tums-off.
There are different types of forced commutation circuits depending upon the way they are connected.
3.1.3.2 Classification of Forced Commutation
Forced commutation circuits can be classified depending upon whether voltage or current is used for commutation. Similarly the classification can be made based on whether the load resonates or commutation components are separate. Some times additional SCR is used for commutation main SCR. Such techniques are called auxiliary commutation methods. Based on these classifications following are some of the main commutation techniques :
1. Self commutation by resonating load and LC circuit
2. Auxiliary voltage commutation (impulse commutation)
3. Auxiliary current commutation (resonant pulse commutation)
4. Complementary commutation
5. External pulse commutation.
ILC
circuit
Fig. 3.1.5 Principle of forced commutation
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Power Devices and Machines 3 - 5 Phase Controlled Rectifiers (AC/DC Converters)
3.1.3.3 Comparison of Natural Commutation and Forced Commutation
Table 3.1.1 shows the comparison between natural and forced commutation techniques.
Sr.No.
Natural commutation Forced commutation
1. No external commutation components are required.
External commutation components are required.
2. Requires AC voltage at the input. Works on DC voltages at the input.
3. Used in controlled rectifiers, AC voltage controllers etc.
Used in choppers, inverters etc.
4. No power loss takes place during commutation Power loss takes place in commutating components.
5. SCR turns off due to negative supply voltage. SCR can be turned-off due to voltage and current both.
6. Cost of the commutation circuit is nil. Cost of the commutation circuit is significant.
Table 3.1.1 Natural and forced commutation
3.2 Single Phase Half Wave Converter and Effect of Freewheeling Diode
3.2.1 Single Phase Half Wave Controlled Rectifier with Resistive Load
Answer following question after reading this topic
1. Explain the operation o f 1<\> half wave converter with the help o fcircuit diagram and waveforms. Most likely and
ImportantQ uestion
The principle of phase controlled operation can be explained with the help of half wave controlled rectifier shown in Fig. 3.2.1. The secondary of the transformer is connected to resistive load through thyristor or SCR Ty The primary of the transformer is connected to the mains supply. In the positive cycle of the supply, Tj is forward biased. T{ is triggered at an angle a. This is also called as triggering or firing delay angle. Tj conducts and secondary (i.e. supply) voltage is applied to the load. Current i0 starts flowing through the load. The output current and voltage waveforms are shown in Fie. 3.2.2.
Fig. 3.2.1 Half wave controlled rectifier with R-load.
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Power Devices and Machines 3 - 6 Phase Controlled Rectifiers (AC/DC Converters)
Since the load is resistive, output current is given as,
Hence the shape of output current waveform is same as output voltage waveform. At n supply voltage drops to zero. Hence current i0 flowing through 7 becomes zero and it turns off. In the negative half cycle of the supply Tj is reverse biased and it does not conduct. There is only one pulsve of V0 during one cycle of the supply. Hence ripple frequency of the output voltage is,
fripple = 50 Hz ‘-e- supply frequency
Power Devices and Machines 3 - 7 Phase Controlled Rectifiers (AC/DC Converters)
Mathematical analysis
The average value of output voltage is given as,
1 TVo(av) = f ! vc M d o t
0
The period of one pulse of v0 (cot) can be considered as T = 2 n. And v0 (cot) =Vm sin cot from a to jr. For rest of the period v0 (cof) = 0. Hence above equation can be written as,
Vo(av )1 71
7T— f Vm sin cot du>t 2n J
... (3.2.1)
The power transferred to the load will be,
o(av)V „ U )
R
Thus the output average voltage and power delivered by the controlled rectifier can be controlled by phase control (i.e. a). The phase control in converters means to control the delay (or triggering) angle a.
3.2.2 Half Wave Controlled Rectifier with RL LoadNow let us study the operation of single phase half wave controlled rectifier for
inductive (RL) load. Normally motors are inductive load. L is the armature or field coil inductance and R is the resistance of these coils. Fig. 3.2.3 shows the circuit diagram of half wave controlled rectifier with RL load.
The SCR will be forward biased in thepositive half cycle of the supply. HenceSCR is applied with the firing pulses in the
. . . . positive half cycle. The waveforms areFig. 3.2.3 Half wave controlled rectifier 0 0 x u iU
with RL load shown in Fig. 3.2.4. Fig. 3.2.4(a) shows thesupply voltage and Fig. 3.2.4(b) shows the
firing pulses to the SCR.
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Power Devices and Machines 3 - 8 Phase Controlled Rectifiers (AC/DC Converters)
Fig. 3.2.4 Waveforms of half wave controlled rectifier for RL load
When the SCR is triggered, the supply voltage appears across load. We normally neglect small voltage drop in SCR. Hence v0 =vs when SCR is conducting. This is shown in Fig. 3.2.4(c). Observe that output voltage is same as supply voltage after a. Because of the RL load, output current starts increasing slowly from zero. The shape of i0 depends upon values of R and L. At n , the supply voltage becomes zero and i0 is maximum. Due to negative supply voltage after n, SCR tries to turn-off. But energy stored in the load
inductance generates the voltage L -~ . This induced voltage forward biases the SCR and
maintains it in conduction. This is shown in Fig. 3.2.5. The basic property of inductance is that it opposes change in current. At n, the current i0 is maximum. As SCR tries to turn-off due to negative supply voltage, the output current i0 tries to go to zero. Such change in i0 is opposed by load inductance. Hence the energy stored in an inductance tries
to maintain i0. To maintain the flow of i0, inductance generates the voltage with
polarity as shown in Fig. 3.2.5. This voltage is higher than negative supply voltage. Hence Tj is forward biased and it remains in conduction. The output current and supply current
Power Devices and Machines 3 - 9 Phase Controlled Rectifiers (AC/DC Converters)
flow in the same loop. Hence i0 =is all the time. The waveform of i0 is shown in Fig. 3.2.4(d) and is is shown in Fig. 3.2.4 (e). After 7 1, i0 (i.e. is ) flows against the supply. Hence energy is consumed in the supply. i0 flows due to load inductance energy. In other words, the inductance energy is partially fed to the mains and to the load it self. Therefore energy stored in inductance goes on reducing. Hence i0 also goes on reducing as shown in Fig. 3.2.4 (d). At P the energy stored
goes to 3.2.4(c)
in the inductance is finished. Hence i0 zero. Therefore T. tums-off. In Fig.
Fig. 3.2.5 SCR conducts due to inductance voltage after n
observe that v0 is negative from n to p . Because Tj conducts from n to p . Hence whenever Tj conducts v0 =vs .
The SCR is triggered again at 2 71 + a. Hence output voltage remains zero from p to 271+ex. Output current as well as supply current are also zero from p to 2?i+a. At 2n + ar Tj is triggered again and the cycle repeats. Here i0 goes to zero at p. Hence this is called discontinuous conduction.
»>■► Example 3.2.1 : Derive an expression for average value o f output voltage for 1 <j) half wave controlled rectifier with RL load.
Solution : For discontinuous conduction, the output voltage waveform is shown in Fig. 3.2.4(c). The output voltage waveform repeats at the period of T = 2ti . The average value is given as,
T
o(av ... (3.2.2)
In Fig. 3.2.4 observe that,
v0(<at)v$ = Vm sin (oI from a top0 from 0 to a and p to2 n
Hence equation 3.2.2 can be written as,P
V.o(av) = f V sin cof dwt = -—-[-coscof]^ 2 n J m 2 71 «
Vo(av) =y^(cos<x-cosP) (3.2.3)
This is an expression for average value of output voltage.
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Power Devices and Machines 3 - 1 0 Phase Controlled Rectifiers (AC/DC Converters)
3.2.3 Half Wave Controlled Rectifier with Freewheeling Diode
Answer following question after reading this topic
1 . C om pare freew heeling diodes and fe ed b ack diodes.Marks [31. D ec.-2006, 2 0 0 8
Now let us consider the half wave controlled rectifier with freewheeling diode across the RL load. This circuit diagram is shown in Fig. 3.2.6.
The SCR is triggered at firing angle of a in positive half cycle of supply. Hence v0 =vs . The waveform of v0 is shown in Fig. 3.2.7(c). Observe that from a to n , v0 is same as supply voltage vs . The freewheeling diode (Dfvv) is reverse biased, hence it does not conduct. The output current i0 increases from zero as shown in Fig. 3.2.7(d). This is shown
in equivalent circuit-I in Fig. 3.2.7. See Fig. 3.2.7 on next page.
After 7i, the supply voltage becomes negative. Hence SCR tries to turn-off. Therefore i0 tries to go to zero. Observe that i0 is maximum at n. But the load inductance
does not allow i0 to go to zero. The energy stored in inductance generates the voltage
L with polarity as shown in Fig. 3.2.8.
The induced inductance voltage forward biases freewheeling diode as well as SCR. But freewheeling diode (DFW) is more forward biased. Hence it starts conducting.
Fig. 3.2.6 Freewheeling diode in half wave controlled rectifier
Fig. 3.2.8 Freewheeling action in half wave controlled rectifier
Therefore Tj tums-off. The output current now flows through the freewheeling diode. In Fig. 3.2.8 observe that i0 = ifW when freewheeling diode conducts. Here iFW is freewheeling current. Fig. 3.2.8(d) and (e) shown that i0 =iFW when freewheeling diode conducts. The freewheeling current flows only due to energy stored in the load inductance. The output current flows in the load itself. Thus inductance energy is supplied back to the load itself. This process is called freewheeling. If load energy is fed back to the supply (mains), then it is called feedback. The energy of inductance goes on decreasing after n .
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Power Devices and Machines 3 - 1 1 Phase Controlled Rectifiers (AC/DC Converters)
Equivalent circuit - 1 Equivalent circuit - II
Fig. 3.2.7 Waveforms of half wave converter with freewheeling diode
Hence i0 also goes on reducing. At p the inductance energy is finished. Hence i0 becomes zero at p. Thus freewheeling diode conducts from n to p. The output is shorted due to freewheeling diode. Hence v0 - 0 whenever freewheeling diode conducts. This is shown in Fig. 3.2.7(c) also. During freewheeling Tj is off. Hence no supply current flows. Therefore
Power Devices and Machines 3 - 1 2 Phase Controlled Rectifiers (AC/DC Converters)
is = 0 during freewheeling period. T: conducts from a to n . Hence i0 = is from a to n as shown in Fig. 3.2.7.
Comparison between freewheeling diodes and feedback diodes
Sr.No.
Freewheeling diodes Feedback diodes
1. Load energy is utilized in load itself through freewheeling diodes.
Load energy is feedback to the source through feedback diodes.
2. Freewheeling diodes have to carry full load current.
Feedback diodes carry full load current some times.
3. Free wheeling diodes are slower. Feedback diodes should be fast.
Example 3.2.2 : Derive an expression for average value o f output voltage fo r 1 <j> half wave controlled rectifier for RL load and freewheeling diode.
Solution : Fig. 3.2.7(c) shows the output voltage waveform. From this we can write,
vs = Vm sin cot from a to n0 from 0 to a and n to2 n
The period of v0 is 2 n . The average value is given as,T
o(av)1 f 1= f j v0(at) d(0t = Vm sin <0/ d(.ot
0
^ [ “ coscot]
o(av) —j-[l + cosa] ,.(3.2.4)
Here note that average output voltage is same as that of resistive load given by equation 3.2.1. This is because output voltage waveforms are same in both the cases.
Example 3.2.3 : A single phase half wave controlled rectifier is used to supply power to 10 Q load from 230 V, 50 Hz supply at a firing angle o f 30°. Calculate - i) Average output voltage ii) Effective output voltage Hi) Average load current.
Solution : The given data is,
R = 10 Q, V, = 230 V 230 V2
a = 30° =
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Power Devices and Machines 3 - 1 3 Phase Controlled Rectifiers (AC/DC Converters)
i) To obtain average output voltage V0 (av)
The load is resistive. For this load V0tav\ is given by equation 3.2.1 as
V.,Vo(av) = ^ - ( 1 + c o s c x )
1 71
•22,042 ( , k
= ^ r [ 1+cos6
= 96.6 V
ii) To obtain effective output voltage v 0 ,™ s,
The rms value is given as,
Vo(rms) i j Vy (of) diot . 0
From the output voltage waveform of Fig. 3.2.2 we can write,
1 *— J v* sin2 cof dcotV.o(rms)
V i f 1 -co s 2 (0/J -------2-------di0t
m2n
V,m 4 71
/VJ dcot- J cos 2 ( 0 1 dcot
\Vm sin 2wf" n \1 I k a - 2I . a J
V. a sin 2 a 1 — + ——
71 271... (3.2.5)
This is an expression for effective rms value of half wave controlled rectifier. Putting values in above equation,
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Power Devices and Machines 3 -1 4 Phase Controlled Rectifiers (AC/DC Converters)
Example 3.2.4 : A single phase half wave converter is operated from a 120 V, 50 Hz supply and the load resistance R = 10 Q. I f the average output voltage is 25 % o f the maximum possible average output voltage calculate -
i) Delay angle a
ii) The rms and average output currents
Hi) The rms and average thyristor currents
iv) The input power factor.
Solution : Given data
Supply voltage Vs = 120; hence Vm = V 2xl20 = 169.7 V, Load resistance, R = 10 Q
Average output voltage VQ(av) = 25 % of V0 av maximum
i) To obtain delay angle a
The average output voltage of half wave converter is given by equation 3.2.1 as,
160.27 V
iii) To obtain average load current I 0(av)
The I0(av) can be calculated as,
_ K(av)*0(00) ' R
V0(av) be maximum at a = 0. Hence above equation will be,
n
It is given that the average output voltage is 25 % of its maximum valve, i.e.,
V o(av ) = 2 5 % o f V o(av )m ax
= 0.25 x 54 = 13.5 V
o(av )m ax
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Power Devices and Machines 3 - 1 5 Phase Controlled Rectifiers (AC/DC Converters)
Consider again equation 3.2.1
V.Vo(av) r^(l+COSCX)
2k v 7
Putting values in above equation,
169713.52 k
(l + cos a )
Solving above equation for a,
a = 2.09 radians = 120°
ii) To obtain rms and average output currents
Average value of output current is given as,
V.o(av) R
13.510
1.35 A
The rms value of output voltage is given by equation 3.2.5 for half wave converter, i.e.,
Vo(rms). a sin 2a 1 — + —
K 2 k
1697 1 209 | sin(2x2.09) ir 2ti
= 37.718 V
Hence rms output current will be,
o(rms)^o(rms)
R
3771810
3.77 A
iii) RMS and average thyristor currents
The waveforms of half wave converter are given in Fig. 3.2.2. There is only one thyristor and output current flows through this thyristor. Hence thyristor current is same as output current. Therefore rms and average valves of thyristor current will be same as these of output current, i.e.,
T(av) o(av)
and ^T(rms) ^o(rms)
1.35 A
= 3.77 A
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Power Devices and Machines 3 - 1 6 Phase Controlled Rectifiers (AC/DC Converters)
iv) To obtain input pow er factor *
Since the load is resistive, the rms value of load current will be same as rms value of supply current. Note that the same current flows in supply and load, i.e.,
^s(rms) = ^o(rms) = 3 .7 7 A
The total supply power will be,
Total supply power = Vs(rms) Is(rms)= 120 x 3.77 = 452.4 VA
The active load power will be,
V 2 Vo(av)Active load power = K
Power factor =
(13.5)'10
Active load power Total supply power
18.225 452.4
18.225
0.04 (lagging)
»>■► Example 3.2.5 : For a single phase half wave converter having resistive load o f 'R ' and
the delay angle o f , determine
i) Rectification efficiency
iii) Ripple factor
Solution : G iven data71
“ = 2
ii) Form factor
iv) PIV rating o f thyristor INOV.-2007,10 M arksl
o(av)
f e ( l + c o s f ) = 0.159
Vo(av)
o(av)R
0.159 VJ, R
V.o(rms) 2a sin 2 a
1 — + —«-----71 2 7t
/o Sm
K 2 71
= 0.353 VLV
o(nns)o(rm$) _ 0. 353 V„ ~R R
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Power Devices and Machines 3 - 1 7 Phase Controlled Rectifiers (AC/DC Converters)
i) To obtain rectification efficiency
P(tc ’ ^o(av)P V Iac o(rms) o(rm$)
= 0.2028 or 20.28 %
ii) To obtain form factor
V,o(rms)FFV,o(av)
iii) To obtain ripple factor
RF = -J fF 2 -1 = y j { 2 .2 2 ) 2 -1 = 1.982
iv) To obtain PIV rating
Peak value of supply voltage appears across SCR in negative half cycle. Hence
3.3 Single Phase Semiconverters (Half Bridge Converter)
The semiconverter is also called as half bridge converter.
3.3.1 Circuit DiagramFig. 3.3.1(a) shows the circuit
diagram of single phasesemiconverter. Observe that the semiconverter has two SCRs 7 and T2. There are two diodes D1 and D2. The input is 230 50 Hz AC supply. The output V0 o f the semiconverter is DC. The load 'R' is connected across the output.
Fig. 3.3.1 (b) shows isolated cathode configuration. Both the configurations of Fig. 3.3.1 are
Fig. 3.3.1(a) Circuit diagram of 1 (j> semiconverter (Common cathode configuration)
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Power Devices and Machines 3 - 1 8 Phase Controlled Rectifiers (AC/DC Converters)
A d ,
230 V, 50 Hz AC supply
o----------<Tv
Fig. 3.3.1 (b) 1 <}) Semiconverter (isolated cathode configuration)
symmetric. In case of common cathode configuration. The cathods of both the SCRs can be made common. But in case of isolated cathodeconfiguration, the gate drives of both the SCRs should be completely isolated. But both the circuits are functionally same.
3.3.2 Working with Resistive Load
Answer following question after reading this topic.
1. With the help o f circuit diagram and waveforms, explain the operation o f three phase semiconverter for 'R' load for a = 0°, 3 0 6 0 ° and 90°. Marks [10], D ec.-2006, 2008 5t likely and
asked in previous University Exam
Let us consider the working of 1<|> semiconverter having resistive load. In the positive half cycle of the supply, SCR Tj and diode D2 are forward biased. SCR Tj is triggered at firing angle a. Current flows through the load. The equivalent circuit is shown below.
Fig. 3.3.2 Conduction of 7 and D1 in positive half cycle of the supply. Dotted lineshows path of current flow
(i.e. supply voltage)
andV V0 _ YsR R
... (3.3.1)
... (3.3.2)
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Power Devices and Machines 3 - 1 9 Phase Controlled Rectifiers (AC/DC Converters)
Fig. 3.3.3 shows the waveforms of this circuit. The waveform of V0 is same as supply voltage Vs, when Tj -D ? conducts. Since the load is resistive, the output current waveform is same as voltage waveform. This is because,
Fig. 3.3.3 Waveforms of semiconverter with R-load
Thus amplitude of V0 is only reduced by the factor 'R' to give i0. But the shape of the current waveform does not change. In the Fig. 3.3.2 observe that iT1 is the SCR Tj current, and is is the supply current. Basically i0, iT1 and is is the same current. Hence,
i0 = is = iTi (when T7 -D 2 conducts)
These currents are in the same direction and flow in the same loop. The waveforms of these currents are also shown in Fig. 3.3.3. See Fig. 3.3.3 on previous page.
SCR Tj and diode D j conduct till n, at 7i supply voltage is zero. Hence current through SCR Tx drops to zero. Hence tums-off. After ti , the supply voltage is negative and Tx is reverse biased. Hence the output voltage V0 is also zero.
At rc+a, SCR T2 is triggered. It starts conducting, since it is forward biased because of negative cycle of the supply. The current i0 flows through load, T2 and D2. Such equivalent circuit is shown in Fig. 3.3.4.
Power Devices and Machines 3 - 20 Phase Controlled Rectifiers (AC/DC Converters)
Fig. 3.3.4 Conduction of T2-D2 in negative half cycle of the supply.Dotted line shows path of current flow
From the above equivalent circuit observe that positive of Vs is connected to positive of V0. Hence V0 remains positive even if supply polarity (i.e. negative cycle) is reversed. Hence we can write,
V0 = ...(3.3.3)
V Vand i0 = = ...(3.3.4)
In Fig. 3.3.4 observe that current through T2 flows in the same direction as i0. Hence iT2 Similarly i0 and is is the same current, but their directions are opposite as shown in Fig. 3.3.4. Hence,
= - *0
The waveforms of all the currents and voltages are shown in Fig. 3.3.3. At In , the supply voltage is zero. Hence T2 turns off. After 2 n T2 is reverse biased. Then Tj is triggered again at 2n + a and the complete cycle repeats.
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Power Devices and Machines 3 - 21 Phase Controlled Rectifiers (AC/DC Converters)
Example 3.3.1 : For the 1<|) semiconverter having resistive load o f 'R' determine the following :i) Average output voltage V0 av^ii) RMS output voltage V0(rms)
Solution : i) Average output voltage :
The average output voltage is given as,
0
Observe the waveform of output voltage in Fig. 3.3.3. It has a period n. Hence above equation can be written as,
In the above equation V0 (cof) = Vm sin cot from a to n . Solving the above integration we get,
a
... (3.3.5)
ii) RMS output voltage :
RMS output voltage is given as,
(rms)
Putting the values in above equation,
... (3.3.6)
This is the required derivation for rms value of output voltage.
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Power Devices and Machines 3 - 2 2 Phase Controlled Rectifiers (AC/DC Converters)
3.3.3 Working with Inductive (R-L) Load
Answer foUowing question after reading this topic
1. Draw the circuit diagram, voltage and current waveform fora = 60°, RL load o f semi-converter. Marks[8], May-2007
Most likely andasked In previous
University Exam
Normally the semiconverters are used to drive the DC motors. These motors are basically inductive (R-L) load. Hence it is necessary to consider the working of semiconverter with R-L load also. With the inductive load, the three modes are possible :
i) Continuous load current
ii) Discontinuous load current
iii) Continuous and ripple free current for large inductive load.
3.3.3.1 Continuous Current Mode
Answer following question after reading this topic.
1. How freewheeling is present inherently in the semiconverters?
In this mode, the current flows continuously in the load because of inductive effect. The waveforms of load current and load voltage are shown in Fig. 3.3.5. In these waveforms observe that SCR T. and diode Dj conducts from a to it. Since the load is inductive current keeps on increasing (saturating) and it is maximum at k . At n, even though the supply voltage is zero, current doesnot go to zero. This is because load inductance opposes this sudden change of current. The load inductance generates a large voltage so as to maintain load current. This current flows through T, and D2 . The equivalent circuit of this operation is shown in Fig. 3.3.6. The SCR Tj conducts even after
n, since it is forward biased due to voltage induced in the load inductance i.e. L —. Diode
D2 is also forward biased due to this voltage. Hence current does not flow through supply i.e. is when freewheeling action takes place. Thus the energy stored in the load inductance is fedback to load itself in freewheeling action.
SCR T2 is triggered at rc+a and the output current starts increasing. Since the current i0 is continuous, it is called continuous current mode of semiconverter. The similar
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Power Devices and Machines 3 - 23 Phase Controlled Rectifiers (AC/DC Converters)
Fig. 3.3.5 Waveforms of 1<J> semiconverter for continuous load current
operation takes place when T2 and D2 conducts in negative half cycle of the supply. Fig. 3.3.5 shows supply current (/..), freewheeling current and other waveforms for inductive load. Note that the output voltage waveform remains same. If there is freewheeling diode in semiconverter, then freewheeling current flows through this diode.
Power Devices and Machines 3 - 2 4 Phase Controlled Rectifiers (AC/DC Converters)
^ D2|___j ^ D1
Fig. 3.3.6 Freewheeling action takes place through T y D2
Average value of output voltage with inductive load
Compare the output voltage waveforms of Fig. 3.3.3 (resistive load) and Fig. 3.3.5 (inductive load). The voltage waveforms are same. Hence average and RMS values of output voltage are also same. i.e. for inductive load,
From equation 3.3.5
From equation 3.3.6
V.V,
o(av) + COS « ) ... (3.3.7)
l\ V 2V - 1 m o(rms) 1 2 7l 7 i-a + i sin 2 a r ... (3.3.8)
3.3.3.2 Discontinuous Current ModeIn this mode, the current through the load becomes zero for some duration. Hence it is
called discontinuous current mode. Fig. 3.3.7 shows the waveforms of discontinuous current mode of semiconverter. (See Fig. 3.3.7 on next page).
As shown in above waveforms, Tj -D ? conducts from a to k and the load current i0 goes on increasing. At n supply voltage is zero. But because of inductance, i0 does not go
to zero. The load inductance induces a large voltage L to maintain current in the same
direction. Hence i0 continuous to flow and it goes to zero at p. Since next SCR T2 is triggered at 7i+a (See Fig. 3.3.7), output current is discontinuous. Freewheeling takes place from 7i to p. The freewheeling current flows through Tj and D2- Similar operation repeats in next half cycle.
Observe that the voltage waveform remains same in discontinuous mode also. Hence ^o(av) anc* ^o(kms) are same as tf*at o f resistive load.
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Power Devices and Machines 3 - 25 Phase Controlled Rectifiers (AC/DC Converters)
Fig. 3.3.7 Discontinuous mode of single phase semiconverter
3.3.3.3 Continuous and Ripple Free Current for Large Inductive Load
Answer following question after reading this topic
With the help o f neat circuit diagram, m ode equivalent circuits and waveforms o f supply voltage, supply current, output voltage, output current, explain the operation o f a single phase half controlled bridge feeding a level (highly inductive) load.
Marks [5], D ec.-2000; Marks [10], May-2006
N Most likely and asked in previous
University Exam
As the load inductance increases, the ripple in i0 reduces. When the load inductance is very large, the ripple in i0 will be negligible. And i0 can be treated as continuous and ripple free. Fig. 3.3.8 shows the waveforms of Ity semiconverter for large inductive load. The load current is continuous and ripple free. Observe that the output voltage waveform is same as resistive load. But the current waveforms are different.
The output current is constant DC of amplitude I^ avy The SCRs conduct for n radians. Hence SCR current is square wave. The supply current has the amplitudes of i ^ avy The supply current is zero whenever freewheeling action takes place.
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Power Devices and Machines 3 - 26 Phase Controlled Rectifiers (AC/DC Converters)
Fig. 3.3.8 Waveforms of 1<J> semi converter for highly inductive load
Example 3.3.2 : Derive an expression for output current for RLE load driven by 2<j> semiconverter. Assume continuous conduction.
Solution : Fig. 3.3.9 shows the circuit diagram of 1 <J> semiconverter for RLE load.(Fig. 3.3.9 see on next page).
Normally, the RLE load is motor load. L is the inductance of the motor and R is the resistance of the inductance. E is an induced emf in the motor. The waveforms of this circuit will be similar to those shown in Fig. 3.3.5. From a to n, T|- D1 conducts and
Power Devices and Machines 3 - 2 7 Phase Controlled Rectifiers (AC/DC Converters)
vs = Vm sin
Fig. 3.3.9 A 1<J> semiconverter driving RLE load
supply voltage vs is applied to the load. Hence an equivalent circuit will be as shown below :
Vm sin cot 0
Fig. 3.3.10 Equivalent circuit when T y D 1 or T2- D2 conduct
By KVL in above circuit we get,
di.Awt)Vm sin cof = R io l(<at) + L —2i_— + £
This equation can be solved using laplace transform. The solution is,
<01(cof) = ^ - s in ( w ( - 0 ) + io l(O)e ' L ... (3.3.9)
Here Z = 2 (toil)2
e -
/ol(0) is initial current at cof = a.
From k to ;c+a freewheeling takes place. T^D2 conduct in this duration. The equivalent circuit is shown in Fig. 3.3.11.
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Power Devices and Machines 3 - 2 8 Phase Controlled Rectifiers (AC/DC Converters)
Forji to n + a
Fig. 3.3.11 Freewheeling action in T^D2 or 72D 1
By KVL to this circuit we can write,
dio2((*t)R io2((ot) + L
dt+ E = 0
This equation can be solved using laplace transform. The solution is,
Eio2( «>t) = 1 —n ( l~ e L )o2 R (3.3.10)
Here iQ2(0) is the initial current at cof = n . In the waveforms of Fig. 3.3.10 and Fig. 3.3.11 observe that,
and /oj(0)=/o2(coi=a)... (3.3.11)
Putting the above two conditions in equation 3.3.9 and 3.3.10 we can get the initial values. Then two currents /’ol(cof) and io2((ot) are separately expressed for semiconverter.
Example 3.3.3 : For a h j> half bridge converter having highly inductive load, derive the follozving :i) Fourier series for supply currentii) RMS value o f nth harmonic o f supply current.iii) Fundamental component o f supply currentiv) RMS value o f supply current. |Nov.-2007, 8 Marks; May-2008, May-2006, 6 Marksl
Solution : i) To determine Fourier series
The general expression for Fourier series is given as,
CO'S(“ 0 = h ta v ) + X c n sin(mof+ <(>„)
;»= 1
where ctJ = yja* +b%
and = tan -1 n
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Power Devices and Machines 3 - 29 Phase Controlled Rectifiers (AC/DC Converters)
Here an y J is (o t) COSMof rfcof
2n:
2kJ is (iot) cosncof if cof
From the supply current waveform of Fig. 3.3.8 we can write,
2 n_2_
2n | l <n<w) cosnmt dwt+ | (-/ ^ Jco sM w f rfco/
2nJ coswof rfcof- J COS H(Ot d(Ot
° ai sin ?;a(l -co snit)tin
-21 o(av) .s in H a
77710
for /i=l, 3,5,
for H = 2 ,4 ,6 ,
The above equation shows that an is zero for even harmonics of supply current.
bn is given as,
~ t
j j 's(w0 sin/7cof d o to
Putting values of T = 2n and /s (cof) from supply current waveform of Fig. 3.3.8,
2n_2_2k \ ! c(av) sin n(0t d(ot + J ) sin TUot do)t
n + a
o(av) In| sin n ot d o t - J sin n ot d o t
<t<w)7171
21
(1 + COS77a) (1 - COS/171)
tlK0
(1 + cos 77 a) for 77 = 1, 3,5,
for 77 = 2 ,4 ,6 ,,
(3.3.12)
(3.3.13)
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Power Devices and Machines 3 - 3 0 Phase Controlled Rectifiers (AC/DC Converters)
The above equation shows that bn is zero for even harmonics of supply current.
Hence cn = +b%
21o(av) .sm nann
4 l 0(av) na— ------ cos—nn 2
21o(av)
nn (1 + cosna)
for n = 1, 3, 5, ... (3.3.14)
This equation gives peak value of nth harmonic of supply current. And <j>„ can be
calculated as,
<fci = tan ~ ' run
tan
= -tan
-21 c{av) .sm nann
21c(av)
nn (1 + cosna)
na~2 ... (3.3.15)
Observe the supply current waveform of Fig. 3.3.8. It has symmetric positive and negative half cycles. Hence its average value is zero. This can also be verified mathematically as follows.
TI (av) is ((ot)d(ot
Here T = 2n and putting values of i$ (cof) from Fig. 3.3.8,
s (av)_1_2n o(av) dwt
o(av)2n
n 2 nJ dcot - J dcot
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Power Devices and Machines 3 - 31 Phase Controlled Rectifiers (AC/DC Converters)
o(<rc>)271 [7i + a -2 jc-(7c+ a))
0
Thus the average value of symmetric waveform is zero
Thus the Fourier series can be written as,
CO 4/*s(“ f) = Z
o((iv) n a .cos - y sm
tin ( — f )h= 1,3,5,—
ii) RM S value of n lh harm onic of supply current
The rms value of the nth harmonic is given as,t l s M cosm .
j _ _ nn______2_>/2 V2
= 2^ Io(av) cos^“ n = 1, 3, 5........h k z
°*9 l o(av) _ n a _ , 0 c = -----------cos — / n =1,3,5
iii) Fundam ental com ponent of supply current
The fundamental component of supply current is obtained by putting n equation 3.3.17. i.e.,
h i = 0 .9 1 ^ cos |
iv) To obtain rms value of supply current
The rms value is given as,
T
s(rms)
With T = 2 k and putting for is (cof) from supply current waveform of Fig. 3.3.8,
J * $ « * * * + J (’ U a v ))2 ^s(rms) 2 71 rr + a
. (3.3.16)
...(3.3.17)
= 1 in
...(3.3.18)
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Power Devices and Machines 3 - 3 2 Phase Controlled Rectifiers (AC/DC Converters)
/ oiav)'y n ... (3.3.19)
The above equation shows that rms value of supply current depends on a.
)>*► Example 3.3.4 : For a 1$ half controlled converter having highly inductive load, derive the followiitg :
ii) Supply power factor (PF)The supply power factor is given as,
PF = sl cosfys(rms)
Putting the values of /sl (equation 3.3.18), / rws) (equation 3.3.19) from previous example and 4>1 above we get,
i) Displacement factor (DF) ii) Supply power factor (PF)iii) Harmonic factor (HF) iv) Current distortion factor (CDF)
Solution : i) Displacement factor
The displacement factor is given as,DF = cos <(>!
From equation 3.3.15, <(>„ = Hence 4>1
DF = cos - j ... (3.3.20)
... (3.3.21)
iii) Harmonic factor
The harmonic factor (HF) is given as,
Putting values in above equation,
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Power Devices and Machines 3 - 33 Phase Controlled Rectifiers (AC/DC Converters)
SIo(av) 2 « ^ COS 2
H F=8 cos
This is an expression for harmonic factor of supply current.
.. (3.3.22)
iv) Current distortion factor (CDF)
The current distortion factor (CDF) is given as,
ICDF = sl
's(rms)
2V2 /d a v ) a--------- COS —71 2
c{av)In - a
\ n
2V2 COS^
yjn(n-a)...(3.3.23)
))»► Example 3.3.5 : For a 1 4> half controlled bridge having continuous and ripple free current, obtain, i) Active power and ii) Reactive power.
Solution : i) Active power
Active power is given as,
PacUve = V s JS1 c °S<t»j
2 - J l lVs'
aCOS c o s I -
^ Vs Jo(av) 2 a— — 2cos 2
(1 + cos a) (3.3.24)
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Power Devices and Machines 3 - 3 4 Phase Controlled Rectifiers (AC/DC Converters)
ii) Reactive power
Reactive power is given as,
^reactive ~ Vs h i s*n 1
v - . Vs ---------------cos — sinH -?)& Vs ^o(av) . a a= -------------------2 sin — cos—
71 2 2
_ _ VmId ‘>v) s in a ...(3.3.25)71
The negative sign indicates that power is reactive.
Comments
i) Active power is consumed by the load.
ii) Reactive power is not consumed by the load. Hence its sign is negative.
iii) Reactive power fluctuates between load and source.
iv) Total power includes active as well as reactive power.
»*► Example 3.3.6 : Single phase semiconverter is operated from 120 V, 60 Hz supply. The load current with an average value o f In is continuous with negligible ripple content. Turns
ratio o f transformer is unity. The delay angle a=-^. Calculate-
a) Harmonic factor o f input current
b) The displacement factor
c) Input power factor
Solution : The given data is,
Vs = 120 V
71“ = 3
a) Harmonic factor is given by equation 3.3.22 as,
HF =o 2 a8 cos - j
Putting the values in above equation,
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Power Devices and Machines 3 - 3 5 Phase Controlled Rectifiers (AC/DC Converters)
HF
8 cos'
= 0.3108 or 31.08 %
b) The displacement factor is given by equation 3.3.20 as,
DF = cos^
= cos7t/3^
0.866
c) The input power factor is given by equation 3.3.21 as,
I 8 2 aj « ^ a ) COS 2
( n/3
PF
8 2 cos7t 71-
= 0.827 lagging
3.3.4 Asymmetrical Half Bridge Converter
3.3.4.1 Operation with Resistive LoadFig. 3.3.12 shows the two configurations of asymmetrical half bridge converter. Note
that both the configurations are functionally same. In both of these circuits the two SCRs appear on the same link.
(a) (b)Fig. 3.3.12 Single phase controlled rectifier
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Power Devices and Machines 3 - 3 6 Phase Controlled Rectifiers (AC/DC Converters)
When Tj is triggered, current flows through T1 and D j. T2 is triggered in the negative
half cycle. Then current flows through T> and D2.
Fig. 3.3.13 shows the waveforms of half bridge converter given in Fig. 3.3.12. These
waveforms are shown for resistive load and a = Observe that the output current
waveform is similar to output voltage. Since T| and D conduct simultaneously their current waveform is same. Similarly, the current waveform of T2 and D2 is same.
Supptyvoltage
Firing pulses of T .
----
Firing pulses otTj
Outputcurrent
S C R T , &
diode D, current
~—~ r
s c r t 2 &d io d e D2 current
irtt
Fig. 3.3.13 Waveforms of half bridge converter of Fig. 3.3.12
Power Devices and Machines 3 - 37 Phase Controlled Rectifiers (AC/DC Converters)
Since the above output voltage is same as that of single phase semiconverter, the rms and average values of output voltage will be,
VLo(av) (1 + cos a)
Vc{rms) 2nrt - a + - sin 2a
3.3.4.2 Operation of Asymmetrical Half Bridge Converter with Level Load
Answer following question after reading this topic
1. Draw the circuit diagram and wauefrorms o f output voltage, output current, supply current and SCR currents for a single phase asymmetrical half controlled bridge feeding a level load. I >
, Marks [6], M ay-2004.2005 J asked inV University Exam
With the similar circuit diagram of Fig. 3.3.12 but for highly inductive load, the operation of asymmetrical converter will be as follows.
M ode - I ( a < c o t<n )
SCR Tj is triggered in this mode. Hence load current flows through T^DV The waveforms are shown in Fig. 3.3.14. (See Fig. 3.3.14 on next page).
Mode - II (it < cot < n + a)
In this mode, the supply voltage becomes zero at n. Hence Tj is no more forward biased. But due to highly inductive load, the constant current is maintained to flow. This load current flows through D - D2. The equivalent circuit is shown in Fig. 3.3.14. Thus the freewheeling action takes place through D j - 0 2 and supply current as well as output voltage are zero.
M ode - III (71 + a < cot < 2 ;r)
SCR T2 is triggered at 7i+a. Since T2 is more forward biased due to supply voltage, it starts conducting. The load current now flows through T2 -D 2. The equivalent circuits - III in Fig. 3.3.14 shows the current flow.
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Power Devices and Machines 3 - 38 Phase Controlled Rectifiers (AC/DC Converters)
r,,ns~
t tEquivalent Circuit - Equivalent Circuit - II
z f z ; ° z f z;»:
f iEquivalent Circuit - III Equivalent Circuit - IV
Fig. 3.3.14 Waveforms of asymmetrical half controlled bridge converter for level load
Power Devices and Machines 3 - 3 9 Phase Controlled Rectifiers (AC/DC Converters)
Mode - IV (2rc < a>t < 2n + a )
At 2 k, the supply voltage becomes zero. Therefore T2 tums-off. But due to heavy inductive load, the current continuous to flow. This current now flows through Dj -D 2 since they are more forward biased compared to T2 -D 2.
At 271+a, SCR T j is triggered again and mode-I starts. Thus the cycle repeats.
Mathematical analysis
Observe that the waveform of output voltage is same as that of semiconverter. Hence the rms and average values of its output voltage are,
o(av)Vn,= (1 + cos a)
71
and Vo(rms)VL2tc
7 i - a + ^ s in 2a
)>*► Example 3.3.7 : For the single phase asymmetrical half controlled bridge circuit derive expressions for
i) Average output voltage ii) RMS output voltage
iii) RMS value o f the nth harmonic supply current
iv) Supply current distortion factor.[May-2004, 2005, 2008,10 M arks!
Solution : Observe the waveforms of semiconverter (Fig. 3.3.8) and asymmetrical configuration of semiconverter (Fig. 3.3.14). The waveforms of output voltage and supply current are exactly same. Hence the above parameters will be same as that of semiconverter, i.e.,
— (1 + cos a)i) Average output voltage, VQ{av)
ii) RMS output voltage, Vo{rtns) =V 2vm2 k
7 i-a + - s in 2a
4/£
iii) n harmonic supply current, Isn = —jLy/2
2V2iv) Current Distortion Factor, CDF =
o(av) na------ cos —7171 2
V2^^^o(av) na------------------C O S^ r-
7171 2
COSa
J k(k - a)
)>!■► Example 3.3.8 : Draw the circuit diagrams o f symmetrical and asymmetrical single phase
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Power Devices and Machines 3 - 40 Phase Controlled Rectifiers (AC/DC Converters)
half-controlled bridge rectifiers and sketch the SCR and diode current waveforms for each circuit for level loads. From these waveforms, derive an expression for the ratio o f average SCR current to average diode current. [Dec.-2003, 8 Marks]
Solution : The circuit diagram of symmetrical configuration is given in Fig. 3.3.1(a). The waveforms are given in section 3.3.3.3 for level loads.
The circuit diagram of asymmetrical configuration is given in Fig. 3.3.12. The waveforms are given in section 3.3.4.2 for level loads.
SCR and diode currents for symmetrical configuration
Fig. 3.3.15 shows the SCR and diode currents for symmetrical configuration.
Average SCR current will be,
Fig. 3.3.15 Symmetrical configuration of 1<j> HCB, VQ, i T and / 01 waveforms
h(av) = \ \ 'r M d w f = ^ } I0(av)dcof = - ^0 a
Similarly average diode current will be,
i 1 f r j . _ ^o(av)o(au) ~ 2 ^J ! o(av)d<ot-----Y ~
a
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Power Devices and Machines 3 - 41 Phase Controlled Rectifiers (AC/DC Converters)
^o(av)Average SCR current 2 j
Average diode current 0{av)
T ~
SCR and diode currents for asymmetrical configuration.
Fig. 3.3.16 shows the SCR and diode currents for asymmetrical configuration.
i* --1
Fig. 3.3.16 Asymmetrical configuration of 1<j> HCB, VQ, i T and i waveform
Average SCR current will be,
1h ( a v ) = j \
0
a
Average diode current will be,
^D(av) 2n i* ^o(av) 2n0
t n - aAverage SCR current _ ' o(av) 2n _ n - a
Average diode current r n + a n + a° ( a v > 2 k
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Power Devices and Machines 3 - 42 Phase Controlled Rectifiers (AC/DC Converters)
3.3.4.3 Comparison of Symmetrical and Asymmetrical Configurations
Answer following question qfter reading this topic
1. Explain different configurations o f semiconuerter andY compare them. Marks [6J, D ec.-2000, May-2000, 2003■■’TiTTi— \ Most likely and
asked in previous University Exam
Table 3.3.1 shows the comparison between symmetrical and asymmetrical configurations of half controlled bridge.
Sr. No. Symmetrical configuration Asymmetrical configuration
1. One SCR is connected on each link. Both the SCRs are connected on single link.
2. SCRs can be driven with common cathode. SCRs must have isolated cathodes.
3. Freewheeling takes place through on diode and on SCR.
Freewheeling takes place through both the SCRs.
4. Average currents of SCR and diodes are same.
Average currents of diodes are higher than SCR.
5. SCR and diodes conduct for equal durations. SCRs conduct for shorter duration compared to diodes.
Table 3.3.1 Comparison of symmetrical and asymmetrical configuration
Example 3.3.9 : A single phase half controlled bridge rectifier operates from the 115 V, 60 Hz mains and supplies a resistive load o f 250 Cl For firing angles o f 45° and 135°,
Calculate :
i) Average output voltage ii) nns output voltage
iii) Load power iv) rms supply current
v) Peak supply current [Dec.-2004, 18 M arks!
Solution: Given : Half controlled bridge
s(rms) = 115 V, therefore Vm = 42V ${rm) = V 2 x ll5 = 162.6 V
R = 250 0
a l = 45° or K4
3tca 2 = 135° or T
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Power Devices and Machines 3 - 43 Phase Controlled Rectifiers (AC/DC Converters)
i) Average output voltage
For a = |, Vo(av) = ^ - ( l + co sa )
— ~ ( l + cos 1 1= 88.35 V
F o r a « - J , V0(av)
ii) RMS output voltage
o(nns)m
2n7 t-a + ^ sin 2a
For a = V0{rms) 162.6J27C
n 1 . 7t\^ " 4 2 ( 4 J 109.63 V
3ti
* - T * 234.65 V
iii) Load power
For a
For a =
4'
3n7 '
P' =U 2o(ai>)
R
(88.35)2 250
(15.16)2 250
= 31.22 Watt
0.919 Watt
iv) RMS supply current
For 1 half bridge inverter with resistive load,
F o r a = 4 ' hirm s)
h(rm s) ^o(rms)
^o(rtns)R
109.63 250
0.438 A
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Power Devices and Machines 3 - 4 4 Phase Controlled Rectifiers (AC/DC Converters)
« = T ' W ) = W = 01386 A
v) Peak supply current
• The supply current will be maximum, when output current is maximum, i.e.
s(max) o(max)*
• Now the output current will be maximum when output voltage is maximum.
• For a = peak value of output voltage is Vm Hence,
_ Vm _ 162.6ls(peak) - R ~ 250 "
• For a = peak value of output voltage is Vm sin Hence,
t / 3tislrl ~T~ 162.6 x 0.7071 . . . 4
W ) = --------R = --------- 2 5 0 --------- = 0 4 6 A '
Example 3.3.10 : A single phase half controlled bridge rectifier supplies a ripple free load current o f 10 A and operates from the 110 V, 60 Hz mains. I f the average output voltage is 75 V calculate :
i) Firing angle ii) rms output voltage
iii) rms supply current iv) rms 7^ harmonic supply current
v) Supply power factor. [Dec.-2003,16 Marksl
Solution : Given : I0 av) = 10 A ripple free
= 110 V, A Vm =V 2V S = V 2 x 110= 155.56 V
i) Firing angle
= 75 V.
V,Vo(av) = - f - d + cosa)
- - 155.56 n .7d - -------- (1 + cos a)K
a = 1.03 radians or 59°
ii) RMS output voltage
I V 2V = -Io(rms) } 2 JT
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Power Devices and Machines 3 - 4 5 Phase Controlled Rectifiers (AC/DC Converters)
f (155.56)2 f ____ 1 .271
7c -1.03 + 2 sin( 2 x 1.03)
l2
= 99.15 V
iii) RMS supply current
j _ j , ~ ~ ,..-1 .03s(rms) ~ o(av)
= 8.198 A
iv) RMS 7th harmonic supply current
V2dropped since it is rms value,
v) Supply power factor
o{av) 7 a— =-----COS r-7ti______ 2_V2
4x10 (7x1.03)-= — cos-----=-----
' 71 -----= - 1.15 A Here negative sign can be
PF = a ) cos2 ^ ... By equation 3.3.21
8 2f 1.03^7c(ti-1.03) COb [ 2 J
= 0.83
))»■► Example 3.3.11 : A single phase HCB operated from the 230 V, 50 Hz mains feeds a resistive load o f 100 Q. If the firing angle is 60°, calculate,
i) Average output voltage ii) RMS output voltage
iii) Total output power iv) DC output power
v) Load current at instant o f turn-on i.e. cot = a .
iv) Peak load current. lM ay-2003,12 M arks!
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Power Devices and Machines 3 - 46 Phase Controlled Rectifiers (AC/DC Converters)
Solution : Given : l<f> HCB
Vs = 230 V, vm = J l V s = 7 2 x 230 = 325.27 V
R = 100 n , a = 60° or |
i) Average output voltage
ii) RMS output voltge
206.3 V
iii) Total output power
Vlrms) (206.3)2 R 100
= 426 Watt.
iv) DC output power
P,o(DC) R
(155.3)2 100
= 241.18 Watt.
v) Load current at the instant of turn-on
, _ vo(tot)0 R
Vm sin cotR
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Power Devices and Machines 3 - 4 7 Phase Controlled Rectifiers (AC/DC Converters)
325.27 sin ^ n= ------ jog— - by putting cot = a = -j
= 2.816 A.
vi) Peak load current
Since SCRs are triggered at a = - , the supply peak voltage occurs at ot=^. Therefore
load current will be at its peak when cof = « i-e.,
_ o(peak) o(peak) ~ r
V cin 325.27 sin -JVm sin co t __________ 2 = o 9 s AR 100
Example 3.3.12 : A single phase semiconverter operates with 230 V, 50 Hz ac input and supplies level load current o f 10 A, operated at firing angle o f 60°. Calculate :
i) RMS supply current ii) Output voltage
iii) Supply power factor v) RMS value o f third harmonic input current.[Dec.-2000, 8 M arks; Dec.-2006, 10 M arksl
Solution : Given : 1<|> HCB
vs = 230 V, Vm = V2 1/ = V2 X 230 = 325.27 V
'< «> = 10 A' « = 60° or |
i) RMS supply current
ln -a) v 71
= 1071
* " 31 71
8.165 A
ii) Output voltage
o(av)V„<n,A = (1+ cos a)
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Power Devices and Machines 3 - 49 Phase Controlled Rectifiers (AC/DC Converters)
3.4.1 Working with Resistive Load
Fig. 3.4.2 Conduction of 7 and T2 in positive half cycle of the supply.Dotted line shows path of current flow
Let us consider the working of 14> bridge (Full) converter with resistive load. In the positive half cycle of the supply SCRs Tj and T2 are triggered at firing.angle a. Hence current starts flowing through the load. The equivalent circuit for this operation is shown in Fig. 3.4.2.
It is clear from Fig. 3.4.2 that, when T{ and conducts,
V0 = Vs (i.e. supply voltage) ... (3.4.1)
V Vand , '» = i f = T - (3A2)
Fig. 3.4.3 shows the waveforms of this circuit. Observe that load voltage is same as supply voltage from a to n. Since the load is resistive, waveforms of V0 and i0 are same. The supply current i$ and i0 are in the same direction hence i$ =i0. T] and T-, turn off when supply voltage becomes zero at n. In the negative half cycle T3 and T4 are triggered at7c+a.
Fig. 3.4.4 shows the equivalent circuit when T3 and T4 conduct.
In the adjacent figure observe that supply current is and load current i0 flow through the same loop. But directions of i$ and i0 are opposite hence
h = -*o
The supply current waveform is also shown in Fig. 3.4.3. T3 and T4 turn off when supply voltage becomes zero at 2 k . At 2 k + a, Tj and T2 are triggered again and the cycle repeats.
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Power Devices and Machines 3 - 50 Phase Controlled Rectifiers (AC/DC Converters)
Power Devices and Machines 3 - 52 Phase Controlled Rectifiers (AC/DC Converters)
Solution : This is a fully controlled bridge with resistive load of 100 Q in series with the battery of 50 V. Hence output voltage of the converter appears across resistance of 100 Q and battery of 50 V. Hence let us first calculate average value of output voltage. The given data is,
a = 30°
Vs = 220 V /. Vm = 220V2
The average output voltage for resistive load is given by equation 3.4.3 as,
Vo(av) = ~ < 1+C°S«)
= - ( 1 + cos 30°)71
= 184.8 V
This voltage is applied to the load. Fig. 3.4.6 shows the equivalent circuit.
By applying KVL to above circuit,
Vo(av)184.8
o(av)
= '«,(,»)* + 50 = '<,(*>) * l« > +50= 1.348 A
Thus the current through the load is 1.348 A.
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Power Devices and Machines 3 - 53 Phase Controlled Rectifiers (AC/DC Converters)
3.4.2 Working with Inductive LoadThe inductive load means resistance and inductance in the load. Such loads are DC
motors. Because of the inductive (R-L) load, the load current shape is changed. Hence operation of the full bridge converter can be discussed into three modes :
i) Continuous load current
ii) Continuous and ripple free current for large inductive load
iii) Discontinuous load current
3.4.2.1 Continuous Load Current
In the continuous load current, the load or output current i0 flows continuously. The waveforms are shown in Fig. 3.4.7.
Fig. 3.4.7 Waveforms o f 1(}> full converter fo r inductive load having continuous load current
Power Devices and Machines 3 - 5 4 Phase Controlled Rectifiers (AC/DC Converters)
As shown in the waveforms of Fig. 3.4.7, Tj and T2 conduct from a to n . The nature of the load current depends upon values of R and L in the inductive load. Because of the inductance, i0 keeps on increasing and becomes maximum at ti . At k , the supply voltage reverses but SCRs T| and T2 does not turn off. This is because, the load inductance does not allow the current i0 to go to zero instantly. The load inductance generates a large
r dinvoltage Ldt
This voltage forward biases Tj and T2 as shown in Fig. 3.4.8. In Fig. 3.4.8 observe that the load current flows against the supply voltage. The energy stored in the load inductance is supplied partially to the mains supply and to the load itself. Hence this is also called as feedback operation. The output voltage is negative from n to n + a since supply voltage is negative. But the load current keeps on reducing.
from Ti to ti + cx due to inductance voltage At n+ a, SCRs T3 and T4 are triggered.The load current starts increasing. The load
current remains continuous in the load. The similar operation repeats. The ripple in the load current reduces as the load inductance is increased.
3.4.2.2 Continuous and Ripple Free Current for Large Inductive Load
s,
Answer follow ing question after reading this topic
1. Draw the circuit diagram of a single phase fully controlled bridge rectifier and sketch the waveforms o f output voltage, output current, supply current and SCR current for a level (ripple free) load. Marks [5], May-2000. 2001 ; Marks [10]. D ec.-2004
Most likely and asked In previous
University Exam
Now let us consider the case when there is large inductance in the load. Because of the large inductance, the ripple in the load current is very small and it can be neglected. Hence load current will be totally DC as shown in Fig. 3.4.9.
In the waveforms shown in Fig. 3.4.9, there is no effect on output voltage waveform for large inductive load. The supply current waveform (/s) is square wave for large inductive load.
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Power Devices and Machines 3 - 5 5 Phase Controlled Rectifiers (AC/DC Converters)
Fig. 3.4.9 Waveforms o f 1<J> fu ll converter fo r continuous andripplefree load current in case o f large inductive load
))*► Example 3.4.3 : For the 1<)> full converter having inductive load and continuous load current, obtain the following :i) Average output voltage V0 av^ii) RMS output voltage V0 rms [Dec.-2004, 3 Marks]
Solution : i) Average output voltage fo r inductive load
The average output voltage is given as,
1 T
Vo(av) = f J vo (“00
Observe the waveforms of l<f> full converter for inductive load given in Fig. 3.4.7 and Fig. 3.4.9. The output voltage waveform has a period from a to 7t+a ; i.e. n. And vQ (cot) = Vm sin (ot during this period. Hence above equation becomes,
j 7i+aVo(av ) = - J Vm sin d(s>t
a
r .nTi+a= — - COS (0 1K 1 J«
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Power Devices & Machines 3 - 56 Phase Controlled Rectifiers (AC/DC Converters)
2 V„o(av) cos a ... (3.4.5)
This is the expression for average load voltage of l<f> full converter for inductive load.
Plot of V0(av) versus firing angle (a)
Following table lists the values of VQ/av\ with firing angle (a)
a Vo{*v) = Km c o s a
0 2V- f = 0 637 Vm
30° 0.55 Vm
60° 0.318 Vm
90° 0
120° - 0.318 Vm
150° - 0-55 Vm
180° - 0-637 Vm
Table 3.4.1 VC(av,) w ith respect to aObserve that VG (av) is positive for a < 9&. Hence it is called rectification. For a > 0,
V0 (av) is negative. Hence it is called inverting mode of operation. In inverting mode, output energy is fedback to the source.
Fig. 3.4.10 Variation o f VD ^ w ith respect to a
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Power Devices & Machines 3 - 5 7 Phase Controlled Rectifiers (AC/DC Converters)
ii) RMS value o f output voltage for inductive load
The rms value is given as,
J \ vo (<°0 d(ot
2
0
2- | V2 sin2 cof d cof
a
1
a
... (3.4.6)
Thus the rms value of load voltage is same as rms value of the AC supply voltage.
»»► Example 3.4.4 : Draw the circuit arrangement o f a single phase full converter feeding a general load comprising o f R, L and E. Sketch the AC supply voltage o/p voltage and the load current waveforms. Assuming continuous load current operation, derive an expression for DC output voltage.A single phase full converter feeding an RLE load is fed by 230 V, 50 Hz mains.If R = 0.5 Q, L = 8 rnH and E = 50 volts, assuming that conduction is continuous and firing angle is 4 0 find average value o f load current.
Solution : Circuit diagram and waveforms
Fig. 3.4.11 shows the circuit diagram of full converter supplying RLE load.
vs = 230 V. 50 Hz
E = 50
R = 0.5 Q
L = 8 mH
Fig. 3.4.11 14> fu ll converter feeding RLE load
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Power Devices and Machines 3 - 58 Phase Controlled Rectifiers (AC/DC Converters)
The RLE load is normally motor load. 'R' is the resistance and 'L' is an inductance of armature winding of the motor. 'E' is the induced emf of the motor. When the load current is continuous, then waveforms of this circuit will be similar to that of RL load. Hence with small ripple in output current, the waveforms of this circuit will be similar to those shown in Fig. 3.4.7. Note that 'E' is not reflected in the waveforms as long as output current (i0) is continuous.
If output current (iQ) is constant and ripple free, then the waveforms will be similar to those shown in Fig. 3.4.9.
RMS and average output voltage
The output voltage waveform remains same with RL load and RLE load when i0 is continuous. Therefore the rms and average values of output voltage will be same as those derived in previous example for RL load, i.e.,
2VVo(av) = - f - cos a
vV , v = -2L = Vv o(rms) ^ 2 s
Second part : To obtain average load current
The ripple in the load current (i0) depends upon values of R, L and E. If load inductance is small, then iG can become discontinuous. In Fig. 3.4.7, observe that iQ repeats at the intervals of ;r . The waveform of i0 remains same whenever Tj-Tj or T3-T4 conducts. Hence in any interval (i.e. a < cof < a or rc+a < cof < 2n+a) the equivalent circuit will be as shown below.
Fig. 3.4.12 Equivalent c ircu it when TyT2 or r 3-T4 conduct
By applying KVL to above circuit,
Vm sin cof = Ri0 + L + E
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Power Devices and Machines 3 - 6 0 Phase Controlled Rectifiers (AC/DC Converters)
a = 40° = 0.698 radians
and Z = -Jr 2 +(coL )1
= ^(0.5)2 +(2.513)2 = 2.5622
Putting values in equation 3.4.8 we get i0(0) as
50 0.5
= 162.48 A
This is the minimum value of output current. If this value becomes negative, then it indicates discontinuous operation.
Putting values in equation 3.4.7 we get equation for i0(ot). i.e.,
. . . 325.27 . . - ,yr7AA\ 50 = '25622 SUl 0 5
= + jl62.48 + | | - | | ^ s in ( 0 .6 9 8 -1.3744)} el5 l3 (a698_“i)
= 126.95 sin (cot - 1.3744) - 100 + 392.89 e -0.1989<at (3.4.9)
This is the equation for output current from a to n+a. This waveform has period of n and it repeats at rc+a. Hence average value of i0 will be given as,
n + a
; o(at>) = ” | a
.. n+0.698= - j [126.95 sin(cof-1.3744) -100 + 392.89c” 01989 ] <io>;
0.698
1 9 A QR 3 ?39 1 n n 3 ?39 qq 3 839= — ^— J sin((of-1.3744) d o t — — J dot + — ^— J e~°^9S9(0t dot
0.698 0.698 0.698
= 217.28 A
This is the average value of output current.
Example 3.4.5 : I f a freewheeling diode is added across the highly inductive load in l<j> full converter, derive an expression for average load voltage.
Solution : We know that freewheeling action does not take place in 1<J> full converter inherently. In the positive half cycle, Tj and T2 conduct from a to n as usual. But from n to n + a freewheeling diode starts conducting. This is shown in Fig. 3.4.13. The freewheeling
*o(0) =325.272.5622 sin(0.698-1 .3744) 1 + 0.5352
1 -0.5352
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Power Devices and Machines 3-61 Phase Controlled Rectifiers (AC/DC Converters)
diode is more forward biased compared to T| and T2. Hence freewheeling diode conducts. The freewheeling diode is connected across the output V0. Hence Vo =0 during freewheeling. The energy stored in the load inductance is circulated back in the load itself. Fig. 3.4.14 shows the waveforms of this operation. The output voltage becomes zero in the freewheeling periods. Compare the load voltage waveform of Fig. 3.4.13 with that of l<j>full converter with resistive load (Fig. 3.4.3). They are same. Hence the average load voltage can be obtained from equation 3.4.3. i.e.,
Vo(av) = ^ - ( 1 + coso) ...(3.4.10)
Fig. 3.4.13 Freewheeling diode conducts from k to rc+a due to inductive load
Fig. 3.4.14 Waveforms o f 14> fu ll converter fo r highly inductive load and freewheeling diode across the load
Copyright!
Power Devices and Machines 3 - 6 2 Phase Controlled Rectifiers (AC/DC Converters)
))!■► Example 3.4.6 : A single phase fully controlled bridge rectifier is fed from230 V - 50 Hz supply. The load is highly inductive. Find the average load voltage and current if the load resistance is 10 Q and firing angle is 45°. Draw the supply current waveform.
Solution : The rms value o f the supply voltage is,
V5(rms) = 230 V
Hence peak value of supply voltage is,
Vm = Vs(rms) 42= 230 V2
Since the load is highly inductive, the load current can be considered continuous and ripple free as shown in Fig. 3.4.9. For such operation, the average load voltage is given by equation 3.4.5 as,
2VVo(av) = ~ cos a
The firing angle a =45°. Hence above equation becomes
2 x 230 V2 Vo(av) = ------ -------cos 45°
= 146.42 volts
The average load current I0tav\ or l a is given as,
_ Vo(av ) o (av ) r
Putting the values of R =10 Q and V0^ v) = 146.42 volts,
146.42l o (av ) 10
= 14.64 A
The supply current waveform will be a square wave as shown in Fig. 3.4.9. The amplitude of the square wave will be I0 m\ i.e. 14.64 A.
Example 3.4.7 : For a 1 fyfull converter having highly inductive load derive the following:i) Fourier series for supply currentii) RMS value o f nth harmonic o f supply currentiii) Fundamental component o f supply currentiv) RMS value o f supply current
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Power Devices and Machines 3 - 63 Phase Controlled Rectifiers (AC/DC Converters)
Solution : i) To determine Fourier seriesThe general expression for Fourier series is given as,
°0>S ( “ 0 = ^ ( a p ) + Z cn S«'» ( « « » » + ♦ „ )
n= 1
where
and
Here,
+bn
tan-1bn\ n /
J i$ (cof) cos n o t d o t
2 n— J is (o t ) cos n o t d o t
o
From the supply current waveform of Fig. 3.4.9 we can write,
n+a 2n+a
= 2it \ Io(av)cosnu>t db)t+ | ( - /o(<n-))cosncot d<atn+a
o(av) 2n+aj cos n o t d o t - j cos n o t d o t
2 Io (av)n n
-4 I
sin n a [cos « j i - 1 ]
o(av)n n sinn a for w=1,3,5,
0 for n = 0,2,4,.
Similarly, 2bn = = J /s (cof) sin n o t d o t
2n
2 nj is (cof) sin n o t dot
From supply current of Fig. 3.4.9,
2n
2n + af l^ m)Sinn(0td(0t+ \ ( - I 0 v )sin ntot d<at^
, (3.4.11)
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Power Devices and Machines 3 - 6 4 Phase Controlled Rectifiers (AC/DC Converters)
Hence
o(av) Ji+a 2n+aJ s in n a td to t- j sin n atd& t
L a rt+a
2 Io(av)nn
4 I
cosna [1 -cos n ti]
o(av)nn c o s n a for n = 1 ,3 ,5 ,.
0 for n = 0 ,2 ,4 ,6 ,.
= + bn
4 Io(av)n n
V. J
[ sin2 n a + cos2 n a ]
4 /o(av)n n for w = l , 3, 5,
And <j>„ = tan 1 jP~ = -n a from equation 3.4.11 and equation 3.4.12.n
Thus <J>„ = -n aThe average value of supply current is zero. i.e. I$(av) = 0. This is clear from
Fig. 3.4.9.Therefore Fourier series is,
4 Isin (ncot-n a )
7 1 = 1 , 3 , 5 ,.-
nn
Ii) RMS value o f nth harmonic supply current
The RMS value of the n^ harmonic of the supply current its given as,
4 !o(av)1 - _ !L - n nsn n V2
nn
(3.4.12)
(3.4.13)
(3.4.14)
(3.4.15)
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Power Devices and Machines 3 - 65 Phase Controlled Rectifiers (AC/DC Converters)
iii) Fundamental component o f supply current
The r.m.s. value of nth harmonic of supply current is given as,
hn =0.91o(av)
With n = 1 above equation we get r.m.s. value of the fundamental component of supply current i.e.,
h i ~ h(av)
iv) To obtain rms value of supply currentThe rms value is given as,
$ (rms)1 7= J /s2 (cof) d o t
1/2
From supply current waveform of Fig. 3.4.9,
Es(rms) In
2n+a
\ I o(«v)d<at + \ ( - ' o i a v ) ) d (ot
h(rm s) ^o(av)
...(3.4.16)
... (3.4.17)
Example 3.4.8 : For a l<j) full converter having highly inductive load, derive the following :i) Displacement factor (DF)ii) Supply power factor (PF)iii) Harmonic factor (HF)iv) Current distortion factor (CDF)
Solution : i) Displacement factor
The displacement factor (DF) is given as,
DF = cos <j) j ... (3.4.18)
From equation 3.4.14 = -n a ; Hence 4>j = -a .
DF = cos ( - a )
DF = cos a ... (3.4.19)
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Power Devices and Machines 3 - 6 6 Phase Controlled Rectifiers (AC/DC Converters)
ii) Supply power factor (PF)
The supply power factor is given as,
PF sls(rms)
COS
From result of previous example and equation 3.4.18,
2 V2 /o(av)
PF cos a
PF = ------ cos a
iii) Harmonic factor
The harmonic factor (HF) is given as,
HF's(rms)
‘sl-1
/ 2o(av)
' 2 V2 /
HF = 0.4834 or 48.34 %
... (3.4.20)
(3.4.21)
Thus the harmonic factor of supply current is fixed to 0.4834, irrespective of triggering angle.
iv) Current d istortion factor (CDF)
The current distortion factor (CDF) is given as,
CDF sls(rm s)
2V2 1o(av)
o(av)
2>/2 0.9 ...(3.4.22)
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Power Devices and Machines 3 - 6 7 Phase Controlled Rectifiers (AC/DC Converters)
Example 3.4.9 : For a 1 § fully controlled bridge having continuous and ripple free current obtain, i) Active power and ii) Reactive power. [Dec.-2000, 6 Marks]
Solution : i) Active power
Active power is given as,
Active = Vs h l c o s * l
= Vs ------- cos(-a), since <j>j = - a
= 2--------------- cos( a)71
^o{av)= ------------- cos a ...(3.4.23)K
ii) Reactive power
Reactive power is given as,
^reactive = h i sm ^1
v 241 , >= V .-------------- sin(-a)K
= - 2 --------------- sin a71
I V I^ ym 1 o(av) . / oAnA\= ---------------- sm a ...(3.4.24)K
The negative sign indicates that the power is reactive.
Comment
Compare the reactive powers of full converter and half converter. They are as follows :
P *tf*<H C B ) = - ^ < f Z ! sin a
2Vn,‘ « av) .*W ii« (FCB) = --------- Sirl“
From above two equations we have,
P r e c i s e B) = 2 *Preaclive (HCB)
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• Thus half controlled bridge draws 50 % reactive power compared to that of full controlled bridge.
))*► Example 3.4.10 : A single phase full converter is operated front a 120 V, 60 Hz supply. The load current with an average valve of la is continuous, with negligible ripple current. If
the turns ratio of the transformer is unity, if the delay angle is a = Calculate the
i) HF o f input current
ii) DF
iii) PF
Solution : Given data
Supply voltage, Vs = 120
Delay angle, a =
i) Harmonic factor (HF)
For continuous load current, the harmonic factor is fixed for full converter. And it is given by equation 3.4.21 as,
Power Devices and Machines 3 - 68 Phase Controlled Rectifiers (AC/DC Converters)
HF = 0.4834 or 48.34 %
ii) Displacement factor (DF)
For 1 <j> full converter, DF is given by equation 3.4.19 as,
iii) Power factor (PF)
For 1 § full converter, PF is given by equation 3.4.20 as,
nc 2V2PF = ------cos a71
= 0.45
This is lagging PF, since current lags the voltage.
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Power Devices and Machines 3 - 69 Phase Controlled Rectifiers (AC/DC Converters)
Example 3.4.11 : A single phase full converter operates with 220 V, 50 Hz ac input and supplies output load consisting of R-L load with very high inductance drawing level load current 10 A and operated at firing angle of 30°. Find -
i) RMS supply current,
iii) Input displacement factor,
v) Power factorSolution : Given : 1 <|>FCB
ii) Fundamental component o f input current,
iv) Harmonic factor
vi) Output voltage. [M ay-2000,10 M arks]
v: 220 V Vm = 220^/2 = 311.12 V
I0(av) = 10 A, a = 30° or - radians.
i) RMS supply current
ii) Fundamental component o f input current
2V2 /L =
o(ar>)n
2V2xlO 9 A
By equation 3.4.17
By equation 3.4.16
iii) Displacement factor
DF = cos
cos a
= cos— = 0.866
iv) Harmonic factor
HF 0.4834 or 48.34 % By equation 3.4.21
v) Power factor
PF = cos a
— cos | = 0.77971 6
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Power Devices and Machines 3 - 70 Phase Controlled Rectifiers (AC/DC Converters)
vi) Output voltage
2Vm 2x311.12 nVo(av) = ~ ~ ~ ~ cos^ = 171.53 V.
Example 3.4.12 : A single-phase fully controlled bridge converter supplies an inductive load. Assuming that the output current is virtually constant and is equal to ld, determine the following performance measures, if the supply voltage is 230 V and if the firing angle is maintained at (n/6) radians.
i) Average output voltage
ii) Fundamental power factor or displacement factor (DF)
iii) Supply power factor (PF)
iv) Supply Harmonic factor (HF) [May-2007, 8 Marks]
Solution : Given : I ^ av = Id
* W ) = 230 V Hence V>» = ^ = 230 V2
a = y radians 6
i) Average output voltage
Vo(a„) = cosa
= 2x- 230-^ . cos I = 179.337 1 6
ii) Displacement factor (DF)
DF = cos a = cos ^ = 0.866o
iii) Supply power factor
or 2V2 2V2 71 PF = ------cos a = ------ cos -7 = 0.7871 71 6
iv) Supply harmonic factor
HF = 0.4834 for fully controlled bridge.
»■► Example 3.4.13 : What happens ifT 2 shown in Fig. 3.4.15 is shorted due to fault in the positive luilf cycle ? [M ay-2007,4 M arks]
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Power Devices and Machines 3-71 Phase Controlled Rectifiers (AC/DC Converters)
Fig. 3.4.15
Solution : In the positive half cycle and short circuit of T2, the situation will be as shown in Fig. 3.4.15 (a)
• Here observe that Tj is forward biased
but it will start conducting when it is triggered.
• T4 and diode D3 are reverse biased inFig. 3.4.15 (a) C ircuit diagrampositive half cycle.
Fig. 3.4.16 shows the situation in positive and negative half cycles. In positive half cycle the controlled supply will be applied to load. But in negative half cycle, supply is shorted through T4.
Fig. 3.4.16 Waveform of c ircu it o f Fig. 3.4.15(a)
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Power Devices and Machines 3 - 7 2 Phase Controlled Rectifiers (AC/DC Converters)
))»► Example 3.4.14 : Draw voltage and current waveform for circuit shown in Fig. 3.4.17[May-2007, 4 Marks]
Resistive load for a = 30°
Fig. 3.4.17
Solution : Fig. 3.4.18 shows the voltage and current waveforms.
Fig. 3.4.18 Output voltage and current waveforms
• The output voltage for a = 30° is shown in Fig. 3.4.18(b).
• For resistive load, the shape of output voltage and that of output current are same. And,
V.
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Power Devices and Machines 3 - 73 Phase Controlled Rectifiers (AC/DC Converters)
3.4.3 Inversion in 1<|) Full Converter (Second Quadrant Operation)
Answer follow ing questions after reading this topic.
y
1. Explain the inuersion in h j) full converter.2. Write short note on 2 quadrant operation o f LCC.
Marks [5], D ec.-2001. Marks [6]. Dec -2006Most likely arid
asked in previous University Exam
The waveforms of 1 <J> full converter for inductive load are given in Fig. 3.4.9. Observe that the output voltage v0 goes negative for some duration. These intervals are 0 to a, n ton + a, ..... and so on. The output current i0 remains positive always. Thus outputinstantaneous power becomes negative in such intervals. In other words, load power flows to source when v() goes negative. The average output voltage is given by equation 3.4.5. i.e.,
V.o( av)
o(av)
COS a
The variation of Vn/M,\ with respect to a is shown in Fig. 3.4.19. In the Fig. 3.4.19 the
V. is positive from 0 to For a = 90° or - , the V0(av} is zero
V,o(av)
Second quadranti0 - Positive
v0 - Negative
First quadrantv0and iQ both
Positive
(a) Variation of output average voltage with respect to firing angle
(b) Second quadrant operation
Fig. 3.4.19
Fig. 3.4.19 shows the waveform of v0 for a = 90° or The SCRs conduct, current flows
in the load but V. 0. This means power fluctuates between load and the source. No power is consumed by the load. The load inductance stores power from source when v0 is
o(av)
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Power Devices and Machines 3 - 74 Phase Controlled Rectifiers (AC/DC Converters)
positive (i.e. rectification). And this stored power is fed back to the source when v0 is negative (i.e. inversion). When the firing angle is increased above 90°, the average output voltage becomes negative as shown in Fig. 3.4.19. This is called second quadrant operation. The net power is fed from output (load) to the source. But where does this power comes from? Because load inductance cannot supply more power than it stores. At a = 90°, stored power and power supplied to the source are equal. For a > 90°, the stored power is less and more power needs to be supplied to the source. Hence an external DC source is to be connected in the load as shown in Fig. 3.4.21. This DC source maintains the forward bias on the SCRs. Hence they keep on conducting even though a >90°. Such output voltage
Fig. 3.4.20 Inversion in A<\> fu ll converters
Fig. 3.4.21 Inverting operation in 1<j)full converter
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Power Devices and Machines 3 - 76 Phase Controlled Rectifiers (AC/DC Converters)
i) Calculate the value o f the current limiting resistor required for nominal charging current o f 15 A if the firing angle is 30°.
ii) Calculate the maximum and minimum firing angles to maintain the current constant if the mains supply voltage varies by +10 % to -10 %.
iii) The above bridge is now operated in the inverting mode by reversing the battery polarity arid adjusting the firing angle appropriately. Calculate the firing angle such that the battery discharge current is 10 A with nominal mains supply voltage. Also obtain the power supplied by the battery and power fedback to the mains. Neglect all devicedrops.
Solution: Given : Vs = 240 V
Internal resistance (Rbatt) = 0.25 + 0.25 = 0.5 Q
batt 144 V
i) To obtain current limiting resistor
[M ay-2006,16 Marks]
Here a = 30° and Io(av) = Ibatt = 15 A
V,2VL
o(av) cos a
2 x 240x V2cos 30° = 187.127 V
The current limiting resistor is given from Fig. 3.4.22 as,
V. VuR^l 0.5 Q
batt
h a l t
187.127-14415
Rcl = 2.375 Q
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ii) Range of firing angles for Vs ± 10 %
Vs(max) = 240 + 10 % o f 240 = 240 + 24 = 264 V
v s(min) = 240 - 10 % of 240 = 240 - 24 = 216 V
To maintain constant charging current V^av) should remain constant. Hence range of firing angles can be calculated as follows:
Power Devices and Machines 3 - 7 7 Phase Controlled Rectifiers (AC/DC Converters)
2VV.o(av)
m(max)n cos a ,
2 x 264 x V2187.127 = ------------— cos a,
And
a max = 38°
2 Vo(av)
w(min)a cos a,
2x216x V2187.127 = ------------— cosa,
71
“ min = 15-79"
Thus a can be varied from 15.79° to 38° to maintain constant charging current,
iii) To obtain firing angle and powers in inverting mode
From Fig. 3.4.23 wre can obtain
^o(av) as'
^ o { a v ) + ^ 44“ ^b a tt^ b a tt = ®
/. Vo(av) ■144 + 10x0.5 = -139 V
V2Vr
o(av) cosa
2x 240 x V2 -139 = ----------------cosa
I = = 10 A
a = 130°
To obtain the power supplied by the battery
Battery current is 10 A and its voltage is 144 V. Hence power supplied by battery willbe,
Battery power = 10x144 = 1440 W.
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Power Devices and Machines 3 - 78 Phase Controlled Rectifiers (AC/DC Converters)
To obtain power fedback to mains
The combined resistance of the reactor and battery is 0.5 Q. Hence power loss due to this resistance will be (10)2 x0.5 = 50 W. The remaining power is given back to mains, i.e.,
Power supplied to mains = 1440 - 50 = 1390 W.
3.4.4 Comparison of Half Controlled and Full Controlled RectifiersNow let us compare the half controlled and fully controlled bridge rectifiers.
Table 3.4.3 shows this comparison.
Sr.No.
Half controlled converter Fully controlled converter
1. This consists of half number of SCRs and half number of diodes.
This consists of all the SCRs as controlled devices.
2. This operates in only one quadrant.
This can operate in two quadrants.
3. Output voltage is always positive. Output voltage can be negative in case of inductive loads.
4. Inherent freewheeling action is present.
External freewheeling diode is to be connected for freewheeling.
5. Power factor is better. Power factor is poor than half converter.
6. Inversion is not possible. Inversion is possible.
7. Used for battery chargers, lighting and heater control.
Used for DC motor drives.
Table 3.4.3 Comparison o f half and fu lly controlled bridges
))*■► Example 3.4.17 : A single phase fully controlled bridge operates with 230 V, 50 Hz ac input and supplies continuous ripple free output current of 5 A. If bridge is operated at a firing angle of 45°. Find,
i) Average output voltage ii) RMS supply current
iii) Harmonic factor iv) RMS value o f 3rd harmonic o f input current.[M ay-2001, 2008, 6 M arksl
Solution : Given : l<t> FCB.
Vs = 230 V, Vm = 230V2 = 325.27 V
l 0(av) = A, a = 45° or ^ radians.
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Power Devices and Machines 3 - 7 9 Phase Controlled Rectifiers (AC/DC Converters)
i) Average output voltage
Vo(av) = ~ c o s a
2x325.27 n . .------------- cos-r = 146.42 VTC 4
ii) RMS supply current
^s(rms)
= 5 A
iii) Harmonic factor
For 1 FCB with highly inductive load, HF is constant, i.e.,
HF = 0.4834 or 48.34 %
iv) RMS value o f 3rd harmonic
c„ 4 lg(av)/nn
41 41 2^2 I0(m)
nn
SJ 3 n 3 n
))»■► Example 3.4.18 : Show that reactive power input reduces to half due to above converter as compared to full controlled bridge for same firing angle a, feeding a continuous ripplefree constant current load. [Dec.-2000,10 Marks]
Solution : a) Reactive power of semiconverter
P(reactwe) = sin *1
2^2 l 0(av) a . ( a = V ,--------------c o s ^ s m j^ --
w ^ l o(av) . a CL= -Vs ------------- 2 sm - cos —n 2 2
'^ ‘ ^sK(av) . . . . a a .= ------------------sin a smce 2 sm — cos — = sin ati 2 2
^o(av) . . r = . , . .= ------------ sm a since V 2 Vs = Vm
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Power Devices and Machines 3 - 8 0 Phase Controlled Rectifiers (AC/DC Converters)
b) Reactive power o f fu ll converter
P.i(reactive) ^s/siSin<J>i
2V2 7
= V
sin a
21/ /m o{av) . --------------sin a since -Jl Vs = Vm
Result : From the reactive powers of semiconverter and full converter, observe that reactive power of semiconverter is half of full converter.
3.5 Three Phase Semiconverters
3.5.1 Operation with Resistive Load
Answer follow ing question after reading this topic
1. Explain the working o f 3 <j> semiconverter with the help of waveforms.
[Dec.-2 0 0 6 , D ec.-2008]
.v».
V - M o s t l i k e l ^ H H
ImportantS i
We have discussed 1 (j) semiconverter earlier. The 3<j> semiconverter delivers more power. It uses three SCRs T j , T3 and T5 and three diodes D4 , D6 and D2. Fig. 3.5.1 shows the circuit diagram of 3 <t> semiconverter.Fig. 3.5.3 (a) shows the waveforms of supply phase voltages R, Y and B. Note that these are phase voltages. These are the voltages with respect to neutral N. In Fig.3.5.1 (3(f) semiconverter), when any SCR and diode conducts, line voltage is applied to the load.Hence it is necessary to draw the line voltage waveforms.
Load
Fig. 3.5.1 3<t> semiconverter or half bridge converter
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Power Devices and Machines 3 - 81 Phase Controlled Rectifiers (AC/DC Converters)
Fig. 3.5.2 shows the phasor diagram of supply phase and line voltages. In this diagram observe that line voltage RB lags phase R by 30°. This is clear from waveforms of Fig. 3.5.3 (b) also. The phase shift between two line voltages is 60°.
B Y
Fig. 3.5.2 Phasor diagram showing the relationshipbetween phase and line voltages o f 3<j> supply
When a < 60°
Tj is triggered at a =30° (see Fig. 3.5.3 (c)). SCR Tj and diode D6 conducts. Hence line
voltage RY is applied to the load from f- + a j to j . At diode D2 is more forward biased
and hence it starts conducting.
Hence line voltage RB is applied to the load. Tj D2 keeps on conducting till next SCR
T3 is triggered at + . The load voltage waveform for a =30° is shown in
Fig. 3.5.3 (c). The devices conducting are also shown in respective intervals.
Observe that one period of the ripple in output voltage waveform is,
Thus there are three cycles of output ripple in one cycle of the supply. Hence ripple frequency is three times of the supply frequency, i.e.,
frippte = 3x50 = 150 Hz In the Fig. 3.5.3 (c), observe that each SCR conducts for the maximum duration of 120°
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Power Devices and Machines 3 - 8 2 Phase Controlled Rectifiers (AC/DC Converters)
Fig. 3.5.3 Waveforms o f 3<j> semiconverter fo r resistive load
Power Devices and Machines 3 - 83 Phase Controlled Rectifiers (AC/DC Converters)
The current waveform for resistive load will be similar to voltage waveform since,
i = Yo0 R
»»► Example 3.5.1 : Derive an expression for the average output voltage of 3<J> semiconverter having resistive load for a < 60°.
Solution : We know that the average output voltage is given as,
TV.o(av) = f j vo (“ 0 d(at
Observe the waveform of V0 given in Fig. 3.5.3 (c). The period T can be considered5k 2 jifrom [ — + cc to -7- + ct I which is -5-. Hence above equation becomes,
1 6 r72^ J vo M d^
[ 3 ) Z+a
3_2 k
5k
2 6 | VRY (cof)dcof + J VRB( o t ) d o t
K6 + a
... (3.5.2)
The equations for VRy and V'»/, can be written from Fig. 3.5.2 (b) as follows,
VRY (cof) = Vm sin I cof +
Vm ( « 0 = V 3 V m s m L t - Z6)
... (3.5.3)
Here Vm is the peak value of the phase voltage. Putting above expressions in equation 3.5.2,
n 5k
Vo(av) 2k f Vm sin cof + jda>t + J V3 Vm sin ^ cof-^ j dot
3\[3 V„ 2 k
J sm cof + jdcof + J siw^cot—gj dot
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Power Devices and Machines 3 - 8 4 Phase Controlled Rectifiers (AC/DC Converters)
3V 3 V„ 2k
3 V 3 U m2 k
3V 3 Vn 2 k
[-cos ((Of+6
{
( K 71A (7 1 K_ eos( _ + _ ) + cos( _ + a + _
(»<-!)5n~6n2
COS5 k k - + a - z \+ c°s ( K-K
U " 6
( l + cos a ) ... (3.5.4)
This is the expression for average output voltage for a < 60.
When a = 60
Fig. 3.5.3 (d) shows the output voltage waveform for a = 60°. Observe that the voltage waveform is just continuous. When Tj is triggered at a = 60°, the line voltage RB is applied
across the load. Tj and D2 conducts from + to f ~ + a j . Similarly next SCR T3 is
triggered at + and T3 D4 conduct. The current waveform will be similar to voltage
waveform for resistive load. The average output voltage is given by equation 3.5.4, since voltage waveform is continuous.
When u > 60
Fig. 3.5.3 (e) shows the output voltage waveform for a = 90°. SCR T is triggered at
^g + a j . Tj and D2 conducts and line voltage RB is applied across the load. In the
waveform observe that, output voltage becomes zero at In Fig. 3.5.3 (b) observe that
line voltage RB becomes zero at Hence SCR Tj is turned off. Since T3 is not triggered,
the output voltage becomes zero. At + T3 is triggered and T3 D4 conducts. Line
voltage YR is applied across the load. Thus for a > 60, the output voltage is discontinuous. Since the load is resistive, the current is also discontinuous. The current waveform will be similar to voltage waveform.
Example 3.5.2 : Derive an expression for average output voltage of 3<j> semiconverter having resistive load for a > 60°.
Solution : Fig. 3.5.3 (e) shows the waveform of output voltage for a =90° (i.e. a >60°). Observe that the period of ripple cycle is
- -2 k
T
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Power Devices and Machines 3 - 8 6 Phase Controlled Rectifiers (AC/DC Converters)
Vs(line) = 440 V
V - « 0s(Ph) ^ 3
Vm = V2Vs(pf0 =V2. ^ = J | 440 V
We have derived an expression for average value of output voltage earlier. For a < 60, VQtav) is given by equation 3.5.5 as,
373 V„(1 + cos a)o(av) 2 k
Here Vm is peak value of phase voltage. Putting values in above equation,
440, _o(av) 2 71
= 524.7 volts.
(1+ cos 40°)
3.5.2 Operation with Inductive LoadWe have discussed the operation of 3<{> semiconverter with resistive load. Now let us
consider the operation with highly inductive load. Hence the load current i0 can be considered continuous and ripplefree. Fig. 3.5.4 shows the circuit diagram of 3(f> semiconverter driving highly inductive load.
Consider that SCRs are triggered at firing angle of 90°. Fig. 3.5.5 (See Fig. 3.5.5 on next page) shows the waveforms of this converter. Fig. 3.5.5 (a) shows the supply phase voltages R, Y and B. Fig. 3.5.5 (b) shows the supply line voltages. These waveforms are
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Power Devices and Machines 3 ■ 87 Phase Controlled Rectifiers (AC/DC Converters)
Fig. 3.5.5 Waveforms o f 3(j> semiconverter fo r highly inductive load
drawn according to the phasor diagram of 3c|> supply shown in Fig. 3.5.2. Fig. 3.5.5 (c) shows the gate drives of SCRs Tle T2 and T3 for a =90°. Fig. 3.5.5 (d) shows the output
voltage waveform v0. Observe that Tj is triggered at f^ + a j . Line voltage VRB is
Power Devices and Machines 3 - 8 8 Phase Controlled Rectifiers (AC/DC Converters)
maximum at the time of triggering. Hence Ti-D2 conducts and line voltage VRB is applied 7 7Cacross the load. At — , the line voltage = 0, but the output current i0 [Fig. 3.5.5 (e)J is
not zero. The output current waveform is continuous and ripplefree. The load inductance
generates a very large voltage L to maintain i0 continuous. This situation is shown in
Fig. 3.5.6.
Fig. 3.5.6 Freewheeling action in 3<J> semiconverter.Dotted lines shows freewheeling current paths
The freewheeling diode DFW is forward biased by the load inductance voltage L
The freewheeling current ipW is basically i0. Thus the energy stored in the load inductance is fed back to the load itself. Fig. 3.5.5 (g) shows the waveform of freewheeling current.
SCR T-, turns off at -7- as soon as freewheeling diode starts conducting. This is because1 6
freewheeling diode is more forward biased compared to Tj and D after — . If extra
freewheeling diode is not connected, then freewheeling current flows through T-. and D4 . This is shown by 'thin dotted line' in Fig. 3.5.6. Thus freewheeling action is inherent in 3<|> semiconverter.
If we neglect the drop of freewheeling diode, then output voltage during freewheeling period is zero. As shown in Fig. 3.5.5 (0/ no supply current flows during freewheeling period. Compare the output voltage waveform of Fig. 3.5.3 (e) (resistive load) and Fig. 3.5.5 (d) (inductive load). Both the waveforms are same.
Example 3.5.4 : Derive an expression for average output voltage of 3<j> semiconverter having highly inductive load.
Solution : We have seen that the output voltage waveforms of 3<f> semiconverter are same for resistive as well as inductive loads. Hence their average values are also same. Hence from equation 3.5.5 we have,
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Power Devices and Machines 3 - 89 Phase Controlled Rectifiers (AC/DC Converters)
V.o(av)3 a/3 V„
2n (l + cos a ) ... (3.5.6)
This is the expression for average output voltage for resistive as well as inductive loads and 0 < a < n .
Unsolved Examples
1. A 3<j> semiconverter is operated from 3<t> 230 V, 50 Hz supply. The load is 10 ohms in series with large smoothing inductor. Determine output voltage and current if triggering angle is 60°.
!A ns.:V o W = 403.5 V, l 0(i„ )= 40.35 A]
2. Derive an expression for average value of output voltage for 3<J> semiconverter.
3.6 Three Phase Full Converters
Answer foUowing question after reading this topic
1. Draw the circuit diagram, voltage and current waveform for a = 30°, resistive load of three phase full bridge converter.
Marks [8], May-2007, Nov.-2007 Most likely and asked in previous University Exam
Three phase half converters operate only in first quadrant of v0 - i 0. The output voltage v0 is always positive for resistive as well as inductive loads. The output current i0 is also always positive. Hence 3 (j) semiconverter operates in first quadrant only. Three phase full converters can operate in two quadrants. The output voltage of 3<j> full converter can be positive as well as negative. It uses six SCRs as shown in Fig. 3.6.1.
Fig. 3.6.1 3<(>full converter
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Power Devices and Machines 3 - 90 Phase Controlled Rectifiers (AC/DC Converters)
3.6.1 Operation with Resistive Load
Answer follow ing question after reading this topic
1. Draw the circuit diagram of three phase full converter with a resistive load. Explain its working and draw the load voltage,
ninput supply, load current at a = —
3
Marks [6], D ec.-2007; Marks [10], May-2008
andasked in previous
^ j^Jniwrsity Exam
Let us consider the operation of 3 <J> full converter having resistive load. Fig. 3.6.2 shows the waveforms of 3 <(> full converter having resistive load. Fig. 3.6.2 (a) shows the supply phase voltages R, Y and B. Fig. 3.6.2 (b) shows the supply line voltages. These supply voltage waveforms are drawn according to the phasor diagram shown in Fig. 3.5.2. Fig. 3.6.2 (c) shows the gate drives for a =30°. For six SCRs, there are six gate drives. See Fig. 3.6.2 on next page.
In Fig. 3.6.2, observe that in interval-I, gate drives are given to SCRs T6 and T-[. Hence line voltage VRY is applied across the load. The equivalent circuit for this interval is shown in Fig. 3.6.3.
In Fig. 3.6.3 observe that SCRs T6 and Tj (normally written as 6-1) conduct. Hence,
= VRYObserve that output current i0 and R-phase current iR flows in the same direction.
Hence,
*R “ *oSimilarly observe that Y-phase current iXJ and output current i0 are in opposite
directions. Hence,
W = - 'o
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Power Devices and Machines 3 -91 Phase Controlled Rectifiers (AC/DC Converters)
Fig. 3.6.2 Waveforms o f 3c{> full converter having resistive load
Power Devices and Machines 3 - 9 2 Phase Controlled Rectifiers (AC/DC Converters)
The SCR pair T6 -T j conduct from + txj to ^ + a j . Line voltage VRY is applied
during this period. At + a j SCR T2 is triggered (Fig. 3.6.2 (c)). Here note that Te
tums-off, since T2 is triggered. Hence Tj -T2 starts conducting and it is marked as
interval-II. In this interval supply line voltage VRB is applied across the load. At + a j ,
T3 is triggered. Hence Tj tums-off and T2 -T3 starts conducting. Therefore line voltage VYB is applied across the load. It is marked as interval-III.
Load and supply currents
Fig. 3.6.4 Output current and supply current waveforms fo r a = 30°
• Since the load is resistive, the shape of output current waveform will be similar
to that of output voltage. Its amplitude will be i0 =
• Whenever Tj conducts, R-phase current will be positive and whenever T4
conducts, R-phase current will be negative.
The follow ing points are important about 3<J>full converter operation.
i) Only two SCRs conduct in any interval.
ii) Each SCR conduct for 120°.
iii) Each SCR pair conduct for one interval of 60°.
iv) SCRs are triggered in following sequence :
.........T’l - h - h -T4 -T5 -T6 -T, -T2 ............
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= 4 i v v h = 4 i - ^ v
Here firing angle is 45°. Hence the conduction will be continuous for resistive as well as inductive load. Therefore the average DC output is given by equation 3.6.2 i.e.,
3>/3 VL V V ) = - T ^ c o s a
Putting values in above equation ,
400
Power Devices and Machines 3 - 95 Phase Controlled Rectifiers (AC/DC Converters)
3V 3 x -J2 ■V3
V *» ) = --------- cos 45°
= 382 volts
3.6.2 Operation with Highly Inductive Load
Answer follow ing question after reading this topic
i1. Draw the circuit diagram of 3 full converter with a highlyinductive load. Explain it's working and draw the load voltage, K load current and current through SCR waveforms for rectification I y. ^ revious
Let us consider the operation of 3<|> full converter with highly inductive load. The output current will be continuous and ripplefree. In the waveforms of Fig. 3.6.2, observe that voltage waveform is continuous till a =60°. But with inductive load, voltage waveform is continuous for any value of a. Fig. 3.6.5 shows the waveforms of 3 4> full converter for highly inductive load. Fig. 3.6.5 (c) shows the output voltage waveform for a =60°. Observe that this waveform is same as that of resistive load shown in Fig. 3.6.2 (e). Fig. 3.6.5 (d) shows the continuous and ripplefree output current. Fig. 3.6.5 (e) shows supply phase current waveforms iR , iY and iB. Observe that the R-phase current is positive whenever Tj conducts and it is negative whenever T4 conducts. All the three current waveforms are of the same nature (quasi square wave) having 120° phase shift with respect to each other.
Fig. 3.6.5 (f) shows the output voltage waveform for a = 90°. The waveform goes negative for same period, because of inductive load. The load inductance generates a large voltage to maintain the load current in the same direction. Hence SCRs continue to conduct and load voltage becomes negative occasionally. Note that there is no freewheeling in full converter.
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Power Devices and Machines 3 - 96 Phase Controlled Rectifiers (AC/DC Converters)
Fig. 3.6.5 Waveforms o f 3<) full converter fo r highly inductive load
Power Devices and Machines 3 - 97 Phase Controlled Rectifiers (AC/DC Converters)
Inversion Mode o f 3 <j) Full Converter
• Fig. 3.6.6 shows the circuit diagram and waveforms for inverting opeartion of3 <|> full converter.
(a)
(a) C ircuit diagram for 3 o full bridge converter for inversion(b) waveform s
(b)
Fig. 3.6.6 Inversion in 3<j>full converter
• At a = 90°, the output average voltage is zero. For a > 90°, the average value of output voltage is negative but current is positive and constant. Hence power at the output side is negative. This means power is fed back from output side to 3 <t> AC supply.
• The DC source is connected in series with RL load as shown in Fig. 3.6.6 (a). This source forward biases the SCR even if a > 90°.
• The waveforms of inverting operation are shown in Fig. 3.6.6 (b) for a =120°. The inversion takes place and average output voltage is negative.
))■► Example 3.6.3 : Derive an expression for average output voltage of 3<j> full converter having highly inductive load.
Solution : In the Fig. 3.6.2 observe that output voltage waveform is continuous for complete range of a. Hence single expression can be derived. In Fig. 3.6.2 (d), observe that one ripple period of output voltage can be,
71
2 + aK \ 71
6 + “ = 3
During this period line voltage VRY is applied across the load. From Fig. 3.6.3 (b), VRY will be,
RY V3 Vm sin
Here Vm is the peak value of supply phase voltage.
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Power Devices and Machines 3 - 9 8 Phase Controlled Rectifiers (AC/DC Converters)
The average output voltage is given as,
TV.o(av)
■j 2^ 7 3 I VRY (“0
n 6 + a
71
2 + “rfcof
3V 3 V„COS (Of + f l
71
r a
3V 3 V.,cos a
This equation holds for complete range of a.
(3.6.4)
Example 3.6.4 : A 3 full converter operated from 3 <J) -V connected 208 V, 60 Hz, supply with Rl = 10 Q. It is required to obtain 50 % of the maximum possible output voltage. Calculate -
i) Delay angle a
ii) rms and average currents
iii) rms and average thyristor ratings
iv) r| o f rectification
v) PF
Solution : Given data
line
Load resistance,
m
r = ion
208 V. Hence V„h = ~ = 120 Vpn V3 V3
j2 V ph = 7 2 x 1 2 0 = 169.7 V
Output voltage Vo(av) = 50 % of Vo(av)max
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Power Devices and Machines 3 • 99 Phase Controlled Rectifiers (AC/DC Converters)
i) To obtain delay angle a
Average output voltage of 3 full converter is given by equation 3.6.2 as,
V 3V3Vmo (av ) = ------ ^ C O S a
Maximum value of output voltage will be obtained when a = 0. i.e.,
V -o(av )m ax n
3V 3xl697280.68 V
71
Since output voltage is 50 % of its maximum value,
^ o (a v ) = 0 *5 V o(av )m ax
= 0.5 x 280.68 = 140.34 V
Consider the formula for output voltage,
V 373V,„o(av ) = — — " c o s «
Putting values in this equation,
373x1697 140.34 = ----------------cos aK
a = 60°
ii) Average and rms output currents
Average output current is given as,
Vo(av) 140.34 Mav) - — ^--------- fo-----
To obtain rms current, we have to obtain rms output voltage, for a < 60, the output voltage waveform is continuous as shown in Fig. 3.6.2 (d). Consider the period from
■g + a j to ^ + when voltage VRY is applied arrows the load. This period is,
K \ ( 7t ^ 71
2 + a j [ 6 + a J - 3
From Fig. 3.6.2 (b) we can write an equation for line voltage VRY as,
Vry = V3Vm sinf cof + $6
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Power Devices and Machines 3 -100 Phase Controlled Rectifiers (AC/DC Converters)
Here Vm is the peak valve of phase voltage. RMS value is given as,
T
; J V2(cOt)d(Oto(rm s)
71
r °
“ *3 V * J sin2fcof + \diot
dcot
nr a
9V2 2 eyvm fn J
a 1 -C O S
dv)t
... (3.6.5)
Putting values in above equation,
V 0(rms) = 3 x 1 6 9 7 ^ 2n
= 159.17 V
Therefore rms output current will be,
_ Vo(rms) _ 159.17 _ 1 5 9 1 7 aM r m s ) --------- £ ------------Jo ------
iii) rms and average thyris to r ratings
In the waveforms of Fig. 3.6.2, observe that two thyristors conduct at a time. The load current is carried equally by three thyristors in one cycle. These three thyristors are Tx, T , T5 and T * T6, T2. Hence the average output current is shared by these three thyristors. Hence average current of single thyristor becomes,
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lT(av) = ^ - (3-6.6)
= = 4.678 A
This is the average current carried by each thyristor.
The rms current is also shared by three thyristors. Hence we can write,
/ 2 _ 12 , i 2 i / 21 c(rms) ~ 1 T(rms) + 1 T(rms) + 1 T(rms)
Above equation shows the relationship between output rms current and rms current of three thyristors. RMS current of each thyristor is same. Hence,
j 2 _ or 21 c(rms) “ 31 T(nns)
iT(rms) = - (3-6.7)V 3
15917 = = 9.189 AV5
iv) Rectification efficiency (r|)Rectification efficiency is given as,
Average or dc load power rms load power
^o(av) o(av)
Power Devices and Machines 3-101 Phase Controlled Rectifiers (AC/DC Converters)
rms) ^o(rms)
140.34x14.034 n ___ 7 7 _ 0/= iW x lW = 0777 or 777 %
v) To obtain power factor
The active load power is the power consumed in the load. It can be calculated as,
Active load power = /^rwig x R
= (15.917)2 x 10
= 2533.5 W
At any time instant two thyristors conduct. Hence the supply current can be given in terms of thyristor currents as,
j 2 _ o/2l s(rms) T(rrns)
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Power Devices and Machines 3 -102 Phase Controlled Rectifiers (AC/DC Converters)
... (3.6.8)
13 A
For 3 <j> supply, the total supply volt-ampere will be,
supply VA = 3 Vs Is
= 3 x 120 x 13 = 4680 VA
The power factor is given as,
Active load power
Let us first see the advantages of 3 (J) power supply.
Advantages o f 3 (j> supply :
i) Higher power supplying capability.
ii) It is suitable for driving AC loads such as high power induction motors, fans, pumps etc.
iii) Phase shift in 3 phases is useful in many applications.
iv) Power demand on one phase is reduced due to three phase.
v) Even if one phase fails, other two phases supply power to the load partially.
Total supply power(VA)
4680“ = 0 5413 (la8Sing)
3.6.3 Comparison between 3c|> and 1<|> Converters
Answer foUowing questions after reading this topic
1. Give the advantages and disadvantages o f converters.Marks [2], D ec-2007
2. Give the advantages o f 3 § supply over 1 <(> supply.Marks [3], D ec-2008
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Power Devices and Machines 3 -103 Phase Controlled Rectifiers (AC/DC Converters)
Advantages of 3 <J> converters :
The three phase converters have most of their advantages due to 3 <|> supply.
i) 3 4> converters are capable of supplying more power to the load.
ii) The ripple frequency is high (i.e. 150 Hz and 300 Hz). Hence filtering requirement is reduced.
iii) Supply power factor for 3 <{> converters is improved.
iv) 3 <|> converters provide continuous load current because of improved ripple frequency.
Disadvantages
i) Since three or six SCRs are to be controlled, the triggering circuits are complex for3 <j> converters.
ii) It is not suitable for simple low power loads.
Applications
i) High power battery charges.
ii) High power D.C. motor drives.
Comparison of 34> and 1<j> converters
Sr.No. Parameter 1<J> converter 3(j> converters
1. Ripple content in output More Less2. Output power Less upto 5 kW More than 5 kW3. Supply current waveform Square wave for 1<(> full
converterQuasi square wave for 3<{> full converter
4. Ripple frequency 100 Hz 150 Hz and 300 Hz5. Control and complexity Less complex and easy
controlComplex control and implementation
6. Maximum supply power factor 0.9 0.9557. Supply and load derating Higher Less
Table 3.6.1 Comparison of 34> and 1<j> convertersIt shows that it is preferable to use 3<|> converters for better power efficiency. Hence for
higher load power requirement 3<j> converters are always preferred. But for simple and low power applications 1<|> converters are used because of their simplicity of implementation.
Unsolved Example
1. A 3<{> full converter operates from 3<J* 415 V, 50 Hz supply. The load is highly inductive.Determine the triggering angle of the converter to get average output voltage of 300 volts. The loadresistance is R = 10 Cl Determine the load current and power.
I A its .: a = 57.63°, Io(av) = 30 A, Po{av) = 9kW]
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Power Devices and Machines 3 - 104 Phase Controlled Rectifiers (AC/DC Converters)
Summary
Sr.No.
Type of converter Load Parameter
1 <b half wave R
RL
2V„(1 f cosa)
RL with freewheeling diode
'o(av) = ^ (c o s a -c o s p )
^o(av) = C0Sa)
1 <j> semiconverter'o(av) = ^ ( 1 + co s« )
V2vmo(rms) 12n ^sin 2 a j
RL'o(av) = T ( 1+COSa)
RL
w _ i mvo(rms) 1 2k
DF = cos
rc - a + 2 s'n 2a
PF =\ 7 T ( 7 t -
2a j C0S 2
hf = E E E J T ,8cos"
3. 1 <J> full converter'o(av) = - f ( 1 + co s« )
n - a + -^sin 2a j
RL
RL with freewheeling diode
2Vno(av) cos a
v o(rms) ^
'o(av) = - f (1 + c o s c t )
y2’o(rms) ~ 1 2tt 7t — a + 2 sin 2a j
RLDF = cos a
PF = ?^?cosaJlHF = 0.4834 or 48.34 %
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Power Devices and Machines 3 -105 Phase Controlled Rectifiers (AC/DC Converters)
4. 3 <J> half w ave converter
R
RL
3J3 VV 0(av ) = • 2 n W ° ° s a fo r a - 30°
= 1+ co s ( g + a jj for a > 30°
_ ^ vm ^o (av ) 2k 005(1
5. 3 <}> sem i converter R , RL w 3>/3Vm (V o(av) 2?l 0 + c o s a )
6. 3 <j> full converter R
RL
R L - with freew heeling d iode
3 V 3V LV o(av) = — ° ° s a o r a -
v o(av) = „ m[ 1+ « » ( 3 + fo r “ > 6 0 °
3v '3V L v o(av) = „ cosa
^o (av ) 's sam e as with R -load.
7. 1 $ dual converter RL 2VV , . = —- c o s a - 0)(av) ti 1
2V
V « > = C°S«2a i + a 2 = 180°
2V r -i ■cir = S l® lc0sw,- c0s“ i j
□ □ □
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