Power Electronics Rectifiers by Bakshi

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 — --------► semicoi semiconverter *

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 are basically AC to DC converters. The power

Controlled rectifier

transferred to the load is

controlled by controlling triggering angle of the devices. Fig. 3.1.1 shows this operation.

Load

a

Control circuit Fig. 3.1.1 Principle of operation of a controlled rectifier

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•

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 d o you mean by commutation o f SCR ? Give types o f commutations. Explain natural commutation in details. Marks (6), May-2007 I \ J

Most likely and masked In previous niversity E xam

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 u tation : 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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C u rren t com m u tation : The SCR is turned off by reducing its anode current

below holding current. •

V oltage com m u tation : The SCR is turned off by applying large reverse voltage

across it. •

P rinciple o f lin e com m u tation

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

Mains AC

Fig. 3.1.3 A half wave rectifier uses natural commutation to turn off SCR

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 load is resistive,

j::::::::::::::::::: I::::::::::;,:::::::;:::::;:::::::::::: ::::::::::::

.........I..... ................... *......... . •..... .......... .

31113S HIS sstS :!t H :3 S !•t:S!?:!:

SHi !H9 5SI! § 8 : HIE B i:! HHi IS

:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: I:::::::::::::::::;::::::::::::::::::::::::::::::::::::::::::::::::::::;:;::::-.;:

Fig. 3.1.4 W aveforms of half wave controlled rectifier to illustrate natural commutation Copyrighted mate

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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.

I

LC circuit

Fig. 3.1.5 Principle of forced commutation

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.

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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 H alf W ave C onverter 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 1half wave converter with the help o f circuit diagram and waveforms.

M ost likely and

Important Q u estion

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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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

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Mathematical analysis

The average value of output voltage is given as,

1 T

Vo(av) = f ! vc

0

M

dot

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, V o (a v )

1 71

7T—

f Vm sin cot du>t

2n J

... (3.2.1) The power transferred to the load will be, V „U )

o(a v )

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 Load Now 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.

. . . .

Fig. 3.2.3 Half wave controlled rectifier with RL load

The SCR will be forward biased in the positive half cycle of the supply. Hence SCR is applied with the firing pulses in the positive half cycle. The waveforms are 00 x u iU shown in Fig. 3.2.4. Fig. 3.2.4(a) shows the supply voltage and Fig. 3.2.4(b) shows the

firing pulses to the SCR.

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Fig. 3.2.4 W aveform s 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

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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 Fig. 3.2.5 SCR conducts due to in­ ductance voltage after n in the inductance is finished. Hence i0 goes to zero. Therefore T. tums-off. In Fig. 3.2.4(c) 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 fo r average value o f output voltage fo r 1 = 10 A'

« = 60° or |

i) RM S sup ply curren t

ln -a ) v 71 71

= 10

*"3

1 71

8.165 A ii) O u tp u t voltage

V„ 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 V0 = Vs and,

V

V

'» = i f

= T

conducts,

(i.e. supply voltage)

... (3.4.1) - (3 A 2 )

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 at 7c+ 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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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 « )

71

- ( 1 + cos 30°)

= 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) = '«,(,»)* + 50 184.8 = ') * l « > +50 o(av) = 1.348 A Thus the current through the load is 1.348 A.

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3.4.2 Working with Inductive Load The 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

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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

voltage L

din dt 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.

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. from Ti to ti + cx due to inductance voltage

3.4.2.2 Continuous and Ripple Free Current for Large Inductive Load Answ er follow in g question after reading this topic 1. Draw the circuit diagram o f 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], M ay-2000. 2 0 0 1 ; Marks [10]. D ec.-2004

s,

M ost likely an d

asked In previous University E xam

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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Fig. 3.4.9 W aveform s o f 1fu ll converter fo r continuous and ripplefree load current in case o f large inductive load ))*► Example 3.4.3 : For the 1full 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]

S olution : i) Average output voltage fo r inductive load The average output voltage is given as, 1 T

J

Vo(av) = f

vo

0

(“0

Observe the waveforms of l 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+a Vo (a v )

=

J

-

Vm sin

d(s>t

a =

—

r

- COS

K 1

.nTi+a (0 1

J«

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o(av )

Phase Controlled Rectifiers (AC/DC Converters)

cos a

... (3.4.5)

This is the expression for average load voltage of l full converter for inductive load. Plot of V0(av) versus firin g 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 a Observe 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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ii) RMS value o f output voltage fo r inductive load The rms value is given as, 2

J \vo ( fu ll converter feeding RLE load Copyrighted material

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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., Vo(av)

2V

= - f - cos a

v

V , v = -2L = V vo(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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a = 40° = 0.698 radians and Z = -J r 2 +(coL )1 = ^(0.5)2 +(2.513)2 = 2.5622 Putting values in equation 3.4.8 we get i0(0) as 325.27 1 + 0.5352 sin(0.698-1 .3 7 4 4 ) 2.5622 1 -0.5352

*o(0) =

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 . . = '25622 SUl

.

- ,yr7AA\ 50 05

= + jl6 2 .4 8 + | | - | | ^ s i n ( 0 .6 9 8 -1.3744)} el5 l 3 (a698_“i) = 126.95 sin (cot - 1.3744) - 100 + 392.89 e -0.1989) = ”

| a

.. n + 0.698

= -

j

[126.95 sin(cof-1.3744) -1 0 0 + 392.89c” 01989

] ;

0.698

1 9 A QR

= — ^—

3 ?39

J

1 n n 3 ?39

q q 3 839

0.698

0.698

sin((of -1.3744) d o t — — J dot + — ^—

0.698

J e~°^9S9(0t dot

= 217.28 A This is the average value of output current.

If a freewheeling diode is added across the highly inductive load in l full converter, derive an expression for average load voltage. Example 3.4.5 :

S olution : We know that freewheeling action does not take place in 1 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

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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 lfull 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)

Fig. 3.4.13 Freewheeling diode conducts from

...(3.4.10)

k

to rc+a due to inductive load

Fig. 3.4.14 W aveform s o f 14> fu ll converter fo r highly inductive load and freewheeling diode across the load

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))!■► Example 3.4.6 : A single phase fully controlled bridge rectifier is fed from

230 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. S olution : The rm s 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,

2V

Vo(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 I 0tav\ or l a is given as, _ o (a v )

Vo (a v ) r

Putting the values of R =10 Q and V0^ v) = 146.42 volts, 146.42 l o (a v )

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 I 0^m \ i.e. 14.64 A.

For a 1fyfull converter having highly inductive load derive the following: i) Fourier series for supply current ii) RMS value o f nth harmonic o f supply current iii) Fundamental component o f supply current iv) RMS value o f supply current Example 3.4.7 :

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S olution : i) To determine Fourier series The general expression for Fourier series is given as, °0 >S

(“ 0

=

^ (a p ) +

where and

Z

n=1

cn

S«'» ( « « » » + ♦ „ )

+bn tan -1 \bn n/

J i$ (cof) cos n o t d o t

Here,

2n — 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

| ( - /o(t +

J

sm ^cof + ^jdcof +

J

J

V3 Vm sin ^ co f-^ j d ot

siw^cot—gj dot

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Phase Controlled Rectifiers (AC/DC Converters) 5n

3 V 3 V„

(» and 1 Converters Answer foU owing qu estion s after reading this topic 1. Give the advantages and disadvantages o f

converters. Marks [2], D e c -2 0 0 7

2. Give the advantages o f 3 § supply over 1 supply. Marks [3], D e c -2 0 0 8

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.

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Phase Controlled Rectifiers (AC/DC Converters)

Advantages of 3 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 for 3 converters. ii) It is not suitable for simple low power loads. A pplications i) High power battery charges. ii) High power D.C. motor drives. Com parison of 34> and 1converters Sr.No.

Parameter

1converter

3(j> converters

1.

Ripple content in output

More

Less

2.

Output power

Less upto 5 kW

More than 5 kW

3.

Supply current waveform

Square wave for 1full converter

Quasi square wave for 3full converter

4.

Ripple frequency

100 Hz

150 Hz and 300 Hz

5.

Control and complexity

Less complex and easy control

Complex control and implementation

6.

Maximum supply power factor ^ 0.9

0.955

7.

Supply and load derating

Less

Higher

Table 3.6.1 Com parison of 34> and 1converters It shows that it is preferable to use 3converters for better power efficiency. Hence for higher load power requirement 3converters are always preferred. But for simple and low power applications 1converters are used because of their simplicity of implementation.

Unsolved Example 1. A 3 full converter operates from 3