Power Electronics Rectifiers by Bakshi

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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 are

basically 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. Controlled rectifier Load a Control circuit

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 d o you mean by commutation o f SCR ? Give types o f commutations. Explain natural commutation in details.

Marks (6), May-2007 I \ Most likely and J 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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Power Devices and Machines 3 - 3 Phase Controlled Rectifiers (AC/DC Converters) • 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

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 :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

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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.

I circuitLC

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 H alf W ave C onverter and Effect of Freew heeling 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 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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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,

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Pow er Devices and Machines 3 - 7 Phase Controlled Rectifiers (AC/DC Converters) Mathematical analysis

The average value of output voltage is given as, 1 T

Vo(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) 7T— f 1 71Vm 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 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.

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

Fig. 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 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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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 in ductance 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 fo r average value o f output voltage fo r 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 top

0 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

1. C om pare freew heelin g d iod es an d fe e d b a c k diodes.

Marks [31. D ec.-2 0 0 6 , 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 W aveform s o f 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

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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 fo r average value o f output voltage fo r 1 <j> half wave controlled rectifier fo r 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 n

0 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(

0

t = 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) T o obtain av erage output voltage V0 (av)

The load is resistive. For this load V0tav\ is given by equation 3.2.1 as V., V_{o (av )} = ^ - ( 1 + c o s c x ) 1 71 •22,042 ( , k = ^ r [ 1+cos6 = 96.6 V

ii) To obtain effective o utp u t voltage v 0 ,™ s , The rms value is given as,

V_{o(rm s)} i j Vy (of) diot . 0

From the output voltage waveform of Fig. 3.2.2 we can write,

1 *

— J v* sin 2 cof dcot V._o(rms) V i f 1 -c o s 2(0/ J ---2---di0t m 2n V,m 4 71 /V

J dcot- J cos 2 ( 01 dcot

\Vm sin 2wf"n \

1

I k a - 2 I . 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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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. If 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 pow er factor.

Solution : Given data

Supply voltage Vs = 120; hence Vm = V 2 x l2 0 = 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 (a v ) = 2 5 % o f V o (a v )m a x = 0.25 x 54 = 13.5 V

o (a v )m a x

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Power Devices and Machines 3 - 1 5 Phase Controlled Rectifiers (AC/DC Converters) Consider again equation 3.2.1

V.

V_o(av) r^(l+COSCX)

2k v 7

Putting values in above equation, 1697 13.5

2k (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

13.5

10 1.35 A

The rms value of output voltage is given by equation 3.2.5 for half wave converter, i.e.,

V_o(rms) . a sin 2a 1 — + —

K 2k

1697 1 209 | sin(2x2.09) ir 2ti

= 37.718 V Hence rms output current will be,

o(rms)

^o(rms) R 37718

10 3.77 A iii) RM S 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) T o obtain in p u t p ow er facto r *

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 data

71 “ = 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 V o(av) o(av) R 0.159 VJ, R V.o(rms) ₂ 1 — + —«---a sin 2 a 71 2 7t /o Sm K 2 71 = 0.353 VL V o(nns) o(rm$) _ 0. 353 V„ ~R R Copyrighted material

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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 _ac V _{o(rms) o(rm$)}I

= 0.2028 or 20.28 %

ii) To obtain form factor

V,_o(rms) FF

V,_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 Sem iconverters (H alf Bridge Converter) The semiconverter is also called as half bridge converter.

3.3.1 Circuit Diagram

Fig. 3.3.1(a) shows the circuit diagram of single phase semiconverter. 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> sem iconverter (Common cathode configuration)

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A d,

230 V, 50 Hz AC supply

o--- <Tv

Fig. 3.3.1 (b) 1 <}) Sem iconverter (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 cathode configuration, 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 p hase sem iconverter fo r 'R' load fo r a = 0°,

3 0 6 0 ° and 90°. Marks [10], D ec.-2006, 2 0 0 8 5t likely and asked in previous

University E x am

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 o f 7^ and D 1 in positive half cycle of the supply. Dotted line shows path of current flow

(i.e. supply voltage)

and V 0 _ YsV

R 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,

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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.

Fig. 3.3.4 Conduction of T2-D2 in negative half cycle o f the supply. Dotted line shows path o f 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 V

and i0 = = ...(3.3.4)

In Fig. 3.3.4 observe that current through T2 flows in the same direction as i0. Hence i

T2

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 I n , 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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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 fo r a = 60°, RL load o f semi-converter. Marks[8], May-2007

Most likely and asked In previous

University E x am

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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Fig. 3.3.5 W aveform s of 1<J> sem iconverter 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.

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^ 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._o(av) V, + COS_{« )} ... (3.3.7) l

\

V 2 V _o(rms) - 1 m_{1 2} 7l 7 i-a + i sin 2 a

r

... (3.3.8)

3.3.3.2 Discontinuous Current Mode

In 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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Fig. 3.3.7 Discontinuous mode of single phase semiconverter 3.3.3.3 Continuous and Ripple Free Current for Large Inductive Load

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

M ost likely and asked in previous

University E x am

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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Fig. 3.3.8 W aveform s of 1<J> semi converter fo r 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|- D 1 conducts and

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vs = Vm sin

Fig. 3.3.9 A 1<J> sem iconverter 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.A wt) Vm sin cof = R io l(<at) + L —2i_— + £

This equation can be solved using laplace transform. The solution is,

<01(cof) = ^ - s i n ( w ( - 0 ) + i o 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) For

ji to n + a

Fig. 3.3.11 Freewheeling action in T^D2 or 72D 1 By KVL to this circuit we can write,

d io2((*t) R io2((ot) + L

dt + E = 0

This equation can be solved using laplace transform. The solution is, E

io2( «>t) = o2 1 —n ( l ~ e L )_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 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. |Nov.-2007, 8 M arks; 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 t a v ) + X c n sin (m o f+ <(>„) ;»= 1 where ctJ = yja* +b% and = tan -1 n Copyrighted material

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Here _an _yJ 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+ | (-/ ^ Jc o s M w f rfco/

2n

J coswof rfcof- J COS H(Ot d(Ot

°^ai ^ sin ?;a(l -c o snit) tin -2 1 o(av)

.

s in H a 7771 0 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(w0sin/7cof d o t o

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 o t d o t- J sin n o t d o t <t<w) 7171 21 (1 + COS77a) (1 - COS/171) tlK 0 (1 + cos 77 a) for 77 = 1, 3,5, for 77 = 2 ,4 ,6 ,, (3.3.12) (3.3.13) Copyrighted material

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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% 21_o(av) _. sm n a nn 4 l 0(av) na — --- cos—_nn ₂ 21_o(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 = ta n ~ ' r un tan = -ta n -2 1 c{av) . sm na nn 21_c(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.

T

I _(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 n J dcot - J dcot Copyrighted material

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o(<rc>)

271 [7i + a -2 jc-(7 c+ a )) 0

Thus the average value of symmetric waveform is zero Thus the Fourier series can be written as,

CO 4/

*s(“ f) = Z _tino((iv) cos - y smn a . ( — f )

h= 1,3,5,—

ii) R M S valu e of n lh h arm on ic o f su p p ly cu rrent The rms value of the nth harmonic is given as,

t l s M co sm . 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) F un d am en tal com p onen t of su p p ly cu rrent

The fundamental component of supply current is obtained by putting n equation 3.3.17. i.e.,

h i = 0 .9 1 ^ cos |

iv) T o obtain rm s 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)

₂71

rr + a

. (3.3.16) ...(3.3.17) = 1 in ...(3.3.18) Copyrighted material

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/_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 cosfy

s(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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SI

_o(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,

I

CDF = sl 's(rms) 2V2 /_{d a v )} _a --- COS — 71 2

c{av)

_{\ n}In - a 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 l

Vs'

COS a c o s I -^ Vs Jo(av) 2 a — — 2cos 2 (1 + cos a) (3.3.24) Copyrighted material

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ii) Reactive power

Reactive power is given as,

^reactive ~

Vs

h i

s*n

^1 v - . Vs---cos — sin

H -? )

& Vs ^o(av) . a a = ---2 sin — co s— 71 2 2 _ _ VmI d ‘>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 a 8 cos - j

Putting the values in above equation,

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HF

8 cos'

= 0.3108 or 31.08 %

b) The displacement factor is given by equation 3.3.20 as, DF = co s^

= cos 7 t/3 ^

0.866

c) The input power factor is given by equation 3.3.21 as,

I 8 2 a j « ^ a ) COS 2 ( n / 3 PF 8 2 cos 7t 71-= 0.827 lagging

3.3.4 Asymmetrical Half Bridge Converter 3.3.4.1 Operation with Resistive Load

Fig. 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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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.

Suppty voltage Firing p u lses o f T . ----Firing p u lses otTj Output current S C R T , & d iode D, current ~—~ r s c r t2 & d io d e D2 curren t irtt

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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,

VL

o(av) (1 + cos a)

V_c{rms)

2n rt - a + - sin 2a

3.3.4.2 Operation of Asymmetrical Half Bridge Converter with Level Load

1. Draw the circuit diagram and wauefrorms o f output voltage, output current, supply current and SCR currents for a single

p h ase asymmetrical half controlled bridge feedin g a level load. I >

, Marks [6], M ay -2004.2 0 0 5

J

asked in

V 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).

M ode - 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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r,,n

s~

t t

Equivalent Circuit - Equivalent Circuit - II

zf z;°

zf z;»:

f i

Equivalent Circuit - III Equivalent Circuit - IV

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Power Devices and Machines 3 - 3 9 Phase Controlled Rectifiers (AC/DC Converters) Mode - IV (2rc < a>t < 2n + a )

At 2k, 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 V_o(rms) VL 2tc 7 i - a + ^ s i n 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.

[M ay-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 2vm

2k 7 i-a + - s i n 2a

4/ £ iii) n harmonic supply current, Isn = —jL

y/2 2V2 iv) 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 Sym m etrical configuration of 1<j> HCB, VQ, i T^ and / 01 waveforms h (a v ) = \ \ 'r M d w f = ^ } I 0(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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^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,

1

h ( a v ) = j \

0

a Average diode current will be,

^D(av) 2n i* ^o(av)

2

n

0

t n - a

Average SCR current _ ' o(av)

2

n _ n - a Average diode current r n + a n + a

° ( a v > 2k

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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 and

Y com pare them. Marks [6J, D ec.-2000, M ay-2000, 2003

■

’TiTTi—

\ M

ost likely and

asked in

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 Com parison 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 pow er iv) rms supply current

v) Peak supply current [D ec.-2004, 18 M arks!

Solution: Given : Half controlled bridge

^s(rms) = 115 V, therefore Vm = 4 2 V ${rm) = V 2 x ll5 = 162.6 V R = 250 0 a l = 45° or K 4 3tc a 2 = 135° or T Copyrighted material

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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 + c o s a )

— ~ ( l + cos 1 1= 88.35 V

F o r a « - J , V0(av)

ii) RMS output voltage

o(nns) _2nm 7 t-a + ^ sin 2a

For a = V0{rms) 162.6J 27C n 1 . 7t\ ^ " 4 2 ( 4

J

109.63 V 3ti * - T * 2 34.65 V

iii) Load power

For a For a = 4' 3n 7 ' P' = U 2 o(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 Copyrighted material

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« = 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 _ 162.6 ls(peak) - R ~ 250 "

• For a = peak value of output voltage is Vm sin Hence,

t / 3ti

slrl ~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. If the average output voltage is 75 V calculate :

i) Firing angle ii) rms output voltage

iii) rms supply current iv) rms 7^ harm onic supply current

v) Supply pow er factor. [Dec.-2003,16 M arksl

Solution : Given : I 0^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 2

V = -I

o(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) l 2 = 99.15 V

iii) RMS supply current

j _ j , ~ ~ ,..-1 .0 3

s(rms) ~ o(av) = 8.198 A

iv) RMS 7th harmonic supply current

V2 dropped since it is rms value,

v) Supply power factor

o{av) 7 a — =---COS^r-7ti______ 2_ V2 4x10 (7x1.03) -= — co

s---=---' 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 cot R

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325.27 sin ^ n

= --- jo g — - 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 -J

Vm sin co t __________2 = o 9 s A

R 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.

[D ec.-2000, 8 M arks; D ec.-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) R M S su p p ly c u rren t

ln -a

) v 71

=

10 71 * " 3 1 71 8.165 A ii) O u tp u t voltag e

o(av)

V„<n,A = (1+ cos a) Copyrighted material

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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 o f 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 V

a n d , '» = i f = T - (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 2k+ a, Tj and T2 are triggered again and the cycle

repeats.

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S olution : 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.

(51)

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 W aveform s o f 1(}> fu ll converter fo r inductive load having continuous load current