Power Electronics Rectifiers by Bakshi

7/27/2019 Power Electronics Rectifiers by Bakshi

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Phase Controlled Rectifiers______ (AC/DC Converters)

Object ives

Principle of 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 of feedback diode and freewheeling operation.

Harmonic analysis of 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.

Controlledrectifier

Load

a

Controlcircuit

Fig. 3.1.1 Principle of operation of a controlled rectifier

( 3 - 1 )

Copyrighted material


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Power Devices and Machines 3 - 2 Phase Contro lled Rectif iers (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.

Applicat ions :

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

An sw er fo l l ow in g quest i on af ter read ing th is to p ic

1. What do you mean by commutation o f SCR ? Give types ofcommutations. Explain natural commutation in details.

Marks (6), May-2007 I \ Most likely andJ masked In previous

niversity E xam

Definition : 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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Pow er Devices and Machines 3 - 3 Phase Contro lled Rectif iers (AC/DC Conv erters)

Cu rrent com mu tation : The SCR is turned off by reducing its anode current

below holding current.

Voltage com m utation : The SCR is turned off by applying large reverse voltage

across it. Principle of line comm 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 triggeredin 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,commutat ion to turn of f SCR

Mains AC

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

.... .....I..... ...................*......... . ..... .......... .31113S HIS sstS :!t H :3S !t:S!?:!: SH i !H9 5SI! 8: HIE Bi:! HHi IS

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

Fig. 3.1.4 Waveforms of half wave controlled rectif ier to i l lustrate natural

commutat ionCopyrighted 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 Commutatio n

3.1.3.1 Principle of Forced Commutation

Forced commutation is used when the supply is

D.C. A commutation circuit is connected across theSCR 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 commutat ion



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Pow er Devices and Machines 3 - 5 Phase Contro lled Rectif iers (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 arerequired.

External commutation components arerequired.

2. Requires AC voltage at the input. Works on DC voltages at the input.

3. Used in controlled rectifiers, AC voltagecontrollers etc.

Used in choppers, inverters etc.

4. No power loss takes place during commutation Power loss takes place in commutatingcomponents.

5. SCR turns off due to negative supply voltage. SCR can be turned-off due to voltage andcurrent 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 Hal f Wave Con verter and Effect of Freewh eel ing

Diode

3.2.1 Single Phase Half Wave Controlled Rectifier with Resistive Load

An swer fo l l ow in g quest i on af ter read ing th is top ic

1. Explain the operation o f 1half wave converter with the h elp o f

circuit diagram and waveforms.Most likely and

ImportantQuestion

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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Pow er Devices and Machines 3 - 6 Phase Contr olled Rectif iers (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


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Mathematical analysis

The average value of output voltage is given as,

1 T

Vo(av) = f ! vc M d o t0The 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>t2n 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 circuitdiagram 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 areFig. 3.2.3 Half wave co nt ro lled rec tif ier 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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Fig. 3.2.4 Waveforms of half wave controlled rectif ier 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

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 for average value of output voltage for 1


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Pow er Devices and Machines 3 - 1 0 Phase Contro lled Rectif iers (AC/DC Conv erters)

3.2.3 Half Wave Controlled Rectifier with Freewheeling Diode

An sw e r f o l l ow i n g quest i on a f t er r ead i ng t h i s t op i c

1 . Com pare freewheeling diodes and feed ba ck diodes.Marks [31. D ec.-2 006 , 20 08

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 halfwave controlled rectif ier

Fig. 3.2.8 Freewheeling action in halfwave controlled rectif ier

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 theload 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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Pow er Devices and Machines 3 - 11 Phase Controlled Rectif iers (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 tofreewheeling 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 Control led Rect i f iers (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 freewheel ing diodes and feedback diodes

Sr.

No.

Freewheeling diodes Feedback diodes

1. Load energy is utilized in load itself throughfreewheeling diodes.

Load energy is feedback to the sourcethrough feedback diodes.

2. Freewheeling diodes have to carry full loadcurrent.

Feedback diodes carry full load current sometimes.

3. Free wheeling diodes are slower. Feedback diodes should be fast.

Example 3.2.2 : Derive an expression for average value of output voltage for 1 half

wave controlled rectifier for RL load and freewheeling diode.

Solut ion : 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

= fj v0(at) d(0t = Vm sin


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Pow er Devices and Machines 3 - 1 3 Phase Contr olled Rectif iers (AC/DC Con verters)

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.,V

o(av)

= ^ - ( 1 + c o s c x )1 71

22,042 ( , k

= ^ r [ 1+cos6

= 96.6 V

ii) To obtain effective output voltage v0, 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)

Vi f 1 -cos 2 (0/

J -------2-------di0tm

2n

V,m4 71

/V

J dcot- J cos 2 ( 0 1 dcot

\Vm sin 2wf" n \

1 I k a - 2I . a J

V. a sin 2 a1 +

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 Contr olled Rectif iers (AC/DC Conv erters)

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 % of 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 2xl2 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(av)max



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Pow er Devices and Machines 3 - 1 5 Phase Contro lled Rectif iers (AC/DC Con verters)

Consider again equation 3.2.1

V.Vo(av) r^(l+COSCX)2k v 7

Putting values in above equation,1697

13.52 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

13.5

101.35 A

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

Vo(rms). a sin 2a1 +

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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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,

V2Vo(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)

> Exam ple 3.2 .5 : For a single phase half wave converter having resistive load of 'R ' and

the delay angle of , determine

i) Rectification efficiency

iii) Ripple factor

Solut ion : Given data

71

= 2

ii) Form factor

iv) PIV rating of thyristor INOV.-2007,10 Marksl

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

a sin 2 a1 + -----

71 2 7t

/o Sm

K 2 71

= 0.353 VLV

o(nns)o(rm$) _ 0. 353 V~R R



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Pow er Devices and Machines 3 - 1 7 Phase Contr olled Rectifiers (AC/DC Con verters)

i) To obtain rectif ication eff iciency

P(tc ^o(av)

P V Iac o(rms) o(rm$)

= 0.2028 or 20.28 %

i i ) To obtain form factor

V,o(rms)FF

V,o(av)

i i i ) To obtain ripple factor

RF = -Jf F 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 Semico nv erters (Hal f Br idge 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^ andT2. 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) Circu it diagram o f 1 (j>sem icon ver ter(Common cathode conf igurat ion)



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Pow er Devices and Machines 3 - 1 8 Phase Contr olled Rectif iers (AC/DC Converters)

A d ,

230 V, 50 HzAC supplyo----------


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Pow er Devices and Machines 3 - 1 9 Phase Controlled Rectif iers (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


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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 Dj 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 Contr olled Rectif iers (AC/DC Converters)

Fig. 3.3.4 Cond uctio n o f T2-D2 in negative half cycle of the sup ply.

Dotted l ine shows path of current f low

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 shownin 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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Pow er Devices and Machin es 3 - 21 Phase Con tro lled Rectifiers (AC/DC Con verters )

Example 3.3.1 : For the 1


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3.3.3 Workin g with Induc tive (R-L) Load

An swer foUow in g quest i on af te r read ing th i s t op ic

1. Draw the circuit diagram, voltage and current waveform for

a = 60, RL load o f semi-converter. Marks[8], May-2007Most likely and

asked In previousUniversity Exa m

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

An swer fo l l ow in g quest i on af ter reading t h is top ic .

1. How freew heeling 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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Pow er Devices and Machines 3 - 23 Phase Con tro lled Rectifiers (AC/DC Converters )

Fig. 3.3.5 Wavefo rm s o f 1sem ico nv erter fo r c on tinu ou s load cur rent

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 isfreewheeling diode in semiconverter, then freewheeling current flows through this diode.


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Power Devices and Machines 3 - 2 4 Phase Contro lled Rectif iers (AC/DC Converters)

^ D2|___j ^ D1

Fig. 3.3.6 Freewheeling action takes place through T y D2

Aver ag e val ue o f ou tput vo ltag e w ith 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

\V2V - 1 mo(rms) 1 2 7l 7i- 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.



25/99

Pow er Devices and Machin es 3 - 25 Phase Con tro lled Rectifiers (AC/DC Con verters )

Fig. 3.3.7 Discontinuous mode of single phase semiconverter

3.3.3.3 Continuous and Ripple Free Current for Large Inductive Load

An swer fo l l ow in g quest i on af ter r ead ing th is t op ic

With the help of neat circuit diagram, mode equivalent circuitsand waveforms o f supply voltage, supply current, output voltage,output current, explain the operation o f a single phase halfcontrolled bridge feeding a level (highly inductive) load.

Marks [5], Dec.-2000; Marks [10], May-2006

N Most likely andasked in previous

University Exa m

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 andripple 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.

Copyrighted


26/99


Fig. 3.3.8 Wavefo rm s o f 1semi co nv erter fo r hig hly ind uc tive load

Example 3.3.2 : Derive an expression for output current for RLE load driven by 2

semiconverter. Assume continuous conduction.

Solution : Fig. 3.3.9 shows the circuit diagram of 1 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 thiscircuit will be similar to those shown in Fig. 3.3.5. From a to n, T|- D1 conducts and


27/99


vs =Vmsin

Fig. 3.3.9 A 1sem ic onvert er dr iv ing RLE lo ad

supply voltage vs is applied to the load. Hence an equivalent circuit will be as shown

below :

Vmsin cot0

Fig. 3.3.10 Equivalent circuit when T yD1 or T2- D2conduct

By KVL in above circuit we get,

di.Awt)Vm sin cof = R io l(


28/99

Power Devices and Machines 3 - 2 8 Phase Contr olled Rectif iers (AC/DC Converters)

For

j i to n + a

Fig. 3.3.11 Freewheeling action in T^D2or 72D1

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 fo r supply current

ii) RMS value of nth harmonic o f supply current.

iii) Fundamental component of supply current

iv) RMS value of 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 t a v ) + X c n s in (m o f + );=1

where ctJ = yja* +b%

and = tan -1 n



29/99

Pow er Devices and Machines 3 - 29 Phase Con tro lled Rectifiers (AC/DC Con verters )

Here an yJ is(ot) COSMof rfcof

2n:

2k J is (iot) cosncof ifcof

From the supply current waveform of Fig. 3.3.8 we can write,

2n_2_

2n| l


30/99

Power Devices and Machines 3 - 3 0 Phase Contr olled Rectif iers (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 l0(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 can be

calculated as,


31/99


o()

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) = Zo((iv) n a .

cos - y smtin

( f )h= 1,3,5,

ii) RMS value of nlh harmonic 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) _ na _ , 0 c= -----------cos /n =1,3,5


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

h i = 0.9 1 ^ cos |

iv) To obtain rms value of supply current


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 71rr+a

. (3.3.16)

...(3.3.17)

= 1 in

...(3.3.18)



32/99

Pow er Devices and Machines 3 - 3 2 Phase Contr olled Rectif iers (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 thefollowiitg :

i i ) 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 previousexample and 4>1 above we get,

i) Displacement factor (DF) ii) Supply power fac tor (PF)

iii) Harmonic factor (HF) iv) Current distortion fac tor (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)

i i i ) Harmonic factor

The harmonic factor (HF) is given as,

Putting values in above equation,

Copyrighted materi


33/99

Pow er Devices and Machines 3 - 33 Phase Con tro lled 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 distort ion factor (CDF)

The current distortion factor (CDF) is given as,

ICDF = sl

's(rms)

2V2 /d a v ) a

--------- COS 71 2

c{av)I n - 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 = Vs J S1 cS


34/99

Power Devices and Machines 3 - 3 4 Phase Control led Rect i fiers (AC/DC Converters)

i i) Reactive power

Reactive power is given as,

^reactive ~

Vsh i

s*n1

v - .Vs ---------------cos sinH - ? )& Vs ^o(av) . a a

= -------------------2 sin co s71 2 2

_ _ VmId >v) sina ...(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 of transformer is unity. The delay angle a=-^. Calculate-

a) Harmonic facto r 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,



35/99


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 2cos

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



36/99


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.

Supptyvol tage

Firing pulses

of T.----

Firing pulses

otTj

Outputcurrent

S C R T , &

diode D,

current

~~ r

s c r t 2 &

diode D2

current

irtt

Fig. 3.3.13 Waveforms of half bridge converter of Fig. 3.3.12


37/99

Pow er Devices and Machin es 3 - 37 Phase Con tro lled 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

An sw er fo l l ow in g quest i on af t er read ing t h is t op ic

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

phase asymmetrical hal f controlled bridge feedin g a level load. I >, Marks [6], M ay-2 004.2 00 5 J a s k e d i n

V Unive rsi ty 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.

Mode - I (a < c o t


38/99


r,,ns~

t tEquivalen t Circu it - Equivalen t 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


39/99

Pow er Devices and Machines 3 - 3 9 Phase Contro lled Rectifiers (AC/DC Conv erters)

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 -D2

since they are more forward biased compared to T2 -D2.

At 271+a, SCR Tj 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)VL

2tc7 i - a + ^ s in 2a

)>* Example 3.3.7 : For the single phase asymmetrical half controlled bridge circuit derive

expressions f or

i) Average output voltage ii) RMS output voltage

iii) RMS value of the nth harmonic supply currentiv) 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

7i -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------------------CO S^ r-

7171 2

COSa

J k (k - a)

)>! Example 3.3.8 : Draw the circuit diagrams of symmetrical and asymmetrical single phase



40/99

Power Devices and Machines 3 - 40 Phase Contr olled Rectif iers (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 of 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 conf igurat ion

Fig. 3.3.15 shows the SCR and diode currents for symmetrical configuration.

Average SCR current will be,

Fig. 3.3.15 Sym m etri cal c on fig ur atio n o f 1HCB, VQ, i T^and / 01 wav eforms

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


41/99


ô(av)

Average SCR current 2 j

Average diode current ^0{av)

T ~

SCR and diode currents for asymmetrical conf igurat ion.

Fig. 3.3.16 shows the SCR and diode currents for asymmetrical configuration.

i*--1

Fig. 3.3.16 Asy m metri cal conf iguration o f 1HCB, VQ, i Tând i waveform

Average SCR current will be,

1h ( a v ) = j \

0

a

Average diode current will be,

^D(av) 2n i* ô(av) 2n0

t n -aAverage SCR current _ ' o(av) 2n _ n - a

Average diode current r n + a n + a

( a v > 2 k



42/99


3.3.4.3 Comparison of Symmetrical and Asymmetrical Configurations

An swer fo l l ow ing quest i on q f t er read ing th i s t op i c

1. Explain different configurations of semiconuerter andY compare them. Marks [6J, Dec.-2000, May-2000, 2003TiTTi \ Most likelyand

asked in previousUniversityExam

Table 3.3.1 shows the comparison between symmetrical and asymmetrical

configurations of half controlled bridge.

Sr. No. Symmetrical configuration Asymmetric al configuratio n

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 diodeand on SCR.

Freewheeling takes place through both theSCRs.

4. Average currents of SCR and diodes aresame.

Average currents of diodes are higher thanSCR.

5. SCR and diodes conduct for equal durations. SCRs conduct for shorter duration comparedto 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 of 250 Cl For firing angles of 45 and 135,

Calculate :

i) Average output voltage ii) nns output voltage

iii) Load pow er iv) rms supply current

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

Solut ion: 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 orK

4

3tc

a 2= 135 or

T



43/99

Pow er Devices and Machines 3 - 43 Phase Controlled Rectif iers (AC/DC Conv erters)

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)

i i ) RMS output voltage

o(nns)

m

2n

7t-a + ^ sin 2a

For a = V0{rms)162.6J

27C

n 1 . 7t\

^ " 4 2 ( 4J 109.63 V

3t i

* - T * 234.65 V

i i i ) Load power

For a

For a =

4'

3n

7 '

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 ' hirms)

h(rms) ^o(rms)

^o(rtns)

R

109.63250 0.438 A



44/99

Power Devices and Machines 3 - 4 4 Phase Controlled Rectif iers (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/ 3 tislrl ~T~ 162.6 x 0.7071 . . . 4

W ) = --------R = ---------250---------= 0 46 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 ang le ii) rms output voltageiii) 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

i i ) RMS output voltage

I V2V = -I

o(rms) } 2 JT



45/99

Pow er Devices and Machines 3 - 4 5 Phase Contr olled Rectif iers (AC/DC Conv erters)

f (155.56)2 f ____1 .

2717c -1.03 + 2 sin( 2 x 1.03)

l

2

= 99.15 V

i i i ) RMS supply current

j _ j , ~ ~ ,..-1 .03s(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)-= cos-----=-----

' 71 -----= - 1.15 A Here negative sign can be

PF = a ) cos2 ^ ... By equation 3.3.21

8 2f1.03^

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 of 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. lMay-2003,12 Marks!



46/99


Solut ion : Given : lHCB

Vs = 230 V, vm = J lV s = 7 2 x 230 = 325.27 V

R = 100 n , a = 60 or |

i ) Average output vol tage

ii) RMS output voltge

206.3 V

i i i ) 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



47/99


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 cot __________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 of 10 A, operated at firing angle of 60. Calculate :

i) RMS supply current ii) Output voltage

iii) Supply power fac tor v) RMS value o f third harmonic input current.[Dec.-2000, 8 Marks; Dec.-2006, 10 Marksl

Solut ion : Given : 1 HCB

vs = 230 V, Vm = V2 1/ = V2 X 230 = 325.27 V

'< > = 10 A' = 60 or |

i ) RMS supply cur rent

ln-a) v 71

= 10

71

* " 31 71

8.165 A

i i ) Outpu t vo l tage

o(av)V


48/99

Pow er Devices and Machines 3 - 49 Phase Con trol led Rectifiers (AC/DC Converters )

3.4.1 Work ing wi th Resisti ve Load

Fig. 3.4.2 Conduction of 7^ and T2in posit ive half cycle of the supply.

Dotted l ine shows path of current f low

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 Va 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 2 k + a, Tj and T2 are triggered again and the cycle

repeats.



49/99



50/99

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 givendata 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+CS)

= - ( 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)

184.8o(av)

= ',(,)* + 50

= ') * l > +50= 1.348 A

Thus the current through the load is 1.348 A.



51/99

Power Devices and Machines 3 - 53 Phase Contro lled Rectif iers (AC/DC Converters )

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 of 1(}>full converter for inductive loadhaving continuous load current


52/99

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 t i . 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 larger din

voltage 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 isalso 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 ,

Answ er fo llow in g questio n after reading th is topic

1. Draw the circuit diagram o f a single phase fully controlled bridgerectifier and sketch the waveforms o f output voltage, outputcurrent, supply current and SCR current for a level (ripple free)load. Marks [5], May-2000. 2001; Marks [10]. Dec.-2004

Most likely an d

asked In previousUniversity Ex am

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 Waveforms of 1full converter for continuous andripplefree load current in case of 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]

Solution : i) Average output voltage for inductive load

The average output voltage is given as,

1 T

Vo(av) = f J vo (00

Observe the waveforms of lfull 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(av) = - J Vm sin d(s>ta

r .nTi+a= - COS (0 1

K 1 J

pyrighted material


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Power Devices & Machines 3 - 56 Phase Controlled Rectifiers (AC/DC Converters)

2 Vo(av) cos a ... (3.4.5)

This is the expression for average load voltage of lfull converter for inductive load.

Plot of V0(av) versus firing angle (a)

Following table lists the values of VQ/av\ with firing angle (a)

aVo{*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,) with 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 of VD^ with respect to a



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Power Devices & Machines 3 - 5 7 Phase Controlled Rectifiers (AC/DC Converters)

ii) RMS value of output voltage for inductive load


J \ vo (full converter feeding RLE load



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Power Devices and Machines 3 - 58 Phase Cont rol led 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 = Vvo(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 inductanceis 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


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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 05

= +j l62.48 + | | -| | ^ si n (0 .6 9 8 -1.3744)} el5l3 (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) -100 + 392.89c 01989 ] ;

0.698

1 9A QR 3 ?39 1 nn3 ?39 qq 3 839= ^ J s i n ( ( o f-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 : If a freewheeling diode is added across the highly inductive load in l

full converter, derive an expression for average load voltage.

Solution : We know that freewheeling action does not take place in 1 full converterinherently. 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) =325.27

2.5622sin(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 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) . ..(3.4.10)

Fig. 3.4.13 Freewheeling diode conducts from k to rc+a due to inductive load

Fig. 3.4.14 Waveforms of 14>full converter for highly inductive

load and freewheeling diode across the load

Copyright!


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Power Devices and Machines 3 - 6 2 Phase Contro lled Rectifiers (AC/DC Converters)

))! 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.Solution : The rms value of 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 andripple 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 V2Vo(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,

146.42

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 I0 m \ i.e. 14.64 A.

Example 3.4.7 : For a 1fyfull converter having highly inductive load derive the following:

i) Fourier series for supply current

ii) RMS value of nth harmonic of supply current

iii) Fundamental component o f supply current

iv) RMS value of supply current



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Solution : i) To determine Fourier series

The general expression for Fourier series is given as,

0

>S ( 0 = ^ ( a p ) + Z cn S' ( + )

n=1

where

and

Here,

+bn

tan-1

bn\ n /

J i$( c o f ) c o s n o t d o t

2 n

J is (o t) cos n ot d o to

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(


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Hence

o(av)Ji+a 2n+a

J s innatdto t - j sin n atd & tL a rt+a

2 Io(av)

nn

4 I

cosna [1

-cos nt i]

o(av)

nnc 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 nfor w= l , 3, 5,

And = tan 1 jP~ = -n a from equation 3.4.11 and equation 3.4.12.n

Thus = -n a

The 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 of nth harmonic supply current

The RMS value of the n^ harmonic of the supply current its given as,

4 !o(av)

1 - _ ! L - n n

sn 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 Contro lled Rectifiers (AC/DC Converters)


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 current


$ (rms)1 7= J /s2 ( c o f ) d o t

1 /2

From supply current waveform of Fig. 3.4.9,

Es(rms) In

2n+a

\ Io(v)d


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ii) Supply power factor (PF)

The supply power factor is given as,

PF

sl

s(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 distortion factor (CDF)

The current distortion factor (CDF) is given as,

CDF sl

s(rms)

2V2 1o(av)

o(av)

2>/2 0.9 ...(3.4.22)



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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 ------- co s(-a), since 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.

CommentCompare 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. Ifthe turns ratio of the transformer is unity, if the delay angle is a = Calculate the

i) HF of 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 Cont rol led Rectifiers (AC/DC Converters)

HF = 0.4834 or 48.34 %

ii) Displacement factor (DF)

For 1 full converter, DF is given by equation 3.4.19 as,

iii) P ower factor (PF)

For 1 full converter, PF is given by equation 3.4.20 as,

nc 2V2PF = ------cos a

71

= 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 factor

Solution : Given : 1 FCB

ii) Fundamental component of input current,

iv) Harmonic factor

vi) Output voltage. [May -2000,10 Marks]

v: 220 V Vm = 220 /2 = 311.12 V

I0(av) = 10 A, a = 30 or - radians.

i) RMS supply current

ii) Fundamental component of 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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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 inductiveload. 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.


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 radians6


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 71PF = ------cos a = ------ cos-7 = 0.78

71 71 6

iv) S upply 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 ? [May-2007,4 Marks]



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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 inFig. 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) Circuit diagram

positive 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 circuit of Fig. 3.4.15(a)



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Power Devices and Machines 3 - 7 2 Phase Controll ed Rectifiers (AC/DC Converters)

)) Exam ple 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.

Copyrighted mater


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Power Devices and Machines 3 - 73 Phase Contro lled Rectifiers (AC/DC Converters)

3.4.3 Inversion in 1


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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 Afull converters

Fig. 3.4.21 Inverting operation in 1


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i) Calculate the value of the current limiting resistor required for nominal charging

current of 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 fir ing 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 device

drops.

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

187.127-144

15

Rc l = 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:


2VV.o(av)

m(max)

ncos a ,

2 x 264 x V2187.127 = ------------ cos a,

And

a max = 38

2 V

o(av)w(min)

acos a,

2x21 6x V2187.127 = ------------ co sa ,

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 ) + ^ 4 4 ^ b a tt ^ b at t =

/. 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 will

be,

Battery power = 10x144 = 1440 W.



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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 Rectifiers

Now 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 ofSCRs and half number of diodes. This consists of all the SCRs ascontrolled devices.

2. This operates in only onequadrant.

This can operate in two quadrants.

3. Output voltage is always positive. Output voltage can be negative in caseof inductive loads.

4. Inherent freewheeling action ispresent.

External freewheeling diode is to beconnected 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, lightingand heater control.

Used for DC motor drives.

Table 3.4.3 Comparison of half and fully 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 of45. Find,i) Average output voltage ii) RMS supply current

iii) Harmonic factor iv) RMS value o f 3rd harmonic o f input current.[May-2001, 2008, 6 Marksl

Solution : Given : l FCB.

Vs = 230 V, Vm = 230V2 = 325.27 V

l 0(av) = ^ A, a = 45 or ^ radians.



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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 of 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 ,--------------cos smj --

w ^ l o(av) . a CL= -Vs ------------- 2 sm - cos

n 2 2

'^ ^sK(av) . . . . a a .

= ------------------sin a smce 2 sm cos = sin at i 2 2

^o(av) . . r = . , . .= ------------ sm a since V 2 Vs = Vm



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b) Reactive power of full converter

P.i(reactive) ^s/siSini

2V2 7

= V

sin a

21/ /m o{av) .--------------sin a since -J l 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

Answ er follow in g question aft er read ing th is topic

1. Explain the working of 3 semiconverter with the help of

waveforms.

[Dec.-2006, Dec.-2008]

.v.

V - Mo s t l i k e l ^HH

ImportantS i

We have discussed 1 (j) semiconverter earlier. The 3 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 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 theline voltage waveforms.

Load

Fig. 3.5.1 3semiconverter or half bridgeconverter



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Power Devices and Machines 3 - 81 Phase Controlled Rectif iers (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 of 3supply

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

Copyrighted mater


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Fig. 3.5.3 Wav

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