CHAPTER 11

11.1 (i) The waveforms of VT1 and VT2 in Fig 11.3 (b) reveals that the greatest forward or reverse

voltage would appear across either of the thyristor when .2pa ≥

Magnitude of these voltages, Em = 2 sE

= 2 .230 = 325.27 V

(ii) RMS value of the output voltage from Eq. (11.3) is given by

Eorms = ( ) 1/ 21 sin 2–2sE ap a

pÈ ˘+Í ˙Î ˚

(a) for a = 120 /∞ , Eorms = 135.10V

\ max forward or reverse voltage = 135.10 ¥ 2 = 191 V.

(b) for a = 60∞ , Eorms = 188.616 V

\ maximum forward or reverse voltage = 266.74 V.

(iii) From Eq. (11.8)

A5= ( )( )

( )( )

2 230 sin 5 1 / 4 sin 5 – 1 / 4–5 1 5 – 1

p pp¥ +È ˘

Í ˙+Î ˚

= – 17.29

From (11.9).

B5 = ( ) ( )4 4cos 6 – 1 cos 4 12 230 –

6 4

p p

p◊ -¥ È ˘

Í ˙Î ˚

= 34.48 From Eq. (11.10), peak amplitude of 5th harmons

output voltage, E5 = ( ) ( )22 25 5 17.29 34.48A B+ = - +

= 38.57

Phase fn = – 1 5

5tan

BA

= 1 34.48tan17.29

--

= - 63.44∞

56 Power Electronics

11.2

(a) For controlling the load, the minimum value of firing angle a = load phase angle,

f = 1tan WLR

-

= 1 43tan 53.13 .- = ∞

Maximum possible value of a is 180∞ .

\ Firing angle control range is 53.13 180a∞ £ £ ∞ .

(b) At 53.13a f= = ∞ , the maximum value of rms load current occurs. But, at this value of firing angle, the power circuit of ac voltage controller behaves as if load is directly connected to ac source. Therefore, maximum value of rms load current is given by

I0 = ( )2 2 22

230 230 46 .3 4

AR WL

=++

(c) Maximum power = 2 20 46 3 6348 .I R W= ¥ =

Power factor = 2 2

0

5 0

46 3 0.6.230

I RE I

¥= =

(d) when a = f, average thyristor current is maximum and conduction angle ° = p.

from Fig. 11.5 (c),we can write

\ ( ) ( )Tavg1I sin

2Em t d tz

a p

aw f w

p+

= -Ú

2 2

2 230 20.707. . 3 4

Em Azp p

¥= =¥ +

.

Similarly, maximum value of thyristor current is

( ){ } 121 sin

2TMEmI t d tz

a p

aw a w

p+È ˘= -Í ˙Î ˚Ú

2 230 32.527 .2 2 5Em A

Z¥= = =¥

(e) Maximum value of diodt

occurs when

a = f. From eq. (11.25),

( ). cos cot 0dio w Emdt z

f- = .

Its value is maximum, when cos(w t – f) = 1.

4

max

2. 230. 2 . 50 2.0437 10 / .5

dio A Secdt

pÊ ˆ\ = = ¥Ë ¯

Solution Manual 57

(f) For a = 0, Fig 11.6 shows that conduction angle r is 180°

For a = 120° and 53.13°,

Fig 11.6 gives a conduction angle of about 95°.

11.3 Given: R = 12 W, L = 24 mH,

Em = 2 240 339.41 ., 50 .V f Hz¥ = =

Z = ( ) ( ) ( )2222 312 2 50 24 10R WL p -+ = + ¥ ¥ ¥

= 14.17 W

f = ( ) 11 1 2 50 24 10tan tan

12WLR

p -- - Ê ˆ¥ ¥ ¥= Á ˜Ë ¯

= 32.13 0.56 .rad∞ =

(a) With zero firing delay, normal steady state theory applies.

Load impedance, Z = 14.17 32.13∞W

Load current = ( )sinEm tz

w f-

= ( )339.41 sin 2 50 0.5614.17

tp -

= ( )23.95 sin 2 50 0.56tp - A.

Load power = ( )240 23.95 cos 32.132

¥ ¥

= 3441 W.

Note that a train of firing pulses will be required as the thyristor will not turn-on until after 32.13°.

(b) Taking t = 0 at the instant of firing ( )90a - ∞ then

i = ( ) ( ) 1223.95sin 2 .50 0.56 23.95 sin 0.562 2 0.024

t tp pp Ê ˆ+ - - - - Ë ¯

= 23.95 sin(2p.50.t + 1.01) – 20.29 e500t A.

Current will cease when i = 0 at t = 7.33 ms/, that is, at an angle of 221.9° on the voltage wave.

Mean load power ( ) ( )0.00733

500

0

1 240 2 sin 2 50 23.95sin 2 .50 1.01 20.290.01 2

tet t dtpp pÈ ˘= + -Î ˚Ú

= 1370 w

(c) Similarly, for a = 120°,

mean load power = 177.2 W.

58 Power Electronics

11.4 From Eq. (11.36), rms output voltage is given by ( ) 12

01 sin 22

2 2sE E ap ap

È ˘= - +Í ˙Î ˚

(i) a = 0°

E0 = ( )1

21. 2 0 02sE pp

È ˘- +Í ˙Î ˚

E0 = Es

If R is the resistance of heater element (load) and the maximum power or rating of heater, p = 1 kW. Therefore,

R = ( )22

3230 52.9

1 10EP

= = W¥

\ Output power or power dissipated by the heater element at a = 0°,

Po = 2 2 2

0 230 1 .52.9

E Es kWk k

= = =

(ii) a = 180∞

Eo = ( ) 121 sin 22

2 2Es pp p

pÈ ˘- +Í ˙Î ˚

= [ ]120.5 0.707s sE E=

Po = ( ) ( )2 22 0.707 0.707 230 50052.9

Eo Es wR R

¥= = @

(iii) a = 70∞

Eo = ( ) 121 sin 1402 70

2 2Es p

p∞È ˘- ∞ +Í ˙Î ˚

= 0.926 Es

Po = ( ) ( )2 20.926 0.926 230 856.71R 52.9

sEw¥=

11.5 (a) Rms live current= 24000 33.39 A.3 415

=¥

Hence, required triac rms current rating = 33.39 A.

(i) For 16 kW,

1624

= ( )12 sin 2p a a

p- +

Solving the above equation, we get

\ a = 74.49°

(ii) For 8 kW,

Solution Manual 59

824

= ( )12 sin 2p a a

p- +

Solving the above equation

we get a = 105.29°.

(b) If thyristors were used, they would control for only one half-cycle of the sine wave, have their required rms

current rating = 33.392

= 23.61 A.

The voltage requirement would be identical to that for the triac.

11.6 Given e1 = 110 V, e2 = 110 V, Ep = 220 V, E6 = 170 V, R = 8 W,

a = 72°.

(a) Rms current of thyristor T1 & T2 can be obtained from eq. (11.38), as

\ I1 = ( ) 12110 110 1 sin 14472

22 8p

p+ ∞È ˘- ∞ +Í ˙Î ˚¥

= 16.19 A.

(b) Rms current of thyristors T3 and T4 can be obtained from (Eq. 11.33), as

I2 = ( ) 12110 1 sin 14472

22 8 p∞È ˘∞ -Í ˙Î ˚¥

= 5.382 A.

(c) The Rms current of second (top) secondary winding is Iw2 = 2 I1

= 22.90 A

Rms current of first (lower) secondary winding, which is the total

Rms current of thyristors T1, T2, T3 & T4 is

Iw1 = ( ) ( )1

22 21 22 2 .I IÈ ˘+Î ˚

= ( ) ( )1

22 22 16.91 2 5.382È ˘¥ + ¥Î ˚

= ( ) ( )[ ]12571.896 57.931+

= 25.095 A.

The volt-ampere rating of primary or secondary,

EA = e1 . Iw1 + e2 . Iw2

= 110 25.095 110 22.90¥ + ¥

= 52.79.45

load power, po = 2

oER

60 Power Electronics

= ( )2170 3612.5

8=

\ The power-factor, pf = poEA

= 3612.55279.45

= 0.684 (lagging)