MD Singh Power Electronics Solution Manual to Chapter 12

CHAPTER

12

Lr = 30 mH, Cr = 5.2 mf, 12.1 Given: Edc = 150 V, R=4W (i) For series resonant inverters, wr =

F0 = 5 kHz,

1 30 ¥ 10

-6

¥ 5.2 ¥ 10

-6

tq = 20 ms,

4¥4

-

4 ¥ 30 ¥ 30 ¥ 10 -12

= 44,384.68 rad /s. But

wr = 2 p fr

\ fr = 7.064 kHz

wo = 2p fo = 2p ¥ 5103 = 31415 ◊ 92 rad / s.

and

Toff > tq =

\

=

p p wo wr p p 31415 ◊ 92 44384 ◊ 68

= 28.94 ms. (ii)

Maximum possible frequency is given by 1 1 = Fmax = p 2 (tq + p / wr ) 2 Ê 20 ¥ 10-6 + ˆ ÁË ˜ 44,384.38 ¯ = 5.496 kHz (iii) Capacitor voltage from Eq. (12.33) is given by 4p È ˘ Í 2 ¥ 30 ¥ 10-6 ¥ 44384 ◊ 68 + 1 ˙ ˙ Ec = 150 Í e Í ˙ Í ˙ 4p -1 ˙ Íe 6 Î 2 ¥ 30 ¥ 16 ¥ 44384.68 ˚ = 152.671 V

(iv) From Eq. (12.36), rms load current is given by ILrms = 25.58 A (v)

Output power,

Po = I 2 Lrms ¥ R = (25.58)2 ¥ 4 = 2617.35 W

(vi)

Rms thyristor current, Irms (SCR) =

25.58 2

= 18.08 A

62

Power Electronics

from Eq.(12.34), average load current is given by IL(avg) = 18.92 A.

\

SCR average current = Iavg(SCR) =

1

2

IL(avg) = 9.46 A.

12.2 Given:- Edc = 150 V, Lr = 25 mH, R = 1W, tq = 25 ms, C1 = C2 = 6 mf. In half-bridge circuit, Ceq = C1 + C2 = 12 mf. (a) From Eq. (12.38), 1 R2 Lr .ceq . 4 Lr 2

wr =

1

=

25 ¥ 10

-6

¥ 12 ¥ 10

-6

-

1 4 ¥ 25 ¥ 25 ¥ 10-12

= 54129.47 rad/s. fr = 8.62 kHz

\ (b)

From Eq.(12.40), maximum resonant frequency fr (max) =

1 2 ¥ 25 ¥ 10-6

= 20 kHz

f0 = 0.3 ¥ 20 ¥ 103 = 6 kHz.

Now,

Period for resonant waveform, tr = Delay time

1 = 0.167 msec. fo

=

Output period T0

td =

1 20 ¥ 103

= 0.05 msec.

To - tr . 2

Ê 0.167 ¥ 10-3 ˆ - 0.05 ¥ 10-3 ˜¯ = 33.5 msec. = ÁË 2 Since, delay time is greater than zero, therefore, this inverter circuit would operate in non-overlapping mode. (c)

Now,

tp =

=

1 Ê 2 wr . Lr .ˆ tan -1 Ë wr . R ¯

Ê 2 ¥ 54129.47 ¥ 25 ¥ 10-6 ˆ 1 tan -1 ÁË ˜¯ 54129.47 1

= 1.847 ¥ 10-5 ¥

69.72 ¥ p = 22.48 usec. 180

Now, 150

Ec1 = e

1¥ 2 ¥p +1 2 ¥ 25 ¥ 10-6 ¥ 54129.47

=e

150 + 1 = 13.39 V 2.322

Solution Manual 63

\

150 + 13.39 Ip = sin ( 2p ¥ 8.62 ¥ 103 ¥ 22.48 ¥ 10-6 ) - e 54129.47 ¥ 25 ¥ 10-6

22.48 ¥ 10-6 2 ¥ 25 ¥ 10-6

= 72.39 A (d)

Average SCR current, - R Lr ˘ ( Edc + Ec1 ) fo ÈÍ 4. Lr + Iav CSCR = 1 e Î ˚˙ ÈÎwr 2 + ( R 2 Lr )2 ˘˚ Lr.

(e)

= 18.96 A. Rms current through the SCR is given by

È

Irms(SCR) =

(f)

Rms load current

( Edc + Ec1 ) Í 1 wr . Lr .

(

- R.tr

1- e Í Í 2.To ÎÍ

2 Lr

)

Ï ÔÔ Lr R 1 Ì - . Ô R Lr ( 2 w )2 + R r ÓÔ Lr

( )

= 24.64 A = 2. Irms (SCR) Io = 2 ¥ 24.64 = 49.28 A

(g)

Output power,

P0 = Io2.R = (49.28)2 ¥ 1

= 2428.52 W P 2428.52 Average supply current (Is) = o = 150 Edc = 16.19 A 12.3 Given: Cr = 6 mf, Lr = 30 mH, R = 3W, tq = 18 ms. Edc = 180 V, (a)

1 Ê 1012 9 ¥ 1012 ˆ Resonant frequency = 2p ÁË 30 ¥ 6 4 ¥ 30 ¥ 30 ˜¯

\

fr = 8.76 kHz wr = 2p fr = 55 kHz fo = 0.3 fr = 0.3 ¥ 8.76 ¥ 103 = 2.628 kHz

\

1

2

To =

1 = 0.38 msec. fo

td =

To 0.38 ¥ 10-3 1 - tr = = 76 ms. 2 2 8.76 ¥ 103

Thus the mode of operation is non overlapping. (b) Maximum operating frequency 1 1 = 27.78 kHz fr(max) where fr(max) = 2tq 2 fo(max) = 13.89 kHz.

fo(max) =

\ (c)

Instant at which the current reaches the peak, tp = Now, initial capacitor voltage is given by

(

)

1 2 wr . Lr = 0.015 msec tan -1 wr R

¸˘ ÔÔ˙ ˙ ˝ 2 Ô˙ ˛Ô˚˙

64

Power Electronics

Ec1

- 3p È - Rp ˘ 6 ¥ 55 ¥ 103 È 30 10 ¥ Í1 e Í1 - e Lr .wr ˙ Í = Edc Í = 180 - Rp ˙ - 3p Í ÍÎ1 + L .w ˙˚ Í 3 6 e r r ÍÎ1 + e 30 ¥ 10 ¥ 55 ¥ 10

˘ ˙ ˙ ˙ ˙ ˙˚

= 178.81 V - R tp

\

Peak current,

Ip =

Ip =

Edc + Ec1 2 Lr . sin ( wr .t p ) e wr.Lr 180 + 178.81 55 ¥ 103 ¥ 30

-3 ¥ 0.15 ¥ 10-3 -6 2 ¥ 30 ¥ 10-6

¥ 10 e

sin (55 ¥ 103 ¥ 0.015 ¥ 10-3 )

Ip = 75.18 A. (d)

Average SCR current: Iav (SCR) =

( Edc + Ec1 ) fo È 2 Ê R ˆ2 ˘ Íwr + ÁË ˙L . 2 Lr ˜¯ ˚ r Î

- R tr È ˘ ÍÎ1 + e 4 Lr ˙˚

= 5.98 A.

(e)

È Ï – R tr Í Ô Ê ˆ Ê E + Ec1 ˆ ÍË - 2 Lr ¯ Ê f o ˆ ÔÊ Lr ˆ - Ê R 1 e Irms (SCR) = Á dc Ë 2 ¯ ÌË R ¯ Á L Ë wr . Lr ¯˜ Í r Ô Á Í Ô Ë Î Ó

¸˘ Ô˙ 1 È ˆ Ô˙ 2 ˜ ˝˙ Í Í ( 2 w )2 + ÊÁ R ˆ˜ ˜ Ô˙ Ë Lr ¯ ¯ ˛Ô˚ ÍÎ r

= 25.34 A (f)

Rms load current

Io = 2 ¥ 25.34 = 50.68 A

(g)

Output power,

po = Io2 . R = 2568.46 W

\

Average supply current (Is) =

2568.46 = 14.26 A 180

12.4 Given: Lr = 0.1 mH, Cr = 10 mf, R = 1W, tq = 12 usec. Resonant frequency,

fr =

=

1 Ê 1 R2 ˆ 2 p ÁË Lr . Cr 4 Lr 2 ˜¯

Maximum possible frequency

2

1 Ê 1 1 ˆ Á 3 6 2p Ë 0.1 ¥ 10 ¥ 10 4 ¥ 0.1 ¥ 10-12 ¥ 0.1˜¯

= 4.97 kHz

\

1

1

2

1

2

Solution Manual 65

Fmax =

1 1 = p p 2 (tq + wr ) ˆ 2 ÊÁ12 ¥ 10-6 + Ë 2p ¥ 4.97 ¥ 103 ˜¯

= 4.44 kHz 12.5 Given: Edc = 12 V, R = 8W, Fs = 15 kHz (i) Source inductance from Eq.(12.62) is given by 4 4 Ldc = 0.4 ¥ = 0.4 ¥ = 33.95 mH. ws 2p ¥ 15 ¥ 103 (ii)

Resonant capacitor from (12.63), C1 =

2.165 2.165 = = 22.97 mF R.Ws 1 ¥ 2p ¥ 15 ¥ 103 Ws . Lr , Assume Q = 7 R

Now,

Q =

\

Lr =

Now,

Ws – Lr –

\

QR ˆ 1 Ws ÊÁ = 0.353 R ˜ Ë ws ¯ ws . cr

8¥7

7 ¥8 -

1 = 0.353 R Ws . Cr

1 = 0.353 ¥ 8 2p ¥ 15 ¥ 103 ¥ cr

Cr = 0.2 mF.

\ 12.6 Given:

Edc = 24 V,

E0 = 15 V,

(i)

p0 = E0.I0

\

I0 = From Eq. (12.75), Zr £

\ (ii) is

= 594.18 mH.

2p ¥ 15 ¥ 103

po = 14 W,

fr = 1.2 mHz

14 = 0.933 A. 15

Edc 24 = 24.17 W = 0.933 Io

Also, from (12.70),

Zr =

and from (12.71),

fr =

Lr = 24.17. Cr 1 2p Lr . Cr

=

\ 1.2 ¥ 106 =

1 2p Lr . Cr

Lr = 3.44 mH, Cr = 0.0052 mF From (12.73), peak switch current (p) = 0.933 + 24

0.0052 ¥ 10-6 3.44 ¥ 10-6

= 1.863 A

66

Power Electronics

(iii) From (12.74), peak voltage rating of capacitor is given by Vc (peak) = 2 Edc = 2 ¥ 24 = 48 V 12.7 Given: Edc = 200 V, E0 = 30 V, I0 = 4A fs = 100 kHz, Deff = 0.55 (i) Calculation duty ratio from (12.113), effective duty ratios is given by 0.5 =

2. n. 30 \ n = 1.67. 200

from (12.114),

Deff = D – 0.25 D

\

0.5 = 0.75 D

(ii)

\ D = 0.67.

Selection of resonant inductor Lr: Here

DD = 0.25 D Lr =

from (12.112), effective inductance

Lreff

0.25 ¥ 0.67 ¥ 100

= 17.48 mH 4 ( 41.67 ) ¥ 100 ¥ 103 = Lr + transformer leakage inductance

= 17.48 mH + 3.46 mH

@ 22 mH (say) from (12.110), minimum load current is given by I0(min) = where

1.67 ¥ 200 1.5 ¥ 500 ¥ 10-12 2 22 ¥ 10-6

c = 500 pF. I0(min) = 0.975 A.

\ Inductor current

\

ILr = I0 =

Io 4 = 2.4 A = n 1.67

VLr = I Lr X Lr = 2.4 ¥ 2 ¥ p ¥ 100 ¥ 103 ¥ 22 ¥ 10-6 = 33.18 V

(iii) Transformer selection. E0 = Deff. Es \ Es =

30 = 60 V 0.5

Ep = n . Es = 1.67 ¥ 60 = 100.2 V

\

VA = 100.2 ¥ 2.4 = 240 V