CHAPTER 8 Additional Problems Solved Problems 8.1 A current commutated chopper is fed from a d.c. source of 236 V. Its commutating components are L = 20 mH and C = 50 mF. If the load current of 200 A is assumed constant during the commutation process, compute the following: (i) turn-off time of main thyristor. (ii) total commutation interval. (iii) turn-off time of auxiliary thyristor. Sol. (i) Peak commutating current is given by ICP = Edc c / L \ 230 \ \ x= I cp Io = 50 = 363.66 A 20 363.66 = 1.82. 200 Turn-off time, tq = [p – 2sin–1(1/1.82)] 20 ¥ 50 ¥ 10-12 = 62.52 usec (ii) ÊI ˆ Ê 200 ˆ q1 = sin–1 Á o ˜ = sin–1 Á Ë 363.66 ˜¯ I Ë cp ¯ = 33.365°. Total commutation interval is given by Ê 5p ˆ ÁË 2 - q1 ˜¯ \ Lc + 2 C ES ◊ sin 2 q1/2 . Io Ê 5p 33.365 ¥ p ˆ Ê 1 - cos 33.365 ˆ 1000 ¥ 10-6 + 50 ¥ 10-6 ¥ 230 Á ÁË 2 ˜ ˜¯ ¯ Ë 180 200 = 229.95 ¥ 10–6 + 9.48 ¥ 10–6 = 239.43 usec. (iii) Turn-off time of auxiliary thyristor is given by (p – q1) Lc Solution Manual 2 Ê p - 33.365 ˆ –6 ÁË ˜¯ 1000 ¥ 10 = 80.93 usec. 180 8.2 A type a chopper operating at 2 kHz from a 100 V d.c. source as a load time constant of 6 ms and load resistance of 10 W. Find the mean load current and the magnitude of current ripple for a mean load voltage of 50 V, Also, compute the minimum and maximum values of load current. Sol. Load time constant, L = 6 ¥ 10–3 s. R R = 10 W. \ L = 6 ¥ 10–3 ¥ 10 = 60 mH. Chopping period, T = 1/f = 1 ¥ 1000 = 0.5 ms. 2000 Average or mean load voltage, E0 = a Es. a= \ 50 = 0.5 1000 Ton = 0.5 ¥ 05 = 0.25 ms. Toff = 0.25 ms. \ As chopping period T = 0.5 ms is much less than the load time constant = 6 ms, the current variation from minimum current Iomin to Iomax (maximum current), must be taken as linear. \ During Ton period, Es – Eo = L \ \ (Iomax – Iomin) = Iomax - Iomin Ton ( Es - Eo ) Ton L = ˘ Eo È Es - 1˙ Ton. Í L Î Eo ˚ = Eo È 1 ˘ - 1˙ Ton Í L Îa ˚ = Eo L Iomax – Iomin = ÈT ˘ - 1˙ Ton Í Î Ton ˚ Eo Toff = DI, current ripple. L \ Magnitude of ripple current, DI = Iomax – Iomin = = Mean load current, Io = Eo . Toff. L 50 ¥ 0.25 ¥ 10–3 = 0.21 A. 60 ¥ 10-3 Iomax + Iomin 2 3 Power Electronics = Eo 50 = = 5 A. R 10 Maximum value of load current Iomax is given by E DI = Io + o . Toff 2 2L Iomax = Io + 50 ¥ 0.25 ¥ 10-3 2 ¥ 60 ¥ 10-3 = 5+ = 5.104 A. Minimum value of load current DI = 5 – 0.104 = 4.896 A. 2 Iomin = Io – 8.3 For a type-A chopper feeding an RLE load, show that the maximum value of rms current rating 3/ 2 ÈE Ê Eˆ ˘ s Í for free wheeling diode is given by 0.39 1 - ˜ ˙ . Consider load current to be ripple Es ¯ ˙ Í R ÁË Î ˚ free. Sol. Average load current for chopper with RLE load is given by Io = Eo - E a ◊ Es - E = R R During period Toff, current flows through freewheeling diode. Therefore, rms value of freewheeling diode current, when Io is ripple free is given by 1/ 2 Toff È T - Ton ˘ ◊ Io Í ˙ t Î t ˚ Iof = È a ◊ Es - E ˘ .Í ˙ R Î ˚ = 1 (a ◊ Es . - E ) R = 1 È ◊ a 1 - a Es - 1 - a ◊ E ˘˚ R Î = 1 È 2 ◊ a - a 3 ◊ Es - 1 - a ◊ E ˘˙ ˚ R ÍÎ ( 1-a ) (i) This current will have maximum value when ( da (2a - 3a ) E = 2 or 2 a -a (2a - 3a ) 2 or 3 1-a a 1-a ) 2 È 1 Í 1 2a - 3a Es 1 E = ◊ + 2 3 R Í2 2 1-a a a Î d I Df s = – E 1 - a. -E Es . ˘ ˙ = 0. ˙ ˚ Solution Manual 4 or 3a – 2 = Ê Eˆ E or a = 1/3 Á 2 + ˜ Es ¯ Es Ë Substituting this value of a in Eq. (i), we get maximum value of rms current rating IDfm of freewheeling diode as, 1 If Dm = R = 1 R 1/ 2 È1 Ê ˘È 1Ê Eˆ E ˆ˘ Í Á 2 + ˜ ◊ Es - E ˙ Í1 - Á 2 + ˜ ˙ Es ¯ Es ¯ ˙˚ ÍÎ 3 Ë ˙˚ ÍÎ 3 Ë 1/ 2 È 1 Ê 2 Es + E ˆ ˘ È 2E + E ˘ ◊ Es - E ˙ Í1 - s Í Á ˙ ˜ 3 Es ˚ ˙˚ Î ÎÍ 3 Ë Es ¯ 1/ 2 1 È 2 Es + E - 3 E ˘ È 3 Es - 2 Es - E ˘ = ˙ ˙Í R ÍÎ 3 3 Es ˚Î ˚ 1/ 2 1 È 2 Es - 2 E ˘ È Es - E ˘ = ˙ Í 3E ˙ R ÍÎ 3 ˚Î s ˚ = 1 2 Es . Es ◊ R 3 3 Es . = Es 3 3 R 2 ◊ È E˘ Í1 - ˙ Î Es ˚ È E˘ Í1 - ˙ E s˚ Î E È E˘ = 0.39 s ◊ Í1 - ˙ R Î Es ˚ 8.4 = 1 3 3/ 2 ◊ ÈÎ Es - E ˘˚ R 2 1 3 Es 3/ 2 3/ 2 3/ 2 A current commutated chopper is operating on a 150 V d.c. supply. The maximum load current is 50 A. The turn-off time of main SCR is 15 msec. If the value of the commutating L is 10 mH and C = 4.4 mF., find out if it is possible commutate the main SCR. If not, calculate the values of L and C to commutate the main SCR successfully. Sol. For current commutated chopper, L= ( 3Edc Loff + Dt ) 4p tom \ Circuit turn-off time tq = (toff + Dt) = L ¥ 4p I om 3 Edc = 10 ¥ 10-6 ¥ 4p ¥ 50 = 13.96 msec. 3 ¥ 150 \ Circuit turn-off time tq < toff. \ This commutation circuit cannot commutate the main SCR successfully. In order to have successful commutation, tq = 2 toff = 2 ¥ 15 = 30 ms. 5 Power Electronics \ L= and, C= 3 ¥ 150 ¥ 30 ¥ 10-6 = 21.48 mH. 4 ¥ p ¥ 50 3 Iom ¥ tq p ◊ Edc = 3 ¥ 50 ¥ 30 ¥ 10-6 p ¥150 C = 9.54 mF For successful commutating, L = 21.48 mH and C = 9.54 mF. 8.5 A voltage commutated chopper is operating with an input d.c. voltage of 100 V. Commutating components are L = 15 mH and C = 4.7 mF. Determine:(i) Minimum duty cycle of the chopper. (ii) Peak capacitor current (iii) Peak current through the device if the load current is 28 A and operating frequency is 400 Hz Sol. (i) Minimum duty cycle, Dmin = = p p Lc T ( Lc ◊ ¥ f = p ¥ 15 ¥ 10-6 ¥ 4.7 ¥ 10-6 1/ 2 ) ¥ 400 = 1%. (ii) Peak capacitor current Icpeak = Edc c/L 4.7 /15 = 55.97 A. = 100 (iii) Peak thyristor current = Iom + Icpeak = 28 + 55.97 = 83.97 A 8.6 A parallel capacitor type A chopper operates at 600 Hz from a 72 V battery and feeds to a load of 10 A. If the turn-off time of main SCR is 45 ms and that of auxiliary SCR is 60 ms. Determine the values of commutating components. Assume a 100% safety margin for turn-off times of main and auxiliary SCRs. Sol. (i) Capacitor C= = (ii) ( I om ◊ toff + Dt ) Edc 10 ¥ 90 ¥ 10-6 = 12.5 mF. 72 L£ C Edc I om £ 12.5 ¥ 10-6 72 10 £ 0.65 mH. (iii) L ≥ 0.01 T 2 Edc T = 1/600 = 1.66 ms. p 2 ◊ C. Solution Manual 6 2 ( ) 2 -6 0.01 ¥ 1.66 ¥ 10-3 \ L ≥ \ L ≥ 16.21 mH. p ¥ 12.5 ¥ 10 ¥ 72 (iv) The time for which SCR2 remains reverse biased is approximately half of the time (t1 – t0). But (t1 – t0) = p Lmin ◊ C. \ (t1 – t0) = p 0.65 ¥ 10-3 ¥ 12.5 ¥ 10-6 = 282.74 ms. \ Time for which SCR2 is reverse-biased = 282.74 2 = 141.37 ms. This is larger than 2 ¥ 60 = 120 ms. \ The value of commutating components calculated in (i), (ii), and (iii) can ensure successful commutation of the SCRs.