# HartChapter10solutions.doc

April 26, 2018 | Author: Carolina de Melo | Category: Mosfet, Bipolar Junction Transistor, Capacitor, Manufactured Goods, Electrical Engineering

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CHAPTER 10 SOLUTIONS 3/20/10

10-1) a) For the elementary MOSFET drive circuit, losses can be determined from the energy absorbed by the transistor. In Probe, the integral of instantaneous power is obtained by entering the expression S(W(M1)) to get the energy absorbed by the transistor. For turnoff losses, restrict the data to 2.5 µs to 4.3 µs. The energy absorbed is 132 µJ. For turnon losses, restrict the data to 5 µs to 5.6 µs. The energy absorbed by the MOSFET is 53.3 µJ. Power is determined as T

1 1   5 s f s 20000 W 132  J   26.4 W . T 5 s W 53.3  J    10.7 W . T 5 s

Pturn off  Pturn on

For the emitter-follower drive circuit restrict the data to 2.5 µs to 2.9 µs, giving 21.3 µJ for turn-off. Restrict the data to 5 µs to 5.3 µs, giving 12.8 µJ for turn-on. Power is then W 21.3  J   4.26 W . T 5 s W 12.8  J    2.56 W . T 5 s

Pturn off  Pturn on

b) For the first circuit, peak gate current is 127 mA, average gate current is zero, and rms gate current is 48.5 mA. For the second circuit, peak gate current is 402 mA (and -837 mA), average gate current is zero, and rms gate current is 109 mA. 10-2) For R1 = 75 Ω, toff ≈ 1.2 μs, ton ≈ 0.6 μs, and PMOS ≈ 30 W. For R1 = 50 Ω, toff ≈ 0.88 μs, ton ≈ 0.42 μs, and PMOS ≈ 22 W. For R1 = 25 Ω, toff ≈ 0.54 μs, ton ≈ 0.24 μs, and PMOS ≈ 14 W. Reducing drive circuit resistance significantly reduces the switching time and power loss for the MOSFET.

10-3) The values of Vi, R1, R2, and C must be selected for the BJT base drive circuit. First, select Vi: let Vi=20 V. then, the value of R1 is determined from the initial current spike requirement. Solving for R1 in Eq. 10-1,

R1 

Vi  vBE 20  1   3.8  I B1 5

The steady-state base current in the on state determines R2. From Eq. 10-2,

R2 

Vi  vBE 20  1  R1   3.8  34.2  I B2 0.5

The value of C is determined from the required time constant. For a 50% duty ratio at 100 kHz, the transistor is on for 5 µs. Letting the on time for the transistor be five time constants, τ = 1µs. From Eq. 10-3, 

R1 R2   C  (3.42)C  1  s  R1  R2 C  0.292  F

  RE C  

10-4) The values of Vi, R1, R2, and C must be selected for the BJT base drive circuit. First, select Vi; let Vi = 20 V. then, the value of R1 is determined from the initial current spike requirement. Solving for R1 in Eq. 10-1,

R1 

Vi  vBE 20  1   6.33  I B1 3

The steady-state base current in the on state determines R2. From Eq. 10-2,

R2 

Vi  vBE 20  1  R1   6.33  25.3  I B2 0.6

The value of C is determined from the required time constant. For a 50% duty ratio at 120 kHz, the transistor is on for 4.17 µs. Letting the on time for the transistor be five time constants, τ = 1 µs. From Eq. 10-3, 

R1 R2   C  5.06 C  0.833  s  R1  R2 C  0.165  F

  RE C  

10-5) a) From Eq. 10-5 through 10-7 for t < tf,   t t  iQ  I L  1    4  1   4  8(10) 6 t 6   t f 0.5(10)    I t 4t iC  L   8(10)6 t tf 0.5(10) 6 I Lt 2 4t 2 vC (t )    8(10)13 t 2 6 6 2Ct f 2(0.05)(10) (0.5)(10) For tf < t < tx, iQ  0 iC  I L  4 vC 

I t IL (t  t f )  L f  8(10)7 (t  0.5(10) 6 )  20 C 2C

Time tx is defined as when the capacitor voltage reaches Vs (50 V.):

vC (t x )  Vs  50  8(10)7 (t x  0.5(10)6 )  20  t x  0.875 s b) With tx > tf, the waveforms are like those in Fig. 10.12(d). c) Turn-off loss is the switch is determined from Eq. 10-12,

PQ 

I L2t 2f f 24C

42 [0.5(10)6 ]2 (120000)  0.4 W . 24(0.05)(10) 6

Snubber loss is determined by the amount of stored energy in the capacitor that will be transferred to the snubber resistor:

1 0.05(10) 6 (50) 2 (120000) PR  CVs2 f   7.5 W . 2 2 10-6) Switch current is expressed as

  t t  iQ  I L  1    4  1   4  8(10)6 t 6   t f 0.5(10)    I t 4t iC  L   8(10)6 t 6 tf 0.5(10) vC (t ) 

I Lt 2 4t 2   4(10)14 t 2 6 6 2Ct f 2(0.01)(10) (0.5)(10)

Capacitor voltage at t = tf = 0.5 µs would be 100 volts, which is greater than V s. Therefore, the above equations are valid only until vC reaches Vs:

vC (t x )  Vs  50  4(10)14 t x2  t x  0.354  s For tx < t < tf, iQ  4  8(10)6 t iC  0 vC  Vs  50

b) With tx < tf, the waveforms are like those of Fig. 10.12(b). Equation 10-12 is not valid here because tx < tf. Switch power is determined from T

T

T

1 1 PQ   p (t )dt   iQ vQ dt  f  iQ vC dt  T 0 T 0 0 

 6       4  8 10 t 4(10) t dt  4  8 10 t (50) dt   1.84 W .            0  tx

120000 

tf

tx

6

14 2

Snubber loss is determined by the amount of stored energy in the capacitor that will be transferred to the snubber resistor:

1 0.01(10 6 (50) 2 (120000) PR  CVs2 f   1.5 W . 2 2 10-7) C

I Lt f 2Vs

10(0.1)(10) 6  3.33 nF . 2(150)

ton D / f 0.4 /100000    240  5C 5C (5)3.33(10) 9 1 PR  (3.33(10) 9 (150) 2100000  3.75 W . 2 I 2t 2 f 102 [0.1(10) 6 ]2105 PQ  L f   1.25 W . 24C 24(3.33)(10) 9 R

10-8) C

I Lt f 2V f

10(0.1)(10) 6  6.67 nF . 2(75)

ton D / f 0.4 /100000    120  5C 5C 5(6.67)(10) 9 1 1 PR  CVs2 f  (6.67)(10)9 (150) 2100000  7.5 W . 2 2 2 I t f 102 [.1(10) 6 ]2105 PQ  L f   0.625 W . 24C 24(6.67)(10) 7 R

10-9) C

I Lt f 2Vs

7(0.5)(10) 6  10.3 nF . 2(170)

ton D / f 0.4 /125000    62.2  5C 5C 5(10.3)(10) 9 1 1 PR  CVs2 f  (10.3)(10) 9 (170) 2125000  18.6 W . 2 2 2 2 I t f 102 [0.5(10) 6 ]2125000 PQ  L f   6.2 W . 24C 24(10.3)(10) 9 R

10-10) C

I Lt f 2V f

7(0.5)(10)6  14.0 nF . 2(125)

ton D / f 0.4 /125000    45.7  5C 5C 5(14)(10) 9 1 1 PR  CVs2 f  (14)(10)9 (170) 2125000  25.3 W . 2 2 2 2 I t f 102 [0.5(10) 6 ]2125000 PQ  L f   4.56 W . 24C 24(14)(10) 9 R

10-11) Using the snubber circuit of Fig. 10-12(a), Eq. 10-12 yields I L2t 2f f

52 [0.5(10) 6 ]2 200000 C   52.1 nF . 24 PQ 24(1) ton D / f 0.35 / 200000    6.72  5C 5C 5(52.1)(10) 9 1 1 PR  CVs2 f  (52.1)(10) 9 (80) 2 200000  33.3 W . 2 2 R

10-12) Using the snubber circuit of Fig. 10-12(a), Eq. 10-12 yields C

I L2t 2f f 24 PQ

62 [1(10) 6 ]2100000  75 nF . 24(2)

ton D / f 0.3 /100000    8.0  5C 5C 5(75)(10) 9 1 1 PR  CVs2 f  (75)(10) 9 (120) 2100000  54 W . 2 2 R

10-13) a ) From Eq. (10-16), TJ  TA  P  R , JA   30  2  40   110 C b) P 

TJ  TA 150  30  3W R , JA 40

10-14) a ) From Eq. (10-16), TJ  TA  P  R , JA   25  1.5  55   107.5 C b) P 

TJ  TA 175  25   2.73 W R , JA 55

10-15) TJ  P  R , JC  R ,CS  R , SA   TA  10  1.1  0.9  2.5   40  85 C

10-16) TJ  P  R , JC  R ,CS  R , SA   TA  5  1.5  1.2  3.0   25  53.5 C

10-17) TJ  P  R , JC  R ,CS  R , SA   TA R , SA 

TJ  TA 110  40  R , JC  R ,CS   0.7  1.0  2.19 C / W P 18

10-18) From Fig. 10.24 using the bottom curve for a single pulse, Z , JC  0.013 C / W for a pulse of 10 5 sec. TJ  Pdm Z , JC   500 W   0.013 C / W   6.5 C

10-19) a) For 50 kHz and D = 0.1, the pulse width is 2s. From Fig. 10.24, Z , JC  0.11 C / W . TJ  Pdm Z , JC   100   0.11  11 C b) Using R  , JC  1.05 C / W , TJ  Pavg R , JC   Pdm D  R , JC  100  0.1 1.05  10.5 C. Note that the value of Z from the graph is very rough, and more precise evaluation in (a) is closer to the 10.5 of part (b).