Hart Chapter 10 solutions

Prévia do material em texto

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 turn-off losses, restrict the data to 2.5 µs to 4.3 µs. The energy absorbed is 132 µJ. For turn-on 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
	
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
	
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,
	
The steady-state base current in the on state determines R2. From Eq. 10-2,
	
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,
	
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,
	
The steady-state base current in the on state determines R2. From Eq. 10-2,
	
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,
	
10-5) 
a) From Eq. 10-5 through 10-7 for t < tf,
	
For tf < t < tx,
	
Time tx is defined as when the capacitor voltage reaches Vs (50 V.):
	
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,
	
	
Snubber loss is determined by the amount of stored energy in the capacitor that will be transferred to the snubber resistor:
	
10-6) 
Switch current is expressed as
	
Capacitor voltage at t = tf = 0.5 µs would be 100 volts, which is greater than Vs. Therefore, the above equations are valid only until vC reaches Vs:
	
For tx < t < tf,
	
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
	
Snubber loss is determined by the amount of stored energy in the capacitor that will be transferred to the snubber resistor:
	
10-7)
	
10-8)
	
10-9)
	
10-10)
	
10-11) 
Using the snubber circuit of Fig. 10-12(a), Eq. 10-12 yields
	
10-12) 
Using the snubber circuit of Fig. 10-12(a), Eq. 10-12 yields
	
10-13)
	
10-14)
10-15)
	
10-16)
	
10-17)
	
10-18)
	
10-19)
	
_1332947809.unknown
_1332947817.unknown
_1332947821.unknown
_1332947825.unknown
_1332947827.unknown
_1332947829.unknown
_1332947831.unknown
_1332947832.unknown
_1332947830.unknown
_1332947828.unknown
_1332947826.unknown
_1332947823.unknown
_1332947824.unknown
_1332947822.unknown
_1332947819.unknown
_1332947820.unknown
_1332947818.unknown
_1332947813.unknown
_1332947815.unknown
_1332947816.unknown
_1332947814.unknown
_1332947811.unknown
_1332947812.unknown
_1332947810.unknown
_1332947805.unknown
_1332947807.unknown
_1332947808.unknown
_1332947806.unknown
_1332947803.unknown
_1332947804.unknown
_1332947802.unknown