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.
