Type:
Solution Manual
Resource:
Power Electronic Circuits
Edition:
1st Edition
Author(s):
Issa Batarseh
Exercise-Chapter 2
Chapter 2: Review of Switching Concepts and Power Semiconductor Devices Exercise Solutions Exercise 2.1 Repeat Example 2.2 for the switching waveforms shown in Fig. E2.1. υsw
V off I on
ι sw
I off V on
t t=0
T s− (t d +t rise) T s− t d
t d t d +t fall
Ts
Ts
Fig. E2.1 Waveforms for Exercise 2.1
Solution: a)
Voff Ion t=0 P(t)
t=Ts
td tfall
t
trise t d
t
Power Electronic Circuits
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Exercise-Chapter 2
b) 0 V I off on t td t − td Voff I on 1 − t fall P (t ) = 0 Voff I on (t − T + t + t ) rise d trise t − T + td Voff I on 1 − td 0
when
t<0
when
0 ≤ t < td
when
td ≤ t < td + t fall
when
td + t fall ≤ t ≤ T − td − trise
when
T − td − trise ≤ t < T − td
when
T − td ≤ t < T
when
t ≥T
c) 1 Pave = Ts
=
=
t fall t rise td t d Voff I on ′ ′ ′ ′ 1 Voff I on t t dt ′ + v(t )i (t )dt = t dt + Voff I on 1 − t ′′dt ′′ + Voff I on 1 − dt ′′′ t fall Ts td trise td 0 0 0 0 0
T
∫
∫
∫
∫
∫
2 t 2fall t 2 Voff I on td2 rise + t − t d t + + − d fall Ts 2td 2t fall 2trise 2td
Voff I on 2Ts
(2td + t fall + trise )
d) Pmax =
Voff I on 2Ts
Power Electronic Circuits
( 2t d + t fall + t rise )
= max
Voff I on 2Ts
2Ts = Voff I on
2
Exercise-Chapter 2
Exercise 2.2 Find the efficiency of the circuit in Fig. 2.1(d) assuming the switching characteristics for S are shown in Fig. 2.4(a) with tON=100ns, tOFF=150ns, Ts=10µs, IOFF=0, VON =0 and D=0.5.
IL +
S RL
Vin
vo _
Fig. 2.1(d) Solution: From example 2.2, we have: V Vin * in 2 Voff I on ton + toff R ton + toff = Vin t + t = Psw, ave = RT on off Ts Ts 6 6 s
(
)
Ts −t off t on Ts T − t V 2 ton + toff t 1 s Pin, ave = dt + I onVin dt + Vin I on dt = in 1 − Vin I on toff RTs Ts ton T s ton T t − 0 s off
∫
η= 1+
∫
∫
1 = 80% ton + toff Ts
Power Electronic Circuits
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Exercise-Chapter 2
Exercise 2.3 Repeat Example 2.3 for (a) vs=12sinωt, and (b) vs is a square wave with peak-topeak voltage equal to ±12V. Solution: a-a)
VS (t ) VS = 12V
π
π
2
3π 2
ωt
2π
Vo (t )
ωt − 12V
iS (t )
ωt −
VS R
a-b) 2π
2π
π 1 1 6 12 Vo ,ave = V t d t td t t ( ) = 12 sin = cos = − = −3.82V ω ω ω ω ∫ ∫ 2π 0 2π π π π 2π
2π
Vo ,rms =
1 V 2 (t )dωt = ∫ 2π 0
Power Electronic Circuits
2π
2π
1 12(12) (12 sin ωt ) 2 dωt = (1 − cos 2ωt )dωt = 6V ∫ 2π π 4π π∫
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