EENG441 SOLVED PROBLEMS P1-) The transistor Q in the chopper shown below below has a maximum maximum (dv (dv/dt /dt ) rating of 50 V/s. It is used
to supply a resistive load of R L = 100 . Find the snubber capacitance C s required to protect the transistor for dv/dt /dt , when it is turned off. Assume that the transistor is turned off at t=0 , and vc(0)= 0.
+
vc
_
R s= 5
Cs
v
+
_
+
Q Vs = 500 V
R L
_
Solution:
The solution for the capacitor voltage after Q is turned off: vc (t ) Vs (1 (1 e t / )
ic
Cs
dv
V s
dt
dvc dt
V s R s
RL
R L C s ( Rs C s
RL )
2
e
e
t /
t /
( Rs RL )Cs
RL t / e Vs 1 Rs RL
v (t ) Vs RLic
RL 9.8765 dv V 6 V/μs 50 10 1 0 V/s 50 V/μ s dt 2 Cs ( Rs R L ) Cs max
0.1975 μF
Alternatively, the voltage across across Q can be solved directly from the eqn. x(t ) x( ) x(0) x( ) et /
v() v c () Vs
v (0 ) Rsic (0 ) Rs
V s R s
RL
( Rs RL )Cs
R s RL e t / v(t ) V s Vs Vs e t / Vs 1 R s RL Rs RL
P2-) In the circuit shown below, transistor Q is turned off at t = 0. The transistor current falls linearly after it is turned off, as shown below. The fall time is t f = 2 s. Diode Ds is ideal. (Note: the freewheeling diode DF does not turn on until the capacitor voltage reaches Vs).
(a) Find and sketch the voltage v across the transistor until steady state is reached. (b) Find the maximum dv/dt across the transistor. Ds
C s= 0.2 F V s= 200 V I0=10 A
Cs R
v
+
_
iQ
+
iQ
Q Vs
I0
DF
I0 _
0
t f
Solution:
(a)
After Q is turned off, the capacitor current is equal to I 0-iQ :
ic
I 0 iQ
v( t)
t t f
t I 0 t f I 0 Ct f
v( t)
t
I0
t'. dt' 2Ct 0
v( t f)
C
ic
1
t
2
dv dt
0
v( t)
t t f
v( t0) v( t f)
f
t
I dt C 0
I 0
2C
t f
t f+
I 0
( t t f ) 50 C
1 C I 0
2C
t
i dt' c
0
t f 50 V
50(t 2) V
v (V) 200 150 100 slope = I 0 /C 50
0
1
2
3
t f
(b) The maximum dv/dt across the transistor is
dv I 0 10 50 V/ s dt 6 max C 0.2 10
4
5
t ( μs)
; t in s
t
P3-) In the circuit shown below, transistor Q is turned off at t = 0. The transistor current falls linearly after it is turned off, as shown below. The fall time t f = 2 s. Diode Ds is ideal.
(a) Find and sketch the voltage across the transistor until steady state is reached. (b) Find the maximum dv/dt across the transistor. (c) Find R so that the initial current through Q is limited to 40A when it is turned on again. Ds
C s= 0.2 F V s= 200 V I0=10 A
Cs R
v
+
_
iQ
+
iQ
Q Vs
I0
DF
I0 _
t f
t
Solution:
(a) Same as in P1. (b) Same as in P1. (c) When Q turns on, its current is the sum of the l oad current and the snubber capacitor discharging current iC . Let Q be turned on at t = 0+. Then, Ds Cs R +
v
iC
_
iQ
+
Q Vs I0 _
iQ (0 ) I 0 iC (0 )
iQ (0 ) 10
200 R
iC (0 )
40
vC (0 )
R
200
R
R 200 30
6.67
P4-) In the circuit shown below, transistor Q is turned off at t = 0. The transistor turns off with its current falling to zero instantly. Diode Ds is ideal. (Note: the freewheeling diode DF does not turn on until the capacitor voltage reaches Vs).
(a) Find the capacitance Cs so that the maximum dv/dt across the transistor is 100 V/s. (b) With Cs = 0.25 F, find and sketch the voltage v across the transistor for 0 t 5 s . Indicate all time and voltage values. Ds Cs
Vs= 200 V R
v
+
I 0= 20 A
_
iQ
+
Q Vs
DF
I0 _
Solution:
(a)
After Q is turned off, the capacitor current becomes equal to I 0: ic
I 0 Cs
dv dt
dv
dt
I 0
100 V/ s
C s
Cs 0.2 F
(b) v (t )
I 0 C s
t
the capacitor charges linearly. t 1
It becomes equal to Vs at
C sV s I 0
0.25 200 20
2.5 s
v (V) 200 150 100
slope = I 0 /C
50
0
1
2
3
4
5
t ( μs)
P5-) The step-down chopper shown below is operated at the switching frequency f s = 10 kHz. (a) Find the duty ratio k so that the average load current I a = 2 A. (b) Find the range of k in which the load current io is continuous ( i.e. k k 1 , find k 1). S
io
+
+
vo
R D
Vs
V s = 100 V E = 40 V L= 5mH R=5Ω
L + E -
Solution:
(a)
I a
(b)
Va
E
R
Va
5 2 40 50 V
k
V a V s
0.5
Conduction is continuous if I 1 > 0
k
1 z
ln 1
E V s
( e z 1)
z
T
104 3
10
0.1
k 10 ln 1 0.4( e0.1 1) 0.412
P6-) The step-up dc-dc converter shown below is operated at a switching frequency of f s = 20 kHz.
(a) For R = 20 Ω find the duty ratio k so that the average power supplied to the load is Pav = 500 W. (b) For k=0.7 find the maximum value of the load resistance R so that the source current is is continuous. D
L
io
is Vs
Q
C
+
Vs = 40 V vo L = 500 μH
R
f s = 20 kHz
-
Solution:
(a) Pav Vdc .I dc
Vdc2
V dc
R
Vs
1 k
500
V s2 R(1 k ) 2
k 0.6
(b) When the switch Q is on L
di s dt
V s
I 2
0 t kTs
I1
V s L
kT s
is (t ) I1
where
I2
V s L
kTs
where I1 is (0)
is( kT s)
Power balance of the converter: Pin Pout
Vs I s ,av
V dc2 R
V V 1 1 I1 I 2 2I1 s kTs I1 s kT s 2 2 L 2L V Vs2 Vs V V s I1 s kTs I1 s kTs 2 2 2 L R(1 k ) 2L R(1 k )
I s,av
For continuous source current, I1 0
V s R(1 k )
2
Vs 2L
kT s
R
2 fs L k (1 k )
2
500 10 6 317.46 2 (0.3) 0.7
2 20 10
3
P7-) The step-up converter shown below is operated with a duty cycle k = 0.75. The minimum value of the source current is is I1 = 10 A . Assume that is decreases linearly when transistor Q is turned off. Find the switching frequency of the converter. D
L
io
is
+
Vs
Q
C
Vs = 20 V L = 5 mH R = 20 C F
vo
R -
Solution:
i s I2 I1
kT
t
T
Power balance: Pin
Pout
I s ,av
Vs I s ,av
Va2 R
I s ,av
V s R(1 k )
2
16 A
1
( I1 I 2 ) I 2 32 10 22 A 2
In 0 < t < kT
f s
1 T s
L
di s dt
kV s L( I 2 I1 )
V s is (t ) I1
0.75 20 5 103 12
V s L
250 Hz
kTs
where I1 is (0)
I 2
I1
V s L
kT s
P8-) The buck regulator shown below is operated at the switching frequency f s = 10 kHz.
(a) Find the average transistor current I Q,av as a function of the duty ratio k, assuming that the inductor current iL is continuous. (b) Find the maximum duty ratio k max if Q has a maximum average current rating of 4 A. (c) Find the maximum duty ratio k c for which iL is continuous. L
Q +
io iL
iQ
Vs
D
+
C
Vs = 50 V C = 100 μF L = 200 μH R=5Ω f s = 10 kHz
vo
R
-
-
Solution:
(a) Pin Pout Vs .IQ,av
(b) IQ,av 4 A
Vo2
Vo
R
k2
V s R
kVs IQ ,av k 2
4A
V s R
k 0.6325
k max
0.6325
(c) When the switch Q is on, L Vo
IQ ,av
di L dt
V s Vo
kVs
1 2
( I1 I 2 ) k
0 t kTs I2
I1
k2
V s R
V s (1 k ) L
iL (t ) I1 kTs (1)
I1 I 2
For continuous conduction I1 0
k
V s
V o L
where
2Vs R
1k
(1) 2 f s L R
kTs I2
where I1 iL (0)
iL( kTs) I1
Vs R
k (1 k )
k 1
Vs
2 f s L
2 104 200 10 6 5
0.2 k c
P9-) The step-down chopper shown below is operated at the switching frequency f s = 10 kHz.
(a) Find the minimum value of the duty ratio k so that the load current io is continuous. (b) Find the duty ratio k so that the average load current is 20 A. (c) Find the average value ( I Q,av) of the transistor current iQ for k = 0.5.
iQ
+
io
Q
+
vo
R D
Vs
L + E -
V s = 200 V E = 50 V L = 10 m H R=5Ω f s = 10 kHz
Solution:
(a) The minimum value of the load current is I1
e- (1-k ) a e- a E 1- e-a R
V s R
a
;
T
L R
2 ms
T
1
f s
0.1 ms
a
0.1 2
0.05
For continuous conduction I1 0
k
1 a
E
V s
ln 1
1
0.05
( ea 1)
ln 1
50 200
( e0.05 1) 0.2547
(b) Average load current I a
E
Va
R
kVs
E 20 5 50 20 k 0.75
R
200
(c) iQ = io when Q is on. In 0 t kT , I1
I Q,av
io (t ) io () io (0) io () et /
200 e0.50.05 e0.05
5
1
1 e
kT
0.05
50 5
9.75 A
i (t ).dt 30k 20.25 1 e T T o
0
kT /
E Vs E t / e I1 R R
V s
io (t ) 30 (9.75 30). e t /
30 20.25 et /
15 20.25 20 1 e 0.50.05 5.0 A
A
P10-) Below is shown a step-down dc-dc converter with an LC filter at its output. Diode D is ideal. Switch S is operated with duty ratio k = 0.4 at the switching frequency f s = 10 kHz. Assume that the output voltage vo = Va is constant and also the output current io = I a is constant. Given that Va = k Vs , Vs = 100 V, L = 250 H and C = 100 F
(a) Find the minimum and maximum values of the inductor current iL for Ia = 10A. Show all your calculations. (b) Sketch the source current is , and find its average value.
S
f s
L
io Switching function of S:
iL
is D
Vs
+
C
vo 1
Load -
0
kT
t
T
Solution:
(a) v D
0 t kT :
Vs
V s Va L
0 t (1 k )T : i L (t ) I 2
0
i L
i L (t ) I1
kT
Va
t
T
I1
t
t'
(3)&(4)
I1 10
I1 I a 100 4
5 10
L
t
1
2
Va
kVs
V s
k (1 k )T
2 L
(0.6)(0.4)104
I1 I 2
I2
5.2 A
V s L
Ia I 2
k (1 k )T
V s 2L
k (1 k )T
14.8 A
(b) i L
I s ,av
I2 I1
kT
T
t
1 2
Vs
Va L
iL (1 k )T I1 I 2
( I1 I 2 ) I a I1 I 2 2I a
I L,av
I2
iL (kT ) I 2 I1
t
k ( I1 I 2 ) kI a
4A
(3) (4)
kT V a L
(1)
(1 k )T (2)
