Download EENG441 FE Spr 12-13 soln

Eastern Mediterranean University
Department of Electrical and Electronic Engineering
EENG 441 INDUSTRIAL AND POWER ELECTRONICS
FINAL EXAM
Date : 27 May 2013
Duration: 120 min.
STUDENT’s
QUESTION
POINTS
NUMBER
1
2
3
4
25
25
30
25
105
NAME
SURNAME
TOTAL
Rules:
1. Numerical results which do not have physical units (A, V, W, etc.) will be considered wrong.
2. You may lose points for untidy and ambiguous presentation.
3. Do exactly what the questions ask. Misunderstanding of questions will not be accepted.
Q1. In the circuit shown below, transistor Q is turned off at t = 0, its current iQ falling to zero instantaneously.
Snubber diode Ds is ideal.
Ds
(a) Find and sketch the voltage v across the transistor
until the freewheeling diode DF turns on at t  t1 .
Take v(0) = 0 V. Also, determine the value of t1 .
R
Ls
(15 pts)
+
(b) After DF turns on, the capacitor voltage is given by
I
1
vc (t )  Vs  0 sin(0t )
t   t  t1 , 0 
0Cs
Ls Cs
until Ds turns off (this happens when i reverses
direction). Find the maximum voltage across the
transistor and the time t2 at which it occurs. (10 pts)
Ls= 5 
C s= 0.2 F
Vs= 200 V
I0=10 A
Cs
v
+
i
_
iQ
Q
Vs
DF
I0
_
Q2. The three-phase half-bridge inverter shown below feeds a balanced Y-connected purely inductive load
having inductance L per phase, and is operated in the square-wave mode (pole voltages are square waves) at the
frequency f s.
(a) Sketch phase-to-neutral voltage van (show voltage and time values) and find its rms value Van,rms. (10pts)
(b) Find the amplitude of the fundamental component of the line current ia in terms of L and fs. (15 pts)
+
a
Vs
b
c
+
ia
van
n
Q3. In the single-phase half-controlled rectifier shown in Figure 3(a), the source voltage is 240 V rms at
frequency fs = 50 Hz. The load: R = 10Ω, L is very large. Assume that the load current i0=Ia is constant.
(a) Sketch the load voltage waveform (v0) (take the firing angle  = 60) and show that its average is given by
Vdc 
Vmax

(1  cos  )
where Vmax is the peak value of the source voltage. Show all the steps in your derivation. (15 pts)
(b) An input filter is inserted between the rectifier and the AC source with the following transfer function
G( j f ) 
I sf ( j f )
Is ( j f )

1
1   f / f0 
2
where f 0  2 f s , as shown in Figure 3(b). The firing angle is  = 60. The rectifier can be represented by a
current source is with current equal to the source current in (a). Considering harmonics of orders up to 5
(i.e. take f  f s , 3 f s , 5 f s ) calculate the total harmonic distortion (THD) of the filtered source current isf .
(15 pts)
io
+
isf
T3
T1
is
R
+
+
vo
vs
-
vs
L
D2
D4
G(jω)
is
(b)
(a)
Q4. The step-up dc-dc converter shown below is operated at the switching frequency fs = 20 kHz.
(a) For R = 30 Ω find the duty ratio k so that the average power supplied to the load is Pav = 160 W. (10pts)
(b) For k = 0.5 find the maximum value of the transistor current iQ (15pts)
D
L
is
Vs
io
+
iQ
Q
C
R
vo
-
Vs = 40 V
L = 400 μH
fs = 20 kHz
SOLUTION
Q1. (a) When Q turns off, the capacitor starts charging by the load current. Ds turns on and bypasses R.
Therefore
C
I0
t
0  t  t1
C
C V 0.2  200
 t1  s s 
 4 s
I0
10
dv(t )
 I0
dt
 v (t ) 
v(t1 )  Vs
v (V)
250
200
150
100
50
0
(b)
Maximum voltage occurs at
2
t2
t1=4
sin(0t )  1
t (μs)
6
 t 

1
  LsCs  1.57 μs
20 2
 t2  t1  1.57  5.57 μs
Vmax  Vs 
I0
L
5
 Vs  I 0 s  200  10
 250 V
0Cs
Cs
0.2
Q2. (a)
van
2Vs / 3
Vs / 3
1
van  (2va 0  vb0  vc 0 )
3
π/3
2π/3
π
2π
ωt
-Vs / 3
-2Vs / 3
(b) Amplitude of the fundamental component of ia
I a1 
Van ,1
s L
where Van ,1 is the fundamental component of van which can be found using the fund.
1
components of the pole voltages: van ,1  (2va 0,1  vb0,1  vc 0,1 )
3
2Vs
2 
2 


sin  s t 
sin  s t 
 vc 0,1 



3 

3 


2V 
2V
2 
2   2Vs


van ,1  s  2sin(s t )  sin  s t 
sin(s t )  Van ,1  s
  sin  s t 
 
3 
3 
3   



va 0,1 
2Vs
 I a ,1 
sin(s t ) vb0,1 
2Vs
s L

Vs
 2 fs L
2Vs
Q3. (a)
400
300
Vdc 
200

100
0

1
V
 
max
Vmax
-100


sin(t ) d (t ) 
Vmax

  cos( )  (  cos( )) 
  cos(t )
Vmax

1  cos  
-200
-300
-400
0
0.005
0.01
0.015
0.02
is
Harmonic amplitudes of is:
(b)
Ia
β
π+α
π
α
ωt
-Ia
I sn 
4Ia
1

sin  n 
n
2

 I s1 
I sf ,1 
4Ia

where     
sin( / 3) 
2 3

Ia
I s3 
4Ia
sin( )  0
3
1
42 3
8
. I s1 
Ia 
Ia
1  (1/ 2)2
3 
3
 THD 
I sf ,5
I sf ,1

is the pulse-width of the current
I s5 
4Ia
2 3
sin(5 / 3)  
I
5
 a
1
4 2 3
8 3
. I s5 
Ia 
Ia
1  (5 / 2)2
21 5
105
I sf ,3  0
I sf ,5 
402
30(1  k )2
 k  0.4226
3
 2.86%
105
Q4. (a)
Pav  Vdc . I dc 
Vdc2
R
Vdc 
Vs
1 k
 160 
(b) First, it should be determined whether conduction is continouos or not.
When transistor Q is on
L
dis
 Vs
dt
0  t  kTs
 I 2  I1 
Vs
kTs
L
 is (t )  I1 
Vs
kTs
L
where I1  is (0)
where I 2  is (kTs )
Power balance of the converter: Pin  Pout  Vs I s ,av 
Vdc2
R
V
V
1
1
 I1  I 2    2 I1  s kTs   I1  s kTs
2
2
L
2
L

2
V
Vs
Vs
V


 Vs  I1  s kTs  
 I1 
 s kTs
2
2
2L
R(1  k ) 2 L

 R(1  k )
I s ,av 
40
40  0.5  50

 4.08 A
30  0.25
2  400
V
 I 2  I1  s kTs  6.58 A
L
 I1 
 continuous conduction