Download EENG 341 M1 spr 08 09 - faraday

Eastern Mediterranean University
Department of Electrical and Electronic Engineering
EENG 341 ELECTRONICS I MIDTERM EXAM I
Date
Duration
: 30 April 2009
: 100 min.
ANSWER ALL 4 QUESTIONS Q1: (a)15 pts. (b)10 pts. Q2: (a)15 pts. (b)10 pts.
Q3: (a)10 pts. (b)10 pts. (c)10 pts. Q4: (a)10 pts. (b)10 pts.
Figure 1
1-) In the circuit in Figure 1, diode D has a constant voltage drop of VD = 0.7
V when it conducts.
(a) Find and sketch the input-output characteristic of the circuit for the
range - 2 V < vi < 2 V (i.e. vo as a function of vi ).
(b) If the input is the sine wave vi (t )  2sin(100t ) V , sketch the waveform
of the output voltage vo (t ) .
D
+
+
200 Ω
100 Ω
vi
vo
2-) The zener diode in the circuit in Figure 2 has the i-v characteristic shown, where Vzo  7.3 V , rz  20  ,
R = 280  and Vs = 12 V.
R
Figure 2
V zo
+
+
+
Vs
Vz RL
vo
i
v
slope = 1/rz
(a) Find the load voltage vo for RL  1500  . (You may use approximation)
(b) Find the minimum value of RL before voltage regulation is lost (for voltage regulation Vz  Vzo )
3-) In the peak rectifier in Figure 3 diode D is ideal and C=100μF.
Figure 3
The input vs is a 50-Hz sinusoidal voltage with rms value 12 V.
(a) Find the minimum value of RL so that the ripple voltage
D
+
Vr  2 V .
(b) With RL  2 k , calculate the average and maximum
values of the diode current during conduction.
(c) Calculate the average supply current I s ,av .
vs
+
is
C
RL
vo
4-) A voltage amplifier has input resistance Ri = 10 k, output resistance Ro = 100  and no-load voltage gain
A= 200.
(a) Sketch the equivalent circuit of the voltage amplifier. Indicate all relevant quantities.
(b) If the amplifier feeds a load RL = 200  from a voltage source vs having an internal resistance
Rs = 5 k, find the overall voltage gain Av  vo / vs , where vo is the load voltage.
Useful Information:
(1) Peak rectifier equations:
Ripple voltage,
Average diode current during conduction,
Maximum diode current during conduction,
Vr 
Vp
fRL C
;

 I 1  2
;
/V  ;
I D,av  I L 1   2Vp / Vr
I D ,max
L
2Vp
r
SOLUTION (Midterm Exam Spring 2008-2009)
1-) (a) Assume that the ideal diode is on
i1
+
v0  vi  0.7
0.7 V
i2
+
200 Ω
100 Ω
vi
 v0  100i1  0.35  vi  0.7
vo
0.7
 3.5 mA
200
v  1.05
i1  i
 0  vi  1.05 V
100
Also v0  100(i1  i2 ), and i2 

Now assume that the diode is off
+ vD
+
100
2
vi
vD  0.7  vi  v0  vi
300
3
2
 vD  vi  0.7  0  vi  1.05 V
3
0.7 V
200 Ω
100 Ω
vi
v0 
+
vo
vo
The transfer characteristic:
 1
vi  1.05V
 v
v0   3 i
vi  0.7 vi  1.05V
slope=1
slope=1/3
0.35 V
(b)
vo(V)
2.0
1.3
0.35
t
-0.67
2-) (a) Exact solution:
Iz  Is  IL,


Is 
Vs  Vz
R
IL 
Vz
RL
 V  V0 V0 
V0  Vz  Vz 0  rz  s


R
RL 

20
r
7.3 
 12
Vz 0  z Vs
280
R
V0 

 7.52 V
1 
1 1 
 1
1  rz  
 1  20  280  1500 
 R RL 
Approximate solution:
1.05 V
vi
V0  Vz 0
Let
12  7.3
7.3
 16.786 mA
IL 
 4.87 mA
280
1500
 Vz  7.3  0.02  11.92  7.538 V

Is 
 I z =11.92 mA
Second iteration
12  7.538
7.538
 15.936 mA I L 
 5.025 mA
280
1500
Vz  7.3  0.02  10.91  7.52 V
Is 
I z =10.91 mA
(b)
V0  Vz 0

RL 

Iz  0

Is  IL
Vs  Vz 0 Vz 0

R
RL
RVz 0
280  7.3

 434.9 
Vs  Vz 0
4.7
3-) (a) Vp  12 2 V
Vr 
Vp
fRLC
2V

RL 
12 2
 1697 
2  50 100 106
(b)
IL 
V0
RL
Vr 
12 2
 1.697 V
50  2  103  100  106
V0  V p

IL 
12 2
 8.485 mA
2 k

2  12 2 
  127.7 mA
I D ,av  8.485  1  
1.697 



2  12 2 
  246.9 mA
I D ,max  8.485  1  2
1.697 


Average of source current over one period of the source voltage: I s ,av  I L  8.485 mA
(c)
4-) (a)
+
Vi
Ro
Ri
+
AVi
(b)
Rs
Vs
+
+
Vi
Ri
Ro
+
AVi
RL
+
V0
Av 
V0 V0 Vi
 .
Vs Vi Vs
Vo 
RL
AVi
Ro  RL
Vi 
Ri
Vs
Rs  Ri
 200   10 
 Av  200 

  88.89
 200  100   10  5 