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Eastern Mediterranean University
Department of Electrical and Electronic Engineering
EENG 341 ELECTRONICS I – INFE 242 ELECTRONICS
FINAL EXAM
Date
Duration
: 25 Jan 2010
: 120 min.
ANSWER ALL 4 QUESTIONS
(1)
Figure 1
In the MOSFET circuit shown in Figure 1, the transistors
have widths W1 = 10 μm, W2 = 40 μm, and their lengths are
equal, i.e. L1 = L2 = 2 μm. Both transistors have
kn  0.2 mA/V2 , and Vt = 1 V.
Find the voltage Vo and the current ID. (25 pts)
+5 V
Q1
ID
Vo
Q2
+15 V
(2)
Figure 2
5 kΩ
The MOSFET in the common-source amplifier
shown in Figure 2 has the small signal
parameters gm = 2 mA/V and   0 . The
coupling and bypass capacitors are short
circuits at the signal frequencies.
(a) Draw the small-signal equivalent circuit of
the amplifier. (10 pts)
(b) Find the voltage gain Av  vo / vi . (5 pts)
(c) Find the overall voltage gain Gv  vo / vsig .
Rsig = 100 kΩ
vo
CC2
CC1
20 kΩ
+
vsig
Rout
+
_
vi
Rin
2 MΩ
1 kΩ

CS
(5 pts)
(d) Find the input (Rin) and output (Rout)
resistances. (5 pts)
(3)
-15 V
Figure 3
The BJT in the circuit of Figure 3 has   1 and VCEsat = 0.2 V. At
the edge of conduction VBE = 0.5 V. For active operation and
saturation VBE = 0.7 V.
(a) For VB = 0 V, find the collector and emitter voltages VC, VE.
6 mA
VC
2 k
(10 pts)
(b) Find the value of VB at which the transistor enters cut-off.
VB
(7 pts)
(c) Find the value of VB at which the transistor enters saturation.
VE
(8 pts)
2 mA
2 k
(4)
The BJT in the emitter-follower amplifier in Figure 4 has
β = 100. The coupling capacitors may be taken as short
circuits at the signal frequencies. Neglect the output
resistance of the transistor.
200 kΩ
(b) Calculate the small-signal parameters of the transistor.
CC1
15 kΩ
CC2
(a) Calculate the bias value of the emitter voltage VE.
(5 pts)
+10 V
Figure 4
vsig
2 kΩ
(5 pts)
2 kΩ
iout
(c) Draw the small-signal equivalent circuit of the
i
amplifier, and find its current gain Ai  out . (15 pts)
iin
MOSFET:
vo
iin
+
_
BJT:
Small-signal model:
(a) Saturation-region current-voltage equation:
1 W
2
iD  kn  vGS  Vt 
2 L
B
+
rπ
Triode region:
iD  kn
E
gmvπ = βib
gm 
(b) Small-signal model:
+
gmvgs
vgs

S
2I D
,
VGS  Vt
ro 
ro

Small-signal parameters:
VDS  VGS  Vt .
gm 
vπ
W
1 2 
 vGS  Vt  vDS  vDS


L
2
Note: for operation in the saturation region
G
C
gmvπ
1
ID
D
ro
IC
VT
VT  25 mV
r 

gm
ro 
VA
IC
SOLUTION
EENG 341 FINAL EXAM
Fall 2009-2010
1-) Because VDG = 0, both transistors operate in saturation.
1 W
1 W
I D  kn 1 (VGS 1  Vt )2  kn 2 (VGS 2  Vt )2
2 L1
2 L2
By KVL, VDS1  VDS 2  5
Also, VDS1  VGS 1 , VDS 2  VGS 2
 3VGS1  10  Vt  9 V,
VGS1  3 V
ID 
L1 W2
. .(VGS 2  Vt )  2(VGS 2  Vt )
L2 W1
 (VGS 1  Vt ) 
 VGS 2  5  VGS 1
Vo  VGS 2  2 V
1
10
 0.2  (3  1) 2  2 mA
2
2
2-) (a) Small-signal equivalent circuit
G
Rsig = 100 kΩ
D
+
vsig
+
_
2 MΩ
Rin
vo
gmvgs
vgs

Rout
5 kΩ
20 kΩ
S
1 kΩ
vo
v
, vi  vgs  (1k ) gmvgs  3vgs  vgs  i
vi
3
8
8
vo   gmvgs (5k  / /20k )  2  4 vgs  8vgs   vi  Av   V/V
3
3
vo
vo vi
2  8
Overall voltage gain: Gv 


     2.54 V/V
vsig vi vsig 2.1  3 
Rin  2 M,
Rout  5 k
(b)
Av 
Voltage gain:
(c)
(d)
3-) (a)
Figure 3
VC  2(6  I C )  12  2 I C
6 mA
VE  2( I E  2)  2 I E  4
Assume that the transistor is active 
VB  0 V 

VBE  0.7 V
IC
VE  0.7 V
I E  1.65 mA
IC   I E  I E
Let I C  I E  0 (in cutoff)
 active.
 VC  12 V, VE  4 V
VBE  0.5 V  VB  VE  0.5  3.5 V
(c)
At the edge of saturation I C  I E
and
VCE  (12  2 I C )  (2 I C  4)  16  4 I C
 VC  4.1 V, VE  3.9 V

VC
2 k
VB
 VC  8.7 V  VCE  8.7  ( 0.7)  9.4 V
(b)
6-IC
VCE  0.2 V
 I C  3.95 mA
VB  VE  0.7  4.6 V
IE
IE -2
2 mA
VE
2 k
4-) (a) KVL equation for the dc circuit
+10 V
 200k   (1   )(2k )  I B  10  0.7  9.3 V
200 kΩ
9.3 V

IB 
 23.13  A
402k 
 VE  (2k ) I E  (2k )(1   ) I B  4.67 V
IB
+
VBE
_
2 kΩ
(b)
I C   I B  2.313 mA
r 
(c)

gm


gm 
2.313 mA
 92.52 mA/V
25 mV
100
 1.08 k
92.52
Small-signal equivalent circuit
15 kΩ
ib
B
C
iin
vsig
+
_
200 kΩ
Rib
βib
rπ
E
iout
2 kΩ
2 kΩ
1
iout  (1   )ib  50.5ib
2
vi  r ib  (2k  / /2k )(1   )ib  (102.08 k )ib
ib 
200
iin  0.6621iin
200  102.8
Ai  33.44 A/A
 Rib 
 iout  50.5  0.6621iin
vi
 102.08 k 
ib
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