Download EENG 341 FEx fall 13 14 - faraday

Eastern Mediterranean University
Department of Electrical and Electronic Engineering
EENG 341 ELECTRONICS I – INFE 242 ELECTRONICS
FINAL EXAM
Date
Duration
: 27 January 2014
: 120 min.
ANSWER ALL 4 QUESTIONS
1-) In the MOSFET circuit shown in Figure 1, the transistor has the
following parameters
kn
+10 V
Figure 1
4 k
W
 1.0 mA/V2 ,   0 , Vt = 2 V
L
ID
Find the the drain current ID, and the gate-source voltage VGS. State in
which region of its characteristics the transistor operates.
+
V GS
3 k
-10 V
2-) The MOSFET in the common-drain
Figure 2
amplifier shown in Figure 2 has the small
signal parameters
gm = 2 mA/V, ro  150 k and   0 .
Assume that the transistor is properly biased at
a point in its saturation region.
Rsig = 100 kΩ
+15 V
CC1
CC2
+
vsig
+
_
(a) Draw the small-signal equivalent circuit of
the amplifier. (10 pts)
vi
Rin
Rout
vo
1 MΩ
15 kΩ

(b) Find the voltage gain Av  vo / vi , the
overall voltage gain Gv  vo / vsig , and the
-15 V
input (Rin) and output (Rout ) resistances. (15 pts)
+10 V
3-) The BJT in the circuit of Figure 3 has
  100, VBE  0.7 V, VCEsat  0.2 V
For the following values of R, determine whether the
transistor operates in the active mode or in saturation. Then
find the values of the collector current I C , and the voltages
VC and V E . (Hint: If the transistor is in saturation  forced will
be unknown)
Figure 3
IC
R
VC
20 k
VE
2 mA
(a) R  2 k (12 pts)
(b) R  10 k (13 pts)
4-) The BJT in the common-emitter amplifier in
Figure 4 has
+12 V
Figure 4
4 kΩ
  100, VA  50 V .
The coupling capacitors may be taken as short
circuits at the signal frequencies.
(a) Calculate the bias value of the collector voltage
Vc. (5 pts)
(b) Calculate the small-signal parameters ( g m , r , ro )
of the transistor. (5 pts)
(c) Draw the small-signal equivalent circuit of the
amplifier, and find its overall voltage gain
v
Gv  o . (15 pts)
vsig
CC1
10 kΩ
vsig
+
_
2 mA
(a) Saturation-region current-voltage equation:
(a) Active mode:
1 W
2
iD  kn  vGS  Vt 
2 L
IC   I B
W
1 2 
 vGS  Vt  vDS  vDS


L
2
for
operation
 Vt .
in
the
I Csat   forced I B
region
(b) Small-signal model:
B
+
rπ
+
gmvgs
vgs

S
 forced  
VCEsat  0.2 V
saturation
(b) Small-signal model:
G
VCB  0.4 V
Saturation:
Triode region:
VDS  VGS
CE
100 kΩ
BJT:
Note:
vo
6 kΩ
MOSFET:
iD  kn
CC2
Vc
D
ro
gmvπ
vπ

E
gm 
C
ro
gmvπ = βib
V
IC

, r 
, ro  A
VT
gm
IC
I C : bias value of collector current
VT  25 mV
SOLUTION
EENG 341 FINAL EXAM
Fall 2013-2014
1-)
Assume that the transistor operates in saturation
VGS  VG  VS  VS
1
1
1
 1  (VGS  Vt )2 mA  I D  ( VS  Vt )2  (VS  2) 2
2
2
2
2
VS  3I D  10  2 I D  (3I D  8)
ID 
9 I D2  50 I D  64  0 
I D  2 mA
is the physically possible solution
Check for saturation
VS  4 V  VGS  4 V > Vt  2 V
VD  10  4  2  2 V  VDG  2 V  Vt  2 V
 Saturation.
2-) (a) Small-signal equivalent circuit
Rsig = 100 kΩ
Rin
G
+
vsig
+
_
1 MΩ
vi
D
+
vgs

gmvgs
150 kΩ

S
vo
15 kΩ
Rout
(b)
vo  gmvgs (15 k 150 k)  2 13.64 vgs  27.28 vgs
KVL:  vi  v gs  vo  0
vi 
1000
vsig  0.909vsig
1100

Av 
vo 27.28

 0.965
vi 28.28

vi  28.28v gs

Gv  0.909  0.965  0.877
Rin  1 M
For Rout, a test source must be applied after removing the load and turning the input signal off:
100 kΩ
G
+
1 MΩ
vi

D
+
vgs

vG  v gs  vt  0  v gs  vt
gmvgs
150 kΩ
S
it
+
-
vt
KCL at S:
vt
0
ro
vt 
1
  g m   vt
ro 
ro 
v
ro
1
150k
 t 


 498.3 
it g  1 1  g m ro
301
m
ro
 it  g m vt 
 Rout
it  g m v gs 
3-)
(a) Assume that the transistor is active
I C   I E  1.98 mA

VC  10  2  1.98  5.96 V
VB  20 I B  20  0.0198  0.396 V 
 VCE  VC  VE  7.056 V  0.2 V
VE  VB  VBE  1.096 V
 transistor is active.
(b) Assume again that the transistor is active
I C  1.98 mA

VC  10  10  1.98  9.8 V
VE  1.096 V  VCE  VC  VE  8.7 V
 transistor cannot be active. It is in saturation
VCEsat  0.2 V, I Csat   forced I B ( forced is unknown)
I E  (1   forced ) I B  2 mA


IB 
2
mA
1   forced
VE  20 I B  0.7 V
VC  10  10 I Csat  10  10  forced I B
VCE  VC  VE  10  10  forced I B  ( 20 I B  0.7)  0.2 V
10

forced
 20  I B  10.5
2 10 forced  20   10.5 1   forced 

  forced  5.32

I B  0.3165 mA, I Csat  1.683 mA
Check:

VE  20  0.3165  0.7  7.03 V
4-) (a) I C   I E  (100 / 101)  2 mA  1.98 mA
(b) gm 
I C 1.98 mA

 79.2 mA / V
VT
25 mV


r 
;
VC  10  10  1.684  6.83 V

gm
VCE  VC  VE  0.2 V
VC  12  4 I C  4.08 V
 1.262 k
;
ro 
VA
 25.25 k
IC
(c)
10 kΩ
ib
B
C
+
vsig
+
_
100 kΩ
rπ
βib
Vπ
25.25 kΩ
4kΩ//6 kΩ
E
25.25 k 4 k 6 k  2.192 k
vo   g m v (2.192 k)  173.6v
1.262 k 100 k  1.246 k  v 
 Gv  173.6  0.11  19.1 V/V
vo
1.246
vsig  0.11vsig
10  1.246