302, ECE Section 2
Spring 2008
April 11, 2008
Exam III
Printed Name: ________________
KEY
_________________________________
Honor Pledge: I certify that I have neither given nor received unauthorized aid while taking this exam.
Signature: ____________________________________________________
-
This is a closed book, closed notes test.
You may use a calculator
Please show all work to get partial credit. Attach any extra pages that were necessary.
DO NOT write on the backs of the pages.
SHOW ALL UNITS!!!!! Points will be deducted for incorrect or absence of units.
Problem
Possible Points
1
30
2
25
3
30
4
15
Total
100
1
Missed Points
Total Points
Small Signal Model Equations
1
FET
gm
2I
D
= 2K n I
D
V V
TN
GS
ro =
S
I
BJT
gm = C
V
T
+V
I
D
V
ro = A
I
C
S
DS
1
I
D
vgs 0.2V V V
TN
GS
V
r = o T
I
C
v 5 mV
be
VT = 0.025 V at room temperature
Single Stage Amplifier Terminal Equations
Common Collector
Common Emitter
Terminal Voltage Gain
Input Terminal Resistance
Output Terminal Resistance
Terminal Current Gain
Terminal Voltage Gain
Input Terminal Resistance
Output Terminal Resistance
Terminal Current Gain
gR
mL
1+ g m R
Common Base
g mR
L 1
1+ g m R
L
r (1+ gmR )
L
E
r (1+ gm R )
E
gm R
L
1
gm
1 Rth
+
gm o
ro1 + g m R
E
o + 1
o
Common Source
gR
mL
1+ g m R
S
Common Drain
ro1 + gm R
th
o + 1
Common Gate
g mR
L 1
1+ g m R
L
1
gm
ro 1+ g m R
S
2
gm R
L
1
gm
ro1+ gm R
th
+1
1. FET Amplifier Small Signal Analysis (30 points total)
80
io
50
2
ii
100
Q-point: (100A, 6 V)
500
7
VGS = 4 V
53
a. Find gm and ro for the circuit above.
gm = 2K n I = 2 * 50 *100 = 0.1 mS
D
ro =
1
1
=
= 752 k
I
0.0133*100
D
b. Draw the small signal equivalent circuit for the amplifier.
Draw the small signal equivalent circuit for the amplifier.
500
500 k
vg
44.4 k
1M
7k
3
c. For what values of vi is the small signal model valid? Assume ro approaches infinity.
Let vg be the voltage from the gate to Ground, then
vg = vgs + (g m * vgs * RS ) = vgs (1 + g m RS ) = vgs (1 + 0.1m * 7 k ) = 1.7 vgs and
vg =
vi * RG
vi * 1M
2v
=
= i therefore
RI + RG 500k + 1M
3
1.7 vgs =
2vi
which implies vi 2.55 vgs
3
For the model to be valid
v gs 0.2V V = 0.2 4 2 = 0.4 V
TN
GS
substituting in the inequality above,
vi 2.55 * 0.4 = 1.02 V
d. Find the voltage gain ( Av = vo / vi ) for the amplifier. Show all work.
Assume ro approaches infinity.
Av =
vo g m vgs RL
=
using the results from above and substituting circuit values,
vi
vi
Av =
0.1* 44.4k * vgs
2.55* vgs
= 1.74
4
2. FET/BJT Terminal Equation Derivations (25 point total)
a. Draw the small signal equivalent circuit and derive the equation given on the equation sheet for the
Terminal Input Resistance, RiB , of the following AC equivalent circuit. Assume ro approaches infinity.
RI
ib
+
vi
+
r
vbe
+
_
o ib
_
RB
vb
+
RL
vo
_
_
RL = R6 || R3
RiB =
vb r ib + (ib + oib )RL r ib + (1 + o )ib RL
=
=
= r + (1 + o )RL r + o RL
ib
ib
ib
substituting
RiB = r + g m r RL = r (1 + g m RL )
5
and
o = g m r
Draw the small signal equivalent circuit and derive the equation given on the equation
sheet for the Terminal Voltage Gain, Avt , of the following AC equivalent circuit. Assume
ro approaches infinity.
gmvgs
RI
S
D
+
vi
+
_
+
R6
vit = - vgs
RL
vo
_
_
Avt =
vo g m vgs RL
=
= g m RL
vgs
vit
6
G
3. Single Stage BJT/MOSFET Amplifier Problem (25 points)
vb
RC = 100 k
R B = 80 k
RE = 450
R I = 20 k
R3 = 100 k
o = 55
a. For the amplifier above, what is the DC collector current, IC, if the terminal voltage
gain of the amplifier (Avt = vo / vb ) is - 100? Assume ro approaches infinity.
From the equation sheet, Avt = 100 =
Solving, g m =
g m RL
g m * 50k
=
1 + g m RE 1 + g m * 450
100
= 20 mS
50k 45k
From the equation sheet,
I
gm = C
V
T
and VT = 0.025 V
Therefore, I C = g mVT = 20m * 0.025 = 0.5 mA
b. What is r and gm ?
From part a, g m = 20 mS
r =
o
gm
=
55
= 2.75 k
20m
7
4. Multistage Amplifier Problem (15 points)
Stage 1
Stage 3
Stage 2
+V
15 k
4.7 k
910 k
160 k
v3
v2
M3
v1
10 k
1M
vi
1.2 Mk
43 k
9k
3k
1.6 k
250
For the circuit above; gm1 = 2 mS
gm2 = 54 mS, r2 = 1.97 k
gm3 = 1.93 mS
Avt1 = - 4.36
Avt2 = - 237
Avt3 = 0.308
a. What is the overall gain of the circuit above, Av = vo / vi ? Assume ro approaches infinity.
Av =
vo vo v3 v2 v1
1M
1M
=
= Avt 3 Avt 2 Avt1
= (0.308)( 237 )( 4.36 )
= 315
vi v3 v2 v1 vi
10k + 1M
10k + 1M
b. What is the overall gain of the circuit if C4 is removed?
Avt3 will remain unchanged. Avt2 will change. Avt1 will change since the input impedance to the second
stage changes and therefore the load resistance to the first stage changes.
Avt 2 =
g m 2 RL 2
237
=
= 2.71
1 + g m 2 RE 2 1 + 54m *1.6k
Avt1 = g m1RL1 = g m1 (RD1 || RB 2 || (r 2 (1 + g m 2 RE 2 ))) = 2m(15k || 33.9k || (1.97 k (1 + 54m *1.6k ))) = 19.6
Av = Avt 3 Avt 2 Avt1
1M
1M
= (0.308)( 2.71)( 19.6 )
= 16.2
10k + 1M
10k + 1M
8
ECE 302, Section 1
Fall 2009
November 11, 2009
Exam III
Printed Name: ______________
KEY_______________________________
Honor Pledge: I certify that I have neither given nor received unauthorized aid while taking this exam.
Signature: ____________________________________________________
-
This is a closed book, closed notes test.
You may use a calculator
Carry all calculations to 3 significant digits
Show all work to get partial credit. Attach any extra pages that were necessary.
Use the backs of the pages if you need more space.
SHOW ALL UNITS!!!!! Points will be deducted for incorrect or absence of units.
Problem
Possible Points
1
40
2
40
3
20
Total
100
1
Missed Points
Total Points
Small Signal Model Equations
1
FET
gm
2I
D
= 2K n I
D
V V
TN
GS
ro =
S
I
BJT
gm = C
V
T
+V
I
D
V
ro = A
I
C
S
DS
1
I
D
vgs 0.2V V V
TN
GS
V
r = o T
I
C
v 5 mV
be
VT = 0.025 V at room temperature
Single Stage Amplifier Terminal Equations
Common Collector
Common Emitter
Terminal Voltage Gain
Input Terminal Resistance
Output Terminal Resistance
Terminal Current Gain
Terminal Voltage Gain
Input Terminal Resistance
Output Terminal Resistance
Terminal Current Gain
gR
mL
1+ g m R
Common Base
g mR
L 1
1+ g m R
L
r (1+ gmR )
L
E
r (1+ gm R )
E
gm R
L
1
gm
1 Rth
+
gm o
ro1 + g m R
E
o + 1
o
Common Source
gR
mL
1+ g m R
S
Common Drain
ro1 + gm R
th
o + 1
Common Gate
g mR
L 1
1+ g m R
L
1
gm
ro 1+ g m R
S
2
gm R
L
1
gm
ro1+ gm R
th
+1
1. BJT/MOSFET Amplifier Problem (40 points)
+5 Vdc
Kn = 100 A/V2
VTN = 1.5 V
= 0.01 /V
18 k
120
120 k
io
ii
vi
100
10 k
vo
390 k
Q point: (107 A, 2.80 V)
VGS = 2.96 V
8.2 k
ro = 935 k
gm = 0.146 mS
a) Draw the small signal equivalent circuit for the circuit above.
b) Using the terminal equations on the equation sheet, find Av = vo / vi and Ai = io / ii.
Assume ro approaches infinity.
Common Source Amplifier with Rs = 0, so Avt = - gm RL = -(0.146m)(18k||10k) = -0.939
v
91.8k
v
vv
Av = o = o gs = Avt gs = (0.932)
91.8k + 100 = 0.938
vi v gs vi
vi
vo
i
R3
R + RG
91.8k + 100
Ai = o =
= Av I
= (0.932)
= 8 . 61
vi
ii
R3
10k
RI + RG
3
2. BJT/MOSFET Terminal Equation Derivations (40 points)
a. Draw the small signal equivalent circuit and derive the equation given on the equation sheet for the
Terminal Input Resistance, RiE, of the following AC equivalent circuit. Assume ro approaches infinity.
RI
ib
+
vi
+
vbe
+
_
oib
r
_
RB
vb
+
RL
vo
_
_
RL = R6 || R3
+
vbe
-
o ib
Applying a voltage source to the Emitter and shorting the independent source, the circuit for
finding RiE is shown below.
Let
RiE =
Rth = RI || RB
v x vbe + vb
i r + (ib Rth ) r + Rth
r + Rth
=
= b
=
=
o
ib oib
1 + o
ix
ix
4
r
o
+
Rth
o
=
r
R
R
1
+ th =
+ th
r g m o
gm o
b. the Draw small signal equivalent circuit and derive the equation given on the equation sheet for the
Terminal Voltage Gain, Avt, of the following AC equivalent circuit. Assume ro approaches infinity.
gmvgs
RI
S
D
+
vi
+
_
+
R6
vit = - vgs
RL
vo
_
_
Avt =
vo g m vgs RL
=
= g m RL
vgs
vit
5
G
3. BJT/MOSFET Design Problem (20 points)
+5 V
RD
RG2
120 k
3900
RI
In the circuit to the right;
2k
vi
Av = vo / vi = 0.494
RG1
3900
ROUT = 41
R6
68 k
R3
100k
ROUT
a) Redesign the circuit above such that Av > 0.9 and the operating point is unchanged.
RI and R3 cannot be changed.
From the equation sheet ,
Then Av = Avt
Avt 1
Let RG = RG1 || RG 2
RG
RG
= 1*
2k + RG
RI + RG
So the following must be true
RG
> 0 .9
2k + RG
Solving RG > 18 k
Since the operating point cannot change, RG1 must equal RG2 and be greater than 36 k
Any solution where RG1 = RG2 and each is greater than 36 k is acceptable.
b) What value of ROUT for the redesigned circuit? Explain your answer?
ROUT = 1 / gm and since IC is unchanged, gm is unchanged.
Therefore,
ROUT = 41
6
vo
ECE 302, Section 2
Spring 2010
April 7, 2010
Exam III
Printed Name: ________________
KEY_________________________________
Honor Pledge: I certify that I have neither given nor received unauthorized aid while taking this exam.
Signature: ____________________________________________________
-
This is a closed book, closed notes test.
You may use a calculator
Carry all calculations to 3 significant digits
Show all work to get partial credit. Attach any extra pages that were necessary.
Use the backs of the pages if you need more space.
SHOW ALL UNITS!!!!! Points will be deducted for incorrect or absence of units.
Problem
Possible Points
1
20
2
20
3
40
4
20
Total
100
1
Missed Points
Total Points
Small Signal Model Equations
1
FET
gm
2I
D
= 2K n I
D
V V
TN
GS
ro =
S
I
BJT
gm = C
V
T
+V
I
D
V
ro = A
I
C
S
DS
1
I
D
vgs 0.2V V V
TN
GS
V
r = o T
I
C
v 5 mV
be
VT = 0.025 V at room temperature
Single Stage Amplifier Terminal Equations
Common Collector
Common Emitter
Terminal Voltage Gain
Input Terminal Resistance
Output Terminal Resistance
Terminal Current Gain
Terminal Voltage Gain
Input Terminal Resistance
Output Terminal Resistance
Terminal Current Gain
gR
mL
1+ g m R
Common Base
g mR
L 1
1+ g m R
L
r (1+ gmR )
L
E
r (1+ gm R )
E
gm R
L
1
gm
1 Rth
+
gm o
ro1 + g m R
E
o + 1
o
Common Source
gR
mL
1+ g m R
S
Common Drain
ro1 + gm R
th
o + 1
Common Gate
g mR
L 1
1+ g m R
L
1
gm
ro 1+ g m R
S
2
gm R
L
1
gm
ro1+ gm R
th
+1
1. General Single Stage Amplifier Questions (20 points)
a. In the following circuit, select a value of R3 such that Avt = - gm
+5 Vdc
18 k
120
120 M
From the Equation Sheet, Avt = - gm RL when RE = 0.
10
So, Avt = - AVt when RL = 1.
RL = 18k || R3 = 1
R3
vi
390 M
8.2 k
Therefore, R3 = 1
b. Two amplifiers, a common emitter with RE = 0 and a common base, have the same magnitudes of
terminal voltage gain. Both amplifiers have the same 4 four resistor bias network, series source
resistor RI, and parallel load resistor R3. Which amplifier would you expect to have the greater
overall amplifier voltage gain? Why?
AVt is the same for each amplifier, however, the input resistances are not.
AV = AVt [RIN / (RI + RIN)] and the input resistance of a common base amplifier is small compared
to a common emitter amplifier. So, [RIN / (RI + RIN)] is larger for a common emitter amplifier than
for a common base when RI is non-zero. Therefore, AV is larger for a common emitter is expected
to be larger for a common emitter amplifier.
c. What are the input and output terminal for the following amplifiers?
Input
Output
COMMON BASE
__Emitter____
____Collector____
COMMON DRAIN
__Gate______
____Source______
d. In a common gate amplifier, if you increase RI, how will the Output Terminal Resistance change?
Why?
Output Terminal Resistance = ro (1 + gmRth) where Rth= RI||R6. If RI increases, then Rth increases.
Therefore, the Output Terminal Resistance increases.
3
2. Amplifier Small Signal Analysis (20 points)
+10 V
270 k
3.9 M
Qpt = (10.9 A, 5.74 V)
KN = 100 A/V2
= 0.001
5k
1.5 M
200
120 k
a) Draw the small signal equivalent circuit for the circuit above.
ro
RI
S
D
gm v g
vi
-
+
_
s
+
R6
vgs
RL
vo
_
+
G
b) Using the terminal equations on the equation sheet, find the Output Terminal Resistance,
RiD, for the circuit above.
ro = 1 / ID = 1 / (0.001 * 10.9) = 91.7 M
gm = (2 KN ID ) = (2 * 100 * 10.9) = 46.7 S
Rth = 120k || 200 = 200
RiD = ro ( 1 + gm Rth ) = 91.7 M
[1+ 46.7 S * (200))] = 92.6 M
4
2. BJT/MOSFET Terminal Equation Derivations (40 points)
a. Draw the small signal equivalent circuit and derive the equation given on the equation sheet for the
Terminal Voltage Gain, Avt, for the AC equivalent circuit below. Assume ro approaches infinity.
ib
+
RL
o ib
vb
-
AVt =
RL = RC || R3
vo
vo
o ib RL
o ib RL
r g m RL
=
=
=
vb vbe + v RE r ib + ( o + 1)ib RE r ib + oib
r + r g m RE
5
=
g m RL
1 + g m RE
b. Draw the small signal equivalent circuit and derive the equation given on the equation sheet for the
Output Terminal Resistance, RiS, for the AC equivalent circuit below. Assume ro approaches infinity.
D
RI I
R
ib
+
vi
+
_
G
++
vvbe
gs
o b
gm ivgs
r
_
R
RBG
vb
-
S
+
RL
RL
_
_
RL = R6 || R3
Applying a voltage to the Source and setting the input to zero,
D
gm vgs
G
vgs
S
RG
RiS =
v gs
vx
=
=
g m v gs
ix
vo
1
gm
6
3. BJT/MOSFET Design Problem (20 points)
12 V
12 V
1k
BJT
BJT or
MOSFET
Four Resistor
Bias Circuit
68 k
3900
2.7 M
39 k
5k
200
1.5 M
27 k
2200
21.8 k
Qpt = (1.1 mA, 5.27 V)
VA = 1,000 V
o = 80
Qpt = (90 A, 6.58 V)
= 0.001 /V
KN = 100 A/V2
VTN = 1.0 V
The goal of this problem is to construct an Amplifier to fit in to the box at the left such that the magnitude of
the voltage gain, |AV |= |vo / vi|, is greater than 50 using either the BJT circuit or the MOSFET circuit above.
a) Would a Follower circuit (common collector or common drain) be a good choice? Why or why not?
No. The Terminal Voltage Gain is approximately 1 so |Av| will be 1 or less.
b) Would an Inverting circuit (common emitter or common source) be a good choice? Why or why not?
Yes. The Terminal Voltage Gain, Avt, is -gmRL for both amplifiers and each has large RL.
Av = Avt [RIN / (RI + RIN)] and both amplifiers have relatively large RIN so the gain is not reduced
by the input resistance of either transistor.
7
c) Would a Non-Inverting circuit (common base or common gate) be a good choice? Why or why not?
No. The Terminal Voltage Gain, Avt, is gmRL for both amplifiers and each has large RL.
For both amplifiers, the terminal input resistance is 1/gm
g m = 2 K N I D = 2 * 100 * 90 = 0.134 mS
.
I C 1.1m
gm =
=
= 44.0 mS
Vt 25m
For the MOSFET,
For the BJT,
So RIN will be small
Since Av = Avt [RIN / (RI + RIN)] and both amplifiers have relatively small RIN, the gain is reduced
by the input resistance with either transistor.
d) Draw the complete diagram of the circuit (including bypass capacitors) that will meet the criteria
|AV| > 50 and calculate the actual value of AV.
So either BJT common emitter or MOSFET common source are good candidates.
1 39
k
3.9 k
68 k
k
2.7 M
1k
1k
5k
5k
200
27 k
1.5 M
21.8 k
2.2 k
For the Common Emitter Amplifier,
Av = g m RL
RB || RiB
68k || 27 k || 1.82k
= (0.044)(3.9k || 5k )
= 60.1
1k + (68k || 27 k || 1.82k )
RI + RB || RiB
For the Common Source Amplifier with RS = 200,
Av =
g m RL
RG
(0.134m)(39.0k || 5k ) 1.5M || 2.7 M
=
= 0.572
1 + g m RS RI + RB
1 + (0.134m)(200) 1k + (1.5M || 2.7 M )
For the Common Source Amplifier with RS bypassed also,
RG
1.5M || 2.7 M
= (0.134m)(39.0k || 5k )
= 0.590
Av = g m RL
1k + (1.5M || 2.7 M )
R I + RB
So Common Emitter Amplifier is the best choice.
8