ECE-342 Test 3: Dec 1, 2009
6:00-8:00pm, Closed Book
Name :
Solution
Please use VT = kT /q = 25.8 mV at 300K for your calculations
1. Problems 1 through 3 all deal with various aspects of the 2-stage amplifier shown below. Assume that β1 =
β2 = 100, VA1 = 75 V, VA2 = ∞.
Solve for the DC bias conditions of Q1 and Q2 . (You may ignore base-width modulation in determining the
bias conditions.) Determine the DC offset of the output signal VOU T , and provide all small-signal parameters
(gm , ro , rπ , and re ) for both transistors. (Be sure to include units)
Solution:
The current source determines both emitter currents:
IE1 = IE2 = IE = 0.5 mA
Using β1 = β2 = 100 gives:
100
100
IE =
(0.5 mA) = .495 mA
101
101
IE
= IB =
= 4.95 µA
101
IC1 = IC2 = IC =
IB1 = IB2
The base currents, with the 0.7 Vdiode drop assumption gives the
node voltages.
VB1 = VIN + (10 kΩ)IB1 = (4.95 µA)(10 kΩ) = .0495 V
VE1 = VE2 = .0495 + 0.7 = 0.7495 V
VB2 = VE2 + 0.7 = 1.4495 V
VOU T = VC2 = VB2 + (100 kΩ)IB2
= 1.4495 + (4.95 µA)(100 kΩ) = 1.9495 V
The small signal parameters come from the bias solution:
.495 mA
IC
=
= .0192 A/V
VT
25.8 mV
β
100
rπ1 = rπ2 =
=
= 5.212 kΩ
gm
.0192 A/V
α
rπ
re1 = re2 =
=
= 51.6 Ω
gm
β+1
VA1 + VEC1
75.75 V
ro1 =
=
= 153 kΩ
IC
.495 mA
VA1
75 V
(≈
=
= 151.5 kΩ)
IC
.495 mA
VA2
ro2 =
=∞
IC
gm1 = gm2 =
2. For the amplifier of problem 1, assume that the following small-signal parameters have been determined for Q1
and Q2 . (These are not the correct values. Use these parameters only for problems 2 and 3.)
gm1 = 0.0116 Ω−1
gm2 = 0.0116 Ω−1
rπ1 = 10.34 kΩ
rπ2 = 7.76 kΩ
re1 = 85 Ω
re2 = 85 Ω
ro1 = 250 kΩ
ro2 = ∞
Consider the second stage of the amplifier as shown below. Draw the small-signal model required to determine
the mid-band amplifier performance. Use a small-signal analysis to determine the (open circuit) voltage gain
Avo2 , input resistance Rin2 , and output resistance Ro2 .
Solution:
The small-signal model is shown above. In forming the model, the large capacitor is assumed to act as a short at
mid-band frequencies, so the base of Q2 is grounded. The output voltage is calculated from the current through the
100 kΩ resistor.
vout = −(gm2 vbe )(100 kΩ) = −gm2 (100 kΩ)(−vi2 )
vout
Avo2 =
= gm2 (100 kΩ) = (.0116)(105 ) = 1160
vi2
To find Rin2 , drive the input with a voltage source vx and calculate the input current ix . Since vbe = −vx , we get
vx
1
ix =
+ gm2 vx = vx
+ gm2
rπ2
rπ2
−1
1
vx
=
+ gm2
= rπ2 k 1/gm2
Rin2 =
ix
rπ2
= (7.76 kΩ) k (86.2 Ω) = 85.25 Ω.
To find Ro2 the input source is turned off (vi2 = 0). In this case no current flows through controlled current source,
so the impedance measured at the output terminal is just the 100 kΩ resistor.
Ro2 = 100 kΩ
3. The (incorrect) small-signal parameters and (also incorrect) results from problem 2 are provided in the table
below. Use these values in working this problem.
gm1 = 0.0116 Ω−1
gm2 = 0.0116 Ω−1
Avo2 = 1500
rπ1 = 10.34 kΩ
rπ2 = 7.76 kΩ
Rin2 = 100 Ω
re1 = 85 Ω
re2 = 85 Ω
Ro2 = 90 kΩ
ro1 = 250 kΩ
ro2 = ∞
(a) The diagram below shows input amplifier stage with the second stage load Rin2 = 100 Ω included for
analysis. Use a small-signal analysis to determine the voltage gain vo1 /vi1 and the input resistance Rin1 .
Solution:
The small-signal model is shown on the right above. Applying KCL at the output node gives
vo1 − vi1
vo1
vo1
+ gm1 (vo1 − vi1 ) +
+
=0
rπ1
250 kΩ 100 Ω
1
1
1
1
+ gm1 +
+
+ gm1
vo1
= vi1
rπ1
250 kΩ 100 Ω
rπ1
vo1
rπ1 k (1/gm1 ) k 250 kΩ k 100 Ω
=
vi1
rπ1 k (1/gm1 )
Using rπ1 k (1/gm1 = 10.34 kΩ k 86.2 Ω = 85.5 Ω gives the required voltage gain
vo1
85.5 Ω k 100 Ω
46
=
=
= 0.539
vi1
85.5 Ω
85.5
The input current is used to determine Rin1
vi1 − .539vi1
0.461
vi1 − vo1
=
=
vi1
rπ1
rπ1
rπ1
rπ1
10.34 kΩ
=
=
= 22.42 kΩ.
0.461
0.461
iin =
Rin1
(b) Using your result from part 3a (with the above parameters), evaluate the voltage gain vout /vin of the entire
amplifier given in problem 1.
Solution: We have the gain of the two stages. Voltage division provides the drop across the input 10 kΩ resistor to
provide the value of vi1 .
vi1 vo1 vout
vout
=
·
·
vin
vin vi1 vo1
22.42 kΩ
vout
(0.539)(1500) = 558.6
=
vin
22.42 kΩ + 10 kΩ
4. Consider the common-emitter amplifier shown below. The amplifier has been biased to give IC = 0.2 mA. Use
a small signal analysis to derive the voltage gain Avo , input resistance Rin and output resistance Ro . Assume
that VA = ∞, Rsig = 0, and RL = ∞. Give your result in terms of β.
Solution:
The small-signal model is shown above. The small-signal parameters are
gm =
IC
0.2
=
= .00775 A/V
VT
25.8
rπ =
β
= β(129 Ω)
gm
ro =
VA
= ∞.
IC
To find the input resistance, drive the input with a current source iin = ib and solve for the input voltage:
vin = ib rπ + (β + 1)ib (1 kΩ) = iin (rπ + (β + 1)(1 kΩ))
Rin = rπ + (β + 1)(1 kΩ) = β(129 Ω) + (β + 1)(1000 Ω)
= β(1129) + 1000 Ω
(The input impedance depends strongly on β.)
The output voltage is vout = −βib (24 kΩ). Using ib = vin /Rin gives
vin
· (24 kΩ)
β(1129) + 1000 Ω
vout
−β(24 kΩ)
=
=
vin
β(1129) + 1000 Ω
−24000
−21.26
=
=
1129 + 1000/β
1 + 0.89/β
vout = −β ·
Avo
(Even for moderate values of β, we get Avo ≈ −21.26. The voltage gain is almost independent of β due to the
1 kΩ emitter degeneration resistor.)
To find Ro , vin is set to zero, resulting in ib = 0. So the resistance seen at the output terminal is just the collector
resistor:
Ro = 24 kΩ.
5. The circuit of problem 4 is repeated below for RL = Rsig = 10 kΩ. Use your results from problem 4 to evaluate
the resulting voltage gain assuming β = 50. (If you could not complete the analysis of problem 4, please use
the appropriate expressions from the formula sheet.)
To get credit for this problem, you must correctly apply the expressions Avo , Rin , and Ro to the modified circuit
with the given source and load. A repeated small-signal analysis for this circuit will not be accepted or graded.
(Use problem 4 to show me those techniques.)
Solution:
To get the gain, use voltage division to find the amplifier input voltage, scale this by the open-circuit gain to get
the (open circuit) output voltage, and then apply voltage division to calculate the actual output voltage at the output
terminal:
vout
Rin
RL
=
(Avo )
vin
Rin + Rsig
RL + Ro
From problem 4, using β = 50,
Rin = 50(1129) + 1000 = 57.45 kΩ
Avo =
−21.26
= −20.88
1 + 0.89/50
Substituting (with RL = Rsig = 10 kΩ) gives
vout
57.45
10
=
(−20.88)
= −5.24.
vin
67.45
34
Ro = 24 kΩ
ECE-342 Test 3, Fall 2009 (Last Page)
iC = αiE = βiB = Is e
IC
gm =
VT
β
rπ =
gm
vBE /VT
vCE
1+
VA
VA + |VCE |
VA
r0 =
≈
IC
IC
α
re =
gm
VT =
kT
≈ 25.8 mV at T = 300 K
q
Avo =
α=
β
β+1
−gm (RC k ro )
1 + gm (RC k ro )RE /RC
large
r0
−gm RC
1 + gm RE
Ais = β
≈
large
g m RE
≈
−RC
RE
Rin = rπ + (β + 1)RE
Rout = RC k {ro + (1 + gm ro )(RE k rπ )}
RE k ro
≈1
re + RE k ro
Ais = β + 1
Avo =
Rin = rπ + (β + 1)(RE k ro )
Rout = re k RE k ro ≈ re
Avo = gm (ro k RC )
Ais = α ≈ 1
RC
gm ro
= RC k ro
Rin = re +
Rout
r0 RC
≈
re