For the following voltage-sampling, current-mixing feedback amplifier,
Vcc
10K
100K
Vo
Rf
RS
1K
vS
the capacitor is a short circuit at the operating frequency, β = 100 and rπ = 2kΩ. Find the
amplifier’s voltage gain, Av = vo/vs , using
a) circuit analysis,
b) circuit analysis and Miller’s theorem,
c) the feedback method, and
d) the feedback method but adding the source and its Thevenin resistance after doing the analysis.
e) Show that the feedback amplifier models found in (c) and (d) are consistent.
ANSWER
a) the amplifier’s voltage gain, Av = vo/vs , using circuit analysis,
+
1K
rπ
vS
Vo
Rf
100K
RS
vπ
10K
gmvπ
-
)
vo − vπ
−
+ gm vπ 10kΩ
Rf
gm (10kΩ) − 10kΩ/Rf
⇒ vo = −
vπ = −454.45vπ
1 + 10kΩ/Rf
vπ
vπ − vo
+
rπ
Rf
(
)
(
)
1
1
1
vo
vo
1
1
1
vo
vπ
+
+
−
=−
+
+
−
rπ
Rs Rf
Rf
454.45 rπ
Rs Rf
Rf
(
)
1
1
)/454.45 +
−vo ( 12 + 11 + 100
100
(
vo =
vs − vπ
Rs
vs
Rs
vs
1
=
=
=
Av =
vo
= −75.06V /V
vs
b) the amplifier’s voltage gain, Av = vo/vs , using circuit analysis and Miller’s theorem,
Resistor Rf represents a conductance 1/Rf connected between base and collector. Thus in can
be replaced by conductances Gin = (1−AM )/Rf and Gout = (1− A1M )/Rf to obtain the following
circuit
vo
RS
1k
10k||Rout
Rin
vS
where
100
× 10kΩ = −500V /V
2kΩ
100kΩ
= 199.6Ω
=
501
= −gm × 10kΩ = −
AM
Rin =
Rout =
Rf
1 − AM
Rf
≃ 99.8kΩ
1 − 1/AM
from which we get
Av =
199.6Ω∥2kΩ
1kΩ + 199.6Ω∥2kΩ
(
)
100
−
(10kΩ∥99.8kΩ) = −69.8V /V
2kΩ
c) the amplifier’s voltage gain, Av = vo/vs , using the feedback method,
To solve the problem, we must first find the non-feedback amplifier that incorporates the
feedback’s network resistive loads. The feedback network is shown on the left of the following
figure:
Vo
i1
+
v1
-
100k
RS
i2
+
v2
-
1K
feedback network
From the table’s formulae,
R11
vS
non-feedback amplifier
v1
|v =0 = 100kΩ
i1 2
v2
=
|v =0 = 100kΩ
i2 1
R11 =
R22
and
β=
i1
1
|v1 =0 = −
v2
100kΩ
10K||R22
Now we can calculate the gain from the schematic shown on the right in the above figure.
Using the formulae included in the exam’s formula sheet,
Av = −gm Rc
Rin
100
100k || 2k
=−
(10k || 100k)
= −301
Rin + Rs
2kΩ
100k || 2k + 1k
Using this result, we can find the transresistance
RM =
vo
vo
=
= Rs Av = −301kΩ
is
vs /Rs
Finally, the amplifier’s gain according to the feedback method is
RM f =
RM
−301kΩ
=
= −75.06kΩ
1 + β × RM
4.01
To determine the voltage gain, observe that vs = is Rs and Av = RM f/Rs so
Av = −75.06V /V
d) the amplifier’s voltage gain, Av = vo/vs , using the feedback method but adding the source and
its Thevenin resistance after doing the analysis.
First build a model for the amplifier.
Vo
i1
100k
+
i2
+
v2
-
v1
-
feedback network
vo
is
RM f
Ri
Ro
10K||R22
R11
iS
non-feedback amplifier
R11
100
× hf e × (10kΩ∥R22 ) = −
(100)(10k∥100k) = −891.27kΩ
R11 + rπ
102
−891.27kΩ
=
= −89.9kΩ
1 + 891.27/100
100k∥2k
= 197.8Ω
= R11 ∥rπ ⇒ Rif =
9.9127
100k∥10k
= 10kΩ∥R22 ⇒ Rof =
= 917.1Ω
9.9127
= −
Now place the voltage source plus its Thevenin resistance.
Rs
vs +
−
iin
Rif
Rof
+
- R i
Mf in
feedback amplifier
+
vo
-
Av =
RM f
89.9
Rof = −
= −75.05V /V
Rs + Rif
1 + 0.1978
e) Show that the feedback amplifier models found in (c) and (d) are consistent.
′ represent the input resistance after removing R from the amplifier in part (c).
Let Rif
s
Ri = Rs ∥R11 ∥rπ ⇒ Rif =
′
Rif
=
1/165.15Ω
100k∥2k∥1k
= 165.15Ω
4.01
1
= 197.8Ω
− 1/1000Ω
Only a fraction of the source current enters the “new” amplifier (the one with Rs removed):
i′s =
1000
is
1197.8
For the “new” amplifier, the trans-resistance should be re-defined in terms of i′s :
′
RM
f =
vo
1197.8
=
RM f = −89.9kΩ
′
is
1000
Since these are the same values obtained in part (d), the two models are equivalent (one can
be obtained from the other) and the results are consistent.