A voltage amplifier has a gain of 200, an input
resistance Ri of 50 kΩ and an output resistance Ro of
1 kΩ. It is connected as shown below to a signal source
with an output resistance Rs of 25 kΩ, and to a load
with a resistance RL of 500Ω.
Rs
Vout
25 kΩ
Vs
RL
500Ω
Using the amplifier model which includes the effect of
its input and output resistance, calculate the gain of the
system (that is, Vout / Vs). (Note that this is similar to
problem 6.3 in your notes.)
ELEC166 Worked Examples Week 7
1
• With the amplifier model included, the circuit looks
like this. Notice that there are two voltage sources
and two potential dividers.
Rs
Vi
AVi
25 kΩ
Vs
B
Ro
Vout
1 kΩ
Ri
50 kΩ
RL
500Ω
amplifier
• The source voltage Vs is the input for the potential
divider consisting of Rs and Ri. So the voltage Vi at
the amplifier input is given by
Vi = Vs ×
Ri
50kΩ
= Vs ×
= 0.6667 Vs
Rs + Ri
25kΩ + 50kΩ
• The voltage at point B is equal to the amplifier input
voltage times its gain, or AVi. This is the input voltage
for the potential divider formed by Ro and RL.
So the output voltage Vout is given by
RL
Vout = AVi ×
Ro + RL
500Ω
= 200 × 0.6667 Vs ×
= 44.4 Vs
1kΩ + 500Ω
Hence the overall gain is 44.4.
ELEC166 Worked Examples Week 7
2
(QB) An amplifier is connected with a feedback
loop as shown below.
amplifier open
loop gain = 6000
Vin
+
Vout
R1
30 kΩ
R2
1 kΩ
A resistance of R1 = 30 kΩ is connected in the
feedback loop, with a resistance R2 = 1kΩ
between the input and ground. The open loop
gain of the amplifier is Aopenloop = 6000, and its
open loop output resistance is 2Ω.
a) Calculate the feedback factor β.
b) Calculate the closed loop gain.
c) Calculate the closed loop output resistance.
ELEC166 Worked Examples Week 7
3
• (a) The feedback factor β is simply the fraction of the
output voltage which is “fed back” in the feedback
loop; that is, the voltage at the (–) terminal of the
amplifier divided by Vout. Since R1 and R2 form a
potential divider, this fraction is simply
R2
1kΩ
β=
=
= 0.03226
R1 + R 2 30kΩ + 1kΩ
• (b) The closed loop gain is the overall gain with
feedback applied. Here we can simply plug in the
formula
Aclosed loop
Aopen loop
6000
=
=
= 30.84
1 + Aopen loop β 1 + 6000 × 0.03226
• (c) When connected with feedback, the output
resistance of the amplifier (originally 2Ω) is divided by
the factor (1 + Aβ). So the output resistance becomes
Rout =
Rout(no feedback)
2Ω
=
= 0.0103Ω
( 1 + Aβ )
1 + 6000 × 0.03226
ELEC166 Worked Examples Week 7
4