Page 76 - Schaum's Outline of Theory and Problems of Electric Circuits

P. 76

AMPLIFIERS AND OPERATIONAL AMPLIFIER CIRCUITS
65
CHAP. 5]
EXAMPLE 5.1 A practical voltage source v s with an internal resistance R s is connected to the input of a voltage
ampliﬁer with input resistance R i as in Fig. 5-2. Find v 2 =v s .

Fig. 5-2

The ampliﬁer’s input voltage, v 1 , is obtained by dividing v s between R i and R s .
R i
v 1 ¼ v s
R i þ R s
The output voltage v 2 is
kR i
v 2 ¼ kv 1 ¼ v s
R i þ R s
from which

v 2 R i
¼ k ð1Þ
v s R i þ R s
The ampliﬁer loads the voltage source. The open-loop gain is reduced by the factor R i =ðR i þ R s Þ.

EXAMPLE 5.2 In Fig. 5-3 a practical voltage source v s with internal resistance R s feeds a load R l through an
ampliﬁer with input and output resistances R i and R o , respectively. Find v 2 =v s .

Fig. 5-3
By voltage division,

R i
v 1 ¼ v s
R i þ R s
Similarly, the output voltage is
R l R i R l V 2 R i R l
v 2 ¼ kv 1 ¼ k v s or ¼ k ð2Þ
R l þ R o ðR i þ R s ÞðR l þ R o Þ v s R i þ R s R l þ R o
Note that the open-loop gain is further reduced by an additional factor of R l =ðR l þ R o Þ, which also makes the output
voltage dependent on the load.

5.2 FEEDBACK IN AMPLIFIER CIRCUITS
The gain of an ampliﬁer may be controlled by feeding back a portion of its output to its input as
done for the ideal ampliﬁer in Fig. 5-4 through the feedback resistor R . The feedback ratio
2
R 1 =ðR 1 þ R 2 Þ aﬀects the overall gain and makes the ampliﬁer less sensitive to variations in k.

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