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CHAP. 5]

AMPLIFIERS AND OPERATIONAL AMPLIFIER CIRCUITS

65

EXAMPLE 5.1 A practical voltage source vs with an internal resistance Rs is connected to the input of a voltage amplifier with input resistance Ri as in Fig. 5-2. Find v2 =vs .

Fig. 5-2 The amplifier’s input voltage, v1 , is obtained by dividing vs between Ri and Rs . v1 ¼

Ri v Ri þ Rs s

The output voltage v2 is v2 ¼ kv1 ¼

kRi v Ri þ Rs s

from which v2 Ri ¼ k vs Ri þ Rs The amplifier loads the voltage source.

ð1Þ

The open-loop gain is reduced by the factor Ri =ðRi þ Rs Þ.

EXAMPLE 5.2 In Fig. 5-3 a practical voltage source vs with internal resistance Rs feeds a load Rl through an amplifier with input and output resistances Ri and Ro , respectively. Find v2 =vs .

Fig. 5-3 By voltage division, v1 ¼

Ri v Ri þ Rs s

Similarly, the output voltage is v2 ¼ kv1

Rl Ri Rl ¼k v Rl þ Ro ðRi þ Rs ÞðRl þ Ro Þ s

or

V2 Ri Rl ¼  k vs Ri þ Rs Rl þ Ro

ð2Þ

Note that the open-loop gain is further reduced by an additional factor of Rl =ðRl þ Ro 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 R2 . The feedback ratio R1 =ðR1 þ R2 Þ affects the overall gain and makes the amplifier less sensitive to variations in k.

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