0195136047.pdf

228 ANALOG BUILDING BLOCKS AND OPERATIONAL AMPLIFIERS

R′i=R 1 +Ri

The output resistance is the same as the Thévenin resistance seen from terminalsCandD. Turning

off independent sourcevS,vinbecomes zero, and as a consequence the dependent source in the

amplifier block goes to zero. Looking to the left of terminalsCandD, R 2 andRocan be seen to

be in parallel,

R′o=

R 2 Ro

R 2 +Ro

The open-circuit voltage gain of the larger circuit is

A′=

vCD

vAB

With

vCD=

R 2

Ro+R 2

Avin and vin=

Ri

Ri+R 1

vAB

we have

A′=

vCD

vAB

=

AR 2 Ri

(Ro+R 2 )(Ri+R 1 )

In further calculations, the dashed box can simply be replaced by the model with parameters

R′i,R′o, andA′. The reader should note that in generalR′imay depend onRL, andRo′may depend

onRS, even though in this simple example they do not.

Two of the most important considerations that influence amplifier design arepower-handling

capacityandfrequency response. In practice, the performance of an amplifier is limited by its

ability to dissipate heat, known as itspower dissipation. The electric power that is converted to

heat in an amplifier can be calculated when the currents and voltages are known at all its terminals,

including the power-supply terminals.

EXAMPLE 5.1.4

For the amplifier circuit shown in Figure E5.1.4 withRi∼=∞,Ro ∼= 0 ,A= 10 ,RL=

100 ,andvin =1 V, calculate the power dissipated in the amplifier if the voltage at the

power-supply terminalVpsis given to be 20 V andIpsis assumed to be equal toIL.

vin vout

+

−

RL

v = 0

(1)

(2)

(3)

(4)

(5)

(6)

v = vps

+ IL

−

Ips

Ips

Figure E5.1.4