0195136047.pdf

226 ANALOG BUILDING BLOCKS AND OPERATIONAL AMPLIFIERS

By defining thecurrent gain GIto be the ratio of the current throughRLto the current through

vS, one gets

GI=

iL

iS

=

Avin/(Ro+RL)

vin/Ri

=

ARi

Ro+RL

(5.1.2)

Thepower gain GP,defined by the ratio of the power delivered to the load to the power given

out by the signal source, is obtained as

GP=

vL^2 /RL

v^2 S/Ri

=

G^2 VRi

RL

=

A^2 RLRi

(Ro+RL)^2

=GVGI (5.1.3)

Note that, for fixed values ofRoandRi,GPis maximized whenRLis chosen equal toRo, and this

corresponds to maximum power transfer to the load. One should also note that the added power

emerging from the output comes from the power source that powers the amplifier, even though

the power-supply connections are usually not shown on the circuit diagram, and that the existence

of power gain does not violate the law of energy conservation.

EXAMPLE 5.1.2

The constants of an amplifier are given byA= 1 ,Ri=10,000, andRo= 100 . It is driven by

a Thévenin source withvTh(t)=VOcosωtandRTh=20,000. The amplifier output is connected

to a 100-load resistance. Find the power amplification, if it is defined as the ratio of the power

delivered to the load to the maximum power available from the Thévenin source.

Solution

The corresponding circuit diagram is shown in Figure E5.1.2. The instantaneous power delivered

toRLis given by

Ro = 100 Ω

RTh = 20,000 Ω

Avin = vin

vin RL^ =

100 Ω

Ri =

10,000 Ω

Signal source Load

Amplifier

vTh =

VO cos ωt

+

−

+

−

+

−

Figure E5.1.2Circuit diagram.

PL(t)=i^2 L(t)RL=

[

Avin(t)

Ro+RL

] 2

RL

The time-averaged powerPLis obtained by representingvin(t) by its phasorV ̄in,

PL=

1

2

∣

∣V ̄in

∣

∣^2 A

(^2) RL
(Ro+RL)^2
Note that the time average of a product of sinusoids is given by