0195136047.pdf

5.1 THE AMPLIFIER BLOCK 225

Solution

Using the model of Figure 5.1.1(b), we have the circuit configuration shown in Figure E5.1.1(b).

SinceRiandRSare connected in parallel,

vin=IS

RSRi

RS+Ri

Using the voltage-divider formula, one has

vout=Avin

RL

Ro+RL

=

ARLRSRiIS

(Ro+RL)(RS+Ri)

Recall thatAis the open-circuit voltage amplification. Let us consider the circuit shown
in Figure 5.1.2 in order to explainvoltage amplification,orvoltage gain. In this circuit a signal
voltagevSis applied to the input of the amplifier block, whereas the output terminals are connected
to a load resistanceRL. Let us evaluate the ratio of the voltage across the load to the signal voltage
vL/vS, which is known as voltage gainGV,

GV=

vL

vS

=

AvinRL

(Ro+RL)vS

=

ARL

(Ro+RL)

(5.1.1)

If there is no load, i.e.,RL=∞, then it is easy to see that the voltage gainGVwill be equal toA;
hence the justification to callAthe open-circuit voltage amplification. The reduction in voltage
gain due to theeffect of output loadingcan be seen from Equation (5.1.1).

+

−

+

−

iin iout

vin vout

Input Output

(a)

+

−

+

−

iin

Ri

Ro

Avin

iout

vin vout

Input Output

(b)

+

−

Figure 5.1.1Amplifier block.(a)Two-

port device.(b)Circuit model.

Ri

Ro

Avin

vin vout = vL

+

−

RL

iL

iS

+

−

+

−

+

−

vS

Figure 5.1.2Circuit to explain voltage amplification or voltage gain.