Question 11.1: Calculating Amplifier Performance A source with an internal ...

First, we draw the circuit containing the source, amplifier, and load as shown in Figure 11.4.We can apply the voltage-divider principle to the input circuit to write

V_{i}=\frac{R_i}{R_i+R_s}V_s=0.667\mathrm{~mV~rms}

The voltage produced by the voltage-controlled source is given by

A_{voc}V_i=10^4V_i=6.67\mathrm{~V~rms}

Next, the output voltage can be found by using the voltage-divider principle, resulting in

V_o=A_{voc}V_i\frac{R_L}{R_L+R_o}=5.33\mathrm{~V~rms}

Now, we can find the required voltage gains:

A_v=\frac{V_o}{V_i}=A_{voc}\frac{R_L}{R_o+R_L}=8000

and

A_{vs}=\frac{V_o}{V_s}=A_{voc}\frac{R_i}{R_i+R_s}\frac{R_L}{R_o+R_L}=5333

Using Equations 11.3 and 11.5, we find that the current gain and power gain are

A_i=\frac{i_o}{i_i}=\frac{v_o/R_L}{v_iR_i}=A_v\frac{R_i}{R_L}           (11.3)

G=\frac{P_o}{P_i}=\frac{V_oI_o}{V_iI_i}=A_vA_i=(A_v)^2\frac{R_i}{R_L}      (11.5)

A_i=A_v\frac{R_i}{R_L}=2\times 10^9

G=A_vA_i=16\times 10^{12}

Notice that the current gain is very large, because the high input resistance allows only a small amount of input current to flow, whereas the relatively small load resistance allows the output current to be relatively large.