Question 10.1: Calculating Amplifier Performance A source with an internal ...

First, we draw the circuit containing the source, amplifier, and load as shown in Figure 10.4. 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 \text{ mV rms}

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

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

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

V_o = A_{\text{voc}}V_i\frac{R_L}{R_L+R_0} =5.33 \text{ 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 10.3 and 10.5, we find that the current gain and power gain are

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.