Question

A given amplifier is capable of presenting three different input resistance of 200Ω , 50KΩ and 1MΩ. If the amplifier's power gain is 1000 and it's voltage gain is 100 , determine which input to use to provide maximum power output and which to use to provide maximum output voltage for a low impedance microphone of 500Ω and 20mV open circuit voltage determine the power and voltage outputs

Answer 1

Since the open circuit voltage is the total voltage, the total current will be equal to $I_{total}=\frac{U_0}{R_1+R_2}$ , where $U_0$ is the open circuit voltage, $R_1$ is the microphone impedance, and $R_2$ is the resistance of the input load, therefore the input load voltage according to the Ohm law will be $U_2=\frac{U_0}{R_1+R_2}\times R_2$ , and the input power:

$P_1 = (\frac{U_0} {R_1+R_2}\times R_2) ^2\times \frac {1}{R_2}$ $= \frac{U_0^2\times R_2} {R_1^2+2\times R_1\times R_2+R_2^2}$ , an output power is equal to $P_2=P_1\times 1000$ $=\frac{U_0^2 \times R_2}{R_1^2+2\times R_1\times R_2+R_2^2}\times 1000$ , $1000$ is a coefficient of increase in power, it is visible that function $P_2$ $(R_2)$ decreases on $R$ therefore its maximum value will be at $R_2=200 \Omega$

$P_2=\frac{0.02^2\times 200}{500^2+2\times 500\times 200+200^2}\times 1000=0.16$ mW.

The voltage is equal at the input $U_2=\frac{U_0}{R_1+R_2}\times R_2$ , therefore at the output $U_1=U_2\times 100 =\frac {U _ 0} {R _ 1 + R _ 2 }\times R_2\times 100$ , the function $U_1 (R_2)$ increases by $R$ , therefore at $R_2=10^6\Omega$ there will be the highest voltage.

$U_1=U_2\times 100=\frac{0.02}{500+10^6}\times 10^6\times 100=2$ V.