Question

A voltage amplifier with an input resistance of 40k ohms, an output resistance of 100 ohms, and a gain of 300 V/V is connected between a signal source and a 100 ohm load. The signal source has an open-circuit voltage of 10mV and a 10k ohm source resistance.
a) What output voltage results?

b) What is the voltage gain from source to load?

c) What is the power gain from source to load?

d) If the output voltage across the load is twice that needed and there are signs of amplifier saturation, suggest the location and value of a single resistor that would produce the desired output. Choose an arrangement that would cause minimum disruption to an operating circuit (for example, if a circuit is operating and you place an additional component in series with part of the circuit it would be a major disruption and not good)

Answer 1

Answer:

89.45 v/v

Explanation:

Let's take the data:

First draw the amplifier circuit.

After the circuit, the voltage division rule can be used to compute the parameters:

The input section is computed like this: $\frac{R_{in} }{(R_{sig} + R_{in} } (v_{sig})$

The output section is computed like this $R_{L}/ (R_out} + R_{in} )$

The product A$V_{in} V_{out}$ gives

A$V_{in} V_{out}$  = A×$\frac{R_{in} }{(R_{sig} + R_{in} } (v_{sig})$×$R_{L}/ (R_out} + R_{in} )$

Computing gives output voltage = 89.45 v/v