Question

A voltage amplifier with an input resistance of 40 kΩ, an output resistance of 100 Ω, and a gain of 300 V/V is connected between a 10- kΩ source with an open-circuit voltage of 10 mV and a 100-Ω load. For this situation: (a)What output voltage results? (b)What is the voltage gain from source to load? (c)What is the voltage gain from the amplifier input to the 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. (Hint)

Answer 1

a) 2.40 V

b) 120

c) 150

d) 8 kΩ in parallel with the amplifier input

Step-by-step explanation:

The voltage divider between the source resistance and the amplifier input resistance makes the amplifier input voltage be ...

vi = (10 mV)(ri/(rs+ri)) = (10 mV)(40/(10+40)) = 8 mV

(a) The gain of 300 makes the open-circuit amplifier output voltage be ...

Vo = 300(8 mV) = 2400 mV = 2.40 V . . . . open-circuit output voltage

__

(b) The load voltage is the result of voltage-divider action between the output resistance and the load resistance.

Vl = Vo(rl)/(ro +rl) = 2.40 V(100)/(100 +100) = 1.20 V

So, the overall voltage gain from the source is ...

vl/vs = (1.20 V)/(0.010 V) = 120 . . . . voltage gain from source

__

(c) The voltage gain to the load from the amplifier input is ...

vl/vi = 1.200 V/0.008 V = 150 . . . . voltage gain from input to load

__

(d) To reduce the input of the amplifier without breaking the circuit, we could add a resistor in parallel with the amplifier input. It would need to reduce the gain of that voltage divider from 0.8 to 0.4 to cut the output voltage in half. That is, for some parallel resistor R (in kOhms), we want ...

R(Ri)/(R·(Ri +Rs) +Ri·Rs)) = 0.40

40R/(50R + 400) = 2/5

200R = 100R +800 . . . . cross multiply

100R = 800 . . . . . . . . . . . subtract 100R

R = 8 . . . . kOhms

Putting 8 kOhms in parallel with the amplifier input will reduce the overall gain by the required amount.