A buffer amplifier with a gain of 1 V/V has an input resistance of 1 M and an output resistance of 20. It is connected between a 1-V, 200-k source and a 100-Ω load. What load voltage results? What are the corresponding voltage, current, and power gains (in dB)?

The Load Voltage is 0.7 v

the corresponding voltage is - 3.098 dB

the corresponding current gain is 78.48 dB

the corresponding POWER gain is 37.7 bB

Explanation:

Given the diagram;

in the circuit of 1 MΩ and 200 KΩ are in  series

also 20  Ω and 100 Ω are in series;

so

V₀ = [ (1V×1MΩ)/(1MΩ+200KΩ)] × [ (1V×100Ω)/(100Ω+20Ω)]

V₀ = 0.7 v

The Load Voltage is 0.7 v

Now considering the formula to find Voltage Gain

A_v = V₀ / V_i

we substitute

A_v = 0.7 / 1

A_v = 0.7 V/V

to convert to dB

A_v (dB) = 20logA_v

= 20log0.7

= - 3.098 dB

the corresponding voltage is - 3.098 dB

To determine the current gain

A_i = V₀/100Ω × 1.2MΩ/V_i

= 0.7/100Ω × 1.2MΩ/1

= 8400 A/A

to convert to dB

A_I (dB) = 20logA_I

= 20log8400

= 78.48 dB

the corresponding current gain is 78.48 dB

To determine the power gain

P_G = V₀²/100Ω × 1.2MΩ/V_i²

= 0.49/100Ω × 1.2MΩ/1

= 5880 W/W

to convert to dB

P_G (dB) = 10logP_G

= 10log5880

= 37.7 bB

the corresponding POWER gain is 37.7 bB

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explanation:

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explanation: