Answer:
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
