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Various amplifier and load combinations are measured as listed below using rms values. For each, find the voltage, current, and power gains (Av, Ai, and Ap, respectively) both as ratios and in dB:

a. Vi= 100mV; i=100μA ; vo =10V ; RL= 100Ω
b. Vi= 10μV; i=100nA ; vo =1V ; RL= 10kΩ
c. Vi= 1V; i=1mA ; vo =5V ; RL= 10Ω

1 Answer

3 votes

Solution:


$V_i = 100 \ mV, i_i = 100 \ \mu A, v_0 = 10 \ V, R_L = 100 \ \Omega$


$i_L= (v_0)/(R_L) =(10)/(100)$

= 10 mA


$A_V = (v_0)/(v_i) = (10)/(100 * 10^(-3))$

= 100


$A_V(dB) = 20 \log (100)$

= 40 dB


$A_i = (i_L)/(i_i) = (100 * 10^(-3))/(100 * 10^(-6))$

= 1000


$A_i(dB) = 20 \log (1000)$

= 60 dB


$A_p = (p_0)/(p_i) = (v_0 i_L)/(v_i i_i)= (10(100 * 10^(-3)))/(100 * 10^(-3) * 100 * 10^(-6))$

= 100000


$A_p(dB) = 10 \log (100000)$

= 50 dB

b).
$V_i = 100\ \mu V, i_i = 100 \ n A, v_0 = 1 \ V, R_L = 10 \ K \Omega$


$i_0 = (v_0)/(R_L)=(1)/(10 K) = 100 \ \mu A$


$A_v= (v_0)/(v_i) =(1)/(10 * 10^(-6)) = 100000$


$A_v(dB) = 20 \log (100000) = 100 \ dB $


$A_i= (i_0)/(i_i) =(100 * 10^(-6))/(100 * 10^(-9)) = 1000$


$A_i(dB) = 20 \log (1000) = 60 \ dB $


$A_p = (p_0)/(p_i) = (v_0 i_0)/(v_i i_i)= (1 * 1000 * 10^(-6))/(100 * 10^(-9) * 10 * 10^(-6)) $

= 100000000


$A_p(dB) = 10 \log (100000000)$

= 80 dB

c).
$V_i = 1 V, i_i = 1 \ m A, v_0 = 5 \ V, R_L = 10 \ \Omega$


$i_0 = (v_0)/(R_L)=(5)/(10 ) = 0.5 \ A$


$A_v= (v_0)/(v_i) =(5)/(1) = 5$


$A_v(dB) = 20 \log (5) = 14 \ dB $


$A_i= (i_0)/(i_i) =(0.5)/(1 * 10^(-3)) = 500$


$A_i(dB) = 20 \log (500) = 54 \ dB $


$A_p = (p_0)/(p_i) = (v_0 i_0)/(v_i i_i)= (5 * 0.5)/(1 * 1 * 10^(-3) ) $

= 2500


$A_p(dB) = 10 \log (2500)$

= 34 dB

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