Question

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

(a) vI= 100 mV, iI = 100 μA, vO = 10 V, RL = 100Ω
(b) vI = 10 μV, iI = 100 nA, vO =1 V, RL= 10 kΩ
(c) vI =1 V, iI = 1 mA, vO =5 V, RL = 10Ω

Answer 1

The solution to this question can be defined as follows:

Step-by-step explanation:

In point (a):

`v_i= 100 \ mV\\\\ i_I = 100 \mu \ A\\\\ v_O = 10 \ V\\\\ R_L = 100 \ \Omega \\\\i_L = (V_0)/(R_L) = (10)/(100) = 100 \ MA \\\\A_v = (V_0)/(V_i) = (10)/(100 * 10^(-3)) =100 \\\\A_v(db) = 20 \lag (100) =40 \ db \\\\ A_i= (i_L)/(i_i) = (100 * 10^(-3))/(100 * 10^(-6)) =1000 \\\\A_i(db) = 20 \lag (100) =60 \ db \\\\`

`A_p= (P_0)/(p_i) =(v_0 i_L)/(v_i i_i) = ( 10(100 * 10^(-3)))/(100 * 10^(-6) * 100 * 10^(-3)) =100000\\\\ A_p(db) =10 \log (100000) =50 \ db \\\\`

In point (b):

`v_i = 10 \mu V\\\\ i_i = 100 \ nA \\\\ v_O =1 \ V \\\\ R_L= 10 \ k \Omega \\\\i_0 = (V_0)/(R_L) = (1)/(10 \ K) = 100 \ \muA \\\\A_v = (V_0)/(V_i) = (10)/(10 * 10^(-6)) =100000 \\\\A_v(db) = 20 \lag (100000) =100 \ db \\\\ A_i= (i_0)/(i_i) = (100 * 10^(-6))/(100 * 10^(-9)) =1000 \\\\A_i(db) = 20 \lag (1000) =60 \ db \\\\`

`A_p= (P_0)/(p_i) =(v_0 i_0)/(v_i i_i) = ( 1 * 100 * 10^(-6)))/(10 * 10^(-6) * 100 * 10^(-9)) =100000000\\\\A_p(db) =10 \log (100000000) =80 \ db \\\\`

In point (C):

`v_i =1\ V\\\\ i_I = 1 \ mA\\\\ v_O =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 =13.97 \ db = 14 \db \\\\ A_i= (i_0)/(i_i) = (0.5)/(1* 10^(-3)) =500 \\\\A_i(db) = 20 \log (500) =53.97 \ db = 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) = 33.97 \ db = 34 \db\\\\`