Question

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

Answer 1

Solution:

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

$i_L= \frac{v_0}{R_L} =\frac{10}{100}$

  = 10 mA

$A_V = \frac{v_0}{v_i} = \frac{10}{100 \times 10^{-3}}$

     = 100

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

             = 40 dB

$A_i = \frac{i_L}{i_i} = \frac{100 \times 10^{-3}}{100 \times 10^{-6}}$

    = 1000

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

            = 60 dB

$A_p = \frac{p_0}{p_i} = \frac{v_0 i_L}{v_i i_i}= \frac{10(100 \times 10^{-3})}{100 \times 10^{-3} \times 100 \times 10^{-6}}$

     = 100000

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

            = 50 dB

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

     $i_0 = \frac{v_0}{R_L}=\frac{1}{10 K} = 100 \ \mu A$

   $A_v= \frac{v_0}{v_i} =\frac{1}{10 \times 10^{-6}} = 100000$

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

 $A_i= \frac{i_0}{i_i} =\frac{100 \times 10^{-6}}{100 \times 10^{-9}} = 1000$

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

$A_p = \frac{p_0}{p_i} = \frac{v_0 i_0}{v_i i_i}= \frac{1 \times 1000 \times 10^{-6}}{100 \times 10^{-9} \times 10 \times 10^{-6}}$

     = 100000000

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

            = 80 dB

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

     $i_0 = \frac{v_0}{R_L}=\frac{5}{10 } = 0.5 \ A$

   $A_v= \frac{v_0}{v_i} =\frac{5}{1} = 5$

  $A_v(dB) = 20 \log (5) = 14 \ dB$

 $A_i= \frac{i_0}{i_i} =\frac{0.5}{1 \times 10^{-3}} = 500$

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

$A_p = \frac{p_0}{p_i} = \frac{v_0 i_0}{v_i i_i}= \frac{5 \times 0.5}{1 \times 1 \times 10^{-3} }$

     = 2500

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

            = 34 dB