electric circuit - 2

[logistic_guy]

Determine the gain \(\displaystyle v_o/v_s\) of the transistor amplifier circuit shown.

 

[logistic_guy]

I will apply \(\displaystyle \text{KVL}\) to the first loop \(\displaystyle \rightarrow\) Let us call it \(\displaystyle I_1\).

\(\displaystyle v_s - 200I_1 - 100(I_1 - I_o) = 0\)

\(\displaystyle v_s - 200I_1 - 100I_1 + 100I_o = 0\)

\(\displaystyle v_s - 300I_1 + 100I_o = 0\)

\(\displaystyle I_1 = \frac{v_s + 100I_o}{300}\)

 

[logistic_guy]

Let us start the second loop.

\(\displaystyle -\frac{v_o}{1000} - 100(I_o - I_1) - 2000I_o = 0\)


\(\displaystyle -\frac{v_o}{1000} - 100I_o + 100I_1 - 2000I_o = 0\)


\(\displaystyle -\frac{v_o}{1000} + 100I_1 - 2100I_o = 0\)

 

[logistic_guy]

The third loop gives us:

\(\displaystyle v_o = 10000(-40I_o)\)

Now I will substitute equation \(\displaystyle 1\) in equation \(\displaystyle 2\).

\(\displaystyle -\frac{v_o}{1000} + 100I_1 - 2100I_o = 0\)


\(\displaystyle -\frac{v_o}{1000} + 100\bigg[\frac{v_s + 100I_o}{300}\bigg] - 2100I_o = 0\)


\(\displaystyle -\frac{3v_o}{1000} + v_s + 100I_o - 6300I_o = 0\)


\(\displaystyle -\frac{3v_o}{1000} + v_s - 6200I_o = 0\)


\(\displaystyle -3v_o + 1000v_s - 6200000I_o = 0\)

I will substitute equation \(\displaystyle 3\) in this equation above.


\(\displaystyle -3v_o + 1000v_s - 6200000\bigg[-\frac{v_o}{400000}\bigg] = 0\)


\(\displaystyle -3v_o + 1000v_s - 62\bigg[-\frac{v_o}{4}\bigg] = 0\)


\(\displaystyle -3v_o + 1000v_s - 31\bigg[-\frac{v_o}{2}\bigg] = 0\)


\(\displaystyle -6v_o + 2000v_s + 31v_o = 0\)


\(\displaystyle 2000v_s + 25v_o = 0\)


\(\displaystyle 25v_o = -2000v_s\)

Determine the gain \(\displaystyle v_o/v_s\)

\(\displaystyle \frac{v_o}{v_s} = -\frac{2000}{25} = \textcolor{blue}{-80}\)