Consider the amplifier circuit shown below. The parameters are g
Consider the amplifier circuit shown below. The parameters are gm = 2 ms, β = 100, r0 = 250 kΩ. Figure shows the Thevenin equivalent faced by load resistance RL.
Thevenin voltage Vth will be approximately:
A. 10 V<sub>in</sub>
B. 90 V<sub>in</sub>
C. V­<sub>in</sub>
D. None of the above
Please scroll down to see the correct answer and solution guide.
Right Answer is: C
SOLUTION
The small-signal equivalent circuit is drawn as:
For calculating Vth, we open the load resistance RL as shown below:
From the circuit, we can write:
\({V_{th}} = (\beta +1)I_b\times r_0\) ---(1)
Applying KVL from the input side to the output side, we get:
\(V_i-270kI_b-I_br_\pi-(\beta +1)I_br_0=0\)
\(I_b=\frac{V_i}{270k+r_\pi +(\beta +1)r_0}\)
Substituting the value of Ib in equation (1), we get:
\({V_{th}} = \frac{{\left( {\beta + 1} \right){r_0}}}{{270k + {r_\pi } + \left( {\beta + 1} \right){r_0}}} \times {V_i}\)
\({V_{th}} \approx \frac{{\left( {\beta + 1} \right){r_0}}}{{\left( {\beta + 1} \right){r_0}}} \times {V_i}\)
Vth ≈ Vi
