Two identical transistors are cascaded as follows with β = 100. The o...
Here the identical BJTs are connected.
First we need to analyse this circuit and find out the DC voltages at first and second stages.
First Stage,
DC Voltage at Base of stage 1 = 10*(R2/(R1 + R2)) = 1.67 V.
VE1 = 1.67 – 0.7 = 0.97 V
IE1 = 0.97 V/4.5 kΩ = 0.22 mA
IC1 = 0.22 mA
VC1 = 10 – 0.22×20 = 5.6 V
Second Stage,
Base Voltage = 5.6 V
VE2 = 5.6 – 0.7 = 4.9 V
IE2 = 4.9 V/10 kΩ = 0.49 mA
VC2 = 10 – 0.49*10 = 5.1 V
Overall Voltage gain is given by,
Total output impedance of stage 1 is R3‖Zin. Where Zin is input impedance of second stage.
Total output impedance is R3‖Zin. 4.85 kΩ
And Input impedance,
Then voltage gain of first stage is
Voltage gain of second stage, since there is no loading effect, Output Impedance
= R5 = 10kΩ
Input impedance was found earlier, re2 = 51.02Ω
Overall Voltage gain is Av = Av1 * Av2 = 8363