Test: Biasing Parameters - Electrical Engineering (EE) MCQ

We know, current gain A_V=h_fe. In π model, h_fe is referred to β.
We know, r_i= β/g_m.
From this, β=r_ig_m.

The circuit is PNP transistor, collector coupled amplifier. The voltage gain is unity for a CC amplifier. Hence on observation, the CC amplifier gives a unity gain.

The change in voltage with temperature can be found by, V(T) = 2.3m(∆T)V_O . In a bipolar transistor at room temperature if the emitter current is doubled the voltage across its base emitter junction thereby doubles.

We know, β/g_m=r_i
= (I_C/I_B)/(I_C/V_T)=V_T/I_B=25m/25µ=1k.

When the collector and base are shorted, the transistor behaves as a normal diode. So, the diode equations imply. I_E=I_O(e^V/V0-1). We get, I_E=49mA.

 The gain for the given circuit can be found by, A_V=R_F/R_S
=100K/10K=10.

A_V=-h_fe R_L^I/h_ie=3*150/3=-150. So, V_O=-150V(t)
But cos20t has low frequency so capacitors are open circuited. Only, the sine component is allowed.
So, V_o =-150Bsin10000t.

The CE amplifier has moderate input and output impedances. The CC amplifier has unity voltage gain. The common ba se amplifier has a unity current gain and high voltage gain.

(I_C – I_CBO)/α=I_E
= (2.945-0.002)/0.98=3mA.
I_E=I_C+I_B . So, I_B=3-2.495=0.055mA=55µA.

 r_o=∆V_CE/∆I_C
=3/0.3m=10kΩ.