Test: Problems on AC & DC Analysis - Electrical Engineering (EE) MCQ

We know, I_C=(V_CC-V_BE)/R_B
By putting the values, we have I_C=5.9mA. I_E=I_C/α. So, I_E=5.99mA.
h_oe= V_CC-R_C(I_C+I_B). We have h_oe=6V.

We know, h_oe=12V
(I_C)_SAT =V_CC/R_L=12/6K=2mA. I_B=10V/0.5M=20µA. I_C= βI_B=1mA. I
h_oe=V_CC-I_CR_L=12-1*6=6V. So, quiescent point is (6V, 1mA).

We know, I_E=V_EE/RE=10/5kΩ=2mA
I_C=α I_E =I_E =2mA
V_CB=V_CC-I_CR_L=20-10=10V. So, quiescent point is (10V, 2mA).

We know, by KVL -12+(I_C+I_B)1K+6+V_BE=0
We have I_E=5.3. I_C= αI_E=5.24mA. From another loop, -12+I_EIK+V_BE=0
We have, h_oe=12-5.3m*1000=6.7V. Hence the Q point is (6.7V, 5.3mA).

The base current I_B is zero when β value is large. So, the h_oe changes to V_CC-R_CI_C. The collector current I_C is changed to (V_CC-V_BE)/R_B from β(V_CC-V_BE)/(1+ β)R_E+ R_B.

We know, I_E=V_EE/R_E=30/10kΩ=3mA
I_C=α I_E =I_E =3mA
V_CB=V_CC-I_CRL=25-15=10V. So, quiescent point is (10V, 3mA).

Sometimes DC current gain of a bipolar transistor is too low to directly switch the load current or voltage, so multiple switching transistors is used. The load is connected to ground and the transistor switches the power to it.

In low frequency region, the small capacitors are open circuited and large capacitors are in active state. In high frequency region, the large capacitors are short circuited and small capacitors are active.

In the mid frequency region, the whole circuit is resistive because the large capacitors are short and small capacitors are open circuited. The gain is constant in this region. So, the circuit is frequency blind as the gain is constant in this region.

At very low frequency, the gain decreases due to the coupling capacitor. When the frequency is decreased, the reactance of the circuit increases and the drop across the coupling capacitor increases. The gain is therefore decreased as the output decreased.