The Common Emitter Configuration - Free MCQ Practice Test with solutions,

The current amplification factor (β) is given by I_C//I_B. When no signal is applied, then the ratio of collector current to the base current is called current amplification factor of a transistor.

 I_C=α I_E +I_CBO =0.995*10mA+0.5µA=9.9505mA.
I_B=I_E-I_C=10-9.9505=0.0495mA. β=α/(1-α)=0.995/(1-0.995)=199
I_CEO=9.9505-199*0.0495=0.1mA==100µA.

Given, I_CBO=10µA, α=0.98 and I_B =0.22µA. I_C=α/ (1-α) I_B+ 1/(1-α) I_CBO
0.01078+0.5=0.51078mA.

 I_C=V across R_L/R_L=5V/5KΩ=1mA.
I_B=I_C/β=1/50=0.02mA.

Here, I_C=0.8/800=1mA
β= α/ (1-α)=0.96/1-0.96=24.
Now, I_B=I_C/ β=1/24=41.67µA.

Here, I_C=0.8/800=1mA.
We know, V_CE=V_CC-I_CR_L
=10-0.8=9.2V.

The subscript ‘CEO’ means that it is collector to emitter base open. It is called as the leakage current. It occurs in a reverse bias in PNP transistor. The total current can be calculated by I_C=βI_B+I_C.

The ac current gain is given by β=∆I_C/∆I_B. When the signal is applied, the ratio of change of collector current to the ratio of change of base current is called ac current gain.

Almost in all the transistors, the base current is less than 5% of the emitter current. Due to this fact, it is generally greater than 20. Usually it ranges from 20 to 500. Hence this configuration is frequently used when appreciable current gain as well as voltage gain is required.