Test: Effect of Various Capacitors on Frequency Response - Electronics and Communication Engineering (ECE) MCQ

There are two capacitors that arise between bases and emitter. One is C_je due to depletion region associated between base and emitter. C_b is another capacitor which arises due to the accumulation of electrons in the base which further results into the concentration gradient within the base of the transistor.

Only one capacitor up between the base and the collector. This is due to the depletion region present between the base and the collector region.

While observing the frequency response of a C.E. stage, we find that all the parasitic capacitances of the B.J.T. end up slowing the speed of the B.J.T. The frequency response of this stage is affected by all the parasitic capacitors.

In the follower stage, the load is present at the emitter. The parasitic capacitors present between the collector and the substrate i.e. C_µ gets deactivated. This is observed from the small signal analysis where both the terminals of this capacitor get shorted to A.C. ground.

The transit frequency is almost inversely proportional to the total capacitance between the base and the emitter of the B.J.T. Hence, the transit frequency will approximately reduce by 2 and the correct option becomes reduces by 2.

The miller effect results in a change in the capacitance seen between the base and the collector. This is why it affects the frequency response of the B.J.T. deeply by changing the poles and affecting the high frequency voltage gain stage.

 If the early effect is considered, the low frequency response of the C.E. stage becomes g_m*(R_l || ro). Thereby, miller approximation shows that the capacitance between the base and the collector, looking from the output side, will be increased by a factor of 1 + 1/g_m*(R_l || ro). Hence the correct option is 1 + 1/ g_m*(R_l || ro).

We have a capacitor from the collector to substrate, C_cs, which comes in parallel to the miller approximation of the capacitance from base to collector. The miller approximation defines the latter as C_µ*(1 + 1/g_m*R₂). Since capacitors gets added, when in parallel, the correct option is C_cs + C_µ*(1+ 1/g_m*R₂).

All the parasitic capacitors of a B.J.T. affect the C.B. stage. None of the parasitic capacitors gets deactivated and they end up behaving as a pole during the frequency response of the C.B. stage.

The transit frequency is directly proportional to the transconductance of the B.J.T. Hence, the correct option is increases. Since it hasn’t been mentioned that whether the transconductance has been doubled or not, we cannot conclude the option “doubles” as an answer.