This set of Microelectronics Multiple Choice Questions & Answers (MCQs) focuses on “Bipolar Amplifiers – Common Emitter Stage”.

1. In absence of early effect, what is the input resistance of the circuit?

a) r_{π}

b) r_{π}+(β+1)*R_{2}

c) (β+1)*R_{2}

d) Infinite

View Answer

Explanation: The above circuit is a degenerated common emitter stage and we invoke the small signal model to find the input resistance.

We use nodal analysis at the emitter node and then we find the emitter voltage. R

_{in}=V

_{in}-V

_{E}/(r

_{π}) and R

_{in}turns out to be r

_{π}+(β+1)*R

_{2}.

2. In absence of early effect, what is the output resistance of the circuit?

a) r_{o}+(r_{π}||R_{2})(1+g_{m}*r_{o})

b) r_{o}

c) r_{π}||R_{2}

d) 1+g_{m}*r_{o}

View Answer

Explanation: We have to invoke the small signal model

Firstly, we need to use KCL at E and find the emitter voltage. Then we use KVL from V

_{in}via r

_{o}and R

_{2}and we derive R

_{out}as r

_{o}+(r

_{π}||R

_{2})(1+g

_{m}*r

_{o}).

3. If the circuit has a very high current gain, find the gain of the following circuit.

a) -g_{m}*R_{C}

b) 0

c) Infinite

d) -R_{C}/(1/g_{m} + R_{E})

View Answer

Explanation: We will have to invoke the small signal model.

We use KCL at the emitter node and find the emitter voltage. Thereafter, we use KVL in the base-emitter loop and the gain is approximately -R

_{C}/(1/g

_{m}+ R

_{E}) since the gain is very high.

4. In absence of early effect, what is the gain of the following circuit?

a) -R_{C}/(1/g_{m} + R_{E})

b) R_{C}/(1/g_{m} + R_{E})

c) -R_{C}*g_{m}

d) R_{C}*g_{m}

View Answer

Explanation: In absence of early effect, the gain of the above circuit is similar to that of a degenerated common emitter stage which is -R

_{C}/(1/g

_{m}+ R

_{E}). The biasing method only places and impact on g

_{m}but the expression of voltage gain is the same.

5. Find the input resistance of the following circuit.

a) R_{2} || R_{3}

b) R_{2} || R_{3} || R_{π1}

c) R_{π1}

d) Infinite

View Answer

Explanation: The input resistance is literally independent of the cascaded structure. We can invoke the small signal model and the result will be the same as that of a simple C.E. stage. The input resistance, R

_{in}, is R

_{2}||R

_{3}||R

_{π1}which concerns the biasing resistors and the resistance in the base-emitter junction.

6. If all the transistors are identical and early effect is neglected, find the gain of the following circuit till Q_{1} . Q_{4} can be assumed to be properly biased.

a) 0

b) g_{m1}*(R || r_{π2})

c) g_{m1} * r_{π2}

d) Infinite

View Answer

Explanation: We draw the small signal model of Q

_{1}& observe that the gain is simply g

_{m1}* R

_{out}. We only need to find Rout which is (R || r

_{π2}). R

_{out}is derived by turning all independent voltage sources to 0. The small signal model of both the transistors reveals that the collector of Q

_{1}is connected to the base-emitter resistance and R. This makes the resistors connected between the same pair of terminals.

7. All the transistors have a high current gain and early effect is absent in each of them. What is the total gain till the collector of Q_{1}?

a) g_{m5}*(4/g_{m4}||r_{π1})*g_{m1}*(r_{π2} + (β+6)*1/g_{m3})

b) g_{m5}*(3/g_{m4}||r_{π1})*g_{m1}*(r_{π2} + (β+1)*1/g_{m3})

c) g_{m5}*(1/g_{m4}||r_{π1})*g_{m1}*(r_{π2} + (β+1)*1/g_{m3})

d) g_{m5}*(7/g_{m4}*r_{π1})*g_{m1}*(r_{π2} + (β+5)*1/g_{m3})

View Answer

Explanation: We approach this circuit by first finding the gain till the collector of Q

_{5}which is -g

_{m5}*(1/g

_{m4}||r

_{π1}). Observe that Q

_{1}is being used as a simple CE stage hence the gain for Q

_{1}is g

_{m1}times the total resistance connected to the collector. This resistance is the input resistance of Q

_{2}which is further degenerated by Q

_{3}. Q

_{3}offers a resistance of 1/g

_{m3}+r

_{π3}which is approximately 1/g

_{m3}and the overall input resistance at Q

_{2}is (r

_{π2}+ (β+1)*1/g

_{m3}). The overall gain is the product of both the gains, becoming a positive quantity. and it turns out to be g

_{m5}*(1/g

_{m4}||r

_{π1})*g

_{m1}*(r

_{π2}+ (β+1)*1/g

_{m3}).

8. If the transistors are identical with high current gain and only Q_{2} and Q_{3} has early effect, what is the output resistance of Q_{2} if a resistor of resistance R is kept in between it’s collector and the supply voltage?

a) R_{C} || r_{o2} + (r_{π2} || r_{o3} || 1/g_{m3})(1+g_{m2}*r_{o2})

b) R_{C} || r_{o2}

c) r_{π2} || r_{o3} || 1/g_{m3}

d) Infinite

View Answer

Explanation: The output resistance of Q

_{5}is given by the total resistance looking its collector in parallel with R

_{C}. We note that Q

_{2}has been degenerated by Q

_{3}, which offers a resistance of r

_{o3}|| 1/g

_{m3}to the emitter of Q

_{2}. Therefore, the resistance looking into the collector is r

_{o2}+ (r

_{π2}|| r

_{o3}|| 1/g

_{m3})(1+g

_{m2}*r

_{o2}) and the total resistance becomes R

_{C}|| r

_{o2}+ (r

_{π2}|| r

_{o3}|| 1/g

_{m3})(1+g

_{m2}*r

_{o2}).

9. Assume that all the transistors are identical with negligible early effect and very high current gain. What is the input resistance at the base of Q_{4}?

a) r_{π4}

b) 0

c) Infinite

d) r_{π4} + (β + 1) * r_{π1}

View Answer

Explanation: We note that Q

_{4}is degenerated by Q

_{5}and Q

_{1}. Q

_{5}offers infinite output impedance while Q

_{1}offers a resistance of r

_{π1}. Note that they are in parallel and the overall resistance becomes r

_{π1}. The overall resistance is r

_{π4}+ (β+1)* r

_{π1}due to degeneration.

10. Assume that all the transistors are identical with negligible early effect and very high current gain. In the following circuit, Q_{4} is behaving as a _____

a) Switch

b) Amplifier

c) Current Source

d) Cut-off

View Answer

Explanation: The input is applied to the base of Q

_{5}while Q

_{4}is simply biased to behave as a current source. It offers a resistance of r

_{o4}||1/g

_{m4}at its emitter which is connected to the collector of Q

_{5}. But it doesn’t behave as an amplifier since the input is connected to the base of Q

_{5}.

11. Assume that all the transistors are identical with negligible early effect and very high current gain. What is the voltage gain till the collector of Q_{2}?

a) g_{m1}*[r_{π2}+(β+1)*(1/g_{m3}||r_{o3})]*g_{m2}*[(r_{π2}||r_{o3}||1/g_{m3})(1+g_{m2}*r_{o2})]||[r_{π4}+(β+1)*r_{o5}]||R_{1}

b) Infinite

c) g_{m1}*[r_{π2}+(β+1)*(1/g_{m3}||r_{o3})]*g_{m2}*[(r_{π2}||r_{o3}||1/g_{m3})(1+g_{m2}*r_{o2})]

d) [r_{π4}+(β+1)*(r_{o5}||1/g_{m5})]||R_{1}

View Answer

Explanation: Firstly, we will calculate the voltage gain to Q

_{1}. This is given by g

_{m1}times the total resistance connected to the collector. This resistance is given by the internal resistance of the degenerated Q

_{2}which is r

_{π2}+(β+1)*(1/g

_{m3}||r

_{o3}). Finally, the gain for Q

_{2}is g

_{m2}times the resistance connected to the collector of Q

_{2}. Observe that the total resistance connected at the collector are R

_{1}, the resistance looking into the base of Q

_{4}which r

_{π4}+(β+1)*r

_{o5}and the output resistance of Q

_{2}which is r

_{o2}+(r

_{π2}||1/g

_{m3}||r

_{o3})(1+g

_{m3}*r

_{o2}). All the resistance come in parallel to each other and the total resistance is r

_{o2}+[(r

_{π2}||r

_{o3}||1/g

_{m3})(1+g

_{m2}*r

_{o2})]||[r

_{π4}+(β+1)*(r

_{o3}||1/g

_{m3})]||R

_{1}. The overall voltage gain is g

_{m1}*[r

_{π2}+(β+1)*(1/g

_{m3}||r

_{o3})]*g

_{m2}*[(r

_{π2}||r

_{o3}||1/g

_{m3})(1+g

_{m2}*r

_{o2})]||[r

_{π4}+(β+1)*r

_{o5}]||R

_{1}.

12. The supply voltage is 5V while the input voltage is 700mV. For a reverse saturation current of 10^{-14}A and R_{1}=1K, calculate the voltage gain without early effect.

a) 19

b) -188

c) 6.5

d) 0

View Answer

Explanation: The voltage gain is simply g

_{m}*R

_{1}and we can calculate gm by finding the collector current first. The collector current is I

_{S}*exp(\(\frac {V_{BE}}{V_T}\)) and that turns out to be 4.9mA. Dividing this by the thermal voltage gives us the transconductance as 188mS and the overall voltage gain is 188.

13. The supply voltage is 5V while the input voltage is 700mV. For a reverse saturation current of 10^{-16}A and R_{1}=1K, calculate the conductance offered due to early voltage of 5V.

a) 750 S

b) 800 S

c) 816 S

d) Infinite

View Answer

Explanation: Firstly, we need to find the collector current which comes out to be 4.9mA. We simply divide this by the early voltage to r

_{o}and its inverse turn out to be 816 S.

**Sanfoundry Global Education & Learning Series – Microelectronics**.

To practice all areas of Microelectronics, __ here is complete set of 1000+ Multiple Choice Questions and Answers__.

**If you find a mistake in question / option / answer, kindly take a screenshot and email to [email protected]**