# Microelectronics Questions and Answers – Bipolar Amplifiers – Common Base Stage

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

1. Assuming that the transistor has a very high current gain and negligible early effect, what is the input resistance?

a) 1/gm
b) 2/gm
c) 3/gm
d) 4/gm

Explanation: The input impedance would be the impedance looking into the emitter of the transistor. We draw the small signal model and simply find the Thevenin’s resistance at the emitter node.

A KCL reveals that v/r + g * v = I, I being the current entering the emitter. Thereafter the input resistance is approximated to 1/gm since gm*rπ=β.

2. Assuming that the transistor has a very high current gain but early effect is present, what is the input resistance?

a) 2/gm||rO
b) 3/gm||2*rO
c) 1/gm||*rO
d) rO

Explanation: Early effect leads to an added conductance between the collector and the emitter in the small signal model.

We use the Thevenin’s theorem at the emitter node and a KCL leads us to the input impedance as (1/(gm+1/rO)) which is 1/gm||*rO.

3. Assuming that the transistor has a very high current gain and early effect is absent, what would be the input resistance of the following circuit?

a) Re || 1/gm
b) Re + 1/gm
c) Re * 1/gm
d) Infinite

Explanation: At the input terminal, we observe that there are two branches spreading out of it containing two resistances. One would be the resistance looking into the input of the emitter which is 1/gm while the other is Re. The two resistance are in parallel since they are both connected between the input node and ground. Note that 1/gm is a representation of resistive behavior offered by the combination of all the elements of that branch while Re is a resistor in general. The representation of the equivalent resistance for two resistances in parallel is Re||1/gm.

4. Assuming that the transistor has a very high current gain and early effect is present, what would be the input resistance of the following circuit?

a) 1/gm || rO
b) Re || 1/gm || rO
c) Re + (1/gm || rO )
d) 0

Explanation: The input node is attached to two branches. One of them offers the resistance seen looking into the emitter of Q2 , undergoing early effect, which is 1/gm || rO and the other is simply Re. They are in parallel to each other and the total input impedance is Re || 1/gm || rO .

5. Assuming that the transistor has a very high current gain and early effect is present, what would be the output resistance of the following circuit?

a) rO+(1+gmrO)(RE||Rπ)||RC
b) RC
c) rO+(1+gmrO)(RE||Rπ)
d) rO+(1+gmrO)(RE||Rπ)+RC

Explanation: The output impedance of the above circuit is same as the output impedance of a common emitter stage with emitter degeneration which is rO+(1+gmrO)(RE||Rπ)||RC. Note that the output impedance is independent of which terminal the input is being given.

6. Assuming that the transistor has a very high current gain and negligible early effect, what is the voltage gain of the following C.B. stage?

a) gm*RC
b) -gm*RC
c) 2gm*RC
d) 0

Explanation: The small signal model of the transistor is used to calculate the gain of the above C.B. stage. In absence of early effect, the gain is simply gm*RC. Note that this stage doesn’t invert.

7. Assuming that the transistor has a very high current gain and negligible early effect, what is the input impedance of the following circuit?

a) Re + (1/gm || rO )
b) 1/gm || rO
c) 1/gm
d) Re + 1/gm

Explanation: We note that Re is in series with the emitter. The total input resistance becomes Re+1/gm where 1/gm is the resistance looking into the emitter. Note that in presence of early effect, the input impedance would have been Re + 1/gm || rO.

8. Assuming that the transistor has a very high current gain and negligible early effect, what is the gain of the following circuit?

a) RC/(1/gm+Re)
b) RC*gm
c) -RC*gm
d) -RC/(1/gm+Re)

Explanation: The gain can be derived by using the small signal model. Firstly, we note that the input voltage at the emitter is undergoing a voltage division between Re and 1/gm. This voltage gets multiplied by a factor of RC*gm and the overall gain becomes RC/(1/gm+Re).

9. Assuming that the transistor has a very high current gain and plausible early effect, what is the output impedance of the following C.B. stage?

a) RC
b) RC+rO
c) rO
d) RC || rO

Explanation: The output impedance of the C.B. stage can be calculated by drawing the small signal model of the transistor. In presence of early effect, we note the resistance come in parallel and hence, the correct answer is RC || rO.

10. Assuming that the transistor has a very high current gain and negligible early effect, we note that Re draw some current from the input and the entire signal doesn’t reach the emitter. What should be done to prevent this loss?

a) Re >> 1/gm
b) Re << 1/gm
c) Re = 1/gm
d) Cannot be improved

Explanation: We will have to maximize Re. The current from the input doesn’t flow much into the resistor but how can we ensure that it doesn’t happen. If the resistance looking into the emitter is less than the Re, all the current flows into Re and hence, Re >> 1/gm is a good way of reducing loss.

11. If VBE=0.7V, find the base voltage such than VIN<=6V can be used but the transistor always stays on. Neglect early effect and assume a high current gain.

a) 4.7
b) 5.3
c) 6.9
d) 4

Explanation: The maximum input voltage is 6V. The base voltage should be greater than atleast 6.7V so that the base-emitter region is always 0.7V and the transistor stays on.

12. If IC=2mA, IS=10-14 find the minimum base voltage such than VIN<=3V. Neglect early effect and assume a high current gain. The supply voltage is 5V.

a) 3.5V
b) 3.4V
c) 3.68V
d) 3V

Explanation: We note that the base-emitter voltage required to generate IC=2mA is VBE=676.56mV and this can be obtained from the exponential relationship between the collector current and base-emitter voltage. The maximum value of VIN=3V which implies that the minimum base voltage to ensure that the collector current is at least 2mA is 3.68V.

13. What happens to the gain of a simple C.B. stage with negligible early effect if the base width of the transistor is doubled?
a) It increases by a factor of 2
b) It decreases by a factor of 2
c) It decreases by a factor of 4
d) It increases by a factor of 4

Explanation: We recall that the gain of a simple C.B. stage is RC*gm and gm decreases by a factor of 2. This is because the collector current is inversely proportional to the base width and thus, the overall gain decreases by a factor of 2.

14. What happens to the gain of a simple C.B. stage with negligible early effect if the emitter area of the transistor is doubled?
a) It increases by a factor of 4
b) It increases by a factor of 4
c) It decreases by a factor of 2
d) It increases by a factor of 2

Explanation: We recall that the gain of a simple C.B. stage is RC*gm and gm increases by a factor of 2. This is because the collector current is directly proportional to the emitter area and thus, the overall gain increases by a factor of 2.

15. Assuming that the transistor has a high current gain and negligible early effect, what is the input impedance of the following stage?

a) R1/(β+1)
b) 1/gm+R1/(β+1)
c) 1/gm
d) 1/gm||R1/(β+1)

Explanation: The small signal model is to be drawn and we use KCL at input node. Using the Thevenin’s theorem, we find that the input resistance is 1/gm+R1/(β+1). This is calculated for negligible early effect. β+1 is approximated to β.

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