This set of Microelectronics Multiple Choice Questions & Answers (MCQs) focuses on “Low Frequency Response of Stages of BJT – Set 2”.
1. In the following C.B. stage, we find that if the magnitude of the coupling capacitor increases, the _______ frequency decreases.
a) Pole
b) Zero
c) Tunneling
d) Auxiliary
View Answer
Explanation: The pole frequency will be found to be inversely proportional to the value of the coupling capacitance. Note that this pole frequency is assumed to be taken less than the lowest frequency and the impedance has to be less than 1/gm+R1.
2. If the base width of the transistor is doubled, by what amount should the magnitude of the coupling capacitor be changed?
a) High amount
b) Low amount
c) No Change required
d) It will be doubled
View Answer
Explanation: If the base width of the transistor is doubled, the current reduces and the transconductance reduces. Note that the previously taken magnitude of the coupling capacitor will still be useful since it is less than the 1/gm+R1.
3. What will be the nature of frequency response for the following circuit?
a) High pass
b) Low pass
c) Band pass
d) Band Stop
View Answer
Explanation: At very low frequencies, the coupling capacitor is open and we need a different design operation. However, as the frequency increases, we find that the gain of the circuit increases until the magnitude of the coupling capacitor decreases and we get the voltage gain of the C.B. stage.
4. In the following circuit, the collector current is 2.6mA. What should be the value of Re if we want all the input current to enter the CB stage? Assume that C2 gets shorted.
a) 1/2
b) 9
c) 2
d) 40
View Answer
Explanation: The transconductance is .1S and its inverse is 10 ohms. Now, we can say that the emitter resistance is required to be much greater than 10 ohms. If we want to reduce the current entering R5, we may take it to be 40 times more than this value. Note that the voltage gain decreases if the emitter resistance increases.
5. What should be the impedance of C2 at a frequency of interest?
a) Less than β/gm
b) More than β/gm
c) Equal to β/gm
d) Equal to 2β/gm
View Answer
Explanation: The impedance of C2 is expected to be low at input voltages of a certain frequency. This shorts VIN to the emitter terminal. It will be less than the input impedance of the CB stage, ie 1/gm. Note that RE will have to be greater than the input impedance of the CB stage. The capacitance may be required to be low for high frequencies.
6. The input impedance of the CB stage is 100. If R5=10k, Check whether the emitter current is drawn towards the ground.
a) Yes
b) Negligible current gets pulled towards the drain.
c) IE/4 goes to ground
d) IE/8 goes to ground
View Answer
Explanation: R5 is 100 times bigger than the input impedance. The impedance is high enough to ensure that negligible current gets pulled towards the drain when the transistor is ON.
7. The following circuit shows a C.E. stage where the voltage divider bias configuration is used to bias a degenerated CE stage, however, the degenerating resistance is bypassed by a capacitor. What is the low frequency response of this stage?
a) High Pass
b) Low Pass
c) Band Pass
d) Band Stop
View Answer
Explanation: We may derive the transfer function for the above circuit by taking into account, the effect of the resistors and the capacitors as VDRAIN/VGATE=-R2/(1/gm+R1||1/C2). We note that there is a zero and a pole, as the frequency of the input voltage increases, the magnitude of the transfer function increases to eliminate the degeneration effect and hence the gain reaches a maximum at the pole which is away from the zero frequency. We may take an inverse laplace transform of the function, however, note that there the input capacitance is to be considered when we make the transfer function b/ the output and the input voltage.
8. What happens to VGATE in the presence of C1?
a) It becomes equal to the gate voltage at high frequency
b) It becomes lesser than the gate voltage at high frequency
c) It becomes greater than the gate voltage at high frequency
d) It becomes equal to the gate voltage at low frequency
View Answer
Explanation: C1 is used to couple the transistor with the previous stage. It needs to short the gate terminal with the input terminal. The transfer function between them will be (R3||R4)/(1/C1s+(R3||R4)) in presence of the coupling capacitance. Now, we note that as the frequency increases, the impedance decreases and the input voltage gets shorted to the gate voltage. Note that the biasing is not altered.
Sanfoundry Global Education & Learning Series – Microelectronics.
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