This set of Electronic Devices and Circuits Multiple Choice Questions & Answers (MCQs) focuses on “A Generalized FET Amplifier”.
1. Ideal maximum voltage for common drain amplifier is _________
a) 0
b) 1
c) 0.5
d) 2
View Answer
Explanation: Similar to the transistor emitter follower, the source follower configuration itself provides a high level of buffering and a high input impedance. The actual input resistance of the FET itself is very high as it is a field effect device. This means that the source follower circuit is able to provide excellent performance as a buffer. The voltage gain is unity, although current gain is high. The input and output signals are in phase.
2. If a certain drain JFET has a transconductance of 4ms. And has external drain resistance of 1.5 ohm than ideal voltage gain will be _________
a) 4
b) 5
c) 6
d) 8
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Explanation: The transconductance, gm is defined as
gm = ΔID / ΔVGS
so gm = Voltage gain / RD
Therefore, voltage gain = gm * RD
=4 * 1.5
=6.
3. Input resistance of common gate of the amplifier is __________
a) zero
b) infinity
c) extremely low
d) extremely high
View Answer
Explanation: For a Common gate amplifier, Current gain is about unity, input resistance is low, output resistance is high a CG stage is a current “buffer”. It takes a current at the input that may have a relatively small Norton equivalent resistance and replicates it at the output port, which is a good current source due to the high output resistance.
4. A FET circuit has a transconductance of 2500 µ seconds and drain resistance equals to 10Kohms than voltage gain will be __________
a) 20
b) 25
c) 30
d) 35
View Answer
Explanation: The transconductance, gm is defined as
gm = ΔID / ΔVGS
so gm = Voltage gain / RD
Therefore, voltage gain = gm * RD
=2500*10-6 * 10 * 103
= 25.
5. Voltage gain of common drain amplifier is always slightly less than _____
a) 0.5
b) 1
c) 1.5
d) 2
View Answer
Explanation: In common drain amplifier
Writing KCL at the source node ;
Gm(vin – vout) – gmbs vout – gds vout = 0
vout vin = Gm / Gm + Gmbs + gds
Therefore gain is less than one.
6. A common gate amplifier has _______
a) low input impedance
b) high input impedance
c) infinite input impedance
d) no impedance
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Explanation: Common gate amplifier just like common base amplifier had a very large voltage gain but input impedance is very low. Also, it gives unity current gain. Hence, the power of the amplified signal will be less.
7. D-MOSFET in case of common source amplifier can operate with gate to source voltage zero at ______
a) Peak positive point
b) Peak negative point
c) Q point
d) Origin
View Answer
Explanation: Q-point” needs to be found for the correct biasing of the JFET amplifier circuit with single amplifier configurations of Common-source, Common-drain or Source-follower and the Common-gate available for most FET devices.
8. A common source amplifier has _______
a) no source resistance
b) no drain resistance
c) no gate resistance
d) low input impedance
View Answer
Explanation: In a common source amplifier the source resistance is connected to the ground(i.e. is grounded) because of which its source resistance is kept zero.
9. The drain of FET is analogous to BJT
a) collector
b) emitter
c) base
d) drain
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Explanation: A common collector amplifier (also known as an emitter follower) is one of three basic single-stage bipolar junction transistor (BJT) amplifier, typically used as a voltage buffer.
In this circuit the base terminal of the transistor serves as the input, the emitter is the output, and the collector is common to both. The analogous field-effect transistor circuit is the common drain amplifier and the analogous tube circuit is the cathode follower.
10. Input signal of common drain amplifier is applied to the gate through ________
a) input resistor
b) coupling capacitor
c) output capacitor
d) transformer
View Answer
Explanation: A common-drain amplifier is also called a source-follower. Self-biasing is used in this particular circuit. The input signal is applied to the gate through a coupling capacitor, and the output signal is coupled to the load resistor through the other capacitor.
Sanfoundry Global Education & Learning Series – Electronic Devices and Circuits.
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