Field Effect Transistor (FET) MCQ
Field Effect Transistor (FET) MCQ, Multiple Choice Questions Field Effect Transistor (FET), Junction Field Transistor (JFET) MCQ, MOSFET MCQ, Engineering MCQ,
Objective Type Questions
Q.1. The JFET is
- a bipolar device
- unipolar device
- current controlled device
- voltage controlled device
Answer: a bipolar & unipolar device
Q.2. The channel of a JFET exists between
- gate and source
- drain and source
- gate and drain
- input and output
Answer: drain and source
Q.3. For low values of VDS, the JFET behaves like a
- resistance
- constant voltage device
- constant current device
- negative resistor
Answer: resistance
Q.4. When VDS reaches the pinch off voltage VP, the drain current becomes
- low
- zero
- constant (saturated)
- reversed
Answer: constant (saturated)
Q.5. In an n-channel JFET, the drain current reaches its maximum value when VGS is
- negative
- zero
- positive
- equal to VP
Answer: equal to VP
Q.6. In an n-channel JFET
- the current carriers are holes
- the current carriers are electrons
- VGS is positive
- the input resistance is very low
Answer: the current carriers are electrons
Q.7. In a p-channel JFET
- the current carriers are electrons
- the current carriers are holes
- VGS is negative
- the input resistance is very small
Answer: the current carriers are holes
Q.8. For an n-channel JFET
- VGS can vary between zero negatively to VGSO
- VGS can vary between zero positively to VGSO
- pinch off occurs for positive VGS
- VDD is negative
Answer: VGS can vary between zero negatively to VGSO
Q.9. A FET cannot operate at VGS = 0 V. The FET is
- JFET
- D-MOSFET
- E-MOSFET
- both JFET and D-MOSFET
Answer: E-MOSFET
Q.10. The transconductance gm of JFET is defined as
- \frac{\Delta I_{D}}{\Delta V_{GS}}
- \frac{\Delta I_{D}}{\Delta V_{DS}}
- \frac{\Delta V_{GS}}{\Delta I_{D}}
- \frac{\Delta I_{DSS}}{\Delta I_{D}}
Answer: \frac{\Delta I_{D}}{\Delta V_{GS}}
Q.11. The amplification factor µ of JFET is given by
- g_{m}\times r_{d}
- \frac{g_{m}} {r_{d}}
- g_{m}\cdot I_{DSS}
- g_{m}\left ( 1-\frac{V_{GS}}{V_{P}} \right )
Answer: g_{m}\times r_{d}
Q.12. A channel is induced in an E-MOSFET by the application of a
- V_{GS}< \left ( V_{GS} \right )th
- V_{GS}> V_{P}
- V_{GS}> \left ( V_{GS} \right )th
- V_{DS}> V_{P}
Answer: V_{GS}> \left ( V_{GS} \right )th
Q.13. The Shockley equation is given by
- I_{D}=I_{DSS}\left [ 1-\frac{V_{GS}}{V_{P}} \right ]^{2}
- I_{D}=I_{DSS}\left [ 1-\frac{V_{GS}}{V_{P}} \right ]
- g_{m}=g_{mo}\left [ 1-\frac{V_{GS}}{V_{P}} \right ]
- I_{D}=C\left [ V_{GS}-V_{GS}(th) \right ]^{2}
Answer: I_{D}=I_{DSS}\left [ 1-\frac{V_{GS}}{V_{P}} \right ]^{2}
Q.14. Ideally, the equivalent circuit of a FET consists of
- a resistance between drain and source
- a current source between the gate and the source
- a current source between the drain and the source
- a current source between the gate and the drain
Answer: a current source between the drain and the source
Q.15. The magnitude of the current source in the AC equivalent circuit of a FET depends on
- the de supply voltage
- Vds
- external drain resistance
- transconductance and gate to source voltage
Answer: external drain resistance
Q.16. A CS amplifier has a drain load RD = 1 kΩ and an external load resistance of RL = 1 kΩ. If gm = 10 mS, the voltage gain will be
- 10
- 5
- 20
- 1
Answer: 1
Q.17. If the external load resistance is removed in Q.16, the voltage gain will be
- decrease
- increase
- remain the same
- become zero
Answer: increase
Q.18. In a CD amplifier, RS = 500 Ω, and the transconductance is 9000 µS. The voltage gain will be
- 0.82
- 1
- 1.22
- 9
Answer: 0.82
Q.19. In a common gate amplifier, the transconductance is gm = 6000 µS, RG = 1 MΩ. The total input resistance will be
- 1666 Ω
- 166.6 Ω
- 1 MΩ
- 100166.6 Ω
Answer: 166.6 Ω
Q.20. In a CD amplifier, IDSS = 5 mA at VGS = 10 V. If the resistor RG is 20 MΩ. The total input of the resistance will be
- 20 MΩ
- 200 MΩ
- 40 MΩ
- 100 MΩ
Answer: 20 MΩ
Q.21. If there is an internal open between the source and the drain in a CS amplifier, the drain voltage will be
- 0 V
- VGS
- VDD
- VDD/2
Answer: VDD