250+ TOP MCQs on Biasing of JFET and MOSFET and Answers

Analog Circuits Multiple Choice Questions on “Biasing of JFET and MOSFET”.

1. Which of the following statements are true?

P: JFET is biased to operate it in active region
Q: MOSFET is biased to operate it in saturation region

A. Both P and Q are correct
B. P is correct and Q is incorrect
C. P is incorrect and Q is correct
D. Both P and Q are incorrect

Answer: C
Clarification: While transistors are biased to work in the active region, to act as amplifiers, FET devices are instead biased in the saturation region to work as an amplifier, whether it be a JFET or a MOSFET. In saturation, current I_DS changes with respect to V_GS, and small changes in V_GS cause proportionate changes in I_DS, and the device can act as an amplifier.

2. In the given situation for n-channel JFET, we get drain-to-source current is 5mA. What is the current when V_GS = – 6V?

A. 5 mA
B. 0.5A
C. 0.125 A
D. 0.5A

Answer: C
Clarification: I_DS = I_DSS(1-V_GS/V_P)²
When V_GS = 0, I_DSS = I_DS = 5mA
When V_GS = -6V, I_DS = 5mA(1 + 4)²
I_DS = 5 x 25 = 125 mA.

3. Consider the following circuit. Given that V_DD = 15V, V_P = 2V, and I_DS = 3mA, to bias the circuit properly, select the proper statement.

A. R_D < 6kΩ
B. R_D > 6kΩ
C. R_D > 4kΩ
D. R_D < 4kΩ

Answer: A
Clarification: In given circuit, V_GS = -5V
V_DS = V_DD – I_DSR_D
To bias properly V_DS > |V_P| – |V_GS|
V_DS > -3
15 – 3mA*R_D > -3
-3mA*R_D > -18
R_D < 6kΩ.

4. Consider the circuit shown. V_DS=3 V. If I_DS=2mA, find V_DD to bias circuit.

A. -30V
B. 30V
C. 33V
D. Any value of voltage less than 12 V

Answer: C
Clarification: V_DS = V_DD – I_DS(10k + 5k)
3 = V_DD – 2(15)
3 = V_DD – 30
V_DD = 33 V.

5. To bias a e-MOSFET ___________
A. we can use either gate bias or a voltage divider bias circuit
B. we can use either gate bias or a self bias circuit
C. we can use either self bias or a voltage divider bias circuit
D. we can use any type of bias circuit

Answer: A
Clarification: To bias an e-MOSFET, we cannot use a self bias circuit because the gate to source voltage for such a circuit is zero. Thus, no channel is formed and without the channel, the MOSFET doesn’t work properly. If self bias circuit is used, then D-MOSFET can be operated in depletion mode.

6. Consider the following circuit. Process transconductance parameter = 0.50 mA/V², W/L=1, Threshold voltage = 3V, V_DD = 20V. Find the operating point of circuit.

A. 20V, 25mA
B. 13V, 22mA
C. 12.72V, 23.61mA
D. 20V, 23.61mA

Answer: C
Clarification: I_DS = [k’W/L(V_GS – V_T)²]/2
V_GS = 20 x 35/55 = 12.72 V
I_DS = 0.25 (9.72)²
I_DS = 23.61 mA.

7. Given V_DD = 25V, V_P = -3V. When V_GS = -3V, I_DS = 10mA. Find the operating point of the circuit.

A. -3.83V, 0.766mA
B. -2.345V, 0.469mA
C. 3.83V, 0.469mA
D. 2.3V, 0.7mA

Answer: B
Clarification: When V_GS = V_P then I_DSS = I_DS = 10mA
Also, in above circuit, V_GS = -I_DSR_S = – I_DSx5k
Thus, I_DS = I_DSS(1-V_GS/V_P)²
Solving we get, I_DS = 0.766mA, 0.469mA
Thus we get V_GS = -3.83V, -2.345V
However, V_GS should lie between 0 and V_P.

8. Consider the following circuit. I_DSS = 2mA, V_DD = 30V. Find R, given that V_P = – 2V.

A. 10kΩ
B. 4kΩ
C. 2kΩ
D. 5kΩ

Answer: B
Clarification: I_DSS = 2mA
I_DS = (V_DD – 15)/50k = 0.3mA
V_GS = V_P[1 – (sqrt{frac{I_{DS}}{D_{SS}}})]
V_GS = -2 x (1 – (sqrt{.15})) = – 1.22V
Thus, V_GS + I_DS x (R) = 0
R = 1.22/0.3mA = 4kΩ.

9. Which of the following statements are true?

A: In a self bias circuit, the current IDS is not stable.
B: Source capacitance, CS, parallel to RS, reduces stability.

A. Both statements are correct and B is the correct reasoning
B. Both statements are correct but B is not the correct reason for it
C. Statement A is correct while statement B is wrong
D. Both statements are incorrect

Answer: D
Clarification: In a self bias circuit, the current I_DS is stable due to the presence of source resistance R_S in the circuit. The source resistance helps provide negative feedback to keep current stable. Capacitor C_S is a bypass capacitor to prevent decrease in voltage gain.

10. For a MOSFET, the pinch-off voltage is -3V. Gate to source voltage is 20V. W/L ratio is 5. Process transconductance parameter is 40μA/V². Find drain to source current in saturation.
A. 0.10 mA
B. 0.05mA
C. – 0.05mA
D. – 50A

Answer: c
Clarification: I_SD = k’W(V_SG – |V_T|)²/2L
I_SD = 20*5*(-20-3)² = 52900μA = 0.05mA.