250+ TOP MCQs on JFET Amplifier and Answers

Analog Circuits Multiple Choice Questions on “JFET Amplifier”.

1. What are the small signal FET parameters?
A. g_m and r_ds
B. g_m and V_gs
C. V_ds and r_ds
D. g_m

Answer: A
Clarification: The small signal model of FET- MOSFET and JFET is obtained from the following equation I_DS = g_mV_gs + V_ds/r_ds
g_m and r_ds are the small signal FET parameters.

2. Find the transconductance when applied gate to source voltage is -2V.

A. 10 Ω^-1
B. 10mΩ^-1
C. 40mΩ^-1
D. 20mΩ^-1

Answer: B
Clarification: V_P = -4V
g_m = g_mo (1 – V_GS/V_P) = 20 (1-2/4) = 20/2 = 10mΩ^-1.

3. Choose the incorrect statement for JFET(s).
A. Maximum transconductance occurs at V_GS=0
B. Transconductance decreases linearly with V_GS
C. Transconductance increases linearly with I_DS
D. Transconductance does not depend on V_DS

Answer: C
Clarification: Maximum transconductance occurs at V_GS = 0, and it lies between 0-g_mo. Transconductance decreases linearly with V_GS according to equation g_m = g_mo(1-V_GS/V_P). However, g_m ∝ (sqrt{I_{DS}}), which is a parabolic increase, not linear.

4. Consider the circuit shown below.

Find the net output resistance given that g_m = 1mΩ^-1 and r_ds = 0.1 MΩ.
A. 10 kΩ
B. 9.09 kΩ
C. 100 kΩ
D. 110 kΩ

Answer: B
Clarification: Net output resistance for the circuit is R = r_ds||R_D = 100k||10k
R = 9.09 kΩ.

5. For an RC coupled common source JFET amplifier without bypass capacitor, find the voltage gain if g_m = 1mΩ^-1, source resistance is 2kΩ, drain resistance is 15kΩ and load is 10kΩ.
A. -2
B. -2.5
C. 5
D. 2

Answer: A
Clarification: A_V = -g_mR_L’/1+g_mR_S
R_L’= 15k|| 10k = 6kΩ
A_V = – 2.

6. Consider the amplifier below.

Find the input resistance and voltage gain of the circuit, given g_m = 0.5mΩ^-1 and r_ds = 0.2MΩ.
A. R_I = 27.27 kΩ, A_V = 5
B. R_I = 44 kΩ, A_V = 1
C. R_I = 27.27kΩ, A_V = 1
D. R_I = 60kΩ, A_V = 100

Answer: C
Clarification: The given circuit is a common drain JFET amplifier.
Its input resistance, R_I = R_G = 60k||50k = 27.27 kΩ
The voltage gain is 1.

7. Which of these is incorrect for a common gate amplifier?
A. It is a current buffer
B. It has ∞ output resistance
C. Its input resistance is high
D. It is used as a high-frequency amplifier

Answer: C
Clarification: A common gate amplifier can be used as a current buffer since its current gain is 1. It has very high output resistance (∞) and low input resistance. It is often used as a high-frequency amplifier.

8. In a high-frequency model of a JFET, which of these capacitances is present?

A: Gate-to-source capacitance
B: Gate-to-drain capacitance
C: Drain-to-source capacitance

A. A and B
B. A and C
C. B and C
D. A, B and C

Answer: D
Clarification: In an HF model of a JFET, C_GS and C_GD are present, in the range of 1pF to 10 pF. C_DS is the internal capacitance of the channel, ranging from 0.1pF to 1pF.

9. Which of these is false for a CS amplifier without a bypass capacitor compared to a CS amplifier with a bypass capacitor?
A. Voltage gain magnitude decreases
B. Input resistance remains same
C. The output resistance decreases
D. The output is 180° out of phase with respect to the input applied

Answer: C
Clarification: For a CS amplifier without a bypass capacitor, the input resistance is unchanged and output resistance increase.
R_I = R_G for both.
R_O = r_ds with bypass capacitor and r_ds = r_ds + (1+μ) R_S without a bypass capacitor.
Output is still 180° out of phase compared to applied input, but the gain decreases.

10. Which of these has an output which follows input?
A. CS amplifier with a bypass capacitor
B. CD amplifier
C. CG amplifier
D. CS amplifier without a bypass capacitor

Answer: B
Clarification: For a CD amplifier, the gain A_V = g_m(R_S||r_ds)/1+g_m(R_S||r_ds)
But since g_m(R_S||r_ds)>>1, gain A_V≈1, that is, it acts as a voltage follower. It is also called a source follower.

11. In a CS amplifier, given that r_ds=0.5MΩ and g_m=5mΩ^-1, the load is 10kΩ, source resistance is 44 kΩ. Calculate the internal amplification factor for the small signal model.
A. 2500
B. 8100
C. 9800
D. 7700

Answer: A
Clarification: When the current source in small signal mode, g_mV_GS is converted into a voltage source, the source is equal to μV_GS where μ=gmrds is the amplification factor of the circuit.
Hence, amplification factor = 2500.