250+ TOP MCQs on Hybrid Equivalent Model and Answers

Analog Circuits Multiple Choice Questions on “Hybrid Equivalent Model”.

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1. Which of the following statement is incorrect?
A. Output of CE amplifier is out of phase with respect to its input
B. CC amplifier is a voltage buffer
C. CB amplifier is a voltage buffer
D. CE amplifier is used as an audio (low frequency) amplifier

Answer: C
Clarification: The output of the CE amplifier has a phase shift of 180^o with respect to the input. The CC amplifier has A_V≅1, thus it is a voltage buffer. However, the CB amplifier has a large voltage gain, and its current gain A_I≅1, thus it is a current buffer. CE amplifier has an application that has an audio amplifier.

2. Consider the following circuit. __________ provides DC isolation. _____________ prevents a decrease in voltage gain. _____________ is used to control the bandwidth.

A. C3, C1, C4
B. C4, C1, C2
C. C2, C3, C2
D. C4, C3, C2

Answer: B
Clarification: Capacitor C3 and C4, are the blocking capacitor and coupling capacitor respectively, both providing DC isolation to biasing circuit. Capacitor C1 is the emitter bypass capacitor, to prevent decrease in voltage gain by avoiding negative feedback. Capacitor C2 is the shunt capacitor, used to control the bandwidth, wherein the bandwidth is inversely proportional to C2.

3. Given h_fe = 60, h_ie=1000Ω, h_oe = 20μ Ω^–, h_re = 2 * 10^-4. Find the current gain of the BJT, correct up to two decimal points.

A. – 58.44
B. -59.21
C. – 60.10
D. – 60.00

Answer: A
Clarification: Current gain, A_I = – h_f/(1 + h_oR_L’) where R_L’ = 2kΩ||4kΩ
R_L’ = 1.33kΩ.
Thus A_I = – 60/(1 + 0.0266) = -58.4453.

4. Consider the circuit. Given h_fe = 50, h_ie = 1200Ω. Find voltage gain.

A. – 278
B. -277.9
C. – 300
D. – 280

Answer: A
Clarification: Voltage gain = A_V = -h_feR_L’/h_ie
R_L’ = 20k||10k = 6.67kΩ
A_V = -50 * 6.67k/1.2k = -277.9 ≅ – 278.

5. Given that I_B = 5mA and h_fe = 55, find load current.

A. 28mA
B. 280mA
C. 2.5A
D. 2A

Answer: B
Clarification: In given circuit, which is an emitter follower, current gain = 1 + h_fe
I_L = I_B (1+h_fe)
I_L = 5mA(56) = 280 mA.

6. Consider the following circuit, where source current = 10mA, h_fe = 50, h_ie = 1100Ω, then for the transistor circuit, find output resistance R_O and input resistance R_I.

A. R_O = 0, R_I = 21Ω
B. R_O = ∞, R_I = 0Ω
C. R_O = ∞, R_I = 21Ω
D. R_O = 10, R_I = 21Ω

Answer: C
Clarification: Since hoe is not given, we can consider it to be small; i.e 1/h_oe is neglected, open circuited. Hence output resistance R_O = ∞.
Input resistance = h_ie/(1 + h_fe) = 1100/51 ≅ 21Ω.

7. For the given circuit, input resistance R_I = 20Ω, h_fe = 50. Output resistance = ∞. Find the new values of input and output resistance, if a base resistance of 2kΩ is added to the circuit.

A. R_I = 20Ω, R_O = ∞
B. R_I = 20Ω, R_O = 2kΩ
C. R_I = 59Ω, R_O = ∞
D. R_I = 59Ω, R_O = 2kΩ

Answer: C
Clarification: R_I = 20k = h_ie/(1+h_fe) = h_ie/51
h_ie =1020 Ω
Hence, after adding base resistance, R_I’= (h_ie+R_B)/(1+h_fe) = (1020+2000)/51 ≅ 59Ω
There is no change in output resistance or current gain due to an extra base resistance. R_O’ = ∞.

8. Consider its input resistance to be R1. Now, the bypass capacitor is attached, so that the new input resistance is R2. Given that h_ie = 1000Ω and h_fe = 50, find R1-R2.

A. 112.2Ω
B. 0Ω
C. 110Ω
D. 200Ω

Answer: A
Clarification: For the circuit, CE amplifier without bypass capacitor, input resistance, R1=h_ie + (1+h_fe)R_E
R1 = 1000 + 51*2.2 = 1000 + 112.2 = 1112.2Ω
With a bypass capacitor attached, input resistance, R2 = h_ie = 1000Ω
Thus R1 – R2 = 112.2Ω.

9. Given that for a transistor, h_ie = 1100Ω, h_fe = 50, h_re = 2*10^-4 and h_oe = 2μΩ^-1. Find CB h-parameters.
A. h_fb = 1, h_ib = 22, h_ob = 3μΩ^-1, h_rb = -1.5×10^-4
B. h_fb = -0.98, h_ib = -21.56, h_ob = 0.03μΩ^-1, h_rb = 1.5×10^-4
C. h_fb = -0.98, h_ib = 21.56, h_ob = 0.03μΩ^-1, h_rb = -1.5×10^-4
D. h_fb = 1, h_ib = -21.56, h_ob = 0.03μΩ^-1, h_rb = -2×10^-4

Answer: C
Clarification: h_fb = -h_fe/(1+h_fe) = -50/51 = -0.98
h_ib = h_ie/(1+h_fe) = 21.56Ω
h_ob = h_oe/(1+h_fe) = 0.03 μΩ^-1
h_rb = (h_ieh_oe/1+h_fe) – h_re = -1.5×10^-4.

10. If source resistance in an amplifier circuit is zero, then voltage gain (output to input voltage ratio) and source voltage gain (output to source voltage ratio) are the same.
A. True
B. False

Answer: A
Clarification: When a source resistance R_S is present, the voltage gain with respect to source becomes
A_VS = A_VR_I’/(R_S+R_I’), where A_V is voltage gain with respect to transistor input. However when R_S=0 then A_VS = A_V.