# Analog Circuits Questions and Answers – Hybrid Equivalent Model

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This set of Analog Circuits Multiple Choice Questions & Answers (MCQs) focuses on “Hybrid Equivalent Model”.

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

Explanation: The output of the CE amplifier has a phase shift of 180o with respect to the input. The CC amplifier has AV≅1, thus it is a voltage buffer. However, the CB amplifier has a large voltage gain, and its current gain AI≅1, thus it is a current buffer. CE amplifier has an application 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

Explanation: 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 hfe = 60, hie=1000Ω, hoe = 20μ Ω, hre = 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

Explanation: Current gain, AI = – hf / (1 + hoRL’) where RL’ = 2kΩ||4kΩ
RL’ = 1.33kΩ.
Thus AI = – 60 / (1 + 0.0266) = -58.4453.

4. Consider the circuit. Given hfe = 50, hie = 1200Ω. Find voltage gain. a) – 278
b) -277.9
c) – 300
d) – 280

Explanation: Voltage gain = AV = -hfeRL’/hie
RL’ = 20k||10k = 6.67kΩ
AV = -50 * 6.67k/1.2k = -277.9 ≅ – 278.

5. Given that IB = 5mA and hfe = 55, find load current. a) 28mA
b) 280mA
c) 2.5A
d) 2A

Explanation: In given circuit, which is an emitter follower, current gain = 1 + hfe
IL = IB (1+hfe)
IL = 5mA(56) = 280 mA.

6. Consider the following circuit, where source current = 10mA, hfe = 50, hie = 1100Ω, then for the transistor circuit, find output resistance RO and input resistance RI. a) RO = 0, RI = 21Ω
b) RO = ∞, RI = 0Ω
c) RO = ∞, RI = 21Ω
d) RO = 10, RI = 21Ω

Explanation: Since hoe is not given, we can consider it to be small; i.e 1/hoe is neglected, open circuited. Hence output resistance RO = ∞.
Input resistance = hie/(1 + hfe) = 1100/51 ≅ 21Ω.

7. For the given circuit, input resistance RI = 20Ω, hfe = 50. Output resistance = ∞. Find the new values of input and output resistance, if a base resistance of 2kΩ is added to the circuit. a) RI = 20Ω, RO = ∞
b) RI = 20Ω, RO = 2kΩ
c) RI = 59Ω, RO = ∞
d) RI = 59Ω, RO = 2kΩ

Explanation: RI = 20k = hie/(1+hfe) = hie/51
hie =1020 Ω
Hence, after adding base resistance, RI’= (hie+RB)/(1+hfe) = (1020+2000) / 51 ≅ 59Ω
There is no change in output resistance or current gain due to an extra base resistance. RO’ = ∞.

8. Consider its input resistance to be R1. Now, the bypass capacitor is attached, so that the new input resistance is R2. Given that hie = 1000Ω and hfe = 50, find R1-R2. a) 112.2Ω
b) 0Ω
c) 110Ω
d) 200Ω

Explanation: For the circuit, CE amplifier without bypass capacitor, input resistance, R1=hie + (1+hfe)RE
R1 = 1000 + 51*2.2 = 1000 + 112.2 = 1112.2Ω
With a bypass capacitor attached, input resistance, R2 = hie = 1000Ω
Thus R1 – R2 = 112.2Ω.

9. Given that for a transistor, hie = 1100Ω, hfe = 50, hre = 2*10-4 and hoe = 2μΩ-1. Find CB h-parameters.
a) hfb = 1, hib = 22, hob = 3μΩ-1, hrb = -1.5×10-4
b) hfb = -0.98, hib = -21.56, hob = 0.03μΩ-1, hrb = 1.5×10-4
c) hfb = -0.98, hib = 21.56, hob = 0.03μΩ-1, hrb = -1.5×10-4
d) hfb =1, hib = -21.56, hob = 0.03μΩ-1, hrb = -2×10-4

Explanation: hfb = -hfe/(1+hfe) = -50/51= -0.98
hib = hie/(1+hfe) = 21.56Ω
hob = hoe/(1+hfe) = 0.03 μΩ-1
hrb = (hiehoe/1+hfe) – hre = -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

Explanation: When a source resistance RS is present, the voltage gain with respect to source becomes
AVS = AVRI’/(RS+RI’), where AV is voltage gain with respect to transistor input. However when RS=0 then AVS = AV.

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