250+ TOP MCQs on Low Frequency Response and Miller Effect Capacitance and Answers

Analog Circuits Multiple Choice Questions on “Low Frequency Response and Miller Effect Capacitance”.

1. In Miller’s theorem, what is the constant K?
A. Total voltage gain
B. Internal voltage gain
C. Internal current gain
D. Internal power gain

Answer: B
Clarification: The constant K=V₂/V₁, which is the internal voltage gain of the network.
Thus resistance R_M=R/1-K
R_N=R/1-K^-1.

2. When applying miller’s theorem to resistors, resistance R1 is for node 1 and R2 for node 2. If R1>R2, then for same circuit, then for capacitance for which the theorem is applied, which will be larger, C1 or C2?
A. C1
B. C2
C. Both are equal
D. Insufficient data

Answer: a
Clarification: Given R1>R2
R/1-K > R/1-K^-1, and so 1-K^-1>1-K
Thus K²>1, K>1, K<-1 (correct)
Thus, C1=C(1-K) and C2=C(1-K^-1)
Hence C1>C2.

3. Find net voltage gain, given h_fe = 50 and h_ie = 1kΩ.

A. 27.68
B. -22
C. 30.55
D. -27.68

Answer: D
Clarification: Apply millers theorem to resistance between input and output.
At input, R_M=100k/1-K = R_I
Output, R_N=100k/1-K^-1 ≈ 100k
Internal voltage gain , K = -h_feR_L’/h_ie
K = – 50xR_c||100k/1k = – 50x4x100/104 = – 192
R_I = 100k/1+192 = 0.51kΩ
R_I’ = R_I||h_ie = 0.51k||1k = 0.51×1/1.51 = 0.337kΩ
Net voltage gain = K.R_I’/R_S+R_I’ = – 192 x 0.337/2k + 0.337k = -27.68.

4. Given that capacitance w.r.t the input node is 2pF and output node is 4pF, find capacitance between input and output node.
A. 0.67 pF
B. 1.34pF
C. 0.44pF
D. 2.2pF

Answer: A
Clarification: C1=C(1-K), C2=C(1-K^-1)
C1=2pF
C2=4pF
C1/C2=1/2=1-K/1-K-1
K = -2
C1 = C(1+2) = 3C
C = C1/3 = 2/3pF = 0.67 pF.

5. Consider an RC coupled amplifier at low frequency. Internal voltage gain is -120. Find the voltage gain magnitude, when given that collector resistance = 1kΩ, load = 9kΩ, collector capacitance is 0. is 0.1μF, and input frequency is 20Hz.
A. 120
B. 12
C. 15
D. -12

Answer: C
Clarification: A_V = -120
f_L = 1/2πC_C(R_C+R_L) = 1/2π*0.001 = 1000/2π = 159.15Hz
A_V’ = (frac{|AV|}{sqrt{1}} + (frac{f_L}{f})^2)
A_V’ = 120/8.02 ≈ 15.

6. Find the 3-dB frequency given that the gain of RC coupled amplifier is 150, the low frequency voltage gain is 100 and the input frequency is 50Hz.
A. 50.8 Hz
B. 55.9 Hz
C. 60Hz
D. 100Hz

Answer: B
Clarification: A_VM = 150
A_VL = 100
f = 50Hz
100 = (frac{150}{sqrt{1}}+(frac{f_L}{50})^2)
1+f²/2500 = 1.5²
f² = 2500*1.25 = 3125
f = 55.90 Hz.

7. Given collector resistance = 2kΩ, load resistance = 5kΩ, collector capacitance = 1μF, emitter capacitance = 20μF, collector current = 2mA, source resistance = 2kΩ. If the effect of blocking capacitor is ignored, find the applicable cut-off frequency.
A. 22.73 Hz
B. 612 Hz
C. 673Hz
D. 317 Hz

Answer: B
Clarification: R_C = 2kΩ, R_L = 5kΩ, C_C = 1μF, C_B = 10μF, C_E = 20μF, R_S = 2 kΩ
h_ie = 1kΩ, I_C = 2mA
f_L1 = 1/2πC_C(R_C+R_L) = 22.73 Hz
f_L2 = g_m/2πC_E = I_C/2πC_EV_T = 612 Hz
Since f_L2 > 4f_L1, hence f_L2 is the correct answer.

8. Consider the circuit shown.

h_fe = 50, h_ie = 1000Ω. Find magnitude of voltage gain at input frequency 10Hz.
A. 100
B. 133
C. 166
D. 220

Answer: B
Clarification: Net load = 10k||10k = 5kΩ = R_L’
A_VM = -h_feR_L’/h_ie = -50×5/1 = -250
f_L = 1/2πC_C(R_C+R_L) = 15.9 Hz
A_VL = (frac{AVM}{sqrt{1}} + (frac{fL}{f})^2) = 133.

9. What is the phase shift in RC coupled CE amplifier at lower 3dB frequency?
A. 180°
B. 225°
C. 270°
D. 100°

Answer: B
Clarification: Total phase shift = 180°+ tan^-1(f_L/f)
At 3dB frequency f_L/f = 1
Total phase shift = 180° + 45° = 225°.

10. Consider that the phase shift of an RC coupled CE amplifier is 260°. Find the low frequency gain when the voltage gain of the transistor is -150.
A. 100
B. 26
C. 40
D. 55

Answer: B
Clarification: 180° + tan^-1(f_L/f) = 260°
f_L/f = tan(80) = 5.67
A = (frac{150}{sqrt{1}}) + 5.672 = 26.05.