Test: Miller Effect Capacitance - Electronics and Communication Engineering (ECE) MCQ

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.

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.

 A_V = -120
f_L = 1/2πC_C(R_C + R_L) = 1/2π x 0.001 = 1000/2π = 159.15Hz
A_V’ = 120/8.02 ≈ 15.

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.

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

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.

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.

A_VM = 150
A_VL = 100
f = 50Hz
1 + f²/2500 = 1.5²
f² = 2500*1.25 = 3125
f = 55.90 Hz.

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 = 133.

180° + tan^-1(f_L/f) = 260°
f_L/f = tan(80) = 5.67
A = 20.3 + 5.672 = 26.05.