Analog Electronics - 2 - Electrical Engineering (EE) MCQ

V₁ = 1V

V₂ = -12V

R₁ = 15 k

R₂ = 100 k

= -0.7V

V_BB­ < 0, the Base emitter Junction is reversed biased

Hence Transistor is in cut off mode

I_E = (1 + β) I_β

3 × 10^-3 = (1 + β) × 0.33 × 10^-3

β = 89.9

So the operating point is (5.43 mA, 5.89 V)

The miller effect causes increase in the Input capacitance increase in the Input capacitance of Common – Emitter Amplifier

Upper cutoff frequency 

Therefore, there is decrease in upper-cut off frequency & Bandwidth of CE Amplifier

DC analysis from voltage divider

Now using appropriate approximations.

Z ≅ β_re = 150 × 18.2 = 2.718 kΩ

Input impedance = R_B || Z_B = 9.09 || 2.718 = 2.092 kΩ

Form DC Analysis

Since β value is large

Collector resistance = R_C’ = R_C ||R_L

= 2.2 || 10

= 1.8 kΩ

Voltage gain = 

Assume transistor is in active region

KVL in base emitter loop

0 + I_B (10) + 0.7 + I_E (1) = 5

I_E = (1 + β)I_B

I_B (10 + 1 + 30) = 4.3

I_C = βI_B = 30 × 0.105 = 3.15 mA

V_E = 5 – 1 (3.255)

= 1.745 V

V_B = 1.745 – 0.7 = 1.045 V

V_C = -5 + 10 (3.15) = 26.5 (not possible)

Transistor is not in active region it is  in saturation region

V_EB = 0.7

V_E = V_B + 0.7

V_EC = V_B - V_C = 0.2

V_C = V_B + 0.5

 

= (0.1V_B + 0.55) mA

I_E = I_B + I_C

4.3 - V_B = 0.1 V_B + 0.1 V_B + 0.55

V_B = 3.13

The AC equivalent circuit is:

Replacing BJT with its h-parameter model

i_b using current division

i_B = 0

Magnitude = |-11.65| ≈ 12

Given R_C = 4.7 kΩ, R_L = 60 kΩ, R_e = 4.7 kΩ, I_C = 1.5 mA

Hence α = 0.983 and β = 59