Analog Electronics - 5 - Electrical Engineering (EE) MCQ

The lower 3 dB frequency ω_L 

For a class B amplifier with V_CC = 24 V and V_O = 22 V efficiency η (in %) is given by

For class A the angle of excursion is 360⁰

For class B the angle of excursion is 180⁰

For class AB the angle of excursion is >180⁰ and < 360⁰

For class C the angle of excursion is less than 180⁰

P_iDC = P_oac + P_Tr

P_ooc = 10 W

η = 50% for class-A transformer coupled

Apply ac analysis to calculate gain

∴ Option (4) is correct.

Given

P_AC(secondary) = 0.48 W

I_CQ = 140 mA

R_L = 8 Ω

P_AC(sec) = P_AC(primary)

P_AC(primary) = 0.48 W

V_1(rms) = 1.96 × 3 = 5.88 V

output DC power

P_DC = V_CC I_CQ

= 10(140 × 10^-3)

= 1.4 W

The cut-off frequency due to source capacitance

 

Where R_S = source resistance

R_i = Input impedance

Where R_i = R₁ // R₂// βr_e

In DC bias values

Taking KVL in base-emitter path

V_B - V_BE - I_E (1.2 Ω) = 0

I_E = 0.917 mA

∴ R_i = R₁ // R₂ // βv_e

= 10 kΩ // 68 kΩ // (120 × 28.36) Ω

≅ 2.44 k Ω

= 103. 87 Hz

For class A transformer coupled amplifier

η = 50%

The Voltage Gain of the common emitter amplifier is equal to the ratio of the change in the input voltage to the change in the amplifiers output voltage. Then ΔV_L is V_out and ΔV_B s Vin. But voltage gain is also equal to the ratio of the signal resistance in the Collector to the signal resistance in the Emitter and is given as:

Voltage gain 
∴ If R_E increase then voltage gain will decrease and –ve feedback increase

Assume that Q₂ in an. It fallows that current will flow from ground through the 1kΩ Resistor into the emitter of Q₂. Thus the base of Q₂ will be at –Ve voltage and base current will be flowing out of the through the 10 kΩ resistor and into +5V supply. This is impossible, since if the base is negative. Current in the 10 kΩ resistor will have to flow into the base. Thus we conclude that our original assumption that Q₂ is ON is incorrect. It follows that Q₂ will be off and Q₁ will be ON.