Analog Electronics - 4 - Free MCQ Test with solutions for GATE EE and Digital

The input is current, current is added in parallel (Hence shunt).

From the output, voltage is sampled (shunt)

Hence shunt series feedback topology.

The collector currents of both Q₁ and Q₂ will be identical due to same base current.

Since Q₃ is in saturation

9 – 0.2 – 5 × 10³ I_C - V_O = 0

V_o = 3.6 V

The ideal closed-loop voltage transfer function

The input resistance
The output resistance is

Given circuit is widlar current source which is particularly suitable for low value of current.

Here I_out ≠ I_ref

Objective approach

The second stage is emitter follower which has voltage gain of unity.

The overall voltage gain = gain of CE stage

r_C1 = effective impedance seen by collector of transistor Q₁

Z_i(base)Q₂ at the base of transistor Q₂.
The impedance Zi (base) Q₂ at the base of transistor Q_2­ is
Z_i(base)2 = βR_E2
= 80 × 3 k

Z_i (base) 2 = 240 kΩ

Therefore:

= 10 k||240 k

r_C1 = 9.6kΩ

Overall voltage gain

A_V− = 3 84
 

Subjective approach:

Draw the equivalent h-parameter model

 r_π = β r_{e = 80 x 25 = 2k}


Apply KVL in the base-emitter loop of transistor Q₂

Apply KCL at node V_B2

800 i_b1 = -255  i_b2

Concept:

For multistage amplifier consisting of n identical amplifying states, the lower cut off f_L (multi) is expressed as

 

Where f₁ = lower cut off frequency of individual state.

Upper cut off frequency is given by:

 

= 20 × 0.643

= 12.87 kHz

Bandwidth = 

= 12714.6 Hz.

Procedure to identify the type of Feedback:

(1) Identify the element responsible for the Feedback.

(2) If Feedback element is directly connected to the output node, it indicates voltage sampling otherwise it indicates current sampling.

(3) If Feedback element is directly connected to the Input node it indicates shunt Mixing otherwise it indicate series mixing.

∴ It is voltage current Feedback.

The base of Q₁ is biased with negative voltage hence

Q₁ → OFF

Q₂ → ON

 

I_C1 = 0

V₀₁ = 12 V

V_E2 = -0.7 V

V₀₂ = 12 – (1 mA) (2k)

= 12 – 2

= 10 V

V₀₁ – V₀₂ = 12 – 10 = 2 V

Also we know IC = βI_B

Since base currents are the same, we have