# Analog Electronics - 2

## 10 Questions MCQ Test GATE ECE (Electronics) 2023 Mock Test Series | Analog Electronics - 2

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Attempt Analog Electronics - 2 | 10 questions in 30 minutes | Mock test for Electronics and Communication Engineering (ECE) preparation | Free important questions MCQ to study GATE ECE (Electronics) 2023 Mock Test Series for Electronics and Communication Engineering (ECE) Exam | Download free PDF with solutions
QUESTION: 1

### For the transistor circuit shown below: If V1 = 1V and V2 = -12V R1 = 15k and R2 = 100k Rc = 2.2 k Vcc = 12V Then the Region of operation of transistor is

Solution:

V1 = 1V

V2 = -12V

R1 = 15 k

R2 = 100 k

= -0.7V

VBB­ < 0, the Base emitter Junction is reversed biased

Hence Transistor is in cut off mode

*Answer can only contain numeric values
QUESTION: 2

### Calculate the value of β for the following circuit

Solution:

I= (1 + β) Iβ

3 × 10-3 = (1 + β) × 0.33 × 10-3

β = 89.9

QUESTION: 3

### In the DC fixed bias circuit shown below the operating point is

Solution:

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

QUESTION: 4

The Miller effect in the context of a Common Emitter amplifier explains

Solution:

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

*Answer can only contain numeric values
QUESTION: 5

For the network shown in the figure. Find the Input independence Zi (kΩ) [Use appropriate approximation]

Solution:

DC analysis from voltage divider

Now using appropriate approximations.

Z ≅ βre = 150 × 18.2 = 2.718 kΩ

Input impedance = RB || ZB = 9.09 || 2.718 = 2.092 kΩ

QUESTION: 6

The voltage gain of the amplifier shown if β = 250 is approximately

Solution:

Form DC Analysis

Since β value is large

Collector resistance = RC’ = RC ||RL

= 2.2 || 10

= 1.8 kΩ

Voltage gain =

*Answer can only contain numeric values
QUESTION: 7

For the pnp transistor shown, the collector current IC is_______ mA.

(Assume VEB(ON) = VEB(Sat) = 0.7 V)

VEC(Sat) = 0.2 V

Solution:

Assume transistor is in active region

KVL in base emitter loop

0 + IB (10) + 0.7 + IE (1) = 5

IE = (1 + β)IB

IB (10 + 1 + 30) = 4.3

IC = βIB = 30 × 0.105 = 3.15 mA

VE = 5 – 1 (3.255)

= 1.745 V

VB = 1.745 – 0.7 = 1.045 V

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

Transistor is not in active region it is  in saturation region

VEB = 0.7

VE = VB + 0.7

VEC = VB - VC = 0.2

VC = VB + 0.5

= (0.1VB + 0.55) mA

I= IB + IC

4.3 - VB = 0.1 VB + 0.1 VB + 0.55

VB = 3.13

QUESTION: 8

For the circuit shown in the figure if RC = 4 kΩ, RL = 4 kΩ, RB = 20 kΩ, Rs = 1 kΩ and transistor parameter are magnitude of hie = 1.1 kΩ, hfe = 50
The current gain is:

Solution:

The AC equivalent circuit is:

Replacing BJT with its h-parameter model

ib using current division

iB = 0

Magnitude = |-11.65| ≈ 12

*Answer can only contain numeric values
QUESTION: 9

For the Common emitter configuration circuit shown below, if IC = 1.25 mA, then the magnitude of voltage gain is_______

Solution:

Given RC = 4.7 kΩ, RL = 60 kΩ, Re = 4.7 kΩ, IC = 1.5 mA

QUESTION: 10

The reverse leakage current of the transistor when connected in CB Configuration is 0.3 μ A and it is 18 μA when the same transistor is connected in CE Configuration the value of α and β of the transistor for a base current of 40 mA will be respected.

Solution:

Hence α = 0.983 and β = 59

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