Analog Electronics - 2 for GATE EE and Digital - Free MCQ Test with solutions

Analog Electronics - 2 - Question 1

For the transistor circuit shown below:

If V₁ = 1V and V₂ = -12V R₁ = 15k and R₂ = 100k R_c= 2.2 k V_cc = 12V Then the Region of operation of transistor is

A.
Active
B.
Saturation
C.
Cut off
D.
Reverse Active

Detailed Solution: Question 1

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

*Answer can only contain numeric values

Analog Electronics - 2 - Question 2

Calculate the value of β for the following circuit

Detailed Solution: Question 2

I_E= (1 + β) I_β

3 × 10^-3 = (1 + β) × 0.33 × 10^-3

β = 89.9

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Analog Electronics - 2 - Question 3

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

A.
5.43 mA, 5.89 V
B.
6.38 mA, 5.21 V
C.
5.03 mA, 5.89
D.
6.38 mA, 5.43 V

Detailed Solution: Question 3

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

Analog Electronics - 2 - Question 4

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

A.
an increase in the low-frequency cutoff frequency
B.
an increase in the high-frequency cutoff frequency
C.
a decrease in the low-frequency cutoff frequency
D.
a decrease in the high-frequency cutoff frequency

Detailed Solution: Question 4

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

Analog Electronics - 2 - Question 5

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

Detailed Solution: Question 5

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

Analog Electronics - 2 - Question 6

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

A.
73.5
B.
85.2
C.
103.5
D.
142.4

Detailed Solution: Question 6

Form DC Analysis

Since β value is large

Collector resistance = R_C’ = R_C ||R_L

= 2.2 || 10

= 1.8 kΩ

Voltage gain =

*Answer can only contain numeric values

Analog Electronics - 2 - Question 7

For the pnp transistor shown, the collector current I_C is_______ mA.
(Assume V_EB(ON) = V_EB(Sat) = 0.7 V)
V_EC(Sat) = 0.2 V

Detailed Solution: Question 7

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

Analog Electronics - 2 - Question 8

For the circuit shown in the figure if R_C = 4 kΩ, R_L = 4 kΩ, R_B = 20 kΩ, Rs = 1 kΩ and transistor parameter are magnitude of h_ie = 1.1 kΩ, h_fe = 50
The current gain is:

A.
50
B.
25
C.
12
D.
35

Detailed Solution: Question 8

The AC equivalent circuit is:

Replacing BJT with its h-parameter model

i_b using current division

i_B = 0

Magnitude = |-11.65| ≈ 12

*Answer can only contain numeric values

Analog Electronics - 2 - Question 9

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

Detailed Solution: Question 9

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

Analog Electronics - 2 - 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.

A.
0.912 and 79
B.
0.983 and 59
C.
0.887 and 89
D.
0.983 and 89

Detailed Solution: Question 10

Hence α = 0.983 and β = 59

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Analog Electronics - 2 - Free MCQ Test with solutions for GATE EE and Digital

MCQ Practice Test & Solutions: Analog Electronics - 2 (10 Questions)

For the transistor circuit shown below:

If V₁ = 1V and V₂ = -12V R₁ = 15k and R₂ = 100k R_c= 2.2 k V_cc = 12V Then the Region of operation of transistor is

Calculate the value of β for the following circuit

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

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

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

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

For the pnp transistor shown, the collector current I_C is_______ mA.
(Assume V_EB(ON) = V_EB(Sat) = 0.7 V)
V_EC(Sat) = 0.2 V

For the circuit shown in the figure if R_C = 4 kΩ, R_L = 4 kΩ, R_B = 20 kΩ, Rs = 1 kΩ and transistor parameter are magnitude of h_ie = 1.1 kΩ, h_fe = 50
The current gain is:

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

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.

Analog and Digital Electronics

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Analog Electronics - 2 - Free MCQ Test with solutions for GATE EE and Digital

MCQ Practice Test & Solutions: Analog Electronics - 2 (10 Questions)

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

Calculate the value of β for the following circuit

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

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

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

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

For the pnp transistor shown, the collector current IC is_______ mA.(Assume VEB(ON) = VEB(Sat) = 0.7 V)VEC(Sat) = 0.2 V

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:

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

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.

Analog and Digital Electronics

Analog and Digital Electronics

Top Courses for Electrical Engineering (EE)

About this Electrical Engineering (EE) Practice Test

Explore more Electrical Engineering (EE) Study Resources

For the transistor circuit shown below:

If V₁ = 1V and V₂ = -12V R₁ = 15k and R₂ = 100k R_c= 2.2 k V_cc = 12V Then the Region of operation of transistor is

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

For the pnp transistor shown, the collector current I_C is_______ mA.
(Assume V_EB(ON) = V_EB(Sat) = 0.7 V)
V_EC(Sat) = 0.2 V

For the circuit shown in the figure if R_C = 4 kΩ, R_L = 4 kΩ, R_B = 20 kΩ, Rs = 1 kΩ and transistor parameter are magnitude of h_ie = 1.1 kΩ, h_fe = 50
The current gain is:

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