Bipolar Junction Transistor NAT Level -2


10 Questions MCQ Test Topic wise Tests for IIT JAM Physics | Bipolar Junction Transistor NAT Level -2


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*Answer can only contain numeric values
QUESTION: 1

Calculate VCE in the figure (in volts).


Solution:

First check to see if Rin (Base) can be neglected.

=150 x kΩ = 150kΩ
Since 150kΩ is more than 10 times R2. The condition  is met and Rin(base) can be neglected.

= 6.88 V
Then, VE = VB + VBE
= 6.88 + 0.7
VE = 7.58V

 lE = 2.42 mA
From IE we can determine lc and VCE

and VC = ICRC = 2.42 mA x 2.2kΩ
= 5.32

= 2.26V
The correct answer is: 2.26

*Answer can only contain numeric values
QUESTION: 2

Calculate the minimum value of C2 (in μF) if the amplifier operate over a frequency range of 2kHz to 10kHz?


Solution:

RE = 560Ω
XC of the bypass capacitor C2 should be

The capacitive value, at the minimum frequency of 2kHz

The correct answer is: 1.42

*Answer can only contain numeric values
QUESTION: 3

If the Q-point is 2 Vand 1 mA with the help of a feedback resistor RB. β = 100, then what is the value of RB in? (Take VBE = 0.7V).


Solution:



The correct answer is: 130

*Answer can only contain numeric values
QUESTION: 4

Calculate the value of VCE (in Volts) for the circuit shown below. (β = 100 and VBE =0.7V).


Solution:

Let us use Thevenin’s theorem


Now, applying KVL to Base circuit

= 9.45 μA
lC = 0.945mA
Now, Vcc = IcRc
- VCE


The correct answer is: 6.22

*Answer can only contain numeric values
QUESTION: 5

A npn transistor circuit has  ∝ = 0.985. If IC = 2mA then the value of lB (in mA) is?


Solution:


= 0.03 mA
The correct answer is: 0.03

*Answer can only contain numeric values
QUESTION: 6

Calculate the collector voltage VC of the transistor circuit should be in the figure :

(given = 0.96, lCBO = 20μA , VBE = 0.3V, RB = 100/kΩ)


Solution:




Similarly,



The correct answer is: 4.72

*Answer can only contain numeric values
QUESTION: 7

In the below figure, the minimum value of lB (in mA) required to saturate this transistor if  [Neglect VCE(sat)].


Solution:

Since,VCE(Sat) is neglected (assume to 0V)


The correct answer is: 50

*Answer can only contain numeric values
QUESTION: 8

In a certain oscillator, Av = 50. The attenuation of the feedback circuit must be?


Solution:

*Answer can only contain numeric values
QUESTION: 9

An amplifier of gain 1000 is made into a feedback amplifier by feeding 9.9% of its output voltage in series with the input opposing. If fl = 20Hz and fh = 200kHz for the amplifier without feedback then due to the feedback what is the new frequency (in kHz)?


Solution:


The correct answer is: 20000

*Answer can only contain numeric values
QUESTION: 10

In the transistor below, base current is 35 μA, VBE = 0.7V, what is the value of RB in kΩ?


Solution:

Using KVL,

The correct answer is: 237.1