In the given figure, given that VBB supply can vary from 0 to 5.0 V, V_{CC} = 5 V ,\beta _{dc} = 200, R_{B}= 100 k\Omega , R_{C} = 1 k\Omega \ and \ V_{BE} = 1.0 V , The minimum base current and the input voltage at which the transistor will go to saturation, will be, respectively : 

Option 1) 25 \mu A and 3.5 V

Option 2) 25 \mu A and 2.8 V

Option 3) 20 \mu A and 2.8 V

Option 4) 20 \mu A and 3.5 V

Answers (1)

Relation between emitter current ,Base current,collector current -

I_{E}= I_{B}+I_{C}
 

- wherein

I_{E}=Emitter\: Current

I_{B}= Base \: Current

I_{C}= Collector \: Current

At saturation V_{CE} = 0

V_{CE} = V_{CC}-I_{C}R_{C}

\Rightarrow V_{CC}=I_{C}R_{C}

I_{C} = \frac{5V}{1\times 10^{3}} = 5\times 10^{-3}A

I_{B} = \frac{5\times 10^{-3}}{200} = 25\mu A

We known that

V_{BB} = I_{B}R_{B} + V_{BE}

R_{B} = 100K\Omega

V_{BE} = 1V

So V_{BB} = 25\times 10^{-6}\times 100\times 10^{3}+1 = 2.5+1 =3.5V

So  V_{BB} =3.5V \text{ and }I_{B} = 25\mu A

Preparation Products

Knockout JEE Main April 2021

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 22999/- ₹ 14999/-
Buy Now
Knockout JEE Main April 2022

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 34999/- ₹ 24999/-
Buy Now
Test Series JEE Main April 2021

Take chapter-wise, subject-wise and Complete syllabus mock tests and get in depth analysis of your test..

₹ 6999/- ₹ 4999/-
Buy Now
JEE Main Rank Booster 2021

This course will help student to be better prepared and study in the right direction for JEE Main..

₹ 13999/- ₹ 9999/-
Buy Now
Knockout BITSAT 2021

It is an exhaustive preparation module made exclusively for cracking BITSAT..

₹ 4999/- ₹ 2999/-
Buy Now
Boost your Preparation for JEE Main 2021 with Personlized Coaching
 
Exams
Articles
Questions