Use app×
Join Bloom Tuition
One on One Online Tuition
JEE MAIN 2025 Foundation Course
NEET 2025 Foundation Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
488 views
in Physics by (121k points)
closed by
In a CE transmitter amplifier, if the amplification factor is 150 and the collector voltage is 4 V and resistance is 2 kΩ, what should be the value of RB, given that the dc base current is 10 times the signal current?

(a) 5 kΩ

(b) 10 kΩ

(c) 15 kΩ

(d) 20 kΩ

1 Answer

0 votes
by (120k points)
selected by
 
Best answer
Right answer is (d) 20 kΩ

Best explanation: Here, the output voltage is 4 V.

So, ic = 4/2000

= 2 mA

The signal current through the base, IB = ic/β

= 0.013 mA

The dc base current = 0.13 mA

Assuming here, that VBE = 1.4 V, we get

RB = (4.0-1.4)/0.13

= 20 kΩ.

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

Categories

...