Electronics and Communication Engineering - Analog Electronics - Discussion

31. 

A certain transistor has adc of 0.98 and a collectors leakage current of 5 μA. If IE = 1 mA, the base current will be

[A]. 35 μA
[B]. 25 μA
[C]. 15 μA
[D]. 5 μA

Answer: Option C

Explanation:

No answer description available for this question.

Mallli said: (Jan 31, 2015)  
I want formulae.

Srinivas said: (Aug 25, 2015)  
Alpha = 0.98, so Beta = 49.

Ic = alpha Ie + Ico = 985 uA.

Ib = (Ic/Beta) = 20 uA.

Gowri said: (Dec 27, 2016)  
25μa B is the correct answer.

Ankit Kumar said: (Feb 12, 2017)  
But how? Explain.

Aphro said: (Mar 3, 2017)  
Why 15μA? Explain.

Prof.Arungandhi said: (Jul 31, 2017)  
Yes, @Srinivas. Here I will give another formula to find the answer.

Ie= (1+beta) Ib;also beta=alpha/1-alpha.
So Ib= (1+beta) /Ie;sub we get 20 microAmpere.

Lobot Kigol said: (Jun 20, 2019)  
It's 15uA because of leakage current.

Post your comments here:

Name *:

Email   : (optional)

» Your comments will be displayed only after manual approval.