Electronics and Communication Engineering - Electronic Devices and Circuits - Discussion
Discussion Forum : Electronic Devices and Circuits - Section 1 (Q.No. 5)
5.
A transistor has a current gain of 0.99 in the CB mode. Its current gain in the CC mode is
Answer: Option
Explanation:
Current gain = 1 + β = 100.
Discussion:
30 comments Page 1 of 3.
Rakesh said:
1 decade ago
Sir I don't know about CB and CC please tell me briefly?
Kamran saifi said:
1 decade ago
CB is Common Base & CC is Common Collector mode. The current gain in CB mode is Ic/Ie & for CC mode it is Ic/Ib.
(1)
Nandan Kumar said:
1 decade ago
Please tell me about CB mode is Ic/Ie & CC mode is Ic/Ib.
R.Gayathri said:
1 decade ago
It was good give brief explanation.
Dilip said:
10 years ago
CB = ic/ie.
CC = ic/ib.
CC = ic/ib.
(1)
Hussain said:
10 years ago
Ic/ie = Alpha.
Ic/ib = beta, b = a/(1-a).
A given = 0.99.
So b = 99 = 1+99 = 100.
Ic/ib = beta, b = a/(1-a).
A given = 0.99.
So b = 99 = 1+99 = 100.
Billz said:
10 years ago
Why are you adding 1 to 99?
AJAY P ROY said:
9 years ago
In CB config: CURRENT GAIN (ALPHA, a) : Ic/Ie=.99 (given).
In CE config: CURRENT GAIN (BETA, b) : Ic/Ib.
In CC config: CURRENT GAIN (gamma, g) : Ie/Ib.
And b = a/(1 - a).
Then, g = 1 + b.
Therefore: g = 1 + (a/(1 - a)).
g =1/(1 - a) = 1/(1 - .99) = 1/.01 = 100.
In CE config: CURRENT GAIN (BETA, b) : Ic/Ib.
In CC config: CURRENT GAIN (gamma, g) : Ie/Ib.
And b = a/(1 - a).
Then, g = 1 + b.
Therefore: g = 1 + (a/(1 - a)).
g =1/(1 - a) = 1/(1 - .99) = 1/.01 = 100.
(4)
Gouarv said:
9 years ago
Very helpful, Thanks @Ajay P Roy.
Sai kumar kunchala said:
8 years ago
Excellent explanation Thanks, @Ajay.
Post your comments here:
Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers