Electronics and Communication Engineering - Analog Electronics - Discussion
Discussion Forum : Analog Electronics - Section 21 (Q.No. 31)
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
Discussion:
7 comments Page 1 of 1.
Mallli said:
1 decade ago
I want formulae.
SRINIVAS said:
1 decade ago
Alpha = 0.98, so Beta = 49.
Ic = alpha Ie + Ico = 985 uA.
Ib = (Ic/Beta) = 20 uA.
Ic = alpha Ie + Ico = 985 uA.
Ib = (Ic/Beta) = 20 uA.
Gowri said:
9 years ago
25μa B is the correct answer.
Ankit Kumar said:
9 years ago
But how? Explain.
Aphro said:
9 years ago
Why 15μA? Explain.
Prof.Arungandhi said:
8 years ago
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.
Ie= (1+beta) Ib;also beta=alpha/1-alpha.
So Ib= (1+beta) /Ie;sub we get 20 microAmpere.
Lobot Kigol said:
6 years ago
It's 15uA because of leakage current.
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