Electronics and Communication Engineering - Analog Electronics - Discussion
Discussion Forum : Analog Electronics - Section 10 (Q.No. 3)
3.
A certain transistor has adc of 0.98 and a collectors leakage current of 5 μA. If IE = 1 mA, the collector current will be
Discussion:
3 comments Page 1 of 1.
S naveena said:
9 years ago
ie = ic + ib.
Ic = B ib (where B is beta).
Ib = ic/B.
B = adc/1 - adc (where adc is alpha).
B = 0.98/1 - 0.98 = 49.
Ic = ie * B/1 + B.
1 * 49/1 + 49 = 0.98 ma (where ma is mills ampere).
Ic = B ib (where B is beta).
Ib = ic/B.
B = adc/1 - adc (where adc is alpha).
B = 0.98/1 - 0.98 = 49.
Ic = ie * B/1 + B.
1 * 49/1 + 49 = 0.98 ma (where ma is mills ampere).
(1)
B T Madhav said:
2 years ago
Ic = Alpha times Ie + Icbo.
Rajashekar balya said:
8 years ago
Here, a=Ic÷ Ie.
Post your comments here:
Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers