Electronics and Communication Engineering - Electronic Devices and Circuits - Discussion

Discussion Forum : Electronic Devices and Circuits - Section 4 (Q.No. 19)
19.
In a bipolar junction transistor adc = 0.98, ICO= 2 μA and 1B = 15 μA. The collector current IC is
635 mA
735 mA
835 mA
935 mA
Answer: Option
Explanation:

Discussion:
5 comments Page 1 of 1.

GlennSid said:   3 years ago
Hence Ic =Ib*β

Saurabh kumar said:   5 years ago
Ic = b * ib+(1+b)ico.
ic = 49 * 15 + 50 * 2,
Ic = 0.835ma.
(1)

Weeknd2020 said:   5 years ago
Ic = Beta (Ib) + Ico.

Sandy said:   8 years ago
The correct answer is C in micro amp.

Amit kumar said:   10 years ago
Ic = B*Ib+(1+B)*Ico.

If y apply this formula why will get 835 mA.

So correct answer is 'C'.

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