If $$\alpha $$ = 0.98, $${I_{{\text{CO}}}} = 6\mu {\text{A}}$$ and $${I_\beta } = 100\mu {\text{A}}$$ for a transistor, then the value of $${I_{\text{C}}}$$ will be
A. 2.3 mA
B. 3.2 mA
C. 4.6 mA
D. 5.2 mA
Answer: Option D
Solution (By Examveda Team)
$$\eqalign{ & {I_{\text{C}}} = \frac{{{I_{{\text{CO}}}}}}{{1 - \alpha }} + \frac{\alpha }{{1 - \alpha }} \times {I_\beta } \cr & = \frac{6}{{1 - 0.98}} + \frac{{0.98}}{{1 - 0.98}} \times 100 \cr & = 5.2\,{\text{mA}} \cr} $$This Question Belongs to Electronics And Communications Engineering >> Analog Electronics
β = α/(1-α) ; Ic = βIc + (1 + β)Ico = 5.2 mA
there is a mistake in the formula current formula is Ic = β* Ib + (1 + β)Ico
FORMULA???