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)
$$\eqalign{ & {I_{\text{C}}} = \frac{{{I_{{\text{CO}}}}}}{{1 - \alpha }} + \frac{\alpha }{{1 - \alpha }} \times {I_\beta } \cr & = \frac{6}{{1 - 0.98}} + \frac{{0.98}}{{1 - 0.98}} \times 100 \cr & = 5.2\,{\text{mA}} \cr} $$Join The Discussion
Comments ( 3 )
Related Questions on Analog Electronics
The action of JFET in its equivalent circuit can best be represented as a
A. Current controlled Current source
B. Current controlled voltage source
C. Voltage controlled voltage source
D. Voltage controlled current source
In a p+n junction diode under reverse bias, the magnitude of electric field is maximum at
A. The edge of the depletion region on the p-side
B. The edge of the depletion region on the n-side
C. The p+n junction
D. The center of the depletion region on the n-side
To prevent a DC return between source and load, it is necessary to use
A. Resistor between source and load
B. Inductor between source and load
C. Capacitor between source and load
D. Either A or B
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???