MCQ
For the transistor circuit shown below, if $\beta = 100,$ voltage drop between emitter and base is $0.7\ V$ then value of $V_{CE}$ will be.....$V$
- A$10$
- B$5$
- ✓$13$
- D$0$
✓
Answer
Correct option: C.
$13$
c
$i_{b}=\frac{5-0.7}{8.6}=0.5 m A \Rightarrow I_{c}=\beta I_{b}=100 \times 0.5 \mathrm{mA}$
$i_{b}=\frac{5-0.7}{8.6}=0.5 m A \Rightarrow I_{c}=\beta I_{b}=100 \times 0.5 \mathrm{mA}$
By using
$V_{C E}=V_{G G}-I_{C} R_{L}=18-50 \times 10^{-3} \times 100=13 V$
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