Modern Physics Ques 253
- A common emitter amplifier circuit, built using an $n-p-n$ transistor, is shown in the figure. Its DC current gain is $250$ , $R _C=1 $ $k \Omega$ and $V _{C C}=10 $ $V$. What is the minimum base current for $V _{C E}$ to reach saturation?
(Main 2019, 8 April II)
(a) $40 $ $\mu A$
(b) $10 $ $\mu A$
(c) $100 $ $\mu A$
(d) $7 $ $\mu A$
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Answer:
Correct Answer: 253.(a)
Solution:
Formula:
- For a common emitter $n-p-n$ transistor, DC current gain is
$ \beta _{DC}=\frac{I _C}{I _B} $
At saturation state, $V _{C E}$ becomes zero.
$ \begin{aligned} \therefore \quad & V _{C C}-I _C R _C=0 \\ & I _C \approx \frac{V _{C C}}{R _C}=\frac{10}{1000}=10^{-2} A \end{aligned} $
Hence, saturation base current,
$ I _B=\frac{I _C}{\beta _{DC}}=\frac{10^{-2}}{250}=40 $ $\mu A $
BETA
