Modern Physics 6 Question 9
9. In the figure, given that $V _{B B}$ supply can vary from 0 to $5.0 V$, $V _{C C}=5 V, \quad \beta _{DC}=200, \quad R _B=100 k \Omega, \quad R _C=1 k \Omega$ and $V _{B E}=1.0 V$. The minimum base current and the input voltage at which the transistor will go to saturation, will be, respectively
(Main 2019, 12 Jan II)
(a) $25 \mu A$ and $2.8 V$
(b) $25 \mu A$ and $3.5 V$
(c) $20 \mu A$ and $3.5 V$
(d) $20 \mu A$ and $2.8 V$
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Solution:
- Transistor saturation occurs when $V _{C E}=0$.
Now, for closed loop of collector and emitter by
Kirchhoff’s voltage rule, we have
$$ \begin{aligned} & V _{C E}=V _{C C}-I _C R _C \Rightarrow 0=V _{C C}-I _C R _c \\ \Rightarrow \quad & I _C=\frac{V _{C C}}{R _C}=\frac{5}{1 \times 10^{3}}=5 \times 10^{-3} A \end{aligned} $$
Now, $\beta _{DC}=200$ (given) $\Rightarrow \frac{I _C}{I _B}=\beta _{DC}=200$
$$ \begin{array}{ll} \Rightarrow & I _B=\frac{I _C}{200}=\frac{5 \times 10^{-3}}{200} \\ \Rightarrow & I _B=2.5 \times 10^{-5}=25 \mu A \end{array} $$
Now, we apply Kirchhoff’s voltage rule in base-emitter closed loop, we get
