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In the given figure, given that V_BB supply can vary from 0 to 5.0 V, $V_{CC} = 5 V ,\beta _{dc} = 200, R_{B}= 100 k\Omega , R_{C} = 1 k\Omega \ and \ V_{BE} = 1.0 V$ , The minimum base current and the input voltage at which the transistor will go to saturation, will be, respectively :

Option 1)
25 $\mu A$ and 3.5 V
Option 2)
25 $\mu A$ and 2.8 V
Option 3)
20 $\mu A$ and 2.8 V
Option 4)
20 $\mu A$ and 3.5 V

Answers (1)

best_answer

Relation between emitter current ,Base current,collector current -

$I_{E}= I_{B}+I_{C}$

- wherein

$I_{E}=Emitter\: Current$

$I_{B}= Base \: Current$

$I_{C}= Collector \: Current$

At saturation $V_{CE} = 0$

$V_{CE} = V_{CC}-I_{C}R_{C}$

$\Rightarrow V_{CC}=I_{C}R_{C}$

$I_{C} = \frac{5V}{1\times 10^{3}} = 5\times 10^{-3}A$

$I_{B} = \frac{5\times 10^{-3}}{200} = 25\mu A$

We known that

$V_{BB} = I_{B}R_{B} + V_{BE}$

$R_{B} = 100K\Omega$

$V_{BE} = 1V$

So $V_{BB} = 25\times 10^{-6}\times 100\times 10^{3}+1 = 2.5+1 =3.5V$

So $V_{BB} =3.5V and I_{B} = 25\mu A$

Option 1)
25 $\mu A$ and 3.5 V
Option 2)

25 $\mu A$ and 2.8 V

Option 3)

20 $\mu A$ and 2.8 V

Option 4)

20 $\mu A$ and 3.5 V

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