In the figure, given that \[{{V}_{BB}}\]supply can vary from 0 to \[5.0V,\]\[{{V}_{CC}}=5V,{{\beta }_{dc}}=200,\]\[{{R}_{B}}=100k\Omega ,\]\[{{R}_{C}}=1\,k\Omega \]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
[JEE Main Online Paper Held On 12-Jan-2019 Evening]
A) \[20\mu A\]and 3.5 V
B) \[25\mu A\]and 3.5V
C) \[20\mu A\] and 2.8 V
D) \[25\mu A\] and 2.8 V
Correct Answer: B
Solution :
Applying KVL at output and input circuit \[{{V}_{CB}}={{V}_{CC}}-{{I}_{C}}{{R}_{C}}\] \[{{V}_{BB}}={{I}_{B}}{{R}_{B}}+{{V}_{BE}}\] At saturation,\[{{V}_{CB}}=0\Rightarrow {{V}_{CC}}={{I}_{C}}{{R}_{C}}\] \[\Rightarrow \]\[{{I}_{C}}=\frac{{{V}_{CC}}}{{{R}_{C}}}=\frac{5v}{1k\Omega }=5mA\] Base current,\[{{I}_{B}}=\frac{{{I}_{C}}}{\beta }=\frac{5}{200}=25\mu A\] Using equation (i), \[{{V}_{BB}}=(25\times {{10}^{-6}})(100\times {{10}^{3}})+1=2.5+1=3.5V\]
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