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), \[{{V}_{BB}}=(25\times {{10}^{-6}})(100\times {{10}^{3}})+1=2.5+1=3.5V\]You need to login to perform this action.
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