Refer to the common emitter amplifier circuit shown below, using a transistor with \[\beta =80\] and \[{{V}_{BE}}=0.7volt\]. The value of resistance\[{{\operatorname{R}}_{B}}\] is |
A) \[330\Omega \]
B) \[330k\Omega \]
C) \[220\Omega \]
D) \[220k\Omega \]
Correct Answer: D
Solution :
\[4-{{I}_{B}}{{R}_{B}}-{{V}_{BE}}=0\]\[\Rightarrow \]\[{{I}_{B}}{{R}_{B}}=4-0.7=3.3volt\] ...(i) |
\[6-{{I}_{C}}=\left( 2.5\times {{10}^{3}} \right)-3V=0\]\[\Rightarrow \]\[{{I}_{C}}=\frac{3}{2.5\times {{10}^{3}}}=1.2\times {{10}^{-3}}\] |
\[=-\frac{{{I}_{C}}}{{{I}_{B}}}=B\Rightarrow \frac{1.2\times {{10}^{-3}}}{{{I}_{B}}}=80\]\[\Rightarrow \]\[{{I}_{B}}=\frac{1.2\times {{10}^{-3}}}{800}\]\[\Rightarrow \]\[{{I}_{B}}=1.5\times {{10}^{-5}}A\] |
Substituting in eq. (i) |
\[1.5\times {{10}^{-5}}\times {{R}_{B}}=3.3\] |
\[{{R}_{B}}=\frac{3.3}{1.5\times {{10}^{-5}}}=\frac{33}{15}\times {{10}^{5}}\] |
\[=2.2\times {{10}^{-5}}\Omega \] |
\[{{\operatorname{R}}_{B}}=220k\Omega \] |
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