Login/Sign Up
Home » For the transistor circuit shown in Fig.14.19, evaluate VE, RB, RE given IC = 1 mA, VCE = 3V, VBE = 0.5 V and VCC = 12 V, β = 100.
For the transistor circuit shown in Fig.14.19, evaluate VE, RB, RE given IC = 1 mA, VCE = 3V, VBE = 0.5 V and VCC = 12 V, β = 100.
CBSE | Class 12 | NCERT Exemplar | Physics | Semiconductor Electronics: Materials, Devices and Simple Circuits
For the transistor circuit shown in Fig.14.19, evaluate VE, RB, RE given IC = 1 mA, VCE = 3V, VBE = 0.5 V and VCC = 12 V, β = 100.

Answer:

{{I}_{C}}={{I}_{B}}+{{I}_{E}}

{{I}_{B}}<<<{{I}_{C}}

\therefore {{I}_{C}}={{I}_{E}}

{{I}_{C}}=1mA(given)

\therefore {{I}_{C}}={{I}_{E}}=1mA

By Kirchhoff’s loop law, in loop DGHJ

{{I}_{C}}{{R}_{C}}+{{I}_{E}}{{R}_{E}}+{{V}_{CE}}+{{V}_{CC}}

{{I}_{C}}({{R}_{C}}+{{R}_{E}})+{{V}_{CE}}=12

1\times {{10}^{-3}}\times \left( {{R}_{C}}+{{R}_{E}} \right)=12

{{R}_{C}}+{{R}_{E}}=\frac{9}{{{10}^{-3}}}

7.8\times {{10}^{3}}+{{R}_{E}}=9000

{{R}_{E}}=9000-7800=1200\Omega

{{V}_{E}}={{I}_{E}}\times {{R}_{E}}={{I}_{C}}\times {{R}_{E}}={{10}^{-3}}\times 1200

{{V}_{E}}=1.2V

{{V}_{BD}}={{V}_{E}}+{{V}_{BE}}=1.2+0.5=1.7V

I=\frac{{{V}_{B}}}{20000}=\frac{1.7}{20000}=0.085mA

{{R}_{B}}=\frac{{{V}_{BD}}}{I+{{I}_{B}}}

{{R}_{B}}=\frac{{{V}_{CC}}-{{V}_{BD}}}{\left[ 1+\left( \frac{{{I}_{C}}}{\beta } \right) \right]}

{{R}_{B}}=\frac{12-1.7}{\left[ 0.085+\frac{1}{100} \right]\times {{10}^{-3}}}=\frac{10.3\times {{10}^{3}}}{0.085+0.01}

{{R}_{B}}=108\times {{10}^{3}}\Omega

{{R}_{B}}=108k\Omega

i

More Material