Numerical Problems-1 - Chapter 9 - Semiconductor Electronics - 12th Physics Guide Samacheer Kalvi Solutions

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Numerical Problems
Question 1.
The given circuit has two ideal diodes connected as shown in figure below. Calculate the current flowing through the resistance $\mathrm{R}_1$.
Answer:
Diode $D_1$ is reverse biased so, it will block the current and Diode $D_2$ is forward biased, so it will pass the current.
Current in the circuit is
$
\begin{aligned}
& \mathrm{I}=\frac{V}{R_s}=\frac{10}{2+2}=\frac{10}{4}=2.5 \mathrm{a} \\
& \mathrm{I}=2.5 \mathrm{~A}
\end{aligned}
$

Net circuit Voltages $=4-(0.7+0.7)=41.4$
$
\mathrm{V}=2.6 \mathrm{~V}
$
Total circuit resistance $=1+18+1$
$
\mathrm{R}=20 \Omega
$
$\therefore$ Circuit Current $\mathrm{I}=\frac{V}{R}=\frac{2.6}{20}$
Question 3.
Assuming $\mathrm{V}_{\mathrm{CEsat}}=0.2 \mathrm{~V}$ and $\beta=50$, find the minimum base current $\left(\mathrm{I}_{\mathrm{B}}\right)$ required to drive the transistor given in the figure to saturation.
Solution:

$\begin{aligned}
\beta & =50 \text { and } V_{E B}=600 \mathrm{mV} \\
\mathrm{V}_{\mathrm{EB}} & =\mathrm{V}_{\mathrm{E}}-\mathrm{V}_{\mathrm{B}} \\
\mathrm{V}_{\mathrm{B}} & =\mathrm{V}_{\mathrm{E}}-\mathrm{V}_{\mathrm{EB}}=3-0.6 \\
\mathrm{~V}_{\mathrm{B}} & =2.4 \mathrm{~V} \\
\mathrm{I}_{\mathrm{B}} & =\frac{\mathrm{V}_{\mathrm{B}}}{\mathrm{R}_{\mathrm{B}}}=\frac{2.4}{60 \mathrm{k}}=40 \mu \mathrm{A} \\
\mathrm{I}_{\mathrm{C}} & =\beta \mathrm{I}_{\mathrm{B}}=50 \times 40 \mu \mathrm{A}
\end{aligned}$

Question 4.
A transistor having $\alpha=0.99$ and $\mathrm{V}_{\mathrm{BE}}=0.7 \mathrm{~V}$, is given in the circuit. Find the value of the collector current.
Solution:

$
\begin{aligned}
\mathrm{V}_{\mathrm{BE}} & =\mathrm{V}_{\mathrm{CC}}-\mathrm{I}_{\mathrm{B}} \mathrm{R}_{\mathrm{B}} \\
0.7 & =12-\mathrm{I}_{\mathrm{B}}(10 k) \\
\mathrm{I}_{\mathrm{B}} & =\frac{12-0.7}{10 k}=1.13 \mathrm{~mA} \\
\beta & =\frac{\alpha}{1-\alpha}=\frac{0.99}{1-0.99}=99 \\
\beta & =\frac{\mathrm{I}_{\mathrm{C}}}{\mathrm{I}_{\mathrm{B}}} \Rightarrow \mathrm{I}_{\mathrm{C}}=\beta \mathrm{I}_{\mathrm{B}}=99 \times 1.13 \mathrm{~mA} \\
\mathrm{I}_{\mathrm{C}} & =111.87 \mathrm{~mA}
\end{aligned}
$
Question 5.
In the circuit shown in the figure, the BJT has a current gain ( $\beta$ ) of 50 . For an emitter - base voltage $\mathrm{V}_{\mathrm{EB}}=600 \mathrm{mV}$, calculate the emitter - collector voltage $\mathrm{V}_{\mathrm{EC}}$ (in volts).
Solution:

$\begin{aligned}
\beta & =50 \text { and } \mathrm{V}_{\mathrm{EB}}=600 \mathrm{mV} \\
\mathrm{V}_{\mathrm{EB}} & =\mathrm{V}_{\mathrm{E}}-\mathrm{V}_{\mathrm{B}} \\
\mathrm{V}_{\mathrm{B}} & =\mathrm{V}_{\mathrm{E}}-\mathrm{V}_{\mathrm{EB}}=3-0.6 \\
\mathrm{~V}_{\mathrm{B}} & =2.4 \mathrm{~V} \\
\mathrm{I}_{\mathrm{B}} & =\frac{\mathrm{V}_{\mathrm{B}}}{\mathrm{R}_{\mathrm{B}}}=\frac{2.4}{60 \mathrm{k}}=40 \mu \mathrm{A} \\
\mathrm{I}_{\mathrm{C}} & =\beta \mathrm{I}_{\mathrm{B}}=50 \times 40 \mu \mathrm{A} \\
\mathrm{I}_{\mathrm{C}} & =2 \mathrm{~mA} \\
\mathrm{~V}_{\mathrm{C}} & =\mathrm{R}_{\mathrm{E}} \cdot \mathrm{I}_{\mathrm{C}}=500 \mathrm{I}_{\mathrm{C}}=500 \times 2 \mathrm{~mA}=1 \mathrm{~V} \\
\therefore \mathrm{V}_{\mathrm{EC}} & =\mathrm{V}_{\mathrm{E}}-\mathrm{V}_{\mathrm{C}}=3-1 \\
\mathrm{~V}_{\mathrm{EC}} & =2 \mathrm{~V}
\end{aligned}$