Numerical Problems-1 - Chapter 9 Semiconductor Electronics 12th Science Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated On May 15, 2024

By SaraNextGen

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:

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