Numerical Problems-2 - Chapter 9 - Semiconductor Electronics - 12th Physics Guide Samacheer Kalvi Solutions

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Numerical Problems
Question 1.
If the energy of a photon of sodium light $(\lambda=589 \mathrm{~nm})$ equals the band gap of a semiconductor, calculate the minimum energy required to create hole-electron pair.
Solution:
$
\begin{aligned}
\mathrm{E}=\mathrm{E}_g=\frac{\mathrm{hc}}{\lambda} & =\frac{6.6 \times 10^{-34} \times 3 \times 10^8}{589 \times 10^{-9}} \mathrm{~J} \\
& =\frac{6.6 \times 3 \times 10^{-17}}{589 \times 1.6 \times 10^{-19}} \mathrm{eV} ; \quad \mathrm{E}=2.1 \mathrm{eV}
\end{aligned}
$
Question 2.
In PNP transistor circuit, the collector current is $10 \mathrm{~mA}$. If $90 \%$ of the holes reach the collector, find emitter and base currents.
Solution:
Here, $\mathrm{I}_{\mathrm{E}}=10 \mathrm{~mA}$
As $90 \%$ of the holes reach the collector, so the collector current.
$
\begin{aligned}
& \mathrm{I}_{\mathrm{C}}=90 \% \text { of } \mathrm{I}_{\mathrm{E}} \\
& \mathrm{I}_{\mathrm{C}}=\frac{90}{100} \mathrm{I}_{\mathrm{E}} \\
& \mathrm{I}_{\mathrm{E}}=\frac{100}{90} \mathrm{I}_{\mathrm{C}}=\frac{100}{90} \times 10 \\
& \mathrm{I}_{\mathrm{E}} \simeq 11 \mathrm{~mA}
\end{aligned}
$
Base current, $\mathrm{I}_{\mathrm{B}}=\mathrm{I}_{\mathrm{E}}-\mathrm{I}_{\mathrm{C}}=11-10$ $\mathrm{I}_{\mathrm{B}}=1 \mathrm{~mA}$.

Question 3.
In the circuit, the value of $\beta$ is 100 . Find $\mathrm{I}_{\mathrm{B}}, \mathrm{V}_{\mathrm{CE}}, \mathrm{V}_{\mathrm{BE}}$ and $\mathrm{V}_{\mathrm{BC}}$, when $\mathrm{I}_{\mathrm{C}}=1.5 \mathrm{~mA}$. The transistor is in active, cut off or saturation state?
Solution:
$
\begin{aligned}
& \beta=100 ; I_C=1.5 \mathrm{~mA}=1.5 \times 10^{-3} \mathrm{~A}, \mathrm{~V}_{\mathrm{CC}}=24 \mathrm{~V} \\
& \beta=\frac{I_C}{I_B} \\
& \mathrm{I}_{\mathrm{B}}=\frac{I_C}{\beta}=\frac{1.5 \times 10^{-3}}{100}=15 \mu \mathrm{A}
\end{aligned}
$
To calculate $\mathrm{V}_{\mathrm{CE}}$, We apply Kirchhoff's rule to loop CEFDC, therefore
$
\begin{aligned}
& \mathrm{V}_{\mathrm{CC}}=\mathrm{I}_{\mathrm{C}} \times 4.7 \mathrm{k} \Omega+\mathrm{V}_{\mathrm{CE}} \\
& 24=1.5 \times 10^{-3} \times 4.710^{-3}+\mathrm{V}_{\mathrm{CE}} \\
& \mathrm{V}_{\mathrm{CE}}=24-7.05=16.95 \mathrm{~V}
\end{aligned}
$

Again, applying Kirchhoff's rule to loop ABEFDCA, We get,
$
\begin{aligned}
& \mathrm{V}_{\mathrm{CC}}=\mathrm{I}_{\mathrm{B}} \times 220 \mathrm{k} \Omega+\mathrm{V}_{\mathrm{BE}} \\
& \mathrm{V}_{\mathrm{CC}}=15 \times 10^{-6} \times 220 \times 10^{-3}+\mathrm{V}_{\mathrm{BE}} \\
& \mathrm{V}_{\mathrm{BE}}=24-3.3 \\
& \mathrm{~V}_{\mathrm{BE}}=20.7 \mathrm{~V}
\end{aligned}
$
Going along loop $\mathrm{ABCA}$, we get
$
\begin{aligned}
& \mathrm{I}_{\mathrm{B}} \times 220 \mathrm{k} \Omega+\mathrm{V}_{\mathrm{BC}}=\mathrm{I}_{\mathrm{C}} \times 4.7 \mathrm{k} \Omega \\
& 15 \times 10^{-6} \times 220 \times 10^3+\mathrm{V}_{\mathrm{BC}}=1.5 \times 10^{-3} \times 4.7 \times 10^3 \\
& \mathrm{~V}_{\mathrm{BC}}=7.05-3.3=3.75 \mathrm{~V}
\end{aligned}
$
As $\mathrm{V}_{\mathrm{CE}}<\mathrm{V}_{\mathrm{BE}}$, both the junctions are forward biased. So, the transistor is in the saturation state.
Question 4.
A transistor has $\alpha=0.95$. If the emitter current is $10 \mathrm{~mA}$, what is (a) the collector current, (b) the base current and (c) gain $\beta$ ?
Solution:
Here, $\alpha=0.95, \mathrm{I}_{\mathrm{E}}=10 \mathrm{~mA}$
(a) $\alpha=\frac{I_C}{I_E} \Rightarrow \mathrm{I}_{\mathrm{C}}=\alpha \mathrm{I}_{\mathrm{E}}=0.95 \times 10=9.5 \mathrm{~mA}$
(b) $\mathrm{I}_{\mathrm{B}}=\mathrm{I}_{\mathrm{E}}-\mathrm{I}_{\mathrm{C}}=10-9.5=0.5 \mathrm{~mA}$
(c) $\beta=\frac{\alpha}{1-\alpha}=\frac{0.95}{0.05}=19$.
Question 5.
For a BJT, the common-base current gain $\alpha=0.98$ and the collector base junction reverse bias saturation current $\mathrm{I}_{\mathrm{C} 0}=0.6 \mu \mathrm{A}$. This BJT is connected in the common emitter mode and operated in the active region with a base drive current $\mathrm{I}_{\mathrm{B}}=20 \mu \mathrm{A}$. Find the collector current $\mathrm{I}_{\mathrm{C}}$ for this mode of operation.
Solution:
$
\alpha=0.98 \text { and } \mathrm{I}_{\mathrm{C} 0}=0.6 \mu \mathrm{A}
$
Collector current,
$
\beta=\frac{\alpha}{1-\alpha}=\frac{0.98}{1-0.98}=49
$
Thus,
$
\begin{aligned}
& \mathrm{I}_{\mathrm{C}}=(49 \times 20)+(50 \times 0.6)=980+30=1010 \mu \mathrm{A} \\
& \mathrm{I}_{\mathrm{C}}=1.01 \mu \mathrm{A} .
\end{aligned}
$

Question 6.
An NPN BJT having reverse saturation current $\mathrm{I}_{\mathrm{S}}=10^{-15} \mathrm{~A}$ is biased in the forward active region with $\mathrm{V}_{\mathrm{BE}}=700 \mathrm{mV}$ and the current gain ( $\beta$ ) may vary from 50 to 150 due to manufacturing variations. What is the maximum emitter current (in $\mu \mathrm{A}$ )
Solution:
$
\begin{aligned}
& \mathrm{I}_{\mathrm{S}}=10^{-15} \mathrm{~A} \\
& \mathrm{~V}_{\mathrm{BE}}=700 \\
& \mathrm{~V}_{\mathrm{T}}=25 \mathrm{mV}
\end{aligned}
$
$\beta$ range from 50 to 150
$
\begin{aligned}
& \mathrm{I}_{\mathrm{C}}=\mathrm{I}_0 \mathrm{e}^{\left(\mathrm{V}_{\mathrm{BE}} / \mathrm{V}_{\mathrm{T}}\right)} \\
& \mathrm{I}_{\mathrm{E}}=\frac{\beta+1}{\beta} \mathrm{I}_{\mathrm{C}} \\
& \mathrm{I}_{\mathrm{E}}=\frac{\beta+1}{\beta} \mathrm{I}_{\mathrm{S}} \mathrm{e}^{\mathrm{V}_{\mathrm{BE}} / \mathrm{V}_{\mathrm{T}}} \\
&
\end{aligned}
$
$\mathrm{I}_{\mathrm{E}}$ will be maximum when $\beta$ is 50
$
\begin{aligned}
& =1.02 \times 10^{-15} \times \mathrm{e}^{700 \times 10^{-3} / 25 \times 10^{-3}} \\
& \mathrm{I}_{\mathrm{E}}=1475 \mu \mathrm{A}
\end{aligned}
$