Modern Physics 6 Question 2
2. The transfer characteristic curve of a transistor, having input and output resistance $100 \Omega$ and $100 k \Omega$ respectively, is shown in the figure. The voltage and power gain, are respectively,
(Main 2019, 12 April I)
(a) $2.5 \times 10^{4}, 2.5 \times 10^{6}$
(b) $5 \times 10^{4}, 5 \times 10^{6}$
(c) $5 \times 10^{4}, 5 \times 10^{5}$
(d) $5 \times 10^{4}, 2.5 \times 10^{6}$
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Solution:
- Given curve is between $I _c$ and $I _b$ as output and input currents, respectively
So, it is transfer characteristics curve of a common emitter (CE) configuration.
In $CE$ configuration,
Current gain, $\beta=\frac{I _{\text {out }}}{I _{\text {in }}}=\frac{I _c}{I _b}$
Voltage gain,
$$ A _V=\frac{V _{\text {out }}}{V _{\text {in }}}=\frac{I _c \times R _{\text {out }}}{I _b \times R _{\text {in }}}=\beta \times \frac{R _{\text {out }}}{R _{\text {in }}} $$
and power gain
$$ A _P=\frac{P _{\text {out }}}{P _{\text {in }}}=\frac{I _c^{2} \times R _{\text {out }}}{I _b^{2} \times R _{\text {in }}}=\beta^{2} \times \frac{R _{\text {out }}}{R _{\text {in }}} $$
Given, $R _{\text {in }}=100 \Omega$ and $R _{\text {out }}=100 \times 10^{3} \Omega$
From Eq. (i), we get
$$ \begin{aligned} \beta & =\frac{5 mA}{100 \mu A} \text { or } \frac{10 mA}{200 \mu A} \text { or } \frac{15 mA}{300 \mu A} \text { or } \frac{20 mA}{400 \mu A} \\ \Rightarrow \quad \beta & =\frac{5 \times 10^{-3}}{100 \times 10^{-6}}=50 \ldots \text { (iv) } \end{aligned} $$
From Eqs. (ii) and (iv), we get
Voltage gain, $A _V=\beta \times \frac{R _{\text {out }}}{R _{\text {in }}}=50 \times \frac{100 \times 10^{3}}{100}$
$\Rightarrow$
$$ A _V=50000=5 \times 10^{4} $$
From Eqs. (iii) and (iv), we get
Power gain, $A _P=\beta^{2} \times \frac{R _{\text {out }}}{R _{\text {in }}}$
$$ \begin{aligned} & =(50)^{2} \times \frac{100 \times 10^{3}}{100} \\ & =2500 \times 1000 \\ \Rightarrow \quad A _P & =2.5 \times 10^{6} \end{aligned} $$
