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:

  1. 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, $A _P=\beta^{2} \times \frac{R _{\text {out }}}{R _{\text {in }}}$

$$ \begin{aligned} & =(50)^{2} \times \frac{100 \times 10^{3}}{100} \\ & =2500 \times 1000 \\ \Rightarrow \quad A _P & =2.5 \times 10^{6} \end{aligned} $$



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