In a bipolar transistor at room temperature, if the emitter current is doubled, the voltage across its base-emitter junction

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Solution

Ans : (c)

$I_{e} = I_{o} [e^{V_{b e} / η V_{T}} - 1]$
$\frac{2 I_{1}}{I_{1}} = \frac{I_{0} [e^{V_{b e_{2}} / η V_{T}} - 1]}{I_{0} [[e^{V_{b e_{1}} / η V_{T}} - 1]}$
$\frac{2}{1} = \frac{e^{V_{b e_{2}} / η V_{T}} - 1}{e^{V_{b e_{1}} / η V_{T}} - 1}$ , $e^{V_{b e_{2}} / η V_{T}} >> 1$
$e^{V_{b e_{1}} / η V_{T}} >> 1$ $\frac{e^{V_{b e_{2}} / η V_{T}}}{e^{V_{b e_{1}} / η V_{T}}} = 2$
$e^{(V_{b e_{2}} - V_{b e_{1}}) η V_{T}} = 2$
$\frac{V_{b e_{2}} - V_{b e_{1}}}{η V_{T}} = ln 2$
$V_{b e_{2}} - V_{b e_{1}} = η V_{T} (0.693)$
Take $η = 1$ $V_{b e_{2}} - V_{b e_{1}} = 1 \times 0.026 \times 0.693$
$V_{b e_{2}} - V_{b e_{1}} = 18 m V$
$V_{b e_{2}} - V_{b e_{1}} ≅ 20 V$

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Similar questions

α = 0.98

I_{B} = 25 μ A

I_{C B O} = 200 n A

β = 100.

V_{B E} = - 0.7 V,

A Bipolar junction transistor has α = 0.98

I_{B} = 25 μ A and I_{C B O} = 200 n A

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