First slide

Diodes

Question

The saturation current density of a P-N junction germanium diode is $250 m A / m^{2}$ at 300 K. Find the voltage ( in V ) that would have to be applied across the junction to cause a forward current density $10^{5} A / m^{2}$ to flow. ( $K = 1.38 \times 10^{- 23} J / K$ )

Very difficult

Solution

$T h e c u r r e n t i n t h e j u n c t i o n d i o d e i s g i v e n b y I = I_{S} (e^{\frac{e V}{η K T}} - 1) F o r G e η = 1$
$D i v i d i n g b y c r o s s - s e c t i o n a r e a, w e g e t J = J_{S} (e^{\frac{e V}{K T}} - 1) \to (1) P u t$
$J_{S} = 250 m A / m^{2} = 250 \times 10^{- 3} A / m^{2} J = 10^{5} A / m^{2}, T = 300 K$
$A f t e r s o l v i n g (1), \frac{J}{J_{s}} + 1 = e^{\frac{e V}{K T}}$
$\Rightarrow \ln (\frac{J}{J_{s}} + 1) = \frac{e V}{K T} \Rightarrow V = \frac{K T}{e} \ln (\frac{J}{J_{s}} + 1)$
$\Rightarrow V = \frac{1.38 \times 10^{- 23} \times 300}{1.6 \times 10^{- 19}} \ln (\frac{10^{5}}{250 \times 10^{- 3}} + 1)$
$\Rightarrow V = 258.75 \times 10^{- 4} \times 12.899 = 0.33 V$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App