At $300$ K temperature the saturated current in Germanium diode is $10$ mA. Find the:
(i) Forward voltage when diode current is $100$ mA.
(ii) Diode current at backward biased voltage $001$ volt.

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Solution

Given,
Satured current $I_{s} = 10 μ A$
$= 10 \times 10^{- 6}$ A
$= 10^{- 5}$ A
Potential barriers for germanium diode $= 0.3$
(i) $V = R (I + I_{s})$
$V = 0.3 [(100 \times 10^{- 3}) + 10^{- 9}]$
$V = 0.3 [\frac{1}{10} + \frac{1}{10}]$
$V = 0.3$
(ii) $V = R (I + I_{s})$
$1 \times 10^{- 2} = 0.3 (I + 10^{- 5})$
$- 10^{- 5} + \frac{10^{- 2}}{0.3} = I$
$I = 0.33$ .

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At 300K temperature the saturated current in Germanium diode is 10mA. Find the:(i) Forward voltage when diode current is 100mA.(ii) Diode current at backward biased voltage 001 volt.

Given, Satured current Is=10μA=10×10−6A=10−5APotential barriers for germanium diode =0.3(i) V=R(I+Is)V=0.3[(100×10−3)+10−9]V=0.3[110+110]V=0.3(ii) V=R(I+Is)1×10−2=0.3(I+10−5)−10−5+10−20.3=II=0.33.

At $300$ K temperature the saturated current in Germanium diode is $10$ mA. Find the:
(i) Forward voltage when diode current is $100$ mA.
(ii) Diode current at backward biased voltage $001$ volt.