Question

the voltage across diode at room temperature of 300 k is 0.71v when 2.5 ma current flows through it .if the voltage increases to 0.8v .calculate the new diode current

Answer 1

∆v=(0.8-0.71)=0.09v

T=300k

$I_0=2.5mA$

$I=I_0e^\frac{e∆V}{KT}$

Put value

$$I=2.5\times10^{-3}(e^\frac{1.6\times10^{-19}\times0.09}{2\times1.38\times10^{-23}\times300}-1)$$

$$I=2.5\times10^{-3}(e^{1.76}-1)\\I=2.5\times10^{-3}(5.69-1)=11.7mA$$