# Diode Voltage in Temperature

Discussion in 'Homework Help' started by matan, May 21, 2009.

1. ### matan Thread Starter New Member

May 21, 2009
1
0
I'm a little bit confused about the Vd Vs. Temperature.

The Diode Equation : [1] Id=Is(e^(Vd*q/n*k*T)-1),
so => [2] Vd=[k*T*n*ln((Id/Is)+1)]/q
Lets :
n=1
k=1.38e-23
T(°k)=273+T(°c)
q=1.6e-19
Id=1mA
Is=1e-12A

Vd/T=86.25e-6*20.72=1.79e-3

Vd(at 25°c)=1.79e-3*(273+25)=0.5326V
Vd(at 26°c)=1.79e-3*(273+26)=0.5343V

so ΔVd=Vd(26°c)-Vd(25°c)=0.5343-0.5326=1.7mV

In the litterature and in all the world at high temperature the Diode Voltage shouled be -2mV/°c ! and i see "+".

why the Vd at 26°c is higher then Vd at 25°c?
In the formula i need to write "-", because "Electron Charge" (q) ?
where is my misteke?
someone have an articale or link that can explain in simple?

10X

2. ### Ron H AAC Fanatic!

Apr 14, 2005
7,012
681
I think the reason is that Is is also a function of temperature. Have a look at this site, and go down to "temperature dependence".
See this site also. It is probably easier to understand.