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We owe a debt to Bell Labs with Shockley, Bardeen and Brattain for inventing the semiconductor. All have now passed on. I remember seeing my first transistor radio in the early 60's. Shockley developed the very important diode equation. I was reading through one of my good old engineering books where the dynamic resistance of a diode is derived. I'm trying to understand this because it's basic to many applications for transistors, Current mirrors, Long tailed pairs etc. The dynamic resistance is used to get the volt drop over the small signal ac resistance. They give an equation :

1) $I =Is(e^{eV/kT}-1)$

I is the diode current

V is the diode voltage

k is Boltzmann's contstant

T is the room temp in Kelvin

Is is the reverse leakage current

They then derive the dynamic resistance (as dV/dI) and get an answer as roughly 1/40I.

I can't get the same answer that they do. Instead, I have to use this equation from Wikipedia to get the same answer.

2) $I=Is(e^{V/Vt}-1)$

V is the diode voltage

Vt is the thermal volts. 25mV at room temperature.

Are they using the wrong equation in (1) , or have I worked it out wrong?

Thanks in advance.

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They're the same thing....… – Chris Gerig Jun 10 '12 at 8:56
As Chris says, if you calculate kT/e at 298K it's 25mV i.e. the same as $V_t$ in your second equation, so your two equations are exactly the same. If you get different answers with the two equations it must be a slip of the pen. Can you expand your post to show your working - we might be able to spot the problem. – John Rennie Jun 10 '12 at 19:10
Hi John, Vt=kT/q =1.38x10^-23x298/1.6x10^-19volts =25mV – johann Jun 11 '12 at 23:53
but kT/e is in joules 1.38x10^-23 x 298/2.718 joules not the same answer ? – johann Jun 12 '12 at 0:02
e here is not 2.718..., but the electron charge: 1.602E-19 coulomb. – Art Brown Jun 16 '12 at 6:46

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