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February 21, 2012

Comments

John Dunn

I should have mentioned that this was how I once made the gain control portion of a couple of "leveller" designs using 1N914 diodes. (The 1N4148 didn't yet exist back then.) Needless to say, I suppose, is that I didn't push the 1N914 diodes to one ampere!

Francisco

A note of caution when using this formula at high forward voltages.The dynamic resistance does not decrease to zero exponentially with applied voltage, as this simple formula would suggest. A more accurate equation for the diode current would be: I=Is*(exp(q/kT*(V-I*R))-1), where R is the parasitic series resistance. When differentiated, this equation indicates that the dymamic resistance tends to a non-zero constant value equal to R.

George Storm

Hmm:

A simple rearrangement gives:
V = k.T.ln(I+Is)/q
which differentiates to give the low frequency value as:
Rdynamic = dV/dI = k.T/q/(I+Is)

However, even at low currents the resistance this would calculate is half the realistic value for a 2N914 (or 4148), and the error increases at higher currents. This is because the emission coefficient (N) of these diodes is approximately 2, and all real components have series resistance (Rs).

So we can write the forward Voltage Vf as:
V = k.T.N.ln(I+Is)/q + I.Rs
and the dynamic (low-frequency) resistance:
Rdynamic = dV/dI = k.T.N/q/(I+Is) + I.RS

More accurate, more intuitive, and no need for a program.
(and that's before we decide that the operating region allows us to omit Is)

Victor Koren

One note that is missing is to make clear that this is usefull only for small signals (really small).
You could add a calculation of the signal distortion vs. signal amplitude.

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