Just wondering how with DC measurements you could measure a diodes saturation current and the Emission coefficient factor N. I was thinking a basic DC circuit varying the voltage taking voltage measurements, then plot the V-I and then try fit the diode equation to the graph?


Two measurements should be enough for the basic Shockley model.

  1. Measure current with fairly large reverse bias, to give you Is. (for example, -20V for a 1N4148)

  2. Measure forward voltage with fairly large forward current (for example, 20mA for a 1N4148) and calculate the emission coefficient:

    \$n = \frac{V_F}{V_T \cdot ln(I/I_S)}\$


Vf is measured forward voltage

I is the test current

Is is the saturation current from step 1

Vt is the thermal voltage calculated from kT/q where T is the junction temperature in Kelvin, q is the charge of an electron and k is the Boltzmann constant.

Very small diodes (or larger diodes at high current) will have a significant resistive term (not modeled by Shockley) that may become significant, in which case you can plot n vs. I over a range to eliminate that effect (or make at least one more measurement and eliminate it mathematically).

  • \$\begingroup\$ is it ln(I/Is + 1)? Thankyou by the way \$\endgroup\$ – JS60 Nov 4 '14 at 23:34
  • \$\begingroup\$ The 1 can be ignored, but leave it in and see what difference it makes. \$\endgroup\$ – Spehro Pefhany Nov 5 '14 at 2:20

I was going to suggest measuring Vf at several current's with decade type steps. 100nA, 1uA, 10uA, 100uA... at higher currents there is a resistive term that creeps in. And plot that against the diode equation.
(at higher currents the diode can heat up too.)

I've never had any luck trying to measure the Is in reverse, I mean in theory it should be independent of the reverse voltage.. and then at some point the diode breaks down.


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