currentdiodesvoltage
I'm having some trouble working with the Shockley's Ideal Diode Equation. I'm trying to generate a model plot for a 1N4001 type semiconductor diode, but the resulting graph seems incorrect.
Is (Reverse Saturation Current) = 5 * 10^-6 A
Vt (Thermal Voltage) = 25 * 10^-3 V
n (Ideality Factor) = 1
Id = Is * (e^(Vd/(n*Vt) - 1)
Nearly everything I've read states that the "Turn On" voltage should be closer to 0.7V. I'm inclined to believe that I'm doing something wrong. Any ideas?
Try looking for eg
tunnel diode transfer function
You are unlikely to get a single tidy expression due to the non linear nature of the beast and its "discontinuous" stable transfer function - which is what it's all about.
Some equations here near 1st diagrams.
The "magic" part is the negative resistance region - shown in red below.
Below is the classic Esaki diode / Tunnel diode transfer function and you can try to fit formulae to that if you think it useful. Unlike most other devices you cxan have the same current at 3 different voltages - with two being stable, and the intermediate voltage being a transient one due to it being in a region of negative resistance.
You could model this as something like y=x^3 with some offsets and shape adjustments but that would be missing the main point of the device.
Related:
Useful comment in [oscillator applications](http://www.powerguru.org/crystal-oscillator- design-and-negative-resistance/)
Here is a completely made up example.
The formula is shown below the graph.
This was made "by eye" solely on the basis of approximating an Esaki diode V/I curve. By adjusting parameters you could probably get a "close enough" fit to a real diode.
I doubt if this equation is especially useful but it does allow you to get a rough model of a real device. Axis units are semi arbitrary - adjust parameters to suit.
If you increase voltage above the maximum seen here you'll get a nasty shock due to the (5-Volts) term - no suggestion that this occurs in real life :-)
Off the Wall:
Plot an energy versus velocity curve for a mass approaching and exceeding light speed.
For mass use standard
M_relativistic = Rest_mass/(1- V^2/C^2)
Use Energy = 0.5 m x v^2.
Ignore (for now) complex-imaginary aspect for V>C.
eg assume for now that jE = E.
Plot energy against V.
Wow!
Ponder.
-> You get a tunneling type curve with two velocities > C that have the same energy as a point for V < C for values of V only above a certain % of light speed. E goes to infinity at V=C as expected. BUT the curve is suggestive of a tunnel effect across C with two velocities > C with the same energy. There's more, but that's enough OTW already :-).
Implications of the tachyonic jE are, of course, unknown.
But you aren't modelling an ideal diode, you're modelling a physical diode to first order. An ideal diode has zero 'on' resistance and infinite 'off' resistance.
Your model should look like this:
simulate this circuit – Schematic created using CircuitLab
Best Answer
Your value for \$I_S\$ is way too high. Where did it come from? Note that the ideality factor is usually closer to 2 for rectifier diodes.
Here is a typical model:-
.MODEL D1N4001 D IS=29.5E-9 RS=73.5E-3 N=1.96 CJO=34.6P VJ=0.627 +M=0.461 BV=60 IBV=10U
None of the numbers in the datasheet you linked are directly useful in determining \$I_S\$, however you could use this figure from the datasheet:
and play with different values of \$I_S\$ until you get reasonable agreement at lower current values. I say "lower current values" because real rectifier diodes have a resistive component to their \$V_F\$ which is more important at higher currents, and your ideal equation does not take that into account. Although \$I_S\$ is temperature-dependent, the figure specifies that it is measured with brief pulses of current, so we can assume that the \$T_J\$ is 25°C.