Diode analysis, reverse-bias saturation current

[Bucephalus]

Hi there

I'm quite inexperienced at electronics, but I'm reading about the diode at the moment, and it talks about various ways to model the diode in a simple circuit analysis. PIecewise linear model, iterative and graphical. It says piecewise is less complex but usually good enough, but not as accurate as using the ideal diode equation in the iterative method. The problem is the iterative method uses the ideal diode equation and it's the following:

Id = IsR[e^(Vd/Vt)-1] + Vd

Is = the reverse-bias saturation current, and it says it's assumed to be known for a diode and a few pages earlier it says that the reverse-bias saturation current is normally around 10^-13 to 10^-15 A, very small. In the example in the book they use 10^-13A. I look on data sheets but I can't find this value, so how are we supposed to use the more accurate iterative method when the reverse-bias saturation current is not disclosed on the datasheets?

I'm taking it that most people are use the piecewise linear model for analysis if they do any analysis at all and that's fine, but I'm just posing the question the question to see if anybody is familiar with the problem I'm talking about?

regards
David

 

[BradtheRad]

Reverse-bias is not a normal way to use a diode, and all bets are off as to whether it will change its operating characteristics.

Seems to me, if it were feasible with ordinary diodes, they would be zener diodes, because zener diodes are made to operate this way.

When I have measured reverse current through a good diode, my meter shows nothing.

- - - Updated - - -

If you want to consider some kind of equation, this might do for a rough idea.

Suppose Vapplied is greater than Vthresh (reverse-bias threshold), then:

A = ( .5+ Vapplied - Vthresh ), raised to an arbitrary exponent such as 3 or 4.

Example, if the diode is rated 50 V, and it is exposed to 50.3 V, then it will conduct 0.8 cubed, or 0.5 A.
If exposed to 52 V, it will conduct 2.5 cubed, or 15.625 A.

 

[FvM]

I look on data sheets but I can't find this value, so how are we supposed to use the more accurate iterative method when the reverse-bias saturation current is not disclosed on the datasheets?

You find IS in simulation models, but not in datasheets. The value is strongly temperature dependent, by the way. But if you like to do calculation based on IS and the Shockley diode equation, you can easily derive it from a forward transfer characteristic.

Reverse-bias is not a normal way to use a diode, and all bets are off as to whether it will change its operating characteristics.

I didn't get the point. Diodes are often operated in reverse bias, e.g. when used as a rectifier. You probably mean to say that the actual reverse bias current isn't exactly specified.

You should consider that the said theoretical IS value doesn't describe the reverse bias characteristic completely. According to the Shockley equation, the reverse bias current would be constant above a few multiples of VT. As we all know, there will be an additional voltage dependent reverse bias current.

 

[crutschow]

If you want to more accurately model a diode in a circuit you should use a Spice type simulator, for example the free LTspice program from Linear Technology. It includes models for many common diodes and utilizes the full diode equation to determine the diode performance in a circuit using many small iterative calculations. Otherwise you just look at the diode's data sheet to determine the approximate forward voltage drop for a given forward current.

 

[BradtheRad]

I wrote: Reverse-bias is not a normal way to use a diode, and all bets are off as to whether it will change its operating characteristics.

FmM wrote:

I didn't get the point. Diodes are often operated in reverse bias, e.g. when used as a rectifier. You probably mean to say that the actual reverse bias current isn't exactly specified.

Yes, we reverse-bias a diode when using it to block current. However if we reverse-bias it so much that it carries current, I think that means we exceeded its safe parameter of operation.

I'm not 100 percent certain that would ruin it...
but it could ruin it, and I'm too economical-minded to risk ruining a good diode even if I did scavenge it for free from a junk circuit board.
:^)

 

[FvM]

Yes, we reverse-bias a diode when using it to block current. However if we reverse-bias it so much that it carries current, I think that means we exceeded its safe parameter of operation.

Any diode in reverse bias will "carry current". As far as we are discussing diode modelling, there's no principle difference between pA or A reverse bias current.

 

[Bucephalus]

Thankyou all for your information. I have learned a few things here. In fact the example in the book, was working out a forward bias voltage drop but using Is.
I'm really more concerned with understanding than doing, i.e. I would prefer to learn how to do it on paper than in spice program.
Question: is the Shockley Diode equation the ideal diode equation that we are talking about?
I will also google this and the forward transfer characteristic to learn more.

Thankyou.

 

[phongphanp]

It's hard on real model to simulate ? It's good for expect some noise, Not only study the mode of operations of equiment. Try more pspice models is the better.

 

[BradtheRad]

Question: is the Shockley Diode equation the ideal diode equation that we are talking about

The Shockley equation is also called the Shockley ideal diode equation. Speaking for myself, I can't use it. I've never seen examples of what specific values to insert for the symbols, depending on the type of diode.

I wanted to determine what equation to use in my homebrew analog circuit simulator. Years ago I did tests on real diodes with a meter and variable power supply. I plotted the results and tried to fit an equation to the graph.

For ordinary silicon diodes I came up with:
A = (V * 1.25) ^ 13

Example, if you apply 0.5 V, you get 2.2 mA.
Or, to get 100 mA, apply 0.67 V.

This formula is only an approximation. Nevertheless it works better than the assumption so often used, that a diode always drops 0.6 or 0.7V.

For led's it is a different formula. For germanium type it is something else. For high power diodes it is something else. Etc.