dear god why cant i understand diodes

I've always thought it was 0.7V for typical diodes. I usually double check with the spec sheet though since this is not always the case.

Can't remember but open any text book on semi-conductor electronics and it will show you how to calculate p-n junction voltage. Lots of calculus here. I remember my professor going through the math on the white board resulting in the approxmation of 0.7V. At which point my brain decided to forget the math and remember the 0.7V.

You draw a load line. That is, on the same graph as the diode's behavior, draw the current through the resistor vs. the voltage across the _diode_: If should have I=0 at the open circuit voltage (5V), and 5V/your resistor value at V=0.

Since the parts are in series, they carry the same current, and hence where the two curves intersect you can read off the current through both of them and the diode voltage (...leaving 5V-V_{diode} as the resistor voltage, of course).

This can be easily extended to two non-linear components in series. For three components you could, at least conceptually, draw a 3D graph and find the intersection of all three curves, although by that point you might as well sit down and get a computer to iteratively obtain the solution.

You could sit down and fit the graph to the various parameters in the diode's equation, but personally before I did that I'd be trying to get a SPICE model out of the vendor which will be somewhat more accurate anyway.

---Joel

Ironically enough, I was never made to use a load line in circuits classes, but I was made to use one in an electric machines class where you were trying to match a motor and generator or somesuch and they wanted to include various non-linear effects!

A precise solution will require either a simulator, or that you be handy with Newton's Method of Successive Approximations (which you can easily set up on a hand-held calculator).

...Jim Thompson

One way is graphically- draw the "load line" across the V-I curve for the diode. This method was used all the time in the evil old days.

Another way is to write out your nonlinear equations and solve them, numerically or otherwise. Or shove it into Spice with a suitable diode model and let the program do the heavy lifting (numerical solution).

You can manually iteratively solve using your initial guess of 0.6V or whatever. For example, if you are using Vf = k*T/q*ln(I/Is) for an ideal diode, find the approximate current from 5V-0.6V, then find the forward voltage, correct for a better approximation of the current from 5V-Vf, rinse and repeat. The initial guess does not have to be very accurate if the supply voltage is high relative to the diode forward voltage. You might also not be using an ideal diode, so the results might not match reality (a diode-connected transistor is closer to an ideal diode than a diode typically is)>

Best regards, Spehro Pefhany

In order to calculate a diodes current at any given point, you must use the third approximation method.

The factor you are missing is "Bulk Resistance" the N/P wafers have a bulk resistance which at low currents is insignificant, but as the current increases the voltage drop Vbr becomes a factor. This is why the knee voltage is sloped.

So a diodes voltage drop is Vf = .7 + Vbr

Most rectifier diodes drop ~1 volt at the max. forward current, so the total bulk resistance would be .3/If

Including bulk resistance in your calculations will give you very accurate values.

Hope this helps,

Ron

Use the lambertW function:

Id = lambertW(l, Is*R/Vt/m*exp(5)*exp(Is*R))/R - Is

(m is the diode 'fudge factor'..., between 1 and 2. Precise enough for you? ;)

Simple, right? How could you have missed that? (xmupad just figured it out for me...)

Dear God, Indeed :-(

...Jim Thompson

am trying

You'll need to know a lot about the diode, including parameters that vary with temperature. And the temperature will vary with ambient and the precise cooling (package to air, through the legs to the PCB). And the parameters vary from one unit to another. And the parameters you need probably won't be in the data sheet, except as graphical curves for a typical unit, for example the fairly ordinary 1N4001:

In other words, there's not a lot of point. Even if you can predict it correctly for the diode you got out of the drawer, the next one will be different. If your design relies on the diode for something crucial, change the design.

Paul Burke

Now, now, Mark! Details, details ;-)

Apparently Kevin, et al, believe looking it up in a table makes it OK ;-)

...Jim Thompson

The closed form solution for the lambert function is?

If it ain't on my cheapy TI calculator it doesn't exist ;-)

...Jim Thompson

a bit too large to fit into the margins of this email...

;)

--- Regards, Bob Monsen

There once was a man from Hornepayne, Who tried to transform the whole plane, It bent a meridian So it wasn't Euclidean, And frustration drove him insane.

- Anonymous

What is the closed form for the ln, cube root, or sin functions again? Your calculator has approximations for them encoded in its little brain. Doing the same for the lambertW function is not beyond the pale, except that it is not nearly as useful, so nobody except the mathcad guys thought of it.

I was reading about Archimedes last night. He managed to bound the solutions to various roots using geometric methods, on his way to bounding the value of PI. He was truly an astonishing fellow, who used the 'method of exhaustion' to prove the various relations between PI and the area of a circle, volume of a sphere, etc. He proved that the volume of a cylinder of height and diameter D is 3/2 the volume of a sphere of diameter D. Also, the surface area of this cylinder is 3/2 times the surface area of this sphere...

Should be simple - why? (I feel your pain!) Would that it were so. Ed

sure!

"I Want" will get you nothing - If this is school work, then spare the sob story and ask the question - or Google/Wiki for "load line".

If you "need" to be able to do this for work, then you are in for some "quality time" i.e. redesigning.

No ;-)