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In high frequency, this plots the input voltage and the current in a simple circuit with a resistor and a diode.

Mathematica graphics

The behaviour I see is that the diode remains forward-biased for longer than an ideal diode would, but as soon as it does it goes back to the usual, perhaps exponentially in a very low time, without any phase shifts.

I have been attempting to (macro) model that particular behaviour of the real diode using an ideal diode and other components (capacitor, resistance, inductor), but so far, failing miserably

Short question is, what could I add to the black box of an ideal diode for it to behave that way?

I would appreciate, should you come up with something, to know how you went about thinking about this, since learning is the only purpose of this question.

Thanks a lot

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Have a look at the Shockley equation for the ideal diode, and the various additions to it. The Wiki diode page is a good place to start, and then follow this up with the page on Diode Modelling –  Oli Glaser Aug 31 '12 at 11:19
    
Thanks @OliGlaser, I am taking a look –  Rojo Aug 31 '12 at 14:22
    
I don't understand. The trace you're showing us comes from a simulator, right? So, by definition, you already have a model for the behavior of the diode. I think you'd learn a lot more by studying that model than by building your own ad-hoc model. –  Dave Tweed Sep 1 '12 at 11:44
    
@DaveTweed on the one hand I am interested in the real diode modeling. On the other hand, and this was the hand I wanted to focus on when I asked, I took this as an exercise: what if I had an ideal diode (perhaps the 0-order model even) and I wanted to generate a response such as the one in the image, for some given set of values and frequencies, by adding capacitors and basic components? I couldn't do it myself after trying. I am trying to get comfortable with circuits with diodes AND components with memory (inductors, capacitors), so I asked to see how people think about these things –  Rojo Sep 1 '12 at 13:16

1 Answer 1

The phenomenon you are seeing is called reverse recovery time. Look that up and you will see it is due to carriers in the junction still being there when the voltage reverses. Until those carriers are "used up", the diode will continue to conduct.

Modeling is all about knowing which characteristics actually matter and ignoring the rest. If you didn't do that, it would be reality instead of a model, but then it would also be too complex to implement.

At first approximation, just assume the diode will conduct in reverse for a fixed amount of time. Diodes meant for applications where this matters will have the maximum reverse recovery time listed in the datasheet. If the purpose of the model is to want to make sure your circuit still works, this is a good model because it represents the worst case conditions.

More accurate models take into account the current immediately before the voltage reversal and look at total charge leaked backwards. There are fancy equations for all that you will have to look them up in semiconductor physics texts if you want this level of detail.

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Thanks +1. I'll be googling reverse recovery time up. Apart from the real diode modelling issue, assuming I "have" an ideal diode and would like with a couple of capacitors and inductors get a response such as the one in the figure, do you see a simple way? –  Rojo Aug 31 '12 at 14:22
1  
@Rojo: You are apparently asking how to simulate a real diode from a ideal diode and additional parts? This makes no sense since you can only physically have real diodes. I can't think of a easy circuit to add around a diode to extend its reverse recovery time. Use a slow diode like the 1N4004. Those are slow enough to basically be useful only in 50 or 60 Hz power circuits. Compare that to a 1N4148 signal diode, which has much faster reverse recovery. Also, Schottky diodes have inherent fast recovery due to being a different junction with less storage capability. –  Olin Lathrop Sep 1 '12 at 12:53

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