# Zener Diode - LT Spice

Discussion in 'Electronic Projects Design/Ideas/Reviews' started by benjamin.grant.du, Nov 19, 2008.

1. ### benjamin.grant.duNew Member

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Hi all,
i'd like to create a zener diode with Vthr = 420 mV, Is = 4uA
and I(zener) = Is*(exp(V(zener)/Vth)-1)... i can't figure out how to do anything like this. I created a custom current soure that I thought obeyed these properties but it seems to output current constantly so that can't be right. Any thoughts on how to approach this?

2. ### RoffWell-Known Member

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I≈Is(e^(V/Vth))
Is=I/e^(V/Vth)
I=4uA, V=.42, Vth≈.026

Is≈3.86e-13

Load the .ASC file below into LTspice and run it.

EDIT: keep in mind that this isn't a zener. It's a diode. The impedance will be high, i.e., if the current changes, the voltage will change significantly. You can run a V-I sweep on it to see what I mean.

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• ###### Diode equation.asc
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Last edited: Nov 19, 2008
3. ### benjamin.grant.duNew Member

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Sorry, I guess I wasn't clear as to what behavior I needed. I essentially have two zener diodes facing opposite direction with different Is and Vs specs, in parallel with a capacitor and essentially a voltage source... However, Is is dependent on the voltage across the zener diode at any given point. That is, I can't just assign a voltage. I guess I can assign Is as an equation instead of the constant that you assigned it to? Attached is the actual problem i'm attempting to solve.

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5. ### RoffWell-Known Member

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I don't see any problem statement on the attachment, and I don't understand your description of the problem. Maybe you can describe it better, or maybe one of the other members can understand what you are saying.

6. ### benjamin.grant.duNew Member

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Circuit Diagram - with description

Sorry, this one should make things clearer. The fact that the current is in units current density isn't significant, it simply mean the resulting voltage is in terms of voltage density, so it's not relevant to the modelling of this circuit.

7. ### RoffWell-Known Member

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Sorry, I don't have a clue as to what the question is.

EDIT: OK, I got it. Let me think on it.

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