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Topic Title: Calculating Is and nVt for Silicon Diode, How?
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Created On: 15 March 2012 09:16 AM
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 15 March 2012 09:16 AM
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Posts: 1
Joined: 15 March 2012

Hello, everyone!

I was asked by my lab instructor for an Electronic course to conduct a diode experiment by which I can get the value of (nVt) and saturation current (Is) for that diode.

I did the experiment by connecting a DC power supply in series with a forward-biased silicon diode (1N4007-MIC), and a 1Kohm resistor. I took voltage reading across the diode starting from 10V DC supply, down to 1V. Now, I have the readings, plus I calculated the diode current (ID) for each -1V difference in DC supply. Also, I graphed a curve of I-V characteristics.

Well, I understand that to get the value of nVt and Is, I have to take two values on that curve, an upper value of VD(V2) on the horizontal axis corresponding to an upper value of ID(I2) on the vertical axis, and a lower value of VD(V1) corresponding to a lower value of ID(I1). I take these values and substitute in the equation:

V2-V1=2.3nVt log(I2/I1) --> From this, I can calculate the only unknown value which is nVt.

I take the resulting value and substitute in the equation:

I=Is e^(V/nVt) --> From this, I can calculate the only unknown value which is Is.

So, my question is, is my understanding to how to get the two values correct? is there any shorter and easier way to get them?

Also, where can I find the the two values in the diode's datasheet? I looked up for it but didn't find any indication of it (this datasheet for example).

 15 March 2012 03:53 PM
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You are broadly correct.
Two comments:

i) The diode equation is I = Is*(e^(V/nVt)-1). In your algebra to arrive at V2-V1=2.3nVt log(I2/I1) you should make it clear that you have assumed that both I1/Is and I2/Is >> 1.

ii) Therefore, to find Is, substitute for nVt in I = Is*(e^(V/nVt)-1) not I = Is*e^(V/nVt).

Note that For I/Is >> 1 you have to also keep I low enough that the resistive drop across the diode bulk resistance (around a couple of Ohms) is negligible compared to the raw diode voltage given by the diode equation. Also note that the power dissipated in the diode will cause self heating and so change the junction temperature. In real device tests, the current through or voltage across the diode is pulsed to reduce this effect by using the thermal inertia of the junction to low pass filter the thermal impulse.

You won't normally find nVt quoted for discrete devices since it is a very process dependent parameter. You may find it quoted in ASIC design and fab work.

Is is basically the reverse leakage current of diode which is given in datasheets.

If you download and install the excellent free and completely unlimited mixed mode LTspiceIV simulator from here:

then you might like to download the zip file, file here:

then extract the two files (into the same folder) and open the .asc file in LTspiceIV.

Andy Fierman


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