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I am trying to get the Rds(on) of a MOSFET. I have tried different methods and I do not know what is the best method. There might be a method that is better than the ones I used.

In practice, it seems that they use a specific device which sends a double current pulse into the MOSFET at different Vgs, but I do not think that I need to do this, as I fixed into the simulation the case temperature. Nevertheless, it could change the junction temperature, so maybe I made an error.

Here are the methods that I used:

1st method (Result 1.43 Ω @ Vgs = 4 V):

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2nd method (Result 2.16 Ω @ Vgs = 4 V): enter image description here

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3rd method: (Result 239 mΩ @ Vgs = 4 V) enter image description here

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None of the three methods gives the same result. What is the correct method?

ocrdu
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Jess
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  • Why did you use 80 V for the 4th pin and 10 V for the supply? Would [this setup](https://i.stack.imgur.com/poUr9.png) be good enough? Also, the 1st method has a series resistance, the 2nd doesn't, and the 3rd has a current source. – a concerned citizen Dec 17 '22 at 08:19
  • Thank you for your comment :) The 4th pin defined the junction temperature. I though it was the case but I took a look on the model and it is said as the junction temperature :) – Jess Dec 17 '22 at 09:37
  • I will take a look on your setup – Jess Dec 17 '22 at 09:39
  • Also for the 1st method, the series resistances was a copy from a previous example that I saw on internet. But I do not know why it will not work ? – Jess Dec 17 '22 at 09:40
  • For the 2nd method and the 3rd method it was a test to know if there was a difference. Does it sounds bad to take a current source ? – Jess Dec 17 '22 at 09:41
  • @aconcernedcitizen So I took a look on your model and I saw that you were looking the value of the Rdson around 2.5V. Does it seems to you a better solution to use a voltage source rather than a current source ? Also do you understand why the results are not the same ? – Jess Dec 17 '22 at 09:56

1 Answers1

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I don't see anywhere any .model so I have to conclude that you're using the basic 4 pin NMOS, which has no such thing as temperature pin. The 4th pin is the bulk and, typically, it's tied to its lowest potential (source here). So, what you did was you forced the bulk to be at a higher potential than the drain. Thank goodness that SPICE has no magic smoke (or is it?). But, if you are using a custom model or subcircuit, please let me know.

Then, you are plotting Vd/Id which, in the first two cases there is a difference of a resistor between them. It's one thing to plot Rds, and another Rds+R.

The third method uses a current source and, as you know, it will deliver that current no matter what. It will also force a different operating point through sheer will.

So, if you want to know the Rds you'll have to first take care of the operating point. The usual way of doing this is with the usual curves which, at its most basic setup, is with two voltage sources:

MOSFET test jig

Two things to note:

  1. I am carelessly using the voltage at the bulk greater than the potential at the source -- this is gross but, it's there for the sake of exemplification
  2. In real life test cases the sources will have a limited impedance and there will be some resistors added, to avoid high currents.

Also, the test is revolving around the threshold voltage, which is set through the .model at vt0=2.5. That's where the sharp change happens so that's where the focus is (and given that, in your pictures, you seem to be interested in the transition).

a concerned citizen
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  • I am sorry. I am using a specific model and the symbol is the same as the nmos4. I "paste" symbol from the "txt" asy file. (at least the part that which interested me) – Jess Dec 17 '22 at 10:19
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    @Jess I hope you didn't forget to set the prefix of the symbol to `X` (since it has a temperature pin, it's most probably a `.subckt`). At any rate, that's the setup that's mostly used. Using "blunt" voltage sources forces the operating point to whatever voltages they're supplying. After all, no matter what `Rds` point you're after, it's dependent on the whole operating point. – a concerned citizen Dec 17 '22 at 10:21
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    "Then, you are plotting Vd/Id which, in the first two cases there is a difference of a resistor between them. It's one thing to plot Rds, and another Rds+R." I do not get it. I still looking for the voltage between the drain and the source and the current flowing through the mosfet so I still measuring the rdson whatever if there is a resistor or not ? – Jess Dec 17 '22 at 10:22
  • Yeah I did it :) I set the prefix of the symbol to X :) – Jess Dec 17 '22 at 10:24
  • Well I agree with you on this point " it's dependent on the whole operating point". So if I used a current source, I forced the current as you said, but if I used a voltage source I forced the voltage across Vds – Jess Dec 17 '22 at 10:27
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    @Jess Oops, you're right! I missed that part about the measurement, you're using `V(d,s)` so it's fine. It still changes the operating point (even if you directly feed the `Vgs`). Don't take my word for it, try it yourself: assign the value `{x}` to the resistor and use `.step param x list 1m 1` when measuring. The difference will only be visible for, well, `Rds_ON` but, you can see the change better if you plot `V(d,s)`. – a concerned citizen Dec 17 '22 at 10:39
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    Thank you for your help ! – Jess Dec 17 '22 at 14:08