# ohmic state vs saturation state drain-source resistance of a MOSFET

If we look at datasheet here we usually see the on drain-source resistance is measured by applying a specific GATE-SOURCE voltage and a specific D-S voltage which will generate a specific current (20A in this case), of course these testing values are such that the MOSFET is at its ohmic region with respect to the prespecified GATE-SOURCE voltage mentioned in data sheet. Now the question is as mentioned in the data sheet at 10V the R_DS (ON) is 3.0 milliohms, does it mean at this 10V GATE-SOURCE voltage if we take the MOSFET get out of ohmic region and and make it saturated by applying sufficient DS voltage can we say even in this saturation state now the drain-source resistance is still 3.0milliohms? if no, is there any way or formula so that we can determine the saturated state ds resistance at a specific GATE voltage? Text books says they usually consider it as a short, but I think that is not precise.

• Figure 1 answers that. At Vgs=10V it's still in the ohmic region at 280A. And the Abs Max continuous drain current is only 220A (page 1). Put those two together and it should be clear that it's not a good idea to get it into saturation at Vgs=10V.
– user16324
Mar 16, 2021 at 19:09
• @BrianDrummond thats not my point, I am not gonna design a practical circuit by this. just to understand theoretically assume its saturated even it is beyond the limit Mar 16, 2021 at 19:14
• You need to be very careful when you talk about the drain-to-source "resistance". That word implies a linear relationship between voltage and current, which no longer exists in saturation. Mar 16, 2021 at 19:22
• My point is that the word "resistance" doesn't have meaning in the saturation mode, so there is no point in asking what its value would be. In saturation the MOSFET acts like a constant current, not a resistance. Mar 16, 2021 at 19:48
• No, you are conflating different things. An ideal current source has infinite parallel resistance. You are asking about series resistance, and saturation does not imply that the "resistance is really low". Mar 16, 2021 at 20:12

if we take the MOSFET get out of ohmic region and make it saturated by applying sufficient DS voltage can we say even in this saturation state now the drain-source resistance is still 3.0milliohms?

Look at the "orange" ($$\V_{GS} = 10 \text{ volts}\$$) characteristic (ignore purple stuff) and estimate a snapshot at 200 amps: - $$R_{DS(ON)} = \dfrac{0.5\text{ volts}}{200\text{ amps}} = 2.5\text{ milli ohms}$$

That's near enough to the quoted typical figure of 2.4 milli ohms at 20 amps. But, do you see the immediate problem of trying to make the current a lot larger so as to possibly take the MOSFET into its saturation region.

The maximum continuous current is 220 amps and it will still be nowhere near saturation and the 2.4 milli ohms will still be valid according to the data sheet graph.

Now, if you were to ask the question but with a gate-source voltage of 3.6 volts, can you see that the on-resistance (purple circle in ohmic part of the characteristic) is: -

$$R_{DS(ON)} = \dfrac{0.5\text{ volts}}{103\text{ amps}} = 4.85\text{ milli ohms}$$

And, if you dragged the MOSFET into its saturation region by forcing a DC voltage of (say) 5 volts, then the current would be about 190 amps. This implies a resistance of 26.3 milli ohms.

But we don't call that $$\R_{DS(ON)}\$$ any more because it's in the saturated region. That name ($$\R_{DS(ON)}\$$) is not applicable any more.

is there any way or formula so that we can determine the saturated state ds resistance at a specific GATE voltage?

Well, that's what I calculated when I came up with 26.3 milli ohms but, as far as I know it's a meaningless number because the device is going to dissipate 950 watts and be fried in about 3 ms: - Of course, if your application required it to operate for about 3 ms and then fry then it is applicable!

as mentioned in the data sheet at 10V the R_DS (ON) is 3.0 milliohms, does it mean at this 10V GATE-SOURCE voltage if we take the MOSFET get out of ohmic region and and make it saturated by applying sufficient DS voltage can we say even in this saturation state now the drain-source resistance is still 3.0milliohms?

No, with constant Vgs = 10V, increasing Vds enough to pull the mosfet out of the ohmic region (linear region) into the saturated region will increase the effective absolute drain-source resistance to much more than 3.0 milliOhms.

By definition, the "saturation region" is the region where the drain current stays approximately constant with changes in the DS voltage, so it doesn't even act like a resistance.

if no, is there any way or formula so that we can determine the saturated state ds resistance at a specific GATE voltage?

Yes, most mosfet datasheets have a graph like this one: (image from User:Krishnavedala at Wikipedia: MOSFET )

From that graph, you can look at the datasheet at the specified gate voltage (GS) and drain voltage (DS) and read out the drain current.

You could use Ohm's law to calculate an effective absolute resistance R = Vds/Id during those conditions.

Since with constant Vgs the current stays roughly constant no matter what Vds is, the value you calculate for R will increase with increasing Vds.

So with constant Vgs, as we increase Vds, the effective absolute resistance will increase to much more than the 3.0 milliOhms.

With physical mosfets, we try to drive the mosfet such that it is either turned off or sufficiently turned on to stay in the ohmic region. If forced into the saturation region, typically the voltage or the temperature will exceed the absolute max guidelines, and then the mosfet self-destructs.

The datasheet says Rds=3[mOhm] max. This means that when the MOSFET is in saturation (aka the Vds is high enough to open the mosfet completely) the Rds is no more than 3[mOhms].

The Rds therefore can be lower than 3[mOhms], but the datasheet does not guarantee it, nor does it provide a graph that shows how can you achieve lower Rds.

if we take the MOSFET get out of ohmic region and and make it saturated by applying sufficient DS voltage

You mean by applying a sufficiend GS voltage? You can achieve saturation by changing Vgs, not Vds.

Thank you @Elliot Alderson for correcting me here, saturation also depends on Vds.

Increasing Vds does lower your Rds, but its just an effect of the mosfet, not something you really 'can control'. If you need a Vds=12V for your system to function, what will you do? decrease it to Vds=5V to get a lower Rds? Probably not. Therefore, driving the Vgs drives the mosfet to saturation.

The more Vgs you apply (without exceeding datasheet's limitations, which here are +- 20V) the lower the Rds, and the higher the Vds, the lower the Rds.

EDIT:

Ok so I now understand that OP wants to keep the Vgs fixed and increase the Vds and see what the Rds on is, untill he/she achieves a specific Rds on.

As mentioned here by @stevenvh

So, yes, $$\R_{DS(ON)}\$$ varies with $$\V_{GS}\$$, and yes, it's higher at higher temperatures.

If your manufacturer can't give you the information and you really need it, move on to another manufacturer.

If you really need this information, find another mosfet/manufacturer that provides it to you. Datasheets are not always perfect.

• no, it was DS , which is to make sufficient so that at a specific GS voltage (here 10V) the MOSFET will go into saturation Mar 16, 2021 at 19:18
• It is increasing drain-source voltage that takes you into saturation. The particular value of DS voltage where you enter saturation is a function of the GS voltage, but it is increasing DS voltage that causes saturation to occur. Mar 16, 2021 at 19:19
• @ElliotAlderson thats what I have written, keeping GS fixed and increasing DS Mar 16, 2021 at 19:20
• Aha! I could not understand that you wanted to keep the Vgs fixed! . You should add this keyword to your question, I think it will make it a bit more clear. Mar 16, 2021 at 19:22
• @Sayan Might comment was for this answer, not for you or your question. I think I was agreeing with you. Mar 16, 2021 at 19:25