MOSFET saturation resistence

[Falesh]

Hi there,

I need to make sure I am getting the right MOSFETs for my project as the delivery time is very high. I am planning to use the IRF640N which has a V_GS(th) of min 2 to max 4. The voltages I will be using are: V_GS 5 (though this could be reduced) and V_DS 5. Since V_GS > V_th and V_DS ? ( V_GS – V_th ) that should mean the MOSFET is saturated. I checked the first graph on the datasheet which seems to indicate, if I am reading it right, that a V_DS of 5 and a V_GS of 5 would let 3A current through, which means its resistance would be about 1.67ohm.

My issue is that I am a complete newbie and am not confident I have done the above correctly. One particular point I don't understand is what the datasheet means by a min - max V_GS(th). Is this something to do with a change in temperature, for instance the min V_GS(th) is when it is cool but it increases to the max when very hot, or vice versa, or is it something completely different?

Many thanks!

[tec5c]

Vgs is the gate to source voltage.

In the datasheet you'll find an absolute term Vgss this is the maximum voltage that can be applied between the gate and the source. Beyond this, you risk damaging the mosfet.

An N channel mosfet is essentially an P type sandwiched between two N type regions.

Party time.

You are hosting a party and inviting all the neighborhood electrons to attend. So you broadcast "PARTY AT MA HOUSE YOLO #SWAG!". Ie, you apply a positive voltage at the gate with respect to the source. Given that your neighbours are some distance away (next door), your broadcast isn't loud enough (You are below Vgs(th)) Once you yell your invite loud enough (ie. Have Vgs at Vgs(th)) you're next door neighbours hear and come party with you.

Vgs(th) is the voltage at which the mosfet channel begins to conduct. At this voltage, a positive voltage, it creates a electric field, which attracts electrons (since our applied voltage is positive, so positive charges on gate). These accumulated electrons near the gate, form a bridge between the source and the drain (which are both n type). Now you have a "continuous" n type path from source to drain.

You only attracted your next door neighbours, so your party is kinda lame. How do you get more people ? You broadcast louder (Increase the range of your electric field - increase your Vgs).

So now you've increased your Vgs to the point that you cannot accept anymore people on your property. You are completely full. You can broadcast your party as much as you want, but there simply isn't enough room to accommodate everyone. You're transistor is now in saturation.

If you start increasing your Vgs to the point of Vgss, well...cops show up and shut you down. Underage drinking, drug use etc.. You go to jail (Your transistor has been damaged).

[Falesh]

Thanks for the replies,

I went with the IRF640N as it seemed suited to my needs, though I have realised I was making false assumptions, and was very cheep for a bunch on eBay. At least they were so cheap that it isn't the end of the world if I need to order different ones.

The mosfets need to be able to handle between 50mA & 2.5A as each is driving a row of LEDs. I will be switching them on for a max duration of about 0.02 milliseconds so I don't mind if there is a delay of something in the region of 400ns or less before the mosfets start letting the full current through. For the speed I was looking at the section with "Turn-On Delay Time" which seemed nice and fast but I didn't realize that the Gate Charge can add a meaningful delay.

I had a search online for a way to calculate the time it would take to fill this capacitance up but I didn't manage to work it out due to the difference between a mosfet and a normal capacitor. Is there a way to get a rough idea how long a mosfet takes to switch from its datasheet and knowledge of the input voltages or do you just have to search for type, e.g. "Logic Level Gate"?

[dmills]

If your fet is showing 1.7 ohms @ your available Vgs, you probably want a different fet if it is to handle 2.5A, 6W or so is a lot of waste heat.

The threshold voltage is where things start to happen, but really you usually want at least twice the threshold voltage to get the thing to switch properly into saturation.

Note that for reasonably fast switching you actually need to supply quite some current to the gate, it does not last long, but there is a reason mosfet drivers can often manage to supply several amps for a few hundred ns at each switching edge.

Keyword is 'logic level mosfet', you can easily get parts that will fully saturate on a 3.3V output.

If you look at the IRF640N datasheet you will see (in the electrical characteristics table) that the threshold will occur somewhere between 2 and 4V, but if you look at at the conditions, you will see that threshold is considered to correspond to 250uA of current....

Further, if you look at the Rds(on) resistance you will see that it is specified at Vds = 10V, so this is not a logic level part, it is probably intended to be driven from a gate driver with a 12V rail.

Compare with something like an IRF7331, much lower working voltage, 20V, but Rds(on) is now specified to 2.5V (0.045 ohms max), so that thing can be driven (slowly) directly from 3.3V logic.

A general observation about mosfets is that as you increase rated voltage or rated current, everything not directly related to that parameter gets worse, you seldom want to use a 200V mosfet in a 12V circuit for example because the 200V rating will hurt you for gate charge, RDS(on) and just about everything else, same way, massively over sizing the thing current wise will hurt you for the same reasons (Not note that maximum steady state Id values become increasingly fictional above 10A or so).

Regards, Dan.

[Falesh]

It is all starting to make sense, though I shall study more to make sure I am really understanding it. On reflection I'll go with the FDC855N which seems to tick all the boxes mentioned. Thanks a lot for the help!