I have an IRF1404 in a circuit that is getting much warmer than it should in theory. The MOSFET is rated at 0.004 ohms Rds when on. With the gate driven at 5v, which is over the 4v max in the spec sheet, the MOSFET still gets very even with a small heatsink attached.

Only about 3A @ 36V is going through it with a 100% duty cycle. So it's basically just on.

watts = current^2 * resistance

So it should be dissipating 0.036w, but it seems like much more, closer to 5w.

 

Doesn't that Mosfet use +20v gate?
According to the Spec sheet?
Why not use a logic level gate type?
Max.

 

It is logic level, spec sheet says gate fully on between 2.0v min and 4.0v max.

 

How come my IR sheet says 20v?
No mention of logic level?
Max.

 

If you are only 1V over the Vth maximum threshold voltage then it is not fully on. You need to look at the Vgs they use when they specify the ON resistance. For a standard MOSFET that's usually 10V. That's the Vgs you need for the MOSFET to be fully on.

Or you need a logic level type MOSFET as Max suggested.

 

That's the threshold voltage, which is irrelevant. Take a look at figure 3 of the IR data sheet. Needs 10V.

[referring to #3]

 

I agree with MAXHEADROOM. The data I see says 10v gate.

Can you show us the spec you are using ?

Me too slow

 

I think I understand that I misunderstood what gate-threshold-voltage really implied.

In figure(3) it does show 50-80A drain-to-source current @ 5v gate, but the on resistance is higher, hence the heat.

Any recommendations for a cheap very low resistance logic level MOSFET that can switch 10A @ 36v with logic level? Or a website where such searching is even possible.

 

Look at the International Rectifier site, the IRL 420/430's etc.
Max.

 

Appears like there is an IRL1404, which is a logic level version of IRF1404. IRL vs IRF, the L standing for logic level. That was easy.

 

It's good practice to characterise power FETs.

You had done most of the work already, by applying a fixed gate voltage and measuring the current (Ids).

All you needed to do was measure the voltage (Vds) and you would have had all the info you needed regarding how well that FET saturated for the given gate voltage and drain current (and known its actual dissipation).

You would also have seen how close it was to the datasheet "ideal" of 0.004 ohms.

 

I measured 5.5v across drain-source on the MOSFET at 3A, 5.5v / 3A = 1.8ohms.

So 3A^2 * 1.8 = 16.2w dissipated. Am I doing this right?

 

I measured 5.5v across drain-source on the MOSFET at 3A, 5.5v / 3A = 1.8ohms.

So 3A^2 * 1.8 = 16.2w dissipated. Am I doing this right?

Yes, but you did it the hard way. You don't have to calculate the resistance to get the power. Since P = V*A = 5.5V * 3A = 16.5W (your calculation has a round-off error since R = 1.83333...Ω.)

 

I measured 5.5v across drain-source on the MOSFET at 3A, 5.5v / 3A = 1.8ohms.

... Am I doing this right?

Yep, and that info (1.8 ohms) shows you the FET is not getting anywhere near enough gate voltage (Vgs) that it needs to turn on properly.

If the gate voltage is higher the FET on-resistance (Rdson) is lower. Obviously to get the datasheet spec of 0.004 ohms Rdson they must have applied a much higher gate voltage than in your test.

So at this point you either need to apply a much higher gate voltage, or change FET type to one like a logic-level FET that will turn on well at a lower gate voltage.