Hi guys

Yet another question. I hope you can answer this. The MOSFET is an IRL540.

I was given the good advice, on this forum, to insert a diode on the mosfet gat, to protect the microprocessor should the
MOSFET give in or something, and direct 30V back through the gate.

I have made one of my famous drawings(see below). When the diode and the 10K resistor is in place, my 5V goes to
4.48V. Will this ensure complete saturation of a logic level MOSFET?

On a less-important note; if I want to connect an LED + resistor from 5V to gnd (before the diode etc) then my current draw
goes from nothing, to 10-20mA...on a pin that has a maximum of 30mA. Will this not affect saturation of the MOSFET?

Thanks for your time!

Nick

 

Modern LED's will have a good brightness if the LED current is >1mA. So increase your led resistor value and Iled around 5mA or lower.
As for the MOSFET we need to know what is the load ?

 

Hi, thanks for your answer!!

The load will be 36-40V

Thanks,

Nick

 

And the load current is??

 

I would think about 5 amps!

 

See "Typical Transfer Characteristics" chart in the datasheet..
and more info here..
https://arduinodiy.wordpress.com/2012/05/02/using-mosfets-with-ttl-levels/

 

Thanks, looking at this graph:

https://arduinodiy.files.wordpress.com/2012/05/irl540ivgs.jpg

...it seems that 4.5V will be good for well above 10A. That should solve my problem then!

/Nicholas

 

Don't use "typical curves" for a design. They don't indicate worst case characteristics.
However the data sheet does shows that, for a Vgs of 4v, the ON resistance is a maximum of 0.11Ω with a drain current of 14A, so you are fine at 5A.

 

The power dissipation in the MOSFET is around Ptot = Id^2*Rds(on) = 5A*5A *2*0.077Ω = 3.85 watts. So, the small heatsink will be needed.

 

The power dissipation in the MOSFET is around Ptot = Id^2*Rds(on) = 5A*5A *2*0.077Ω = 3.85 watts. So, the small heatsink will be needed.

It's only half that value at 1.92W.
You multiplied by the "2" which is used in that formula as the "squared" indicator.
But even for that dissipation you should use a small heatsink.

 

It's only half that value at 1.92W.
You multiplied by the "2" which is used in that formula as the "squared" indicator.
But even for that dissipation you should use a small heatsink.

This "2" in the formula I use because, as you know the Rds(on) resistance will rise with junction temperature and this "2" includes this effects (more or less).
http://www.vishay.com/docs/91300/91300.pdf (Fig 4)

 

This "2" in the formula I use because, as you know the Rds(on) resistance will rise with junction temperature and this "2" includes this effects (more or less).
http://www.vishay.com/docs/91300/91300.pdf (Fig 4)

Okay. You just threw that fudge factor of 2 in there with no explanation so I didn't know.
But that's reasonably conservative since, according to the data sheet, the ON resistance doubling occurs at a junction temperature of about 140°C and you likely don't want to operate much above that junction temperature for good reliability.