I am using an opto-coupled IRFP240 n channel enhanced mosfet to switch 12 volts at 6.0 amps continuous.
Calculated:
P=I^2 * Rds (.180) = 6.46W
No heat sink yet.
Can't touch it after a 10 seconds. Does this sound correct.
I am using a voltage divider on the source voltage to drive the gate.

Thanks for any insights.

 

6.5W is a LOT of heat to get rid of through a fairly small package. Look at how big are 5W resistors, and those would be well above 100°C at full rated power.
Try using a better suited mosfet, basically the lower voltage it is rated for, the lower Rdson usually is. For example IRFP064N has 0.008 ohm, rated for 55V and 110A.

 

yes that sounds correct..
You need a significant heatsink to dissipate 6.5W on a typical mosfet.
TO-247AC package right?

Rja (thermal resistance junction to ambient) is 40 deg C per watt
So with 6.5 watts you are subjecting the junction to approx 260deg C temp rise over ambient. which equals WAY..WAY too hot.

 

I am using a voltage divider on the source voltage to drive the gate.

Why? You want at least 12V on the gate to be sure the MOSFET is operating at its minimum Rdson. Otherwise, they get hot.

 

Thank you for the help. Apparently no one has any problem with my using the power source for the load on the gate? My thought was that the ground for the load source and the gate have to be shared, wasn't sure about the power source for the gate. I will try the different mosfet.
Many thanks

 

Maybe if you posted a schematic we would actually have some idea about how and why you have it connected.

 

I'm a little confused about what you are asking. Can you provide a schematic?

The MOSFET's source pin should be at power-supply ground. The gate voltage must be limited below the device's rating, 20V for that MOSFET. I'd aim for ~12V. If your power supply is less than ~15V, then you can just use the power supply voltage directly applied to the gate.

 

New at forums. I'll try to supply schematic.
I think you have answered my question very well. The source pin is connected to the power supply ground. I guess my question was a theoretical one as I didn't think applying a sufficient positive voltage to the gate without it having a common ground with the source pin would turn it on. I am running into this problem when I switch 160 VDC from rectified mains and want to separate mains ground from signal or gate ground. Have to use second transformer from same power source.
Many thanks again.

 

Yikes, this is not as simple as it first sounded. I think you might want to look into an opto-isolator so that you can maintain isolation. But by all means post a schematic so the pros here can give you a safe solution.

 

Here is the schematic. Hope this works.

 

Here it is

 

Yeah, you can (and should) lose the pot as long as R2 is adequate to limit current through the transistor.

 

The optoisolators I have seen only have a transfer ratio of about 50% (current for current). You have 13.2ma going through the LED in the opto so R2 should be at least 1800 ohms. You did pretty good, but you can waste less power by changing R2 to 18,000 ohms and connect the +12V directly to the collector of the opto. That way the opto won't struggle to get the gate voltage high enough, and that will reduce the heat.

and, while I'm here, rectifying the mains to 160 volts is a no-no on this site. Too dangerous. If you talk about rectifying the mains here, the Moderators will shut you down.

 

A little question turned into a big one. I can understand if you don't want to respond further. You have all been very helpful.
Thank you very much.

Your correct. I should have added rectifying mains after going through a 1:1 transformer. It is too dangerous otherwise. Really appreciate your help.

I can see the pot is not needed as long a I limit the current on the transistor. Do I need to limit the current on the gate?

 

Yeah, you can (and should) lose the pot as long as R2 is adequate to limit current through the transistor.

Isn't R2 a pull down, to make sure gate turns off? Not to control gate current. If used to control gate current, wouldn't it go between opto and gate?

With out R2, the gate would be slow turning off.

 

Mosfet is controled by voltage, so there is no current flowing through the gate. The gate is basicly a capacitor with a few nF.

 

That's what I intended it to do. Provided the gate has a very high impedence I could see where it might possibly current through the transistor. However I think it is most effective turning off the gate.

 

The main function of R2 is to limit current thru the opto's transistor, but it also provides a pull-down for the gate.

 

Spot-on, wayneh. I give a, "Thanks" for posts that give me an LOL, but in this case, it's for posting exactly what I was thinking and saving me the time it would take to type out the answer.

 

thank you again. A lot of circuits put another resistor where I have the pot to limit the current on the opto-isolator. Is that not necessary or should it be added to R2 for a total R? Apparently the resistance doesn't have to be on the collector side?
Learning a lot of my first forum.