Help for Microcontroller isolation!

lmarklar

Joined Apr 23, 2015
24
He's done it so much he just doesn't say the steps that he does in his head automatically :p

Rds@5v * Load Amperage = Voltage Drop across FET
Voltage Drop * Load Amperage = Wattage the FET needs to dissipate

http://www.mouser.com/Search/Produc...505virtualkey57370000virtualkey942-AUIRLI2505

It's a bit expensive but has a crazy low Rds on. For your application with a 5v gate source

0.01 ohms * 5v = 0.05 voltage drop across
0.05v * 5a = .25w

Should be able to run that without a heat sink, but I would still recommend at least a small one to help out.

I just had that datasheet handy because I happened to be getting a couple for a high amperage job. You can probably find even better than that if you only need to go up to 5a
 
Last edited:

joeyd999

Joined Jun 6, 2011
5,236

joeyd999

Joined Jun 6, 2011
5,236
...Should be able to run that without a heat sink, but I would still recommend at least a small one to help out.
To help what out?

The junction to case thermal resistance is 65C/watt. 1/4 watt power dissipation will lead to a junction temperature increase of ~16C over ambient. Maximum junction temperature is 175C. Therefore, this part could be run at 1/4W at up to 159C ambient without a heatsink with no problems.

Heat sink is definitely not required.
 
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