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Power FETs

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kinarfi

Well-Known Member
I scrolled through Allied Electronics Power FETs and my main criteria was high current and later price comes into play. In choosing a Power FET for my project, are there any certain specs I should be looking for in the spec sheets?
Some of these are rated over a hundred amps continuous in a TO-220AB package and with my background as an electrician, I don't see how that much current can flow through the leads and not heat up severely. I'm thinking of running parallel FETs, Should I? or would it be a waste of board and components and which of the follow would you recommend and Why. The motor I'm using is off a Power Wheels ride on toy and pulls around 30 amps stalled and around 17 loaded but still turning, 3 pole, permanent magnet field. Amp measurement come from voltage across a .1 ohm resistor in series, the highest reading I ever got was 2.7 volt, = 27 amp. On another unit, I expect to see current in the 40 amp range.
Thank you,
Kinarfi
 

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Hero999

Banned
Some of these are rated over a hundred amps continuous in a TO-220AB package and with my background as an electrician, I don't see how that much current can flow through the leads and not heat up severely.
It will heat up severely, that's why you need a huge heatsink and fan.


The motor I'm using is off a Power Wheels ride on toy and pulls around 30 amps stalled and around 17 loaded but still turning, 3 pole, permanent magnet field.
Use Ohm's law to calculate the power dissipation. Generally you can ignore surges unless they're really huge so just use the average current.

Once you have the maximum power, work out the maximum tolerable temperature rise.

What's the maximum ambient temperature?

What's the maximum junction temperature? See the datasheet.

ΔT = Ta - Tj

Now work out how good the heatsink needs to be.

Heatsinks are specified by thermal resistance which is measured in °C/W, a rating of 1°C/W means the heatsink's temperate will increase 1°C per Watt of power dissipated.

Add the thermal resistance of the junction to case to the number you calculated and any insulating tab you've used.

Now select a heatsink with a lower thermal resistance than the calculated value.
 

audioguru

Well-Known Member
Most Helpful Member
The datasheet for most Mosfets show an allowed momentary peak pulse of hundreds of Amps but the small pins are rated for only about 30A continuously.
 

wealth210

New Member
It will heat up severely, that's why you need a huge heatsink and fan.



Use Ohm's law to calculate the power dissipation. Generally you can ignore surges unless they're really huge so just use the average current.

Once you have the maximum power, work out the maximum tolerable temperature rise.

What's the maximum ambient temperature?

What's the maximum junction temperature? See the datasheet.

ΔT = Ta - Tj

Now work out how good the heatsink needs to be.

Heatsinks are specified by thermal resistance which is measured in °C/W, a rating of 1°C/W means the heatsink's temperate will increase 1°C per Watt of power dissipated.

Add the thermal resistance of the junction to case to the number you calculated and any insulating tab you've used.

Now select a heatsink with a lower thermal resistance than the calculated value.
For instance,I use one P-Mosfet GB3318 and the using ambient Ta=25°C,Read from the datasheet,know that Tj=150°C,that means ΔT = |Ta - Tj|=125°C,also know Rthj-a=66°C/W.
My undstanding is this mosfet the maximum power=125°C/66°C/W=1.894W,assume the RDS(on)=150mΩ,Per P=I²R,then 1.894W=I²*0.15Ω,so the reasult I=3.55A(allow constan Amp),is right my understanding?
and I want to know how calculate the size of the heatsink?Per your saying,I have understood the a rating of 1°C/W means the heatsink's temperate will increase 1°C per Watt of power dissipated.but how to determine the size?
 

bailey45

New Member
Rthj-a is the thermal reesistance junction to ambient (No Heatsink). You have it right except use the Rthj-c (Junction to case) and then add the thermal resistance of the heatsink you use to determine maximum power.
 

Hero999

Banned
I want to know how calculate the size of the heatsink?
You don't need to, buy a heatsink with a desirable °C/W rating.
 

kinarfi

Well-Known Member
As I read through this, I come the to conclusion that a low Rds(on) and ohms law P=I²R, combined with thermal resistance determines the amperage of a MosFet and audioguru seems to agree about the ampacity of leads and if I was truly running something at 60 amps for any period of time, it would be good to run parallel FETs for lead heating (both FETs on the same heat sink), but one FET rated at 120 amp continuous would handle the current, even if the lead can't. Yes - No?
Thanks
Kinarfi
 

smanches

New Member
I have been finding that same case more and more in datasheets for fets. They give the theoretical current of the junction, but the package is much more limiting. Have to read the fine print on everything. :(
 

Hero999

Banned
smatches said:
I have been finding that same case more and more in datasheets for fets. They give the theoretical current of the junction, but the package is much more limiting. Have to read the fine print on everything. :(
Yes, what's the point in that?

Why put a die that can supposedly take 180A in a package that can only handle 75A?

Continuous Drain Current, VGS @ 10V (Silicon Limited): 180A
Continuous Drain Current, VGS @ 10V (Package Limited): 75A
http://www.electro-tech-online.com/custompdfs/2009/07/irl1404zpbf.pdf

I'd rather but a cheaper MOSFET that can only handle 75A.
 

tcmtech

Banned
Most Helpful Member
I typically consider the high current capacity to be a overload rating. I have not had any problems with running 60 amps though a TO-220 type case provided the case sat right on the board and the leads were soldered right up to the base of the case itself and the circuit traces were built up with heavier wire and solder overlays.

I think its when you have the leads at full manufactured length then there is a reasonable concern for the actual working amp capacity limits. If the leads are short as possible and the actual thermal limits of the device are kept under check the full capacity can be obtained provided the circuit board traces and related built up solder or heavy gauge wire/sheet can carry the sustained currents to and from them without too much added heat build up themselves.

But a good fabricator and designer would realistically only have the device switching about half its rated current or less.
I would have no problem running a 30 amp motor with 50 amp surges on a 100+ amp rated device. But I would go a bit over sized on the heat sink in regards to calculated thermal load and use a rather generous amount of circuit board build up in order to help pull heat off of the leads.

If there is room on the board and heat sinks and the price is no big difference dividing up the loads between two or more devices doesn't hurt anything though and does give you further protection.
 

wealth210

New Member
Rthj-a is the thermal reesistance junction to ambient (No Heatsink). You have it right except use the Rthj-c (Junction to case) and then add the thermal resistance of the heatsink you use to determine maximum power.
Sorry,I still not understand your means,of course ,I know Rghj-a is the thermal resistance juction to ambient.In my calculation,you say all my understanding are right except using the Rthj-c,Per your saying,I should update: 125/5.5=22.7W,right?
 
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