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### Topic: Rds(on) loss calculation? (Read 6337 times)previous topic - next topic

#### 2632

##### May 20, 2013, 02:46 am
I am trying to determine the amount of power the following FETS will eat up but even after looking at lots of examples I am not confident I am getting the correct numbers.   At this point I just care about the loss in full on. (no switching loss)

13.8v at 100A
Rds(on) = 2.6m?

P = I2 * RDS
1000 * 0.0026 = 2.6watts ??

So out of 1380W only 2.6W are lost to heat in the FET?

Next question is how much power can they safely handle each on a PCB only in still air at 120F?  My though is to just parallel them to get the capacity needed since space is not a problem.

http://www.irf.com/product-info/datasheets/data/irfs3107-7ppbf.pdf

#### mauried

#1
##### May 20, 2013, 03:21 am
Need to work on your maths.
100 amps and 2.6 mohms = 100 * 100 * 0.0026 = 26 watts.
Fet will need a heatsink.

#### oric_dan

#2
##### May 20, 2013, 03:25 am
Quote
P = I2 * RDS
1000 * 0.0026 = 2.6watts ??

Ummm,
P = I2 * RDS
10000 * 0.0026 = 26watts

Those ratings you see in the datasheets for MOSFETs are usually somewhat optimisitc
from a practical perspective. No way you could dissipate 26W in a TO-220 sized MOSFET
without some incredible heatsinking. More like 6-8W with a common heatsink - and this
does not mean a pcb copper area. So, for use of pcb area as heatsink, maybe 20-25A, not
100A.

If I were doing it, I might rather get a TO-220 MOSFET rather than the D2PAK, and use a
real heatsink.

#### AmbiLobe

#3
##### May 20, 2013, 03:30 am
There is a mistake in your P = IIR calculation.
100 * 100 is 10,000 not 1000
So the MOSFET power is 26 watts at 100 amps and .0026 ohms.
That is safe because the spec for  IRFS3107 is 370 watts maximum power dissipation using a big heat sink.

If the load and MOSFET have a total of 13.8 volts across them, yes 1380 watts is the total power and only 26 watts is heating up the MOSFET.

With no heat sink the device will melt at 26 watts after a few minutes.  At 120 degrees F ambient air temp on a PC board with no heat sink and no fan, I can only guess at how many parallel devices would be needed : I guess 11 MOSFETS. My project will use an ice bath on the heat sink for 1000 watts.

#### 2632

#4
##### May 20, 2013, 03:47 am
I knew that was to good to be true!
Yep need another 0... oops

Other than the 100 * 100 being incorrect is the formula correct?

#### sonnyyu

#5
##### May 20, 2013, 07:01 am

I guess 11 MOSFETS. My project will use an ice bath on the heat sink for 1000 watts.

TO-220FP/TO-220F package  will be perfect for this type application. no heat sink is needed and directly into ice bath.

#### oric_dan

#6
##### May 20, 2013, 07:26 am
OP, I don't know about icebaths and whatnot, but one thing you should take note
of in the spec sheet is where it says,
Quote

TJ Operating Junction -55 to + 175 °C

Thermal Resistance
R?JA Junction-to-Ambient (PCB Mount)  --- 40 °C/W

NOTE 7:
When mounted on 1" square PCB (FR-4 or G-10 Material). For recommended
footprint and soldering techniques refer to application note #AN-994.

This tells you how hot the device will get, ie the temperature rise above ambient,
assuming you use the recommended 1"x1" pcb copper heat sink. So,

t.r. = 40 °C/W *   26W = 1040 °C (pop corn)
t.r. = 40 °C/W * 3.75W =  150 °C (darn hot)

#### sonnyyu

#7
##### May 20, 2013, 07:28 amLast Edit: May 20, 2013, 08:23 am by sonnyyu Reason: 1

That is safe because the spec for  IRFS3107 is 370 watts maximum power dissipation using a big heat sink.

It might need enough PCB area, if not available then d2pak heatsink.

further more we could use air cooling(fan), water cooling, ice-water cooling, peltiers, dry ice, liquid nitrogen, mini air conditioner, other type heat pump...

Electronics is art.

change ice cooling to ice-water cooling.

#### MarkT

#8
##### May 20, 2013, 02:50 pm
You would never normally take a MOSFET to anywhere like its nominal current rating at continuous
duty cycle - that figure usually represents the maximum power dissipation with an infinite heatsink.

Take the Rds(on) value and work out the dissipation for your load, then decide if thats OK, if not
select a device with lower Rds(on), or parallel them.

The other thing it might be worth checking is the voltage drop from drain to source at your load
current, V = IR.   Again if the voltage drop is more than you'd like, select a better device or parallel them.
[ I will NOT respond to personal messages, I WILL delete them, use the forum please ]

#### dc42

#9
##### May 20, 2013, 11:47 pm
If you use (say) 5 of those mosfets connected in parallel, then the total power dissipation will be reduced by a factor of 5, and the power dissipation will be shared almost equally between them. So that's just over 1W per mosfet, instead of 26W.

The other loss to be aware of is the loss during switching on and off, particularly if you will be using PWM. To keep the switching loss low, you should use a mosfet driver chip such as MCP1407 when switching such large currents.
Formal verification of safety-critical software, software development, and electronic design and prototyping. See http://www.eschertech.com. Please do not ask for unpaid help via PM, use the forum.

#### oric_dan

#10
##### May 21, 2013, 01:05 am
Quote
If you use (say) 5 of those mosfets connected in parallel, then the total power dissipation will be reduced by a factor of 5, and the power dissipation will be shared almost equally between them. So that's just over 1W per mosfet, instead of 26W.

That works.

#### sonnyyu

#11
##### May 22, 2013, 05:32 am

Quote
If you use (say) 5 of those mosfets connected in parallel, then the total power dissipation will be reduced by a factor of 5, and the power dissipation will be shared almost equally between them. So that's just over 1W per mosfet, instead of 26W.

That works.

LOL, I think everyone know it is dave's typo, but I really like your senses of humor.

#### oric_dan

#12
##### May 22, 2013, 08:06 am
Oh, I thought it was right, but some weeks maths doesn't work, everything
looks fuzzy, calculator looks like slide rule, LOL. Where's the decimal point?
100A / 5 = 20A
Pd = 20A * 20A * 0.0026ohms = 1.04W

#### sonnyyu

#13
##### May 22, 2013, 02:54 pmLast Edit: May 22, 2013, 03:00 pm by sonnyyu Reason: 1

Oh, I thought it was right, but some weeks maths doesn't work, everything
looks fuzzy, calculator looks like slide rule, LOL. Where's the decimal point?
100A / 5 = 20A
Pd = 20A * 20A * 0.0026ohms = 1.04W

Ok, You are right, I drink too much coffee lately.  It is not my fault, once I learn some high school  math teachers here are failed go Dutch after eating because no calculator. I tried to do "all the math" by heart and tried to teach my kids that ways as well.

#### sonnyyu

#14
##### May 22, 2013, 03:44 pmLast Edit: May 22, 2013, 03:49 pm by sonnyyu Reason: 1
The two issues might be worth to clear;-

1.About FET, FET has huge family members;- CNTFET, DGMOSFET, HEMT, IGBT, JFET, MOSFET...
Not all FET has power version, but at least IGBT, JFET does. The one we talk about here should be called MOSFET.

more   https://en.wikipedia.org/wiki/Field-effect_transistor

2. Rds(on) is only valid at logic power mosfet, Power linear MOSFET 2SK1058 for example, the Rds(on) = 15000 m?

check here http://forum.arduino.cc/index.php?topic=164436.msg1227775#msg1227775

Power linear MOSFET for RF application support neither d2pak nor TO-220 as package, plus Rds(on) is invalid.

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