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### Topic: Calculating power loss in a MOSFET (Read 3011 times)previous topic - next topic

#### 2632

##### May 25, 2012, 01:36 am
Ok, I need some help with this one.
I am working on a project that has a battery disconnect.  The current is about 50A (calculate at 70A) @14v dc (continuous)  The switch should almost always be "on"

Mfg: Infineon
Part No.: IPD 70N10S3-12
Rds (on): 11.5m ohms
I d: 70A  (what is the d?)

So how do I calculate how hot this sucker is going to get?

#### RuggedCircuits

#1
##### May 25, 2012, 01:44 am
The 'd' stands for "drain", which probably doesn't help much. It just refers to the amount of current flowing into the drain terminal and out the source terminal.

The power dissipation of a resistance is I^2 * R, so in your case it would be 70*70*0.0115=56.4W. Way too much. It will get hot for about 2 seconds then catch on fire.

In that current domain you can:

1) Get a MUCH lower on-resistance MOSFET (like 1mohm instead of 11.5mohms), and use big thick wires all over the place
2) Put MOSFET's in parallel (e.g., 10 MOSFET's at 11.5mohms each is like 1 MOSFET at 1.5mohms....though you have to be careful to make sure they all switch on and off at exactly the same time)
3) Use an IGBT
4) Use a contactor

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#### 2632

#2
##### May 25, 2012, 03:20 am
Thank you!

I like the idea of a IGBT but when I look at most of them the dissipated power is also way high.  But the screw terminals and form factor would work good for the application.

Ideally I would use a latching relay (big one) since this would not use any power on standby.  We were looking into solid state thinking it would use less power on standby (on) than a contactor, but looking at the number you came up with I may need to reconsider.

Any suggestions on part numbers to look at?

#### RuggedCircuits

#3
##### May 25, 2012, 05:18 am
Forgot to mention the possibility of a really big heat sink and a fan. Together with a 1mohm (or so) MOSFET and it might just work.

Suggestions on part numbers for what? MOSFETs? The NXP PSMN1R8-30PL is a good starting point.

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#### 2632

#4
##### Jun 02, 2012, 03:27 am
So after looking for other options. MOSFETs keep coming up as our best option.  At this point I am thinking paralleling enough to satisfy the current requirement and to keep the ON losses as low as possible is the plan.

Quote
The power dissipation of a resistance is I^2 * R, so in your case it would be 70*70*0.0115=56.4W. Way too much.

So how about 10 of them? (round number)
70A/10 = 7A for each MOSFET
7*7*0.0115 = 0.5635W loss in each MOSFET
Total power loss 5.6W

Is this correct?

Would adding more MOSFETs just keep reducing losses at the expense of space and price?

#### retrolefty

#5
##### Jun 02, 2012, 03:39 am
Quote
Would adding more MOSFETs just keep reducing losses at the expense of space and price?

Well it would keep improving the source/drain composite losses in theory, but keep in mind that the more gates you parallel together the more combined parallel gate capacitance you will have, and that will consume more current during switching transitions and/or cause slower source/drain switching times which can increase main current circuit losses in a major way, including device damage. In robust industrial designs using very high power mosfets or multiple mosfets they tend to use special purpose 'gate driver' chips that can source or sink many amps of current very quickly, just so the main mosfets will switch as fast as needed for the high power circuits. The Arduino output pin current capacity must be analyzed carefully when attempting to use with very high power mosfet switching applications.

Lefty

#### 2632

#6
##### Jun 02, 2012, 03:48 am
This application is a basic on/off switch that will be on almost 24/7.

I was planing on having the Arduino drive a transistor that then drives the MOSFETs.

Primary goal is to have a cool running high current switch.  If the losses are higher than a contactor than the contactor will win out.

#### jackrae

#7
##### Jun 02, 2012, 09:39 am
In simple terms, 2 mosfets in parallel means half the current through each therefore only a quarter of the power loss through each, so potentially less than 1/2 total loss ( 2 x 1/4).  On the basis that power loss is a square law, if you have 4 in parallel, the total power loss will be 1/4  (4 x 1/16)

#### MarkT

#8
##### Jun 03, 2012, 07:15 am
Several things:  if running from 5V you _must_  have logic-level MOSFETs.

With such high currents you need to use a MOSFET driver chip to prevent drain-gate capacitance feedback from putting voltage transients on the gate and to avoid exceeding the safe-operating-area ratings (at 70A you need to avoid being in the linear region for more than a few us - gate drive of several hudred mA is good)

Heat sink and fan is going to be hard to avoid at 70A.  If you parallel a couple of 3 milliohm FETs you'll get 15W dissipation which isn't too bad. 10 devi ces in parallel is overkill and hard to drive fast enough I suspect.

There are high current MOSFET packages available with much higher ratings than TO220, but are expensive - they bolt onto a large heatsink.  High current MOSFET drivers are required for such modules.
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