Mosfets

Discussion in 'General Electronics Discussion' started by Lord_grezington, Jun 15, 2013.

1. Lord_grezington

124
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May 3, 2013
Hello...

I am hoping to use this Mosfet http://www.farnell.com/datasheets/51964.pdf

The Specs are as follows:

Max Current = 61A (38.5A @ 100 degrees)
Max Voltage = 200V
Rds(on) - 0.044 Ohms
Max Power dissipation = 417W (3.3W/degree for derivative above 25 degrees)

Now, I want to draw around 30A (below max current @ 100 degrees) nearly continuously, , so are the calculations below correct?

Max temperature needs to be around 80 degrees, so: (80 - 25) x 3.3 = 181.5W max power dissipation.

Power dissipation is (30 x 30) x 0.044 = 39.6W (which is below Max)

I will only be taking around 48V through the Mosfet, I will also be driving the Mosfet through a driver with a Vgs of 15V to keep losses to a minimum. I will also me running it from 100Khz.

Will this work? if not where am I going wrong?

Thanks

2. (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

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Jan 21, 2010
At 30A and an Rds of 0.041 ohms, the power dissipation will be 30*30*0.041W = 3.69W -- a very small amount of power.

However, your switching losses will be significant. The turn on rise time is 440ns and the fall time is 350ns. Assuming you can achieve that, the switching losses are:

(440*10^-9 * 48 * 30 * 100,000)/2 + (350*10^-9 * 48 * 30 * 100,000)/2 W
= 32 + 25 = 57W

Do the total dissipation will be 61W

If you want the max temperature rise to be 55 degrees, then you need a total thermal resistance of 55/57 which is about 1 degree C per watt.

From this we subtract the case to junction thermal resistance of 0.3 degC per watt and another approx 0.5degC per watt for the losses between the device and the heatsink and you find you have precious little remaining.

To have any hope at all you'd need a VERY good heatsink with forced air cooling at a minimum.

You may wish to consider a lower operating frequency (50kHz would almost halve the power dissipation) or having multiple devices in parallel.

Switching 30A at this frequency you'll have to be very careful about stray inductances which could lead to some embarassingly high voltages appearing across the mosfet as it turns off, leading to even higher dissipation

edit: I'm pretty tired. Repeat the calculations yourself to make sure I've not put things out by an order of magnitude or something silly.

3. KrisBlueNZSadly passed away in 2015

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Nov 28, 2011
I think all your calculations, and your conclusion, are correct, and you're doing everything right. Your power dissipation is way under your limit, so no heatisnk is needed. I guess you had assumed that.

Make sure that your gate drive waveform has a high slew rate. In other words, don't leave the MOSFET in its linear region for very long.

Edit: I hadn't noticed you were switching at 100 kHz. Steve is probably right. I'm also tired and somewhat inebriated so I probably shouldn't have posted. I'll stop now :-/

Last edited: Jun 15, 2013
4. (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

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Jan 21, 2010
30 x 30 * 0.041 is not 3.69. Hahahaha. I was an order of magnitude out.

That just makes it worse with a total power dissipation of 84W (worst case) and an allowable rise in temperature of 55 degrees

5. Lord_grezington

124
2
May 3, 2013
Ok, I am get the logic, but I just want to understand your equations a little better.

Rise time:
440x10^-9 (rise time) * 48 (Voltage) * 30 (Current) * 100,000 (Frequency) / 2 (is this 2 assuming a duty cycle of 50%?)

Fall time same as above...

The first part of your thermal resistance I get as well, why do you subtract the junction to case thermal resistance? should this not be added on along with the additional losses?

I am hoping to parallel the mosfets up as I actually need around 55.5A. I know that when I parallel these mosfets up I need to watch my PCB traces to ensure I get (as close as possible) identical resistances and I need to use separate gate resistors. I am a little worried however to use more than 2 on parallel. Is it OK to say use around 4 of these in parallel without creating some sort of bias due to the tolerances?

thanks - great help so far

6. Lord_grezington

124
2
May 3, 2013
New Question (on top of the one above)

Is it possible to use Mosfets (or IGBT's) @ up to 60Vdc and 60A? I am looking for the lowest cost option. A high switching frequency (100Khz) is preferred to keep additional component costs down (caps and inductors).

I have looked at both IGBT's and Mosfets (even relays), but without coupling at least 4 components in Parallel the dissipated power is too high.

Anyone have any good ideas?

7. (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

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Jan 21, 2010
No, it assumes the power rises from zero to the max and then falls again. If the max energy dissipated is voltage * current * time, and if the rise and fall is linear then the energy dissipation is voltage * current * time / 2 (geometry shows us that for a triangle in a rectangle. f is 1/t, to that gets rid of the extra time value (cancels from rise time)

8. (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

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Jan 21, 2010
Look for devices in a TO-3 package. They have a lower junction to case thermal resistance in many cases.

If you can get by with a "live" heatsink (no insulation between case and heatsink) you will have lower thermal losses.

Also you might be able to find a device which can switch faster.

If this is a SMPS, then your diodes are going to be dissipating a lot of power too.

An offline SMPS is better in these cases because you switch a higher voltage and lower current and use a transformer to produce the lower voltage and higher current.

9. Lord_grezington

124
2
May 3, 2013
Hi Steve

Thank you, very good explanation.

However I am now a little stumped to the gate current. To calculate this do i divide the rise time (440ns) by the gate charge (75nC) to get a current of 5.87A? This seems very high.

Last edited: Jun 16, 2013
10. BobK

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Jan 5, 2010
Q = I * T

So it is the other way around, divide the charge by the time to get the current. In this case 74/440 = .16A.

I think you can find MOSFETS that switch a lot faster than that. This would help a lot.

Bob

11. Lord_grezington

124
2
May 3, 2013
Hi Bob

Sorry, silly mistake, 0.17A sounds much better.

I expect this is how we calculate the gate resistor, 15V gate voltage gives a gate gate resistor value of 88 Ohms?

12. BobK

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Jan 5, 2010
No, you really want to get as much current as is reasonable to make it switch fast. Usually just something like 10 Ohms will do. With gate drivers, you typically do not use any resistance.

Bob

13. Lord_grezington

124
2
May 3, 2013
Is it the Total Gate charge I need to take? Using the total gate charge looks to be giving me some high gate currents. Should I be using the gate source charge or the gate drain charge?

14. (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

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Use total gate charge. Large currents are typically not a huge issue until you hit the limit of maximum gate current or total gate dissipation (And for the latter you'll need to know Rgi, which is not often specified).

Gate currents exceeding several amps are not unheard of. For this MOSFET I doubt you'll need them because it simply can't switch that fast.

If your load is inductive at all, you may need to slow down the switch off. If it is capacitive, you may need to slow down the switch on. Both of these will increase dissipation as the cost of saving the mosfet from nasty voltage or current spikes (respectively). Since you're very close to thermal limits already, I'd be hoping the load is purely resistive and that you minimise wiring inductance.

15. KrisBlueNZSadly passed away in 2015

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Nov 28, 2011
It sounds like you have a lot of experience in this area Steve. Perhaps you might like to combine this advice into a tutorial on MOSFET driving in demanding applications. I would certainly read and save it. Advice like this can be worth its weight in gold!

16. (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

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Jan 21, 2010
Whilst I am more than a babe in the woods, I am experienced only by the number of mistakes I've either made or observed others make.

Gonzo Engineer is the man for this sort of stuff.

I can point out a lot of problems -- He can point out a lot of solutions.

The other thing that makes me seem a lot smarter than I am is that I've read a lot of application notes.

edit: It might be interesting to put together some notes on mosfet driving but it would certainly need to be checked by people with more experience than I

17. KrisBlueNZSadly passed away in 2015

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Nov 28, 2011
I'm like that too Steve. Any information about factors to consider in electronic designs (in general) is more important than the detailed information itself; you can always look that up later, or ask an expert. The important thing is to recognise all the factors that might affect your design. As a former design enginner, I understand this reasoning. Most problems encountered in a production environment (actually, in any process, probably) are unanticipated. Therefore, having strong problem anticipation instincts is important.

This is why, sometimes, my responses to technical questions on the forums contains other questions that begin "Have you considered ...".

And I agree on your comments about application notes. Unlike most electronics documentation on the web, application notes (and any technical material published by component manufacturers) are written by real engineers, and who have the background and the specific experience to give you the best information and advice on that particular topic. They're a powerful way to broaden your knowledge.

Last edited: Jun 17, 2013
18. Lord_grezington

124
2
May 3, 2013
Thanks Steve, you have been a lot of help.

I have now found a new Mosfet http://www.mouser.com/ds/2/196/IPP_B230N06L3_Rev2.2_-70616.pdf this one has a tiny total gate charge (only10nC) which allows a much lower switching current and as its quiet well priced its worth stacking them up in parallel to get the total current i need.

Also I have noticed that if I have a lower gate charge, with a higher gate voltage (15V) the gate charges very quickly which massively reduced switching losses.

However I have worked in reverse to Steves original idea, when stacking the fets up in parrellel its the gate current which cause issues. So I chose a driver with a output current of 1A, spread this over 4 fets so allows 250mA per fet. With a gate charge of 10nC it gives me switching times of 40ns. Check this is slower than the fastest possible switching time (which it is). Now I can calculate the total power losses. I now find that the power loss for each devise (@ 15A each running on 48V and 100Khz) is only around 8.06W. And if I want to improve this I take a higher gate driver to supply 2A giving 500mA per fet and now decreases the total power loss down to 6.62W each.

The current is limited to each fet using the gate resistance. ie, if I need 0.5A with a gate voltage of 15V. Keeping in mind this devise has a gate Resistance of 0.9 Ohms, the required gate resistor is 29.1Ohms.

Is this theory all correct?

19. (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

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Jan 21, 2010
That seems to have a much faster rise and fall time so the switching losses will be much lower.

I haven't done the calculations, but if you're happy with the ones I've done above and followed them with the figures from this mosfet then you should be in the ballpark.

I notice that this device actually specified the internal gate resistance so you can calculate the gate dissipation should you want to.

20. BobK

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Jan 5, 2010
Yes, that is a much better MOSFET for you application. Lower Rdson, lower gate charge and lower rise / fall times -- all of these parameters move in the right direction.

Bob