that is about 2.6 kW at the start of avalanche

Yes, but only at the very very start.

It all comes down to the inductance value. The total energy stored in an inductor passing DC is

W-s = 0.5 x L x I^2 >> 1/2 inductance times current squared, in watt-seconds

If the magnet inductance is 1 mH, then at 75 A the total energy is 5.625 W-s. When the magnet is disconnected from its DC source, this is the energy that must be dissipated in the suppression network and the magnet's coil resistance. Assuming the resistance is zero (worst case), it all goes into the external diode or FET body diode. The discharge current decreases exponentially, so a linear approximation yields a value with safety margin built-in. As long as we're making assumptions, lets make another one - the discharge time to essentially zero current is 100 ms. In this case, the energy in the snubber is 56.25 W for 0.1 s, followed by 0 W until the next discharge event.

Using Watt's Law, the 35 V body diode must be able to handle an average discharge current of 1.61 A for 0.1 s. The peak current is greater than this, but depends on the inductor's resistance.

Note - all of this is based on a 1 mH inductor. Bigger inductor = bigger heat.

Note Note - All of this is based on several approximations. In reality, the inductor energy is flowing through the coil resistance into a 35 V zener diode, a much more complex circuit to analyze for peak current and power. Still, the total heat analysis is valid.

ak

 

To quote myself: "that is about 2.6 kW at the start of avalanche,"

There are multiple limits to what is allowed for avalanche. The most critical spec in most cases is the the total energy. Very high current FETs tend to run in the range of 200-600 mJ, if the starting die temperature is 25°C. Starting die temp is rarely going to be that low.

A 1 mH inductor at 75 A stores 2810 mJ. You will be extremely hard pressed to find any FET in anything other than a chassis-mount package that will come near that. You would be extremely hard pressed to find any other FETs that would survive that with less than about 5 in parallel (and very well laid out) unless you start with 25°C die temperature which is not easy at 75 A.

e.g.
IRFLP3034, 1.7 milliohms max, 327 A (silicon limited) 195 A (package limited) (TO-247, which has a very large chunk of copper)
Maximum single-pulse avalanche energy at starting die temperature of 50°C is only 175 mJ.
At 1% duty cycle that drops to about 140 mJ.

I looked at one sub-millohm FET (TI part, I think) rated for 600 mJ avalanche energy at 25°C starting die temperature

 

Aside from the connections having been corrected, and the gate drive circuit being clarified, the use of a controller board to control switching is just silly! A simple 555timer IC, a flipflop such as a 4013, and probably a couple transistors such as the MPSA13 would deliver identical functionality far better.

 

I agree with epb - having inverting driver transistors means that there is a chance the system will power up with all FETs on. One solution is to change to logic-level FETs and drive then directly from the arduino. I still see no need for gate resistors.

ak

The gate resistors are used to prevent high frequency oscillation during the period the FET is in its linear area. This oscillation at 75 amps could radiate quite a bit of RF into everywhere, possibly causing the arduino to go crazy. This is prevented by a 1k quarter watt resistor in each gate lead. Cheap insurance!

 

It isn't difficult to find FETs that will dissipate on the order of 10 W at 75 A with 4.5 V gate-source (that is, for a single FET). With most logic level FETs you will reduce the ON resistance by 20% or so with Vgs of 6-8 V versus that at 4.5 V. You can actually easily find lower ON resistance in surface mount parts than in through-hole, but SM makes managing current and heat harder.

As I mentioned previously, a free-wheeling diode across the load will slow turn-off, but I suspect that is irrelevant for the application. Without knowing the inductance it is impossible to know if the body diode of the FET(s) can be safely avalanched. If you use a (say) 30 V FET, it will probably avalanche at something around 35 V. At 75 A, that is about 2.6 kW at the start of avalanche, which is rather a lot. A 100 µH inductor would store 281 mJ at 75 A, which is within the rating of many high-power FETs if the die starts out at a very modest temperature.

The FET itself will always avalanche before the reverse diode, thus the avalanche limitations of the FET MUST be adhered to. The FET body diode will not help in any way. Add the kick back diode externally to the FET. As noted above, the kickback energy is dependent on the energy stored in the magnetic field so it can be large or small dependent on the inductance. In most cases the surge rating of the diode is all that is needed to be 75 amps. The average current rating of a diode to get a 75 amp peak current may be less than 10 amps for a suitable diode (costing a dollar or two) so it is not a significant expense to add it.

 

I still assert that using an Arduino is not the best choice because it is a whole lot of excess hardware and must be programmed and costs way more than the CMOS ICs that could do the job far more reliably. Also, the IC devices can all run very well on the 12 volt supply, reducing the complexity and removing the need for the regulator. OR, you can use a CD 4017 to generate the duty cycle with an off-time between the on times, except that since the two switches can be on at the same time with no problem that is over-doing the controls a bit. So really, what is so special about using the arduino to generate a square wave????

 

The FET itself will always avalanche before the reverse diode, thus the avalanche limitations of the FET MUST be adhered to. The FET body diode will not help in any way. Add the kick back diode externally to the FET. ...

Avalanche in a power MOSFET is in the body diode which exists as an intrinsic structure without any option to omit it. The vast majority of modern power MOSFETs are rated for avalanche operation, both single pulse and repetitive, and one is perfectly entitled to use it to discharge an inductive load (as I have done in some small flyback converters where the leakage inductance is discharged by FET avalanche) provided the current, energy and power limits are observed. Modern power FETs are carefully designed to prevent turn-on of the parasitic BJT, which is something you don't want to have happen. (By "modern" I mean those designed in the past couple of decades).

My point was that with "high" inductance at high current it is likely impossible to find a FET with adequate avalanche energy handling capability without greatly exceeding the ratings required for normal switching of the load with decent efficiency. An inductor of 50 µH at 75 A would likely pose no problem at moderate die temperature and diode duty cycle. 500 µH at 75 A (1.4 joules) would likely make avalanche discharge impractical. Unless paralleled FETs are carefully matched and laid out carefully, it is unwise to expect sharing of avalanche current because of unit-to-unit variation in actual avalanche voltage. The FET with the lowest actual breakdown voltage will probably hog almost all of the avalanche current.

Putting a free-wheeling diode across the inductive load such as that under discussion is the typical thing to do because it is cheap and easy, but, as I also pointed out, it greatly increases the time taken for discharge of the inductance in comparison with methods that permit a higher discharge voltage. If you need fast discharge, a single free-wheeling diode will likely be unsuitable (but a small number in series might be satisfactory).

 

Lots of erudite circuit analysis here! But my question is why are you using two magnets above their power ratings rather than using one magnet that can provide the required holding force within its power rating? You are reducing the duty cycle of each magnet by 50% and the implication is that they are OK at that power dissipation but will burn up at 100%. Can't you find/make a magnet that can do the job at 100% duty cycle?

Or am I missing something here in your description of your application?

 

If the FET switches had adequate heat sinks then the switching off time could be a bit slower, greatly reducing the dv/dt and the resulting spike. That would be another option. But external diodes would certainly be an effective choice as well. AND, I would ask the same question, which is why not simply build an electromagnet that is good for the 100% DUTY CYCLE? Then the problem could be solved. Use more iron and higher temperature insulation and possibly cooling fans, or even water cooled coils, like used in some spot welding transformers. That has been done and it worked in automotive production lines for many years. Not simple, not cheap, but very effective.

 

Lots of erudite circuit analysis here! But my question is why are you using two magnets above their power ratings rather than using one magnet that can provide the required holding force within its power rating? You are reducing the duty cycle of each magnet by 50% and the implication is that they are OK at that power dissipation but will burn up at 100%. Can't you find/make a magnet that can do the job at 100% duty cycle?

Or am I missing something here in your description of your application?

Suspect that the OP's application is a "rail gun" so heavy duty and heavy weight components are a limiting factor.

 

Not making a rail gun. Maybe next month though!

What I ended up settling on was a revised circuit using two of either IRLZ44PBF or
PSMN8R0-40BS mosfets driven directly from Arduino @4.5V, bypassing the 2N3904 transistors. From reading your comments about avalanche, I'll probably include the freewheeling diodes. Which mosfet would be the better choice?

Stay tuned for the railgun! Sure got a kick out of that comment.

 

The IRLZ44NPBF would be suitable to drive directly from the Arduino.
The PSMN8R0-40BS would need a gate drive >10V.

I would suggest that you isolate the power supplies for the Arduino and the heavy current "device" using opto-couplers.
If you did that the PSMN8R0-40BS would be the better device to use.

This will minimize the possibility of having spikes on the ground or supply rails of the Arduino when the device is triggered.

 

How many FETs do you intend to parallel?

The IRLZ44 is unsuitable for 75 A with fewer than 4 or 5 in parallel unless you are prepared to accept high losses. 5 in parallel at 75 A will dissipate at total of about 25-30 W.

 

IRLZ44NPBF has lower Id, higher Rds(on), Vgs of 16V? How would that be a better option?

I would ideally like one with the TO220 architecture as I already have heatsinks to fit TO220.

Can anyone find a TO220 that can handle 75A between 2-3 parallel, at a Vgs to drive from arduino without dissipating too much heat?

 

You are reading the maximum Vgs (+/-16V) for the IRLZ44NPBF.
At a Vgs of 4V, the Rds is 35 milliohms (less at 5V).
As ebp suggests, you would need 4 or 5 of those to safely handle the 75A.

What kind of TO220 heatsinks do you have? Those little clip-ons?

The PSMN8R0-40BS has an Rds of only 7.6 milliohms so your power dissipation is going to be way less.
Two should be sufficient. Just drive with a transistor or (my preference) an opto-coupler.

 

The main concern I have with the PSM is that the 2nd terminal (I believe it’s drain) is not possible to connect to. Why would they design it like that, how are you supposed to hook up drain?

 

The main concern I have with the PSM is that the 2nd terminal (I believe it’s drain) is not possible to connect to. Why would they design it like that, how are you supposed to hook up drain?

The "unconnectable" pin is also the tab. It's labeled MB on the datasheet.

 

You could select an alternative TO220 device here:

https://www.infineon.com/cms/en/pro...ax,559_max,559_90_max,297_max,478_min,478_max

 

You could select an alternative TO220 device here:

https://www.infineon.com/cms/en/pro...ax,559_max,559_90_max,297_max,478_min,478_max

Alright... I like this. Say I decided to go with the IRF2805PBF, which is rated at for a max of 175A. How many of these would I need to switch the current of 75A (I'm assuming there's some parameter I have been overlooking as to why I'd need 4-5 of the last mosfet, which was rated for a good bit).

 

As I said upthread (I think, I've said the same thing on multiple threads), when you are selecting power MOSFETs for switching applications it is usually best to select based on ON resistance rather than current rating. Sometimes the manufacturer's current rating will be more limiting than ON resistance, but not often.

The IRLZ44N has an ON resistance of 22 milliohms with 4.5 V gate to source. Power dissipation is current squared time resistance, so
75 A squared times 22 milliohms = 5625 x 0.022 = 123.75 watts, for a single FET
If you used 5 in parallel, you would reduced the ON resistance by a factor of 5, so the total power loss would be about 25 W, or 5 W per FET, though you can't count on exact sharing unless you actually match the FETs. At 5 W, you don't need much of a heatsink. It is actually preferable to mount paralleled FETs on a common heatsink. This helps them stay at approximately the same temperature which improves sharing.

The above calc's ignore switching loss, but at a very low switching rate it can be mostly ignored, even though the instantaneous dissipation will be hundreds of watts.

The IRF2805 has an ON resistance of nominally 4.7 milliohms maximum with 10 volts gate to source, so provided you supply the higher gate voltage, one 2085 is better than four paralleled 'Z44Ns. At 75 A, one would dissipate about 26 W, which would require a moderate heatsink.
Two in parallel, on opposite sides of the same heatsink, might be pretty good overall.

You can find single FETs in through-hole packages with ON resistance of less than 2 millohms with 4.5 Vgs

Since you are controlling 900 watts, 20 or 30 watts of loss isn't bad from an efficiency point of view, it just means you do need moderate heatsinks. Paralleling is a nuisance if it isn't absolutely necessary.

75 A is a lot to put through connecting wires, so that has to be dealt with carefully, Paralleling several FETs does make it easier to attach adequately large wire. At 15 A per FET, 18 AWG wire would be (just minimally) adequate - each FET gets a short length pigtail (all of the same length) to a common terminal for perhaps 6 or 8 AWG to the supply & load. I rather like EPDM-insulated wire if it needs to be moderately flexible, but it is rather expensive & not very common. I like fluoropolymer insulation if the wire is going to run hot, but it is even more expensive.