Going by the schematics on the previous page, Bromikey seems to be using a high side switch between the caps + and the battery +, he would be better served to use a low side switch between the battery negative and the cap negative. That would simplify the driving circuitry.

If he has the mosfet drain at battery voltage he will have problems without a working level shifting driver circuit or something. I'll ask him.

Cheers

I've just recently downloaded the newer manuals for the picaxe chips and it tells me now that the M2 chips can all run at the default clock speed of 4 Mhz or they can be clocked up to 8 -16 and 32 Mhz, the old manuals only stipulated that the M2 chips could be clocked to 8 Mhz. Theoretically the M2 chips at 4 Mhz clock speed can output a 500 Khz PWM signal,
but at 32 Mhz the theoretical PWM signal could be 4 Mhz. Not bad for a cheap 8 pin micro.

What I like about the smaller chips is the price, the small size and low power draw along with many of the functions of the larger chips. For more outputs and inputs the 14M2 seems good and I intend to use those more.

..

An alternative 8 pin part is the Atmel ATtiny85.  64MHz PWM oscillator, ~$1.00 each.  The core runs 8MHz, 8MIPs.

BroMikey's self-education project continues to reveal to him
interesting developments just when he thinks he's got it
figured out.  His education has progressed to the point
where he "knows enough to be dangerous."

Not meant to denigrate BroMikey or his endeavors but it is
a phase that all aspiring electronics technicians work through
as they grow in their understanding.

Six paralleled MOSFETs should be more than capable of handling
40 Ampere pulses, or even more, if properly driven and properly
protected from flyback pulses which are capable of destructive
avalanche of the MOSFET Body diode.

Hopefully BroMikey will either find the answers to his dilemma through
his own research efforts or a knowledgeable tech will provide him with
the elusive technical details he hasn't yet found.

It is possible in his case that his MOSFET gate driver chips aren't properly
positioned as close to the MOSFETs as possible and with sufficient capacitance
to properly drive the Gates.

It is also possible, and very likely the case, that the length of the cables from
his capacitor bank to his battery bank results in quite a lot of stray inductance.
At the instant pulsing current switches off a very substantial flyback pulse will
be generated which, although very brief, will take the MOSFET's into avalanche.
Since avalanche characteristics are quite different amongst the parallel connected
MOSFETs it is probable that one of them will serve as a current hog for the avalanche
current.  In time this will destroy the MOSFET and result in its failure much as has
been the case so far with his setup.

Thankfully, these problems are rather routine in high current switching systems
and the remedies are well established,  BroMikey will find success if he keeps
searching.

Maybe if he is dumping enough capacitance charged to a high enough voltage into a battery he might be exceeding the maximum pulsed current rating for the mosfet, and with several in parallel one mosfet might be taking most of the abuse, as a cap discharge can produce a very high initial peak current.  That is why it's done, isn't it.

And coupled with poor switching and high "on"resistance then "Pop goes the mosfet".

He also mentioned at one stage he w is using quite long cables, like meters in length.

I've made a drawing for him with a store bought transformer run from the wall to give a 12 + 12 volt supply so he can rectify 12 volts for switching supply and 24 volts for charging the caps with a Large inductor between smoothing caps and dump caps to partially isolate the dump cap from the supply when dumping.  Maybe a MOT primary would work there, (just remove the secondary for safety XXX).

..

Maybe if he is dumping enough capacitance charged to a high enough voltage into a battery he might be exceeding the maximum pulsed current rating for the mosfet, and with several in parallel one mosfet might be taking most of the abuse, as a cap discharge can produce a very high initial peak current.  That is why it's done, isn't it.

That's true. A capacitor discharge can produce very high currents, the more capacitance the more current ... into low impedance loads. The max pulsed current for the P460 is 80 amps and that's under pretty narrow pulse parameters. So if he's charging to 80 volts, then dumping the caps through a single mosfet due to poor paralleling, he needs at least one whole ohm of impedance in the circuit after the mosfet, to keep the max current below 80 amps.  R = V / I  so 80/80 = 1 ohm. Oh, wait, the mosfet itself has 0.27 ohms minimum resistance, so the rest of the circuit needs less than three quarters of an ohm impedance (resistance) to keep the maximum surge current below 80 amps. No matter the capacitance.

I'm still having my first coffee of the day, so please check my math and reasoning.

Dumping a cap through a mosfet into a load is a lot less problematic than charging an empty cap through a mosfet switch or crowbarring a cap bank with a mosfet. The empty cap looks like a dead short, so if the supply voltage is there, the surge current in the mosfet can be very high. But dumping a cap into a resistive or inductive load with a mosfet is much less problematic because the load impedance limits the surge current. I think.

And coupled with poor switching and high "on"resistance then "Pop goes the mosfet".

He also mentioned at one stage he w is using quite long cables, like meters in length.

Then he should put the diode I mentioned above, at the load end of the cables, and supplement with an additional similar diode right at the mosfet pins.

I've made a drawing for him with a store bought transformer run from the wall to give a 12 + 12 volt supply so he can rectify 12 volts for switching supply and 24 volts for charging the caps with a Large inductor between smoothing caps and dump caps to partially isolate the dump cap from the supply when dumping.  Maybe a MOT primary would work there, (just remove the secondary for safety XXX).

..

Is the complete actual circuit he is actually presently using right now, available for study? I'd like to take a look but I can't be arsed to sift through page after page of nonsense at EF.

Aye, BroMikey still has a ways to go before he acquires full
and complete technical understanding of what he's up
against with his circuit.

Good education always seems to come with some sort of
pricetag - be it in dollars or burned out MOSFETs.  But the
rewards can be enormous if the "student" has the gumption
to stick with it until he is finally able to make it do what it is
supposed to do reliably and without fear of premature failure.

And know the reasons why...

What would be the peak current of 36 volts in a 200,000 uf cap dumped into a paralleled set of four 12 volt batteries in good condition ?

The peak current depends on the total resistance of the circuit plus load. The capacitance _does not enter into it_.

The peak current into a _direct short_, when switched by the mosfet, is found by applying Ohm's Law: I = V / R. 
Since v = 36 volts and R = 0.27 ohms, the mosfet's minimum on-state resistance, the maximum current is just over 133 amps. Into a dead short, from whatever you are using that provides 36 volts, whether it be a capacitor of infinite capacity or a battery or a hydroelectric power plant.

For _how long_ can that peak current, or any current, be sustained? NOW the capacitance matters. Higher capacitance, more energy, more time to discharge it. As soon as the discharge from the cap begins, the voltage will begin to go down and the current will drop from its peak value, and as the voltage decays exponentially so does the current. More capacitance, longer time constant. More _voltage_ more peak current.

Now let's imagine a more realistic load impedance. Say it is one single ohm. Now solve I = V / R, and find that you can only push a bit over 28 amps through the circuit, no matter what the source is, no matter what the capacitance you are dumping is. That solution to Ohm's Law is the _maximum_ current that 36 volts will push through your circuit.

Now I don't know what other elements he has between the mosfet switch and the load, or what the equivalent internal resistance is of his battery banks, but I will bet that the total circuit resistance, even with all six mosfets properly paralleled, is more than 1.27 ohms. Which means that the _maximum pulse current_ from a 36 volt source will be less than 28 amps.

However in his post he says he's using 80 volts and is attaining 6 amps per mosfet, or 36 amps total. Again, solving Ohm's Law, we find that using those numbers at face value, R = V / I gives us a total circuit resistance of a little over 2.2 ohms.

So going back to the original question, if your capacitor of a million billion Farads is charged to 36 volts, what will be the peak current when it is discharged thru the mosfet (or thru a simple mechanical switch closure) into BroMikey's 2.2 ohm circuit? I = V / R = 36/2.2 = less than 17 amps. A single mosfet will work safely for a 36 volt supply, IF the backspike is handled properly.

BroMikey is struggling with some internalized misconceptions and still
lacks full understanding of some critical details.  He's still in the embryonic
stages of his technical development.  Ways to go yet.

And yes, BroMikey you're reaching some of us.

The good news is that we've all been there; it too will pass. Then the
real understanding will congeal with eye-opening results.

Don't stop the quest!  True understanding is built upon numerous
failures which are temporarily disappointing but serve to fuel the
fires of desire.  The desire to know with certainty.