MOSFETS overheating

[WingZER0]

I had my CA249 setup with 3 mosfets (IRF2804) in parallel of each other. I've used a 100 ohm resistor infront of the gate and a 30,000 ohm resistor behind it. I'm using a 9.6v 1700 mah battery.

Any fire burst longer then a second however heats the mosfets to untouchable levels (60+ degrees) and has already killed one of the mosfets.

No short circuts are in the circut itself: I pulled it off the motor and held the trigger down for 10 seconds, mosfets didn't heat up at all.

It's not short circuting on the mechbox: I tested it connected to just the motor and it still overheats.

I'm completely stumped on this. :s Any input?

[Dynamo]

shouldt you be using only one mosfet?

[Skruface]

Airsoftmechanics.com has a few guy who REALLY know electronics/MOSFET theory. They even have some public domain wiring diagrams and stuff.

[Drake]

Slightly off topic but did you build this setup yourself, or is this a commercial MOSFET (and if so, which one) ?

[m102404]

In addition to the PM I sent...

+1 on going to airsoftmechanics.com...I haven't seen anything on this forum re. mosfets that compares to what they have there.

I bought one of Terry's SW-AB-Long mosfets and it seems to work really well so far. Resetting 20A fuse, active braking, etc... I had it short out in my first build and found that in the wiring channels of the mechbox there were little pins in the right half of the mechbox that I suppose are there to hold the wires in place.

Squeezing 2 16AWG wires and a 24AWG wire in one channel took up too much space and one pin was shorting the system. The mosfet fuse blew and got really, really hot (Terry said it might be in excess of 120deg when tripped). After I rewired the mechbox (18AWG primaries and 24AWG trigger) there was enough space and the system works fine. It's only a 9.6v 1500mah 2/3A battery, pushing a M110 spring with a stock EG1000 motor...so it's not the 10.8 or 12v systems that some guys use.

Haven't fielded it enough to comment on it's practical performance.

Tys

[mcguyver]

Mnay FETs will heat up horrendously under load, even a small load. A heatsink is the solution, to increase the effective mass and surface area of the FET when dealing with heat dissipation.

Using more than 1 FET in parallel on a single-waveform load is fine. The problems with multiple FETs are when you are using them to drive AC loads with 1 FET driving the positive wave half and another driving the negative wave half. If the aren't biased properly (the waveforms meet up well at the crossover point or zero volts), then you get crossover distortion which alters the waveform and can create excess heating. This is not the case here.

If you have no load issues when the motor is disconnected from the FETs, this should be a reasonable inference that you have no damage in the wiring between the FET and the motor. It's not to say that the motor itself is not faulty. Use a DMM with DC-series current capability of a DC clamp meter and test your inrush current and running current with the motor removed from the gun (in other words with no load). If it exceeds 20 amps inrush or 5 amps running at no load, turf your motor and replace it. I've seen bad motors at no-load draw 20 amps or more.

You did not specify whether you had a fuse in the system, but if you had one it likely would have blown already or shown signs of thermal damage.

[WingZER0]

I based my design off of the airsoft mechanics one (thus why I put the 100 ohm and the 30k ohm resister in front of and behind the gate). My P90's actually wired to exactly the same specs as the airsoft mechanics example one.


Yes, the mosfet is a home built one, but I had mosfetted other guns already (although with a somewhat beefier mosfet, not the T220 package) which operates completely fine as far as I know.


There isn't a fuse in the system




And I'll see if I can get my hands on my dad's old DMM and I'll post up what I measured.

[Kos-Mos]

I bet that you have a friend FET in the lot. That will cause 1- extra load on the remaining fets if it burnt open 2- reversed load and to a certain extend short circuit if it is burnt closed.

Remove the unit and try to use the one from an other gun and vice-versa. Then it should confirm that theory.

*edit*

If you have a bog standard multimeter, you can check conductivity between Source and Drain. If it conducts and have a close to null resistance, then it most probably is fried. Used that to check the fets while prototyping back in college. It was an autonomous "sumo" robot. When the PIC controler was reset because the power regulator was not powerfull enought to drive the 2 servos for the claws, the fets where receiving random signals and burnt some times.

[WingZER0]

Added a massive heatsink to the mosfets, and that seems to be working well (old cpu thermal paste FTW). Been meaning to get the DMM but it keeps slipping my mind. <.<

[MadMax]

Originally Posted by WingZER0

Added a massive heatsink to the mosfets, and that seems to be working well (old cpu thermal paste FTW). Been meaning to get the DMM but it keeps slipping my mind. <.<

I have to wonder if MOSFETs actually offer all that much benefit if they have to be significantly heatsunk. It's so easy to say they rock and are more efficient than the usual switch gear because they're so voodoo to most users.

If your FETs heat up enough that they'd melt heatsink made of nylon then they're less efficient than a carbon block between phosphor bronze plates supported by nylon (the traditional switchgear that usually doesn't melt).

Heat = inefficiency. If you need serious heat sinking, then you're expending watts of inefficiency heating FETs that aren't driving the motor. The only time I've used FETs to replace a mechanical switch was for a ludicrous M249 running a 14.4v 4.8Ah pack. The first trigger pull welded the switch closed so it got FETs that were heatsinked to the solid aluminum upper. Obviously FETs were the only switching that would have worked with that AEG because mechanical switches can't be as easily heat sinked and the arcing issues are pretty difficult, but I wonder if they're really beneficial for lower powered setups which draw current well within the current capacity of mechanical switches.

FETs ALWAYS incur a 0.7v drop in the forward bias mode which means you're always burning 7.3% of the energy leaving the battery in your FET cct (0.7v/9.6v). If the voltage drop in a mechanical switch is less than 0.7v then you're doing worse with a FET cct. 0.7v is a considerable voltage drop for our low DC voltage packs.

RC cars and airplanes depend on FET switching to accomplish PWM speed control, so do cordless power tools, but we don't need this feature.

Has anyone made direct measurements comparing the voltage drops over mechanical switches vs. 0.7v?

[MadMax]

Originally Posted by WingZER0

I had my CA249 setup with 3 mosfets (IRF2804) in parallel of each other. I've used a 100 ohm resistor infront of the gate and a 30,000 ohm resistor behind it. I'm using a 9.6v 1700 mah battery.

One other thing comes to mind:

Are you sure you read the colour codes right? Screw up by a few decades and you could be running your FETs in their active region instead of the saturated region.

[Kos-Mos]

hmmm

I don't know what FET you are getting that mesure from, but most recent FETs have a very tiny internal resistance.

The ones I use is 0.033 ohm. This is not a 0.7v drop!!

[MadMax]

The rated resistance is a separate property from the forward bias drop. There are two sources of voltage drop in most transistor devices (Field Effect Transistors included). One is the usual ohmic loss due to path resistance. The other is the 0.7v drop inherent in most semiconductor materials.

Besides 0.033ohm * 10A = 0.33v which means that your FET would be burning:

(0.7v + 0.33v) * 10A = 10.3W

1.03v/9.6v = 10.7% loss in your FET for a 9.6v batt running 10A

[mcguyver]

All FETs have a 0.7v drop. That is the price for a semiconductor (ie. diode). This is the reason we don't have 100% efficiency in anything electronic that requires switching, regardless of the switching speed, pulse-width modulation, etc. The 0.7v drop is independant of current flow (I squared R losses) and the internal DC resistance is not a major factor generally in any internal losses of the FET.

That is why people are still on the quest for room-temperature superconductors.

[MadMax]

I realize that the 0.7v losses are not escapable. What I'm wondering is what is the break even current where a mechanical switch has the same voltage drop as a FET cct.

The switching losses related to non saturated operation aren't important for our application as we don't do very much pulse work, but the 0.7v drop may be considerable for many typical AEG applications compared to mechanical switching losses.

I think Kos is quoting high resistance values. 0.33v drop at 10A current is pretty crappy.