The OP hasn't posted any test results that are conclusive regarding damage. At this point I am treating it as speculation until the OP posts the resistance comparison of a suspect device compared to a brand new unused device. That would be conclusive if the difference is noticable.
Simply saying it won't turn off with out having any further information is not conclusive. There may be some other explanation. There has been no mention of overheating or smoke or exploding devices so it's too soon to tell.
raschemmel:
The OP hasn't posted any test results that are conclusive regarding damage.
Really?
First post:
Then at some point (not always immediately) my MOSFETs, rather than switching off with a gate voltage of 0, seem to continue to conduct current (just as if its a 50hm resistor).
Seems pretty clear he's noticing a degradation over time compared to initial operation.
Second post:
Also disconnected the gate and connected it to ground - same result.
Has tried directly grounding the gate. Still leaks like an open faucet. Assuming it's wired properly, that seems pretty definitive.
Right now, I'm leaning towards the fault being caused by a good toasting, courtesy of Mr Watt.
I still think it's the 324 op-amp not pulling all the way to GND, look at the OPs drawing and imagine 50 mV at the output of the Vfollower, applied to the + input of the second amp would cause the output to rise, forcing the FET to conduct until the voltage at the top of the 2 R resistor and the - input was at 50 mV, causing 25 mA to flow through the sense resistor.
The 324 datasheet states the Vlo voltage at output may be as high as 20 mV @ 5V VCC, what could it be at 12V?
A quick stab with a DMM probe on the gate would tell the tale.
I've solved the issue
Thanks for the good feedback and thoughts, I've done some more testing and come to the conclusion that the cause of the damaged MOSFETS is indeed the soldering, and that my iron is not earthed. I measured the voltage on the tip relative to ground at 70V AC ! This seems to be the culprit not the circuit.
Using an earthed socket for the iron, have replaced the failed MOSFET and everything is now functioning correctly
I've checking the failed and a new MOSFET in a component checker, which seemed to show the failed one as a Depletion Mode device and a new on as an Enhanced Mode device - which would perhaps explain why the circuit was seeming to work but not turning off - no idea how damaging the device changed it but out of interest personal i'll test the failed device further to see if it really does behave like a D rather than E device and if it an indeed be turned off with a -ve input.
Big thanks again to all those helping me out here
You can edit the post title in your original post to read:
"[SOLVED] MOSFET continuously conducts"
SEE REPLY # 21
It is possible that that the MOSFET is being damaged during
soldering.
You really should be using a grounded iron.
It is not real difficult to add a ground wire.
The FET has a reasonably large input capacitance but
they can be damaged.
Even with a grounded tip, install the MOSFET last.
I've checking the failed and a new MOSFET in a component checker, which seemed to show the failed one as a Depletion Mode device and a new on as an Enhanced Mode device - which would perhaps explain why the circuit was seeming to work but not turning off - no idea how damaging the device changed it but out of interest personal i'll test the failed device further to see if it really does behave like a D rather than E device and if it an indeed be turned off with a -ve input.
Depletion_and_enhancement_modes
For N-type depletion-load devices, the threshold voltage might be about –3 V, so it could be turned off by pulling the gate 3 V negative (the drain, by comparison, is more positive than the source in NMOS). In PMOS, the polarities are reversed.
Hi,
Do you have the gnd of the arduino connected to gnd of the load circuit, that is the gnd end of the 2R resistor?
Without the gnd to the arduino your current sense from 2R resistor is not valid.
What is the maximum current you put through the load and does the MOSFET have a heatsink?
As you are not PWM controlling the MOSFET then it will be dissipating all your load energy.
Do you have a fuse in the load circuit?
Tom... 
I am not surprised that the ungrounded iron did the
damage.
At a previous place I worked, we damaged about $1500s worth
of MOSFETs using ungrounded soldering irons. These were nice
Wellers that the assemblers took home and cut the ground pin
off.
MOSFETs come in two types. Some have an input protection
zener diode and some don't.
We had both types in our design. None of the zener input ones
failed. Almost all of the unprotected ones where damaged.
We fixed the plugs and had no more issues.
The other part that I've found to be sensitive to static is
the parts like 4051, 4052 and 4053.
Just handling these part while ungrounded would damage them.
Also, the more sensitive parts should always be the last parts
to be installed. This keeps the unconnected leads from being
static antennas.
Dwight
I'll be convinced by the resistance measurements. (preferably out of circuit)
Still no load in schematic .
schematic still "incomplete"
replying "liad is constant current" does not satisfy schematic requirements
Read Reply# 14
I didn't say it answered all the questions about the circuit, but it answered your question about why the PWM is being filtered. It's a constant current driver. I know that because Dave Jones (EEVBlog) made a video about a circuit using the exact same topology, but with a pot in place of the LPF to adjust the output current.
Plus, you know, the fact that the picture's named "Constant Current Load.jpg" is a pretty big hint.
I agree that the load is important, but the circuit is not working the way I think you think it is.
post complete schematic that shows load and it's connections
We don't know any mire now than we did at Reply # 14.
How can you think the mosfet is "toasted" in absence of ANY mention of device temperature ?
Read OP, Reply # 6, & Reply # 14.
See ANY temp info ? (answer: NO !)You're drawing conclusions without the most imposrtant info (temp ,resistance & load wiring)
Show me the evidence.
Where did I say that I knew it was toasted? I said I suspected it, and was leaning towards it. I very specifically asked in reply #15 about the heat dissipation setup. I asked for the important info.
The MOSFET isn't being switched between saturation and cutoff like when it's used as a digital switch, but is operating in the linear region due to the op amp feedback circuit. This means that, depending on the load, it can dissipate very significant amounts of power.
How can mosfet "work" if it is damaged ?
I am not stupid, and would appreciate not being treated like I am. You could at least give me the courtesy of thinking about my words before ranting.
Semiconductor devices have multiple failure modes, not just "open". They can also fail short, sometimes all at once, sometimes progressively.
Device is rated for 9.2 A @25 degree
Your point? You might as well have said that it has a 100V VDS maximum. That's not the limit I was talking about.
The maximum junction temperature is 175C, and with 62C/W Junction-to-Ambient resistance it only needs about 3W of power to hit that limit and start damaging the junction if there's no heat sink.
Would YOU choose a 9 A mosfet for a 50 mA load ? (answer: NO !)
For a professional production circuit? I wouldn't expect to ever see that.
For a hobbyist project or a development prototype? Wouldn't surprise me.
Also, 50 mA is not the load, it's the fault current when the device is supposed to be off.
So then we can assume load current is 5 A +/- 4A.
You rant at me earlier in this same post about drawing conclusions without evidence, and then you pull this out?
The MOSFET is part of an adjustable dummy current load. It is the driving circuitry that defines what current will go through it. And if you do that math, with a 5V PWM signal the maximum current that will go through is 2.5A.
If the heat sinking is inadequate, this kind of amperage could very easily cause the MOSFET to burn out.
I mentioned thermal damage as one possibility. The others mentioned (gate damage from an ungrounded iron, inductive voltage spikes from the load) are good possibilities too. We don't know yet.
I've solved the issue
Thanks for the good feedback and thoughts, I've done some more testing and come to the conclusion that the cause of the damaged MOSFETS is indeed the soldering, and that my iron is not earthed. I measured the voltage on the tip relative to ground at 70V AC ! This seems to be the culprit not the circuit.
I don't think that's enough evidence to conclude it's the iron. I assume that when you're soldering, the board itself is not earthed but left floating. If the 70 VAC mains frequency you measured was capable of damaging the MOSEFT in this case, it would damage everything that couldn't handle at least 100V.
An earthed iron is necessary because the fragile MOSFET gates are very easily broken by static electricity. You are not going to be able to measure static electricity with a typical multimeter.
I've checking the failed and a new MOSFET in a component checker, which seemed to show the failed one as a Depletion Mode device and a new on as an Enhanced Mode device
An unsurprising result, given the symptoms.
I think it's too early to say that the problem has been fixed. You said in your OP that the problem didn't always show up immediately. Wait for a time and give it an opportunity to get damaged. If it goes for a long enough time to give you confidence that it isn't getting damaged, perfect. If not, we'll need to revisit the inductive spikes of overheating possibilities.
dwightthinker:
These were nice
Wellers that the assemblers took home and cut the ground pin
off.
Did you ever find out why these people vandalized company property?
These were nice
Wellers that the assemblers took home and cut the ground pin
off.
That's messed up.
That's just wrong on so many levels...
The MOSFET is part of an adjustable dummy current load. It is the driving circuitry that defines what current will go through it. And if you do that math, with a 5V PWM signal the maximum current that will go through is 2.5A.
The load current is a function of the two parameters we don't have, the Load voltage and the load current. It has nothing to do with the 5V drive signal.
In conclusion, the damage was NOT heat related so that rules out "toasting".
If the heat sinking is inadequate, this kind of amperage could very easily cause the MOSFET to burn out.
I mentioned thermal damage as one possibility.
You forget we don't know the load voltage (or current) but heatsinking is not an issue because the OP never reported overheating.
And while the gate protection will work against ESD transients, it will not protect against low impedance 70 V AC.
Jiggy-Ninja:
Did you ever find out why these people vandalized company property?
Yes, they did a lot of extra assemble at home for extra pay.
It was a small company, with varying work loads. The houses
they lived in had two pin sockets.
The main boss thought it was fine. Most of the board we made
were burnin boards. These were mostly hundreds of sockets and
resistors.
The boards with the MOSFETs were driver boards that we only
occasionally made.
About 1/2 of the soldering stations had missing ground pins.
I was surprised that the boss felt it was not an issue but then
he wasn't an engineer either.
As for damaging other parts, in most cases the impedance of
parts like TTL are so low that the capacitive 70VAC from the irons
is no longer 70VAC when touched to he parts. The fets where different.
The specifically have high input resistance in the 100s of megohm and
low input capacitance.
The thermal failure is also quite likely for these parts. One wonders
if this was considered for this design.'
Dwight
KeithRB:
And while the gate protection will work against ESD transients, it will not protect against low impedance 70 V AC.
If there's low impedance 70V on an exposed metal surface of your soldering iron (or ANY tool you own), you have a lot more to worry about than the MOSFETs on your circuit board. Like your own life. I can't imagine anything other than the cheapest of the dirt cheap fly-by-night bottom-of-the-barrel Chinese junk would actually have an electrically hot soldering iron tip.
The load current is a function of the two parameters we don't have, the Load voltage and the load current. It has nothing to do with the 5V drive signal.
In conclusion, the damage was NOT heat related so that rules out "toasting".
It's like you're not even trying to understand how the circuit is supposed to function. That MOSFET is not being used as a switch.
You forget we don't know the load voltage (or current) but heatsinking is not an issue because the OP never reported overheating.
He didn't say it wasn't overheating either. he didn't say anything about the temperature. That's why I brought it up as something to check.
. Your comment that I was inaccurate by rounding it down to 5A +/- 4A is not valid since all I am doing is giving the OP the benefit of the doubt that he specified a part such that the "nominal " load current is < 60% of MAXIMUM continuous current rating of the device.
Your "guess" ignores the entire rest of the circuit around the MOSFET, which has a defined way of operating.
When analyzing why it's failing, yes it is important to know the specs of the actual parts used and the applied load. None of those details are necessary to see that the intention of the circuit is to be an adjustable linear constant current regulator with a 0 - 2.5A range, controlled by PWM from a microcontroller.
What are some of the general failure categories of a linear power circuit?
Overvoltage: Steve specified using 5V and 10V on the output, well within the 100V rating of the transistor. That leaves spikes from an inductive load as a cause, which is why we asked about the nature of the load.
Overcurrent: The control loop, if it is properly functioning, should have a maximum output current of 2.5A, well below the 9A limit. Because of the 2 ohm current sense resistor and 12V supply to the controlling op amp, it is basically impossible for this circuit to put more than 6A though the MOSFET, and it will be far less if you take into account the threshold voltage. This one is the cause you can put at the bottom of the list even without knowing anything about the attached load, simply due to nature of the control circuit. The only real way for the MOSFET to conduct above the rated current in this circuit is if the sense resistor gets shorted (there are other ways, but the only ones I can think of involve total failures in the power supply or op amp).
Overtemp: The MOSFET is being operated as a linear element, so there is basically a 100% chance that is is being subjected to a non-trivial amount of heat dissipation (otherwise what's the point of the feedback control circuit?).
And the MOSFET specific failure mode, gate damage caused by overvoltage, either from static or capacitive coupled mains.
Those are the obvious failure causes. We need the details to figure out which one is the actual cause, but coming up with a short list of possibilities can be done without details.
That being said, it is pointless to discuss excessive current or overheating when no such observations were posted by the OP .
That is quite a statement.
"He didn't say it was hooked up to an inductive load, so it's pointless to discuss overvoltage as a cause."
"He didn't say what equipment he used to assemble the board, so it's pointless to discuss if an ungrounded iron damaged it."
Does that really make sense to you?
You need to clarify this because the load current is still a function of load voltage and impedance , both unknown.
Here is what I was responding to:
The load current is a function of the two parameters we don't have, the Load voltage and the load current. It has nothing to do with the 5V drive signal.
If "5V drive signal" is referring to the part of the schematic labeled "Arduino PWM", the bolded part is completely false. Figure out for yourself why, I have been more than patient enough.
raschemmel:
Whether the LM324 output is HIGH or LOW
Neither of those amps are configured as a comparator.
The maximum is based on the load current and voltage which are unknown.
The maximum is based on the control circuit, which is completely known.
that section of the schematic is irrelevant to this post. (now that it has been confirmed that they are damaged)
I only know of the MOSFET being damaged, who said the amps were?
The output of the 324 (running off 12V ) cannot damage the mosfet
Where did I say that it could?
raschemmel:
If it is known then tell us the maximum possible current that the mosfet could be subjected to
I have, in 3 different posts.
Hi,
Post #6 seems to indicate that the 2R resistor in the circuit is the load.
So there is no current feedback it relies on the volt drop across the 2R to adjust the bias on the MOSFET directly and hence current.
There is no gnd of arduino to gnd of load circuit for the analog input to reference against.
I did ask the OP about the gnd connections but he didn't answer the question
We have no code?
I would like to see how the OP thinks his circuit works?
Tom....
If you are referring to the short-circuit current (Vin/R=5V/2 ohms = 2.5A , or 10V/2 ohms= 5A (depending on whether Vin is 5V or 10V), then why not just say so ?
Because I wasn't.
The 2 ohm resistor looks like a current sense resistor
Because it is.
so the mosfets could not have been damaged by excessive current
I also consider that to be the least likely cause of failure.
There could be some heating if he didn't have a heatsink but he didn't mention one.
There will not be some heating, there will be a lot, and it will require substantial thermal dissipation. It's in the nature of this kind of circuit.
Schematic should label the 2 ohm resistor as a Current Limit/Sense resistor
Why? If you see a voltage divider hanging off a LM317 regulator output and connected to the ADJUST pin, does that need to be labelled "feedback network" for you to understand what it's for?
If he had 10V supplied and the arduino analog out was 5V, the
FET would have 5V drop at 2.5 amps. That is the worst case
condition for power when using 10V in.
It would be 12.5W.
When using 5V source, worst case is 2.5V at 1.25A is 3.125W.
Power in the MOSFET makes a big difference as to what the
supply voltage is.
Dwight
dwightthinker:
If he had 10V supplied and the arduino analog out was 5V, the
FET would have 5V drop at 2.5 amps. That is the worst case
condition for power when using 10V in.
It would be 12.5W.
When using 5V source, worst case is 2.5V at 1.25A is 3.125W.
Power in the MOSFET makes a big difference as to what the
supply voltage is.
Dwight
YES! YOU GET IT!
I confess that I did not think the max power point would be at half-voltage like that. I thought that you were incorrectly generalizing from the Maximum Power Transfer Theorem (which deals only with resistive loads), so I did the math myself. I was surprised to find that it was correct.
More generally, maximum power dissipation in the FET occurs when the DC control voltage (from the LPF PWM signal) is half of VIN.