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Topic: [SOLVED} MOSFET continuously conducts (Read 8179 times) previous topic - next topic

KeithRB

And while the gate protection will work against ESD transients, it will not protect against low impedance 70 V AC.

dwightthinker

#31
Sep 29, 2016, 04:36 pm Last Edit: Sep 29, 2016, 04:38 pm by dwightthinker
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

Jiggy-Ninja

#32
Sep 29, 2016, 06:06 pm Last Edit: Sep 29, 2016, 06:06 pm by Jiggy-Ninja
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.
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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.
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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.
Hackaday: https://hackaday.io/MarkRD
Advanced C++ Techniques: https://forum.arduino.cc/index.php?topic=493075.0

Jiggy-Ninja

Quote
. 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.

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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?

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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:
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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.
Hackaday: https://hackaday.io/MarkRD
Advanced C++ Techniques: https://forum.arduino.cc/index.php?topic=493075.0

Jiggy-Ninja

Whether the LM324 output is HIGH or LOW
Neither of those amps are configured as a comparator.

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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.

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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?

Quote
The output of the 324 (running off 12V ) cannot damage the mosfet
Where did I say that it could?
Hackaday: https://hackaday.io/MarkRD
Advanced C++ Techniques: https://forum.arduino.cc/index.php?topic=493075.0

Jiggy-Ninja

If it is known then tell us the maximum possible current that the mosfet could be subjected to
I have, in 3 different posts.
Hackaday: https://hackaday.io/MarkRD
Advanced C++ Techniques: https://forum.arduino.cc/index.php?topic=493075.0

TomGeorge

#36
Sep 30, 2016, 12:29 pm Last Edit: Sep 30, 2016, 12:30 pm by TomGeorge
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.... :)
Everything runs on smoke, let the smoke out, it stops running....

Jiggy-Ninja

Quote
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.

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The 2 ohm resistor looks like a current sense resistor
Because it is.

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so the mosfets could not have been damaged by excessive current
I also consider that to be the least likely cause of failure.

Quote
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.

Quote
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?
Hackaday: https://hackaday.io/MarkRD
Advanced C++ Techniques: https://forum.arduino.cc/index.php?topic=493075.0

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

Jiggy-Ninja

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.
Hackaday: https://hackaday.io/MarkRD
Advanced C++ Techniques: https://forum.arduino.cc/index.php?topic=493075.0

Jiggy-Ninja

that's only because he chose a crappy mosfet with a high RDSon
RDSON has absolutely nothing to do with this circuit. It is an irrelevant metric.
Hackaday: https://hackaday.io/MarkRD
Advanced C++ Techniques: https://forum.arduino.cc/index.php?topic=493075.0

allanhurst

Would your application be happy with PWM ?- saves a lot of dissipation....

regards

Allan

TomGeorge

#42
Oct 01, 2016, 01:05 am Last Edit: Oct 01, 2016, 01:08 am by TomGeorge
Hi,

Quote
Quote

Quote
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?
Hmm..  if the 2R is a current sense resistor, then how is it used without the gnd of the circuit connected to the gnd of the arduino?
Where do you connect the load that you are putting the controlled current through?

Post # 28, can you answer my questions, please.

Please an new circuit diagram showing us all the asked for items.

Tom.. :)
Everything runs on smoke, let the smoke out, it stops running....

Jiggy-Ninja

I have not kept anything meaningful to myself. I have stated repeatedly that the MOSFET is not being used as a switch, the feedback to the op amp driver keeps it in the linear region. The datasheet graph you cite is IRRELEVANT because the MOSFET is not being saturated. Almost everything that I have read that you posted since our feud began has been wrong because of this very fundamental misunderstanding of the circuit.

I have been very consistent in stating the reason why the MOSFET is at risk of dissipating too much power, and I will repeat it here for emphasis:  The MOSFET is not being used as a switch, the feedback to the op amp driver keeps it in the linear region.

The first place I saw this circuit topology was in Dave Jone's EEVblog #102 - DIY Constant Current Dummy Load for Power Supply and Battery Testing video, and there's probably about a million other Youtube videos and blog posts doing variations of the exact same circuit. He brings out a white board with this circuit on it less than two minutes into the video, and after shortly discussing an alternate circuit he analyzes this circuit. OP's is a trivial variation on Dave's circuit that will have nearly identical performance.

I mentioned in reply #30 that EEVBlog made a video about it, and the key words "constant current load" and "dummy current load" were mentioned in other replies of mine. If you aren't familiar with what kind of circuit those words refer to, either say so or do a web search.

The math that I did in response to Dwight's post was merely verifying a particular fact that I found mildly surprising enough to remark about. I did not think it would be important to someone that can't get "the MOSFET is not being used as a switch" into their head, so I didn't feel like wasting time typing it up. If you feel it will be helpful, I have attached my handwritten calculation for your study.

3rd time's the charm: The MOSFET is not being used as a switch, the feedback to the op amp driver keeps it in the linear region.
Hackaday: https://hackaday.io/MarkRD
Advanced C++ Techniques: https://forum.arduino.cc/index.php?topic=493075.0

dwightthinker

#44
Oct 01, 2016, 01:27 am Last Edit: Oct 01, 2016, 01:30 am by dwightthinker
I'll admit I did not look at the specification for the
transistor before making my statement. My bad.
The maximum power would still be at 2.5V, on the
2 ohm load, for the source of 5V.
It would never be able to reach 5V on the load because
the on resistance of the FET is only slightly less
than 2 ohms Looking at Fig2 of TC=175c.
The maximum power is still 3.125W in the FET.
He would never be able to get the full 5v across the
2 ohm resistor because the FET wouldn't go below
about 1 ohm ( 3A at 3V with 5.5V gate ). It would
hit a constant current someplace before the load
went over about 3v.
With 10 volts the compliance of the transistor is no
longer an issue. The FET would only be below
its compliance driving 2.5A.
Now, the power is 5v time 2.5a or 12.5W.
Still, these are typical and the FET may move
from constant resistance to constant current at a
slightly lower drain source voltage or require a
slightly higher gate voltage.
This is all with a gate drive limited to 5.5v.
This is right at the limit of what the LM324 can output
with 12V supply. ( 5.5V gate to source and 5V on the 2 ohm
resistor ).
So, yes, after looking at the specification of the transistor
I would say that my calculations are correct. If the
transistor wasn't compliance limited first,
with 5v source 3.125W or less.
With a 10v source 12.5W.

https://en.wikipedia.org/wiki/Maximum_power_transfer_theorem

In electrical engineering, the maximum power transfer theorem states that, to obtain maximum external power from a source with a finite internal resistance, the resistance of the load must equal the resistance of the source as viewed from its output terminals. Moritz von Jacobi published the maximum power (transfer) theorem around 1840; it is also referred to as "Jacobi's law".[1]

There, I did my homework.
Dwight


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