Overcurrent protection when using mosfet + microcontroller

Hello folks,

I am designing a circuit that will supply 12V to a device that the user will connect - the voltage will be switched by a MOSFET which will be controlled by a microcontroller based on some input.

For example, imagine the user uses this device to switch a bulb on/off based on some input to the microcontroller.

I want to offer over-current / short protection. I know there are specialty devices designed for this purpose, but cost, size and component count are factors... I was considering using a low value resistor in series with the power output and measuring the voltage across it with an ADC in the microcontroller, then I could turn off the power if the current exceeds a preset threshold.

I am not sure if this would be fast enough to stop the MOSFET from being destroyed in the case of something like shorting the output terminal to ground?

The power supply would be a 12V car battery, so it's not easy to implement overcurrent protection on a power supply side. I know there are specialty chips for this purpose, but I would like to expand the device to have 10+ channels in the future, so adding 10 over current protection devices could be expensive and take a lot of space, whereas if I can do it in the micro, it's much easier.


If the above idea is flawed, can someone suggest a more appropriate method for monitoring current draw with a microcontroller and turning off the supply in case of over current?

I want to offer over-current / short protection. I know there are specialty devices designed for this purpose, but cost, size and component count are factors... I was considering using a low value resistor in series with the power output and measuring the voltage across it with an ADC in the microcontroller, then I could turn off the power if the current exceeds a preset threshold.

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Hi Corp666 !
You can do it easier ! with a simple Op-Amp ! the idea is using a hiccup method to limit the current ! ( overload or short circuit protection ! ) .
it is enough that when the current is higher than your authorized value , the GS voltage of your mosfet become zero ( if it is N- Mosfet ) . thus it will protect it ! ( an op amp + a resistor as a current sensor + a bjt transistor )
Best Wishes + Good Luck
Goldsmith

I have done a similar work and I accomplished this by overdimensioning (do one say that or is it called something else?) the component so it could actually handle that very high incurrent rush so the MOFSET doesn't get destroyed.. I would say this is by far the easiest way to accomplish this.

Some would even suggest using a thermistor for that purpose that varies its resistance by the current (more current, higher resistance = limits the current) while others would not like it since it would require a resistor of a bigger size (size is a factor for you) that can handle the heat developing when limiting the current. If this would be a solution for you, make sure to calculate the effect that develops over the thermistor and buy a thermistor that can handle that effect (W).

goldsmith said:

You can do it easier ! with a simple Op-Amp ! the idea is using a hiccup method to limit the current

Click to expand...

Hi Goldsmith,
Is it fast enough to protect Mosfet from a dead short
ani

Abraham900 said:

I have done a similar work and I accomplished this by overdimensioning (do one say that or is it called something else?) the component so it could actually handle that very high incurrent rush so the MOFSET doesn't get destroyed.. I would say this is by far the easiest way to accomplish this.

Some would even suggest using a thermistor for that purpose that varies its resistance by the current (more current, higher resistance = limits the current) while others would not like it since it would require a resistor of a bigger size (size is a factor for you) that can handle the heat developing when limiting the current. If this would be a solution for you, make sure to calculate the effect that develops over the thermistor and buy a thermistor that can handle that effect (W).

Click to expand...

Sorry to reply to the thread so late.... been traveling a bit for work.

The problem I see with using large components so that they won't fail is that the power source is a lead acid battery, so the potential current if the output was shorted could be 2000 amps. I could use an old fashioned fuse, but I don't know if that would be fast enough to protect the FET if someone shorted the output terminals. Also, the user can change to a larger fuse, and they are not auto-resetting.

I've seen devices where they monitor the current and automatically shut off the output, then automatically re-activate as soon as the overcurrent condition goes away - looking to do the same thing.

Thermistor is an option, but they get hot and offer a variable resistance, whereas I'm really looking for a hard cut-off at a specific current level. There must be an easy way since lots of devices do this (the latest generation Harley Davidson motorcycles do this - the 'fuses' are now sold state where the circuit just shuts off in case of over-current and turns back on as soon as the condition is remedied).

goldsmith said:

Hi Corp666 !
You can do it easier ! with a simple Op-Amp ! the idea is using a hiccup method to limit the current ! ( overload or short circuit protection ! ) .
it is enough that when the current is higher than your authorized value , the GS voltage of your mosfet become zero ( if it is N- Mosfet ) . thus it will protect it ! ( an op amp + a resistor as a current sensor + a bjt transistor )
Best Wishes + Good Luck
Goldsmith

Click to expand...

Thanks for the info.

Can you explain "hiccup method"? You're suggesting an op-amp to measure the voltage across a sense resistor so the output of the op-amp would take the gate high on an N-fet and turn it off? And this is fast enough to protect the FET even if I was to short the output to ground (I need to use a P-FET but I can just a BJT to turn off the gate as you describe).

Hi Goldsmith,
Is it fast enough to protect Mosfet from a dead short
ani

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Hi Ani
Why it shouldn't be very fast ? it depends on your design . if you design your circuit , for operating in fast situation , it will handle this purpose as well .

Regards
Goldsmith

Thanks for the info.

Can you explain "hiccup method"? You're suggesting an op-amp to measure the voltage across a sense resistor so the output of the op-amp would take the gate high on an N-fet and turn it off? And this is fast enough to protect the FET even if I was to short the output to ground (I need to use a P-FET but I can just a BJT to turn off the gate as you describe).

Click to expand...

Hi again
Of course . do you know what the word "hiccup " does mean ?
Just put the meaning of this word a side and then think about what i'm going to say while you have that meaning in your mind .
Well and what i want say ! consider you have a PSU ( Power Supply Unit ) . with out put voltage of 10 volts . and you are taking 10 ampere as it's out put current ! and then consider , when your have putted a short circuit instead of the load , it wants to increase the current as large as infinite ! ( but in practice that's not possible ) . so just consider in such a mode you turn your PSU off current will going to be reduced as low as zero but suppose at this time before current be zero you turn it on again . and you do it as a pulse ! ( on / off / on / off .... ) what will happen ? your current can't exceed from a known value.
it means if you cut the current before exceeding from your desired value and then turn it on before decreasing the current from your desired value , your current will be constant in your desired value . so if you increase your load , this section won't do anything because the current will be lower than your mentioned red boundary . this section will do it's best while you have overload issue !
So it is concept of "Hiccup" current limiting method .
A notice : in practice you're not going to turn your supply on and of with your hands ! you'll use a transistor because it can work as a very fast switch !
Best Wishes + Good Luck
Goldsmith

I think, in many case a simple polyfuse may just work. You need to select right size polyfuse. Search for polyfuse ( may be in digikey).

Hi Goldsmith,
By "hiccup " do you mean a current limited power supply

Hi Goldsmith,
By "hiccup " do you mean a current limited power supply

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Hi Ani
I thought you know about Hiccup current limiter ! i offer you to read my post #7 .
Best Wishes
Goldsmith

goldsmith said:

Hi Ani
I thought you know about Hiccup current limiter ! i offer you to read my post #7 .
Best Wishes
Goldsmith

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Thank you Goldsmith can you kindly post an example circuit so that I can study it in detail.
regards ani

Thank you Goldsmith can you kindly post an example circuit so that I can study it in detail.
regards ani

Click to expand...

Hi Ani
Of course . take a look here please :


As you can see base of a Darlington pair has been drove via a 20 volts PSU . so the out put voltage across the R1 should be around 20 volts . and as you can see we've took a sample from out put current via R2 .
When the voltage across the R2 be around or higher than V4 voltage source . thus 2n2222 will be on and it will decrease current through the base of Darlington pair . and thus as you can see , the current has been limited .

Best Wishes
Goldsmith

Thanks Goldsmith once again for a detailed explanation with circuit.

Thanks for the tip on the off-the-shelf parts. I was aware of a few parts (I think they were NXP?) but I did not have much luck with higher current parts. I need to handle about 10-20 amps in a P-channel and the other parts I saw topped out around 3-5amps.

Thanks for the tip on the off-the-shelf parts. I was aware of a few parts (I think they were NXP?) but I did not have much luck with higher current parts. I need to handle about 10-20 amps in a P-channel and the other parts I saw topped out around 3-5amps.

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Hi Corp666
I think i've showed you the idea ! then you can increase it's ability to be able to obtain higher currents . the idea is something like that but for 20 amperes you can easily replace current sensor with everything else e.g : hall effect sensor and it's driver circuit or perhaps clamp current meters or CT before rectification or perhaps a pretty low value resistor !

Could i understand your problem as well ? or perhaps your problem is something else ?
Best Wishes
Goldsmith