Urgent: Current limiter heat dissipation problem

[ekankjatwani]

View attachment 66439

Hi,

I am working on a design to limit the current in case of short circuit
by the user in one of my ic's. The circuit is shown in the image
below:


So here is my problem:
I simulate the circuit and it does everything as I want. It limits the
current to 200 mA right now. I know I can find better mosfet's which
can tolerate higher collector but no matter what mosfet i choose ,
none of them will be able to tolerate a short. With 30 V and 200 mA
dropping across the Q7 transistor , i am dropping about 6 W. I do not
have enough space in the device design to provide for cooling. I
cannot use a resettable fuse because of the cooling time, no current
limiting diodes because they are rated for too less of a current than
a usual load would require . My last choice would be a microcontroller
trying to sense a voltage across a sense resistor. Please let me know
if you have a better solution to my problem.

Thank You

 

[alec_t]

There isn't a Q7 in the schematic
When you say 'short' do you mean Rload = 0, or Q10 collector is shorted to ground

It limits the current to 200 mA right now.

Yes. Q10/M7 current will be limited to ~200mA regardless of the Rload value.

no matter what mosfet i choose, none of them will be able to tolerate a short

If the FET is a modern low-Rds(on) logic-level type it is turned fully on by V16 (if that's 10V), so at 200mA it will have ~mV across it and dissipate negligible power. Q10 dissipates nearly all the power.

Edit: What do you want the circuit to do when the 200mA current limit is reached?
(a) hold the current at 200mA (in which case you will have up to 6W of heat to cope with somehow),
(b) switch the current off until a reset button is pressed, or
(c) switch the current off for a preset time then switch it on again automatically?

 

[ekankjatwani]

Thank you for your response. I apologize, i was talking about Q10 , where I mentioned Q7. When I say short , I mean rload =0 . I might not have gotten explained the problem correctly . The picture below shows my circuit without any kind of protection.

View attachment 66443

So if Rload=0 , there is a lot of current flowing through my FET , no matter which FET i choose will cause it to blow up.
So i just want the FET to survive this Rload=0 situation. My design which i posted earlier helps me limit the current to 200 mA but as you said I drop about 6W across Q10. This would mean my Q10 would heat up and my problem is the the way the device is design , there is absolutely no space for heat sinks. I was wondering if you know of a better design to do this.

If you have any more questions please let me know thank you !

 

[ekankjatwani]

oh and I want to hold the current at 200 mA. Thanks

 

[alec_t]

Then you're stuck with up to 6W to lose somehow . A heatsink is absolutely necessary for the FET/resistor/junction-transistor/etc which is dissipating the 6W or you'll let the magic smoke out.

 

[ChrisP58]

200mA at 30 Volts is 6 watts. No way around that.

But, do you really need to deliver 200mA, even when the output is shorted? Or only when driving the nominal load? If not, you might try using a foldback current limit scheme.

View attachment 66444

 

[ekankjatwani]

I was reconsidering my design and I might be able to get by with only delivering 200 mA when I have a nominal load and may be show some kind of fault in case of a short circuit. I was thinking may be using an led, will you guys have a suggestion for me in that case. If I have to break the circuit temporarily till the short exists and then have it go back to its regular operation when a nominal load returns . One way of doing this would be using a microcontroller , sensing a voltage across a resistor and once it is over what I set it to be , i turn of the mosfet and check after every few seconds to see if the voltage across Rsense has dropped. But if you guys have a solution without using a microcontroller , let me know.

Thank you for replying to my query.

 

[alec_t]

But if you guys have a solution without using a microcontroller , let me know.

See post #6.

 

[Roff]

What is the range of your nominal load?
How much dissipation can Q10 tolerate?

 

[ekankjatwani]

I don't want to dissipate more than 1/2 watt. Also, the range of nominal load varies for different supply voltages. This circuit is rated for 10-30 V and I max=200 mA, so for 10 V supply , the nominal load should be not less than 50 ohm and for 30 V supply, should not be less than 150 ohm .

Lastly, I don't understand foldback current limiting technique after reading the section on it in the Art of Electronics book .Could you tell me what part of the circuit is the regulator and what is the current limiting part ?

Also , does anyone have a simple circuit that will help me understand the foldback current limiting a little better?

Thanks a lot guys

 

[alec_t]

I don't understand foldback current limiting technique

Perhaps this will help
https://www.google.co.uk/url?sa=t&r...3YDwDA&usg=AFQjCNFnzsjeo6dRg4VKi9Yi45zX2WxwsA

 

[dr pepper]

If you could do a total redesign you might be able to accomplish what you want using a switch mode supply, heat dissipation can be managed.

 

[Roff]

Perhaps this will help
https://www.google.co.uk/url?sa=t&r...3YDwDA&usg=AFQjCNFnzsjeo6dRg4VKi9Yi45zX2WxwsA

With all due respect, I think you are oversimplifying the power limiter. IMHO, he needs the voltage analog of a current limiter. If the output voltage tries to exceed a set limit, it will get clamped at that level. Either that, or, above a set voltage, the current is reduced as the voltage rises. I have fiddled with this for a while (simulations only). I designed circuits that worked, but they latched up, and had to be reset or powered down and back up.

 

[alec_t]

The system you suggest is probably best, Roff, but to give the OP an idea of what is meant by foldback here's an example which, when 200mA is exceeded, drops the current down to ~40mA until either the load resistance is increased or the supply voltage is reduced:

 

[Roff]

The system you suggest is probably best, Roff, but to give the OP an idea of what is meant by foldback here's an example which, when 200mA is exceeded, drops the current down to ~40mA until either the load resistance is increased or the supply voltage is reduced:

Alec, you might be interested in my voltage-controlled resistor. It is good for circuits like this where a resistor needs to be time-variant. If you already have one, ignore the rest of this post.

VCR.asy is an hierarchical block. The schematic is VCR.asc. I have VCR.asc in my top level LTspiceIV directory, and VCR.asy is in \lib\sym\misc.
You can right-click on the symbol and it should allow you the opportunity to open the schematic. There you can see the formula. It's very simple. R = (V+)-(V-).
I'm also attaching your circuit, modified to use the VCR.
If you already have one, forgive me.

 

[ekankjatwani]

I want to thank you guys for taking out the time to help me out with this. I think foldback technique makes sense to me or atleast I can follow what is happening in the circuit you gave me. So just to make sure I understand it clearly , the reason ultimately the current drops to 43mA, is because the Vgs in short circuit condition ends up being such that it wont allow more current to flow through right?

I guess what I am trying to get at is if I pick a Mosfet with higher threshold , can reduce the short circuit current to about 20 mA ? Conversely that can help me lower my power dissipation to less than a watt.

Thanks guys ! this is my first project and I feel i am getting to learn a lot .

 

[alec_t]

the Vgs in short circuit condition ends up being such that it wont allow more current to flow through right?

Yup. You can play with R2, R4, R5 to adjust the foldback current. U1b pulls down Vref and the current is then such that Vsense = Vref. The FET turn-on voltage threshold does affect the foldback current too.

Thanks for the VCR files, Roff. I'll have a play.

 

[4pyros]

Seems either way you need a big current limiting or sensing resistor.

 

[alec_t]

Nah. If R= 2 and I=0.2A then watts = 0.08W. R will hardly get warm

 

[4pyros]

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