But I need to provide at least a half amp under normal circumstances.

OK, not so pricy for larger currents...
http://www.digikey.com/product-sear...t=0&page=1&quantity=0&ptm=0&fid=0&pageSize=25

 

What trip current should I use? Under normal conditions my controller can use about a half amp.

Will I lose much efficiency?

What resets them? Returning to normal conditions?

 

What trip current should I use? Under normal conditions my controller can use about a half amp.
Will I lose much efficiency?
What resets them? Returning to normal conditions?

Here is my understanding from some past experience:
I am guessing somewhat over 1/2 amp. They are effected by ambient temperature so you might have to take that into account.
Yes, the fuse has to drop enough voltage to heat up. This means that even at currents lower than the trip current it will be dropping some voltage.
The fuse has to cool down to reset. I think it will stay "tripped" until you either remove the load or power off the circuit long enough for the fuse to cool down.

Remember that the FET has to survive long enough for the fuse to trip. The power supply must supply the trip current for the fuse to work properly.

Since the FET should be hard on when the fault is applied then power dissipation should not be a problem. If the fault is a reversed power supply polarity, the FET's drain-source diode will cause more dissipation than then the FET in normal polarity and this has to be taken into account .

Be sure to check the data sheets for fuses that you think might work. Be warned that fuses are a _lot_ more complicated than you would ever expect.

Note: I once went round and round with UL over a proper fuse selection. My error, not theirs. If I had just read what they were telling me then the problem would have been fixed in just one cycle.

 

As with the constant current limiter circuit, it comes down to what your load circuit can tolerate. The CCL "costs" about 0.7-0.8 V of lost power supply headroom when limiting kicks in. A PTC with a cold resistance of 1.5 ohms will drop about 0.75 V at 0.5 A. Note that a PTC must dissipate power in order to function. The response time for an overload is slower than a fuse, but there is that self-resetting thing. And a PTC does not provide constant protection. With each trip cycle it's response time increases and the precision of its trip point changes. Nothing gives the response time, precision, or adjustability of a constant current or current foldback circuit.

ak

 

As with the constant current limiter circuit, it comes down to what your load circuit can tolerate. The CCL "costs" about 0.7-0.8 V of lost power supply headroom when limiting kicks in. A PTC with a cold resistance of 1.5 ohms will drop about 0.75 V at 0.5 A. Note that a PTC must dissipate power in order to function. The response time for an overload is slower than a fuse, but there is that self-resetting thing. And a PTC does not provide constant protection. With each trip cycle it's response time increases and the precision of its trip point changes. Nothing gives the response time, precision, or adjustability of a constant current or current foldback circuit.

ak

"0.7-0.8 V of lost power supply headroom when limiting kicks in."

Well if limiting kicks in then then that means it is shorted and my lights are out anyway. It probably means some idiot made a mistake on installing the light. So I really don't care if the overhead occurs when tripped.

But the speed concerns me. Will the fuse even be worth it? In other words if I have a short will I zap everything anyway?

 

But the speed concerns me. Will the fuse even be worth it? In other words if I have a short will I zap everything anyway?

Oops. I forgot about what happens to the FET before the fuse trips! Even with the fuse you will need to limit the current. What the fuse gives you, then, is some protection for the FET overheating while it in in its linear range because of the current limiting.

 

To avoid excess dissipation in the FET current limiter you can make it into a fast electronic fuse that would shut the current off until it is reset by turning off either the input signal or the power.

Below is the simulation of such a circuit.
It adds a PNP transistor to act as an SCR type latch and shut off the MOSFET when the current exceeds the threshold current.
Initially, with the MOSFET on, and the V+ power slowly ramps up, it trips and turns off the MOSFET at about .65A as expected for R_Sense = 1Ω.
Then after the input control signal goes to 0V to reset the circuit, it trips as soon as the control signal goes high and the current exceeds .65A.

 

Check out NCV8402 series low side driver from ON Semiconductors they have built in temperature and current limit protection.

Allen

 

You can use a comparator and a very small sense resistor to trigger the FET, this would remove the heating caused during the linear region as it would either be fully on or fully off. This would also let you set the exact trip current that you wanted. As such if you wanted to reuse the circuit but with a different current rating, it would be as simple as choosing different values for your comparator to use.

As for changing the circuit to trigger from an inverted signal, adding a super low power, small, and cheap FET would allow you to control your much bigger power FET.

 

"0.7-0.8 V of lost power supply headroom when limiting kicks in."

Well if limiting kicks in then then that means it is shorted and my lights are out anyway. It probably means some idiot made a mistake on installing the light. So I really don't care if the overhead occurs when tripped.

But the speed concerns me. Will the fuse even be worth it? In other words if I have a short will I zap everything anyway?

If I understand correctly your concern is 12 volts on the drain of the FET.
If so the first simulation Carl posted with just a pull up on the transistor fed back to the micro would do it.
Lets say you used a 0.5 ohm resistor in the source as the sense resistor it would trip at about 1.3 amps. You would loose about .25 volts under normal operation.
The resistor would need to be about 1/2 watt, and the micro would need to respond fairly fast or it would become the fuse.
The .5 ohms would limit the current in the FET to 24 amps which it looks like is okay.

 

...................
The .5 ohms would limit the current in the FET to 24 amps which it looks like is okay.

Don't understand that calculation.
Are you talking about some other circuit than the one I simulated?

 

Don't understand that calculation.
Are you talking about some other circuit than the one I simulated?

Your right. That will teach me not to be lazy.
The current will be lower because the FET can't turn all the way on under the short condition.
But it looks like it is still okay.

 

A fool is quick to open his mouth. Well, maybe I'm being foolish, but what about putting a voltage matching light bulb in series with the circuit? IF you short power then the light bulb will simply light. NO? Correct me if I'm wrong - and I don't mind being corrected. I learn from my mistakes.

If the circuit doesn't draw much power then the light bulb should merely act like a series resistor. Only when shorted will the light bulb light up. At least if I remember basic electronics from High School - OH SO MANY YEARS AGO.

 

A fool is quick to open his mouth. Well, maybe I'm being foolish, but what about putting a voltage matching light bulb in series with the circuit? IF you short power then the light bulb will simply light. NO? Correct me if I'm wrong - and I don't mind being corrected. I learn from my mistakes.

If the circuit doesn't draw much power then the light bulb should merely act like a series resistor. Only when shorted will the light bulb light up. At least if I remember basic electronics from High School - OH SO MANY YEARS AGO.

That would sort of defeat the high efficiency requirement.

 

A fool is quick to open his mouth. Well, maybe I'm being foolish, but what about putting a voltage matching light bulb in series with the circuit? IF you short power then the light bulb will simply light. NO? Correct me if I'm wrong - and I don't mind being corrected. I learn from my mistakes.

If the circuit doesn't draw much power then the light bulb should merely act like a series resistor. Only when shorted will the light bulb light up. At least if I remember basic electronics from High School - OH SO MANY YEARS AGO.

That may work okay if you can tolerate the cold resistance of the bulb in series with the load (which is about 1/10th or less of the hot resistance).

A further thought on that is that is, a properly sized bulb could be used as the sense resistor in the circuits I posted.
That would minimize the loss under normal current operation since the operating resistance would be less than the trip resistance.
The trick is determining the particular bulb needed for the desired current limit.
A low voltage bulb would give the highest resistance ratio between operating and trip points.