Need explaination of this sense resistor datasheet please

[jnz]

I have a 30A relay. Let's say there is a 25A load on it. I want to current sense that and when I see any more than 25A cut the control to the relay. Typical current sense circuit where I'll need a resistor inline to the load.

The issue I'm having is that at 12V, a 25A load is 300W. But there doesn't seem to be a reasonable sized sense resistor out there that'll work for that load. This can't be right because I know there are power supplies in the 1500W range that are determining their own current.

For example,
http://www.mouser.com/ds/2/427/wsbs8518-240930.pdf

This sense resistors is 50-1000 uOhm or so, large, but at 36W apparently I would need 9 of them wired parallel to handle the load properly? Something doesn't seem right. There seems to be more than enough copper even with the small resistance than is in the wire going to the circuit and definitely in the contact at the conntector terminals between the source and the load.

How are current sense circuits in the 50W+ range even remotely feasible? And what I am I missing, because I know they are.

[PSR B1257]

Math problems?

P=I²R=(25A)^2*1mOhm=625mW

[jnz]

Ha.... Um.... yea. Let's say I'm not great at math.

For some reason every time I was looking at P=I^2*R, I was getting mixed up somewhere. Ok, so since the resistor is so tiny in terms of it's resistance, the wattage it sucks up will be very small, yea?

I think I was getting mixed up in looking at the total load of the circuit. So in this case, I can use much much smaller resistors and I'm a dummy. Got it!

[fubar.gr]

The 36 Watt figure is for the resistor itself, not the load. Since the resistance is so small you need something like 180+ Amps to make it dissipate 36 watts.

[suicidaleggroll]

For some reason every time I was looking at P=I^2*R, I was getting mixed up somewhere. Ok, so since the resistor is so tiny in terms of it's resistance, the wattage it sucks up will be very small, yea?

The problem was you were using P=I*V, and for the voltage you were using the full 12V dropped across the load.  12V won't be dropped across the resistor through, if it was, your load wouldn't get anything!  At 1mohm and 25A you'll only get 25mV dropped across the resistor, and the remaining 11.975V will go to the load.

[jnz]

Thanks everyone. Feeling dumb over here

[T3sl4co1l]

FWIW, you will need a sensitive amplifier to read the voltage on that resistor.  It may not be the best choice.  Often, a larger resistor can be used, to trade power dissipation and voltage drop for a larger signal, which results in a more accurate measurement.

Tim

[jnz]

FWIW, you will need a sensitive amplifier to read the voltage on that resistor.  It may not be the best choice.  Often, a larger resistor can be used, to trade power dissipation and voltage drop for a larger signal, which results in a more accurate measurement.

Tim

Right. Now I know that I'll want to find a much larger resistance in order to get a good read off the amp I chose. But with a 25A load at the device I should be able to go up in resistance and down in package size now that I'm being less dumb.