Partial Jag failure?

[techhelpbb]

Originally Posted by Mike Betts

JW,

If you look at the spec sheet for the MX5-A40, you can pull 133A through the circuit breaker for about 3 seconds before it trips. This can be a long time for a resistor to gain heat depending on the thermal resistance of the resistor.

Add to this that the MX5-A40 is an auto-resetting CB meaning that the heating can be cumulative as it repeatedly trips and resets.

Likewise, the 120A can pull much more than 266A for quite a long time...

Regards,

Mike

Edit: I had to back a couple of years to get a spec sheet on the 120A CB but you can see that it can pull 266A for over 20 seconds...

This was why I considered putting a fast blow 120 Amp fuse in series with my design for measuring the current (just like most digital multimeters). I fully expected someone would eventually make the common mistake of putting the circuit itself across the battery (either with no other load to monitor or effectively shorting out the load) and burn out the resistor even at 30 Watts...even if a fan was blowing on it and it was on a heat sink.

So that every one can see why:

E = I x R
E = 140 Amps x 0.002 Ohms
E = 0.28 Volts

P = E x I
P = 0.28 Volts x 140 Amps
P = 39.2 Watts

Now...even at 140 Amps...assuming you don't ever turn off the load...that resistor in my circuit (not the Jaguar) has too small a Wattage. Never let us mind that the 4 Watt resistor in the Jaguar would be toast at this point.

So what would happen if we put the resistor itself across the battery and ignore the wire and current limiting device resistance?

E = I x R
I = E / R
I = 12 Volts / 0.002 Ohms
I = 6,000 Amps

P = E x I
P = 12 Volts x 6,000 Amps
P = 72,000 Watts

So right about that point even the 30 Watt resistor would be destroyed...quickly.
Never mind that the battery might do something unpleasant.

Some fast acting fuses like the CNL125 which are targeted at fork lift application:
http://www.littelfuse.com/products/F...L/CNL125..html

Would probably work...but then...it costs a little more than $20.

The resistor I ended up using is only about $4:
http://www.riedon.com/us/images/stor...f/FPR2T218.pdf

In the end...this is why I was concerned about finding what FIRST might call a 'COTS' resistor for this application. Obviously if someone used a great big piece of copper or even aluminum with a known electrical resistance...one could fabricate a resistor that could survive dissipating a pretty serious amount of heat. However, in doing so you create a situation where the end user (any team using it) might have to be able to measure it and it has a small resistance that makes a simple digital multimeter pretty ineffective as an Ohmeter. Of course one could try to use it as shunt to measure it...but then I fear they could hurt themselves if they aren't careful. This is probably why in the past a coil was used to measure the current (so I've been told)...but the downside with that is that small currents become harder to read and fast changes in current get removed.

I'm open to better solutions for this sort of trade off (perhaps a bunch of slightly larger resistors in parallel dividing the Wattage, even 2 of the 0.002 Ohm resistors I've linked above would increase the rated Wattage to basically 60 Watts)...but if someone wants to discuss them...let's not distract further from this topic.