Main breaker tripping, dead CIM

[shagun.g]

Originally Posted by Ether

Perhaps something like this.

I've been trying to read up on this - I don't quite understand this concept. How you can use an op amp to measure the voltage drop (without a resistance)? How does this work?

[Ether]

Originally Posted by shagun.g

I've been trying to read up on this - I don't quite understand this concept. How you can use an op amp to measure the voltage drop (without a resistance)? How does this work?

There is resistance associated with the main breaker. So therefore there is a voltage drop across the main breaker associated with the current flowing through it.

The resistance of the main breaker is quite small, and thus the voltage drop across the breaker is quite small.

The op amp doesn't measure the voltage drop. It just amplifies it to a level suitable to be digitized by the A/D.

Instead of measuring the voltage drop across the main breaker, you could measure the voltage drop across a length of wire.

[Ether]

Originally Posted by Bill_B

Not legal...

What rule are you looking at?

[Bill_B]

Originally Posted by Ether

What rule are you looking at?

Thanks. Eagle-eye Ether does it again. That is great to be able to do. I think we were called out for an extra connecter on the battery-side of the main breaker once and I just carried the incident forward. I'll be checking specifically next year for the same verbiage. It is very good to have your program know how much current is being drawn real-time. You might even consider alteration of drive power on the supposition that 120+ amps means you're stalled. YMMV of course.

[Oblarg]

Originally Posted by Bill_B

You might even consider alteration of drive power on the supposition that 120+ amps means you're stalled. YMMV of course.

FWIW, with 6 CIMS geared 6.1:1 we were drawing 180 amps while slipping the wheels.

[Al Skierkiewicz]

Bill,
1 foot of #10 or 2 feet of #6 will drop 0.1 volt for every 100 amps. These two are equivalent to 1 mohm resistance.

[brycen66]

Originally Posted by Ether

The op amp doesn't measure the voltage drop. It just amplifies it to a level suitable to be digitized by the A/D.

Instead of measuring the voltage drop across the main breaker, you could measure the voltage drop across a length of wire.

Although your method does make mathematical sense, it will run into an error if you use this on the 120 amp breaker. You forgot to account for the fact that even the best op amps cannot read more than 95% of the reference voltage. This means that even if you used a differential resistor setup, which I assumed from your drawing, you could not use the 5 volt supply from the analog breakout. You would need somewhere around 16 volts, which means this would have to have an independent power supply. You could get around this by using a differential op amp setup on a length of wire attached to your main ground wire, but that adds unnecessary resistance.

A completly diffrent solution would be to use a linear hall sensor such as this part (http://www.jameco.com/webapp/wcs/sto...01_1915940_-1).

Place a steel washer or ring(any ferrous metal works) around one of your main wires. Then cut a slot in this steel washer and place the hall sensor in the slot. We epoxied ours in to keep it intact. The hall sensor outputs voltage which is directly related to the total robot current, and it can be plugged directly into the analog breakout. This is the system we have implemented into our practice robot and the results so far look good.

If you have any questions feel free to PM me.

[Ether]

Originally Posted by brycen66

You could get around this by using a differential op amp setup on a length of wire attached to your main ground wire, but that adds unnecessary resistance.

You don't need to add unnecessary resistance. Just measure the voltage drop across the existing ground wire:

Originally Posted by Ether

Instead of measuring the voltage drop across the main breaker, you could measure the voltage drop across a length of wire.

Or use an isolation amplifier. I'm sure a EE could figure it out.

I like your Hall Effect solution too.


[EDIT] By the way, your link is broken: it has a trailing ")"

[Aren Siekmeier]

Originally Posted by Al Skierkiewicz

Roger,
There is nothing that happens to a breaker that trips for over current that doesn't happen during normal operation.

Except that you are running several hundred amps through it and it is quite hot, while there is minimal current and temperature increase when you manually trip the breaker via the red button. Pat's suggestion that a phase change in the material takes place (some sort of plastic deformation) seems likely, considering also the difference in temperature and current through the conductive element of the breaker. Perhaps the increase in temperature and current brings the lattice above a threshold energy that allows it to plasticly deform more easily than when the red button is pressed.

[Al Skierkiewicz]

Aren,
The breaker is a simple bimetallic device. As the current flows through it (and heat is generated), the contact latch cam deforms due to the expansion of two different metals. Theoretically, the breaker could exist forever at just under the trip temperature without damage. Pushing the red button merely forces the latch cam back to a position that prevents contact.

[Aren Siekmeier]

Originally Posted by Al Skierkiewicz

Aren,
The breaker is a simple bimetallic device. As the current flows through it (and heat is generated), the contact latch cam deforms due to the expansion of two different metals. Theoretically, the breaker could exist forever at just under the trip temperature without damage. Pushing the red button merely forces the latch cam back to a position that prevents contact.

I'm aware of the operational principle. Pressing the red button to manually release the latch cam does not increase its temperature, and perhaps does not even force any deformation (elastic or plastic). We all know that this mechanical action, typically performed with the breaker closer to room temperature, does no harm to the breaker internals, since we have successfully used the breaker for years.

But things change under a trip condition when all the components are much hotter, and perhaps plastic deformation thresholds are different. It's also not clear to me whether there is any material deformation at all when the button is pressed, while there is of course significant deformation due to heating in an over current condition. This difference in conditions could conceivably lead to component damage in a trip condition while there is no damage when the button is pressed.

[Al Skierkiewicz]

Remember that the deformation takes place in the latch cam. The latch is actually ties directly to the flag. It turns as the flag is moved in or out. What change may take place is on the contact area of the two parts. When a high current through an inductive load is interrupted, arcing usually occurs. While the contact is designed to survive multiple interruptions without damage, some pitting will occur with each arc. This effect is worse when the space between the contacts is very small. So the design of this device makes the opening of the contact large and quick when either tripped or opened manually.