Electronic Circuit Breaker

[Badpuppy]

Originally Posted by feufeu

https://www.amazon.fr/EBTOOLS-Disjon...0867SH3XX?th=1

hope this will work?

12 volts, yes. 6 volts uncertain.

[Badpuppy]

Originally Posted by nkaminar

BadPuppy, But the current with a dead short is going to be so great that the circuit breaker (or fuse) will trip in so short of a time that it does not matter where it is 6 or 12 volts.

Breaker should protect above its rating, not necessarily a dead short.

[DanK]

The power to trip the circuit breaker is the current squared times the resistance of the circuit breaker. It doesn't matter if that current comes from a 6 volt or a 12 volt source. Amps are amps.

[Badpuppy]

Originally Posted by DanK

The power to trip the circuit breaker is the current squared times the resistance of the circuit breaker. It doesn't matter if that current comes from a 6 volt or a 12 volt source. Amps are amps.

True statements, but consider the middle one. At 6 volts, 4x current is required to trip a 12V thermal CB.

6v/R creates 1/4 (36/144) the heat of 12v/R, at 1/2 the amps.

Clear as mud, as they say.

[DanK]

You are right about the heat at 1/2 the amps of a 12v circuit but the circuit breaker will not trip until the amps equal that of the circuit breaker rating. Once the amps are equal in a 6v circuit to the amps in a 12v circuit the heat generated in the breaker will be the same.

[nkaminar]

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On the other had, it takes twice as much current to run a 6 volt device as a 12 volt device. Headlight bulbs are an example. The resistance is half for the 6 volt system device which doubles the current. In any case, the circuit breaker is there to protect against a short which is most cases is quite a lot of amps.

[Bob Johnson]

Please note that when the circuit breaker trips it does not see the full 6V or 12V. Most of the voltage drop is across the rest of the circuit. In the example below the breaker trips when when it sees a voltage of 0.6 volts regardless of the system voltage (6V or 12V). The resistance of the circuit breaker has to be quite low because you do not want to have a voltage drop across it. During normal operation you do not want to have a 1V drop across the breaker leaving only 5V to power the car. In the example below I use a resistance for the breaker of 0.02 ohms. If you are not using the headlights you will have under 5 amps which would have a voltage drop of only 0.1V. Below I state that it the number of watts that triggers the breaker watts = volts * amps. Dan said it a different way Amps * Amps * ohms. (note volts = Amps * ohms).




Here is some information:
If you calculate the resistance needed for 30 amps for a 6 and a 12 volt system you get 0.4 ohms for a 12V and 0.2 for 6V. So a dead short across a circuit breaker would require it have those resistance. But that is not how it works. The circuit breaker has a much lower resistance, typicality 0.02 ohms. Thus it will "see" a voltage of 30 X 0.02 = 0.6 volts when it triggers regardless of the overall voltage of the system. The number of watts (power) is what triggers the breaker. In this case it is 18 watts (30A X 0.6V).



Note that when the breaker triggers there is a difference in the "short" resistance for 6V vs 12V. In a 6V system it take a resistance of (6V - 0.6V) / 30 A = 0.18 Ohms for 6V and (12V - 0.6V) / 30 A = 0.38 Ohms for 12V. That would result from a 180 watt load for a 6V system and a 360 watt load for a 12V system. That is where the difference between a 6V and 12V system is seen.




Bob

[feufeu]

reply to #21
i wasn't speaking about part but link :-)
I installed this part in a chrysler 1936 with stock 6v and worked perfectly.