“Tip: An easy way to put a circuit breaker in series with an Innovation FIRST speed controllers is to install the circuit breaker in a junction block and draw +12 Vdc from the protected output tab. This method should be fine for most teams and is recommended by FIRST for simplicity and reliability. However, robots pushing the performance barrier can obtain a slightly higher continuous output current from the speed controller by placing the circuit breaker on the output side in series with the motor.”
I happened to have a 1999 competetion manual around, and I was reading it. When I read this statement, I began to wonder about it. First of all, is this still legal? Second, is the gain worth the trouble?
With the new 40A breakers I doubt it would be very useful for the drive system unless you have horrible gearing. I can’t say for sure about the actual legality of it (though I doubt it is) but I don’t think an inspector would let that pass during the competition.
[quote=Venkatesh]“Tip: An easy way to put a circuit breaker in series with an Innovation FIRST speed controllers is to install the circuit breaker in a junction block and draw +12 Vdc from the protected output tab. This method should be fine for most teams and is recommended by FIRST for simplicity and reliability. However, robots pushing the performance barrier can obtain a slightly higher continuous output current from the speed controller by placing the circuit breaker on the output side in series with the motor.”
I don’t believe this will have the desired results. At anything less than max output from the speed controller (i.e. duty cycle less than 100%), the current in the motor leads will be larger (easily 2X) than the current drawn from the supply. A breaker in the motor leads will trip sooner than a breaker in the power supply lead to the speed controller.[/quote]
the current in the motor leads will be twice the current in the supply leads?!
where do you think the motor leads are getting that current from?
all the victor does is pulse width modulate the supply, when you are going slow it pulses the output on for a short period, then switches it back off for a longer period - full on, full off.
the victor does not have anyway to store up energy, so the current on the output side is = the current on the input side, minus the slight amount the victor uses (or wastes due to it not being 100% effecient)
Thats what I think they were getting at in '99, you might get an extra half amp out of the breaker if its on the output side
but there was a tiny little problem with doing that. When the victors fail they often short internally, +12V right to ground - so if you have a chip of metal fall into it, and the breaker is on the output side, the only thing protecting the victor is the main breaker - which can source enough current to turn the victor into a pyrotechnic display, including the 10 gauge wire that is feeding it.
Which is why its no longer legal - many victors went up in smoke that year.
The fans on the Victors are not there for looks! Those FET get HOT.
I have not measured it but at 100 amps, I would not be surprised if you dropped a half volt across each leg of the H-bridge – that means you drop a full volt over both. Power = IV = 100Amp*1Volt = 100 Watts. This is a significant fraction of the power you are pounding through the motor.
Bottom line: I think that if you are tripping the 40 Amp breaker (or any of the breakers) you will get a noticeably longer time to trip if you put it on the motor side of the Victor. But as it is not legal any more, and, as Ken rightly pointed out, it can be a dangerous situation, don’t do it.