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[brokinspokes78]

Hello folks,
So I have question about solenoid Contactors. If the contactor indicates the make or break amps is 400, does this mean the contactor requires at least 400 amps to operate. Not sure what contactor is best suited for an Alltrax 650 amp controller, 58V battery. I plan to turn the alltrax down to about 450-500 amps max. Should I purchase a 400 or 600 amp contactor? Also is the three amp coil suppression diode sufficient for this set up? Or should I make my own 6 amp diode? Or could I use two 3 amp diodes in series or parallel? Thank you in advance, build on....

[Doug S]

I think you're conflating the two sides of the contactor. On the drive (coil) side, there's a "kickback" diode which handles the inductive spike caused when the drive to the coil ceases. Without that diode, the energy represented in the magnetic field inside of the coil would have no place to go, so it would cause a voltage spike which can be damaging to components. The kickback diode gives that energy somewhere to go, allowing it to ramp down fairly slowly without a big voltage spike. Depending on how the coil side of the contactor is wound, it has a certain current requirement to pull the contactor shut, say 12V @ 100mA, which will be provided by the driving circuitry. It's not related in any way to the contact current. The kickback diode only needs to carry this amount of current, regardless of the contact side current. If a 3A kickback diode is specified for that contactor, it will be adequate for any contact side current.

The "make/break" current is the rating for the contact side of the contactor. You won't actually want to make or break the contactor under 400A conditions (you'll always want to both make and break the contactor at exactly 0 amps), but in an emergency, it's rated to do so without the contacts welding together -- and an emergency is what the contactor is there for. No, the contactor doesn't need 400 amps, that's the maximum it's rated for. It will work fine at any current level lower than it's rated for, but make sure the electronics package is designed such that the system can never exceed that rating, even if (say) a transistor shorts out. It's exactly that sort of failure the contactor needs to work properly in.

[ZeroPointZero]

I think you're conflating the two sides of the contactor. On the drive (coil) side, there's a "kickback" diode which handles the inductive spike caused when the drive to the coil ceases. Without that diode, the energy represented in the magnetic field inside of the coil would have no place to go, so it would cause a voltage spike which can be damaging to components. The kickback diode gives that energy somewhere to go, allowing it to ramp down fairly slowly without a big voltage spike. Depending on how the coil side of the contactor is wound, it has a certain current requirement to pull the contactor shut, say 12V @ 100mA, which will be provided by the driving circuitry. It's not related in any way to the contact current. The kickback diode only needs to carry this amount of current, regardless of the contact side current. If a 3A kickback diode is specified for that contactor, it will be adequate for any contact side current.

The "make/break" current is the rating for the contact side of the contactor. You won't actually want to make or break the contactor under 400A conditions (you'll always want to both make and break the contactor at exactly 0 amps), but in an emergency, it's rated to do so without the contacts welding together -- and an emergency is what the contactor is there for. No, the contactor doesn't need 400 amps, that's the maximum it's rated for. It will work fine at any current level lower than it's rated for, but make sure the electronics package is designed such that the system can never exceed that rating, even if (say) a transistor shorts out. It's exactly that sort of failure the contactor needs to work properly in.

Care to dumb this down for us non-EE's ?  lol   

[brokinspokes78]

 Doug S. Thank you for the response and taking the time out of your day to explain this to me. This will help me from blowing stuff up or welding contact points together, thank you. Knowledge is power.
Cheers

[Doug S]

Care to dumb this down for us non-EE's ?  lol   

Best I can do: A relay (a big one is often called a contactor, but it's the same thing) is a device that allows a fairly small voltage/current signal to switch a much larger voltage/current.  It has two sides: the coil (control), and the load (contact) side. In automotive use, you often see the coil set up for 12V, drawing maybe 100mA (the current can vary widely though). That small current pulls the contacts together, which switches on the high-current side.

How's that? Hopefully you can now pick through my previous post.

[brokinspokes78]

 Update: I ended up going with the SOL-600 Main Contactor 600 Amp. Water resistant
Double-make or double-break contact
Rated load—making & breaking 600 Amps
Voltage:  24V, 36V, or 48V coil
Current:  600 amps continuous
Inrush Current 1000 plus Amps
Contacts:  Silver oxide
Terminals:  1/2" main lug, #10 coil
Mounting:  2-up to 1/4

470 Ohm 10 Watt Pre-charge Resistor
3/8" Ring Terminals
Used on 48 Volt Applications

The coil suppression diode calls for 1 amp when using 200 amp contactors, and 3 amp when using 400 amp contactor. I could Not find any specifics on what size diode for the 650 amp controller.  Alltrax Doc100-068 Spec DCX AXE Only shows Diode sizes for 7-200 amp solenoid contactors and 400 amp solenoid contactor. So calculating that there is 1 amp suppression diode to every 200 amp solenoid contactor, and 3 amp diode to 400 amp Solenoid. This should mean I should use a 5 amp suppression diode on a 600 amp solenoid. is this correct?
 This will be providing power from my 58V battery pack, into an Alltrax 48650(650 Amp) controller, in turn the Agni 95R.

[Doug S]

There's not necessarily any relationship between the coil and contact rating on a contactor, except that bigger (higher current rated) contacts might be heavier and require higher coil current to get the armature to move.

You didn't post specs on the coil voltage and current rating, so I'm assuming the manufacturer doesn't supply them either. You can always just measure it yourself. When you get the contactor, measure the coil resistance with a DC ohmmeter. Divide the coil voltage by that resistance. That will be the amount of DC current the coil will draw when the contactor is activated. If you wish, you can multiply that number by 2, or even 5 if it makes you more comfortable...but keep in mind, the diode will only conduct for a very short period of time when the contactor is deactivated, so you really don't need to derate the diode much if at all. That should give you a warm fuzzy about the diode rating.

In any event, diodes are cheap, and there's no downside in using a higher-current diode than you really need. Maybe just put a 10A diode in there and be sure.

[brokinspokes78]

 Doug S, Thank you again, I'm learning as I go. Your help is much appreciated.

[BrianTRice@gmail.com]

I've tried to digest Doug's explanation into the wiki:
http://zeromanual.com/index.php/Unofficial_Service_Manual#Contactor

[Doug S]

This also seems like a good opportunity to explain the mysterious "precharge" circuitry. First let's discuss why precharge is necessary.

The attached circuit fragment shows the contactor in the yellow box, which (as we've talked about) consists of a control or "coil" side and an electrically isolated load or "contact" side. The dashed lines in the middle represent the iron core of the contactor, which is the electromagnet that pulls the armature of the contactor when it's energized by closing SW1 (more often than not SW1 is a transistor, not a mechanical switch).  As we've also already discussed, this magnetic field contains energy, which must be safely dissipated when the switch opens, or a large voltage spike can result which can destroy things (usually the switch itself). That's the function of the "flyback" or "kickback" diode shown.

But there are transient issues on the load side, as well. In particular, the capacitor shown is typically a very large value "bypass" capacitor which is needed as a reservoir for the battery current. It smooths out and quiets down the voltage transients on the high-power circuitry, which can be very severe when you're driving 500+ amps through an electric motor. Capacitors on your typical printed circuit board range in the sub-microfarad range to maybe a few microfarads, but in applications like this, bypass capacitors are almost always in the millifarad range or may even approach a farad. That's a TON of capacitance, and keep in mind it's designed with high-current wiring designed to handle hundreds of amps. The problem with that is when the cap is discharged, close to 0 volts, and the contactor is closed, applying full battery voltage (~110 volts on my Zero SR) essentially instantaneously. That's a voltage difference between the battery and discharged cap of >100V, which results in a giant, fat blue spark inside the contactor just before the contacts close. That spark can easily represent enough energy to momentarily melt the contact material (copper), and as the contacts slap together, they cool rapidly and actually weld to each other. Needless to say, you do NOT want your contactor contacts welded together.

[Doug S]

To prevent that from happening, we add a power resistor and external switch around the load contacts, to "precharge" the bypass caps. See attached circuit fragment.

The resistor value varies, but a few hundred ohms is typical, and it does dissipate some heat (though only for a short time), so it's a bigger resistor than you'll find in most applications. Before the contactor is activated, SW2 is closed for several seconds, which allows the bypass caps to charge up in a controlled fashion through the resistor, which is called "precharging". Then, when the main contactor is energized, there's no voltage difference between the contacts, so no spark happens. SW2 doesn't have the sparking problem because the resistor limits the current through it to a very small value, even when the caps are at a very different voltage than the battery.

[giacomo]

Very nice and detailed explanation.

I think it is worth to clarify that capacitors are "funny" circuit element. When you apply a voltage change to them, like when you close SW1 and the capacitor was at 0V,
they behave has a very very low resistors, let's say 0.1 Ohm, so as soon as the SW1 is closed you ended up with I = V/R = 110V / 0.1 Ohm = 1100 AMPS!!!, that's a lot of current ...
Numbers are indicative only ... But as soon as the capacitor charged up the "effective" resistance grows up exponentially up to almost infinite (not counting the leakage).

So, if you want a quick way to understand the behavior of a circuit with capacitors, just think that at time zero (when the voltage change happens) they behave like a "short" and that after a "long" time they behave as an open.

Hope it helps ...

Giacomo 

[brokinspokes78]

Thanks again Doug S. very detailed indeed. You have helped me choose my new parts with confidence. I feel I've come a long way in just a weeks time. All the answers are out there. Just have to research, and helpful advice goes a long way. Sure do appreciate it. I hope this will help others with their projects as well.