2
\$\begingroup\$

My uncle has a 48V electric boom lift that sat for awhile and we're trying to get up and running again. We had a lot of stuff replaced and refurbished and we're almost there, but there's one last hurdle. There are two motors that drive the wheels. We sent these away to be refurbished. The brakes are part of these motors. The brakes are engaged by springs (I'm assuming). When you go to move the lift the brakes are energized and they release and as soon as you stop moving the springs once again stop the wheels. For whatever reason the wiring for the brakes is separate from the motors. There's two terminals with 16 AWG wire on each.

One of the two brakes won't release. I thought the two were separate, but I've discovered they're actually receiving 40v (the schematic says 48V) in series from the B+ wire (so going in). Power goes into the terminal on the right brake, a wire comes out the second terminal, into a loom and end up at one terminal at the second brake before coming out the second "return" wire. I've tried every configuration of hookup on these four terminals and the right brake is always the one that's working (I would have thought if I swapped it the left would be the only working one at some point, but that was not the case). If give the left brake 48V from battery power it works. I can get them both working by wiring them in parallel, but then they see 48V when I believe they should see 24v (if they're wired in series, right?). I don't know if that's going to damage them.

So that's where I am right now and I'm not sure how to proceed. Current flows through both, I'd expect them both to work. Maybe the resistance on the left one is higher? But I thought it'd still work, just maybe not as well. Maybe because it's only seeing 40v to begin with (as opposed to 48v) the voltage at the second one is too low to activate? This is pure speculation and it's probably flat out wrong.

There are two sets of four batteries. I believe each set it wired in parallel and then the two are wired in series for 48V input. My plan for if I can't figure this out would be to use two 48V relays to send 24V from one group of batteries to each brake and use the original 48V wire as the trigger.

Obviously my understanding of electricity is lackluster, but I know how to avoid getting zapped. If anyone can shed some light on what's going on I'd greatly appreciate it.

UPDATE: The schematic is on page 96 (3-20):

https://csapps.jlg.com/OnlineManuals/Manuals/JLG/JLG%20Boom%20Lifts/30_35_40_45E/Service_3120743_10-01-01_ANSI_English.pdf

UPDATE 2: Left brake (the problematic one) measures 18.6 ohms, the right one measures 18.9 ohms.

I also forgot to mention I've tried swapping around the terminals to see if maybe they were reinstalled incorrectly, but it didn't make a difference. I tried AB-CD, AB-DC, BA-CD, BA-DC. On a related note, is there polarity? there's no markings. Obviously the original connectors had a wire in mind for each terminal, but those have been dismantled and it moves the same way no matter how its hooked up.

\$\endgroup\$
1
  • 1
    \$\begingroup\$ When you "swapped" them, what did you actually do? If the problem stays with the same physical brake, then there's a mechanical issue with that brake, not an electrical one. When you overpower it with the full battery voltage, you're overcoming whatever extra drag there might be in the mechanism. Is there any way you can move the mechanism by hand in order to judge the mechanical resistance? \$\endgroup\$ – Dave Tweed Aug 3 '18 at 15:03
1
\$\begingroup\$

The schematic shows B+ 48V power applied to pin 7 and both brakes in series with return to pin 8 which is connected to B- (0V) by either fowd or rev contactor.

This means 24V across each brake coil.

If you are seeing 40V, there may be a problem with the wiring or battery condition. Only 1 diode drop should be expected ~ 1V rise above ground on Brake return when active.

Check contactor resistance or trace the voltage drop from Vbat to Brake voltage.

If all else fails, contact motor service company for details on how they validated the brake performance, as there is a problem.

Each brake can be tested for release threshold with a variable 24V supply and ought to be released < 80% of Vbat min. (SWAG)

\$\endgroup\$
3
  • \$\begingroup\$ I'm sorry if I was unclear. The 40V measurement is at the B+ wire with nothing attached. Which might be a problem in itself, but I can't seem to physically find or access the other end of that wire. \$\endgroup\$ – Inclemency Aug 3 '18 at 15:02
  • \$\begingroup\$ What is the specific gravity of each cell? and Voltage for each battery. They should be matched , otherwise suplhation has occured and may need pulse rejuvenation. \$\endgroup\$ – Tony Stewart EE75 Aug 3 '18 at 15:04
  • \$\begingroup\$ B+ should be same as battery. But Brake return is switched thru diode so not the same as B- and may be 1V when active. \$\endgroup\$ – Tony Stewart EE75 Aug 3 '18 at 15:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.