Electric discharge from electric motor controller after unplugging battery.

Thread Starter

Firsttimeposter

Joined Jun 29, 2020
9
Hello!

I was being a bit careless after unplugging a 48v electric bike battery and I dropped the power cord leading motor's controller, it hit the rim and shorted the remaining current from the motor's controller. (I assume the steel rim completed the connection between the prongs and shorted the remaining current.) It gave off a nice pop.

I assume the motor is fine, but I am concerned about the controller board. Could such a discharge damage anything within the controller?<-my real question
What are electronics? and why electricity? Would an old TV work if the person repairing it forgot to wait after unplugging it?

My guess would be no, but I figured it couldn't hurt to be shamed for my carelessness by some circuit experts.

Thanks in advance!
 

DickCappels

Joined Aug 21, 2008
6,463
Can't tell from the information given. It might have done something -the "pop" is a sign that enough energy was involved to hurt something. Probably the best way to find out is to put it back together and test it.
 

Thread Starter

Firsttimeposter

Joined Jun 29, 2020
9
Thank You! I charged and assembled it again this morning and it runs fine under light load.

Apologies, I am mostly wondering in a theoretical sense: are circuit boards hurt in this manner? When I google it, I find articles explaining static discharge.

For example, if I was to remove the battery from my laptop and then short the remaining electricity in the capacitors of the laptop by shorting it (perhaps connecting the laptop's terminals with a wire), would that damage the laptop or any circuitry?

Please let me know if pictures or any specific descriptions would help my idiosyncratic writing.
 

dl324

Joined Mar 30, 2015
11,000
Welcome to AAC!
Apologies, I am mostly wondering in a theoretical sense: are circuit boards hurt in this manner?
As with most things, it depends.

I once shorted a scope ground clip to the power pin on a TTL IC in a refrigerator sized computer. All that happened is that the lead on the chip was vaporized (the 5V supply could supply 100A into a short). Luckily, the power distribution on the board was able to handle the current.
When I google it, I find articles explaining static discharge.
ESD damage can be a double edged sword. If you're lucky, the device will die outright. If you're unlucky, it makes the device unreliable and subject to an early death; but not before potentially causing erratic behavior.
For example, if I was to remove the battery from my laptop and then short the remaining electricity in the capacitors of the laptop by shorting it (perhaps connecting the laptop's terminals with a wire), would that damage the laptop or any circuitry?
That's not likely to cause damage because the capacitors wouldn't be storing much energy. That wouldn't necessarily be the case if high voltage, high capacitance devices were involved.
 

Thread Starter

Firsttimeposter

Joined Jun 29, 2020
9
Welcome to AAC!
As with most things, it depends.

I once shorted a scope ground clip to the power pin on a TTL IC in a refrigerator sized computer. All that happened is that the lead on the chip was vaporized (the 5V supply could supply 100A into a short). Luckily, the power distribution on the board was able to handle the current.
ESD damage can be a double edged sword. If you're lucky, the device will die outright. If you're unlucky, it makes the device unreliable and subject to an early death; but not before potentially causing erratic behavior.
That's not likely to cause damage because the capacitors wouldn't be storing much energy. That wouldn't necessarily be the case if high voltage, high capacitance devices were involved.

It wouldn't be science if there was a definite answer i suppose.

TTL IC in a refrigerator sized computer. <- The feeling of relief must have been immense. haha what was it computing? (if you can say)

Makes sense, a laptop also might have reverse current protection as well?

In my case I am working with a 48v 13s3p 18650 battery and a 86v, 80ah controller however, I got the knockoff version I'm not 100% on there being a PCB in the controller.

So 48v ~6ah and 30-60 instant amps it can produce? If the controller can accept that charge, why would would it become a problem in reverse?
My guess would be that (1) circuitry doesn't appreciate reverse flow, but I am not familiar enough with specifics to make a call. (Electrical engineers are too smart for me) (2) The built up charge can release faster than 48v 60 amps. (But wouldn't it be regulated by the resistors in the system in reverse as well?)

I hope my response isn't too convoluted! (and thanks again!)
 

Thread Starter

Firsttimeposter

Joined Jun 29, 2020
9
Extremely impressive, I appreciate the assistance from such an expert source!

I believe our situations may be slightly different. There was no live current attached to the controller when it shorted. The discharge was leftover from the motor and controller after I unplugged the battery. You mentioned it maybe different on a high voltage application. What would you expect could go wrong? Would I experience the erratic performance you described above? (I guess that makes me unlucky, because it seems to work)
 

Irving

Joined Jan 30, 2016
820
If the controller is half decent it will have some large storage capacitors in there; I suspect they discharged back out of the input connector, which also means there's probably no polarity protection diode either.

However, if it was half decent they would have put bleed resistors across the caps to discharge them safely!
 

Thread Starter

Firsttimeposter

Joined Jun 29, 2020
9
If the controller is half decent it will have some large storage capacitors in there; I suspect they discharged back out of the input connector, which also means there's probably no polarity protection diode either.


However, if it was half decent they would have put bleed resistors across the caps to discharge them safely!
Yes, exactly what happened I found a small weld on my rim where I believe it made contact. It's a 5000w electric bike hubmotor so capacitors are likely larger. (Things I should have put in the original post.)

Some Capacitors when wired reverse current can draw excess energy from the battery. In this case they were being grounded with the battery disconnected so they can't draw any new charge. I can't see how this scenario would risk any damage unless perhaps 2 capacitors of varying capacitance were wired in series (which wouldn't happen? maybe with diodes between them?) That is where my knowledge breaks down. I would imagine discharging out the back would be harmless then?

Saved $50 bucks though! :) I did drop the cord about 3 seconds after unplugging it, if the resistors were extremely resistant, would it take any noticeable time for them to bleed the system?
 

Irving

Joined Jan 30, 2016
820
Normally bleed resistors would be sized to draw off the charge in a few seconds. But there are limits. Say you had 10000uF capacitance on a 48v system. If they were fully charged the charge Q = C . V = 10000e-6 x 48 = 0.48 coulombs. A coulomb is the charge that is equivalent to 1Amp flowing for 1 second. So if your short took 1mS to discharge the capacitors, that was a current of 1000A, easily enough to weld the connector to the wheel rim

If we wanted to discharge the capacitors in 10sec after switch off, to 36% of their initial voltage then this is the time constant t = C . R,
so 10 = 10000e-6 . R, or R = 10/10000Mohm = 1kohm. But that 1kohm is subject to the full 48v when in use, dissipating W = V^2/R = 48*48/1000 W or about 2.3W, needing a 5W resistor. That adds cost, a few cents, uses energy and generates heat, so some manufacturers just leave them off...
 

Thread Starter

Firsttimeposter

Joined Jun 29, 2020
9
Interesting, I would expect then there either aren't bleed resistors or
Normally bleed resistors would be sized to draw off the charge in a few seconds. But there are limits. Say you had 10000uF capacitance on a 48v system. If they were fully charged the charge Q = C . V = 10000e-6 x 48 = 0.48 coulombs. A coulomb is the charge that is equivalent to 1Amp flowing for 1 second. So if your short took 1mS to discharge the capacitors, that was a current of 1000A, easily enough to weld the connector to the wheel rim

If we wanted to discharge the capacitors in 10sec after switch off, to 36% of their initial voltage then this is the time constant t = C . R,
so 10 = 10000e-6 . R, or R = 10/10000Mohm = 1kohm. But that 1kohm is subject to the full 48v when in use, dissipating W = V^2/R = 48*48/1000 W or about 2.3W, needing a 5W resistor. That adds cost, a few cents, uses energy and generates heat, so some manufacturers just leave them off...
Interesting, I couldn't tell you what type/ strength capacitors are being used so I can't say for sure if there are bleed resistors or not. How do we know they are discharged to 36%? (I know over discharge may be a problem, is that simply their "balance" point) I am understanding your formulas (1 small piece at a time :) ), but I don't see where the resistor might stop bleeding?



I'm sorry, although I am getting closer, I haven't been able to answer the original question. I greatly appreciate everyone's assistance, and I am getting to learn a ton of things that I did not understand previously. 3 days ago I had little to no idea how capacitors and resistors worked and now I am at least "familiar" or better.

My original question was if there would be any damage to the motor controller if the battery was not attached and the leads of the power to the controller were connected. Would it just discharge the system at an acceptable/safe rate? or can I expect something to be damaged?
 

Irving

Joined Jan 30, 2016
820
To answer your question - and I assume you meant "if the leads to the controller were shorted together" - is 'probably not', but its impossible to be definitive without knowing what's inside the box. You've already established that the controller stores an appreciable charge and that can be released pretty quickly. And that the charge is there for some significant time after disconnecting the batteries. Its a fair bet there are big capacitors and they don't have bleed resistors. All we can say is that its probably a good idea to avoid shorting them together or to the bike frame. One solution might be, once you've disconnected the battery, to try and turn the bike on and drain the capacitors that way. This may or may not work.

The bleed example I gave above was merely to demonstrate why a manufacturer may choose not to fit bleed resistors for energy saving and/or cost reasons. The figure of 36% is simply to make the maths easy by way of demonstration, of course a bleed resstor will continue to bleed down to 0v, but the lower the level of charge when a short occurs, the lower the risk of damage. A bleed time of several minutes would obviously be less desirable as a short is more likely to happen soon after switching off I would think?
 

Thread Starter

Firsttimeposter

Joined Jun 29, 2020
9
"One solution might be, once you've disconnected the battery, to try and turn the bike on and drain the capacitors that way. This may or may not work."
One solution might be, once you've disconnected the battery, to try and turn the bike on and drain the capacitors that way. This may or may not work.
This will work, I can see the electronics attempting to boot when I power w/o the battery. Everything is running so I'll say I was lucky, it will be some time before I am able to open the box and check for any damage but I will update when once I do. I should probably set it up so I never need to unplug the battery again. With how strong the battery is its not worth the risk to remove the battery for charging.

Based on the seller's storefront, I would assume they went with no resistors. I will try to find a diagram for curiosity sake.

Thank You Irving and everyone who shared their information with me. The knowledge on capacitors, resistors, diodes etc in the last 3 days has been more than I have learned in the last 3 years. I greatly appreciate it!
 

Irving

Joined Jan 30, 2016
820
If it still works then that's good. It's very unlikely you will see any visible damage due to this internally - no magic smoke came out then or subsequently. Of course, some component may have been stressed and might fail prematurely but that could mean in 10y rather than 15y or next week - basically you'll never know for sure.
 
Top