help explain why when battery is switched off it reads 44.5V?

[wakyct]

hi, total beginner here so bear with me if this is obvious. I'm putting together my bike and I noticed that when my new Luna 48V bottle battery is switched off, it reads 44.5V, and when switched on, it reads about 49V (I just got it Monday -- haven't charged it yet). I'm curious how this relates to the function of the switch?

The reason I ask is because I'm planning to hardwire the motor power to the contacts in the battery cradle (to keep things tidy and more simple), thinking that if the battery is switched off then there wouldn't be any problems with sparking when putting the battery on the cradle. However if the battery still has volts at the contacts when the switch is off maybe that won't work? Thanks for any pointers.

edit a couple days later: for any newbie like me who may be curious, there was no problem with sparking with the battery switched off when putting it in the cradle.

[ykick]

If it’s a MOSFET switched output what you’re seeing is “float voltage”. Try connecting to the motor controller and then measure the voltage OFF while connected to the controller. Even the slightest load voltage should drop to nearly 0V in OFF position.

[wakyct]

Thanks, so I take it the MOSFET is in the BMS? Doing further reading I saw a couple references that on bottle batteries the switch typically switches the BMS which would make sense.

[ykick]

Probably a MOSFET switch because a physical hardware switch will not survive arcing which takes place when opening closing circuits at these DC voltages. Back in prehistoric times of eBikes 24V systems could get away with an automotive toggle switch or similar to open/close battery output. 36V began experiencing arcing failures and once the industry adopted 48-52V nothing except a seriously large DC contact switch can survive.

So yes, some battery pack designs incorporate a simple "soft" ON/OFF switch using the BMS power FETs.

Not sure that's what you have but if it is, that would explain why you measure floating voltage that's simply sitting statically on the devices. Any amount of load will dissipate that voltage and should drop to near zero. There may be some voltage leaking through the FETs but it shouldn't be much when connected to a load.

[Tommycat]

I was wondering the exact same thing when I was testing my newly acquired 52 volt 11.5 ah bottle battery! 56.6 vdc with the switch on...29.07 vdc with the switch off. The same at both, charging port and output contacts.

What I've learned...As Ykick says it's just some kind of phantom voltage. Digital voltage meters depending on internal resistance may show voltage potential, but there's no power behind it. Any resistance will bleed the potential voltage off, as in connecting the controller, or a power drain of some sort.

But you must have the switch 'ON' to charge! Other wise my 3 amp mini Luna charger will switch on and off about once a second.
It's also recommended that you charge with out the battery connected to the controller. (...and in a bar-b-q grille....??? lol)

Which for me brought up two interesting points. 1) After the battery gets done charging, the charge will reverse and drain back into the charger at a rate of 3mA. So at this time you have to disconnect the charger right when it's done charging to avoid losing some charge. Is this amount anything to worry about? I'm going to be working on eliminating this when some diodes get here that are on a slow boat from China. :-) 2) It has been recommended that the battery be removed from it's holder, or disconnected from the controller, to be charged. I wanted to be able to just leave it on the bike as is to charge. So I plan on installing Luna's remote switch between the battery output and the controller to allow me to not have to take it on and off. It also gives me a secondary power 'ON/OFF' switch to disable the bike.

Total beginner here too. Ykick....would love to see a wiring schematic of these bottle packs! Got one?

Modern Times... Remote_Switch


Regards,
T.C.

[ykick]

I do not recommend disconnecting charger from a pack too prematurely. Even though the charger may stop current output (LED green, fan stops) it may simply be in a dormant state while pack BMS is in balance/equilibrium mode.

In that mode, once the BMS bleeds off excessive charge for the affected cell groups the charger can/should resume bulk charging until cell group OVDV (over voltage detection voltage) is reached. Rinse and repeat...

I say for new packs at least 12-24 hours connected to a charger plugged into the wall. Then burn off some charge but not too deeply and repeat this conditioning process 3-5 times.

[Tommycat]

Well said Ykick... Would you say bleed off happens ALL internally to the battery pack? (Wondering if my diodes could get in the way...) And of course this conditioning only occurs at 100% charging rate.
Any schematics?

[AZguy]

The big problem a diode in series to the battery will give you is that it will drop voltage and then your pack will not fully charge/balance

[Tommycat]

Not an electronics guy and I appreciate your comment. If I'm reading the spec. sheet right...it looks like it may drop the voltage by .7 vdc. Hopefully there's enough adjustment in P1 on my mini to make up for this!

[AZguy]

Don't do it!

I'm not sure what the P1 is but you'd need to bump the output of the charger up... but this is a big problem too!

If you try and charge with a 48V charger and a diode in series... the charger output will limit at ~54.6V which is what the battery will need to reach full charge and more importantly to balance the cells but the diode will drop about 0.7-1V at 2A (presuming a silicon diode vs. schottkey) at 2A and then as the charger, "thinking" the charge complete will hold ~54.6V and the voltage at the battery will creep very slowly as the current drops and the voltage across the diode drops... while it will very slowly keep on creeping up and possibly even balance and increase the charge, it would take a very long time (think likely days) and still never get fully charged and who knows what the BMS does at the very low currents that are below full charge.

So say you bump the charger voltage up to compensate for the diode drop. Then there's nothing to stop the voltage creep at the end of the charge cycle and then you get yourself in the very dangerous region of overcharging the battery.

I'm not sure what value the diode will bring regardless. Why do feel the need to put it in? I can say for certain it will give you heartburn unless it's only purpose is to *prevent* the battery from charging fully and balancing....

[Tommycat]

Don't do it!

I'm not sure what the P1 is but you'd need to bump the output of the charger up... but this is a big problem too!

You are correct in your assumption, this is what I'm going off of... HERE.

If you try and charge with a 48V charger and a diode in series... the charger output will limit at ~54.6V which is what the battery will need to reach full charge and more importantly to balance the cells but the diode will drop about 0.7-1V at 2A (presuming a silicon diode vs. schottkey) at 2A and then as the charger, "thinking" the charge complete will hold ~54.6V and the voltage at the battery will creep very slowly as the current drops and the voltage across the diode drops... while it will very slowly keep on creeping up and possibly even balance and increase the charge, it would take a very long time (think likely days) and still never get fully charged and who knows what the BMS does at the very low currents that are below full charge.

This is a very good point. On re-visiting the spec sheet I see I was thinking in discrete terms, picking out just the one maximum spec. Where as diode voltage drop is analog depending on amp draw. Good catch! Just a reminder I'm running a 52v battery...

So say you bump the charger voltage up to compensate for the diode drop. Then there's nothing to stop the voltage creep at the end of the charge cycle and then you get yourself in the very dangerous region of overcharging the battery.

Possibility.

I'm not sure what value the diode will bring regardless. Why do feel the need to put it in? I can say for certain it will give you heartburn unless it's only purpose is to *prevent* the battery from charging fully and balancing....

All good input, thank you. The most important at this time is the last line...Why?

My thinking on adding diodes would be to eliminate the need to unplug the charger for a few hours after a full charge to allow the BMS to balance the cell charges. And then re-connecting and re-starting the charger. Perhaps a couple times. Thinking that this is needed because of the battery draining out to the charger at a rate of 3 mA with the charger off. Are these assumptions correct? Also at the 80, and 90 percent charge settings allowing the charger to remain connected and removed at a later convenient time with out any loss of charge. (impossible I know, but get as close as possible) As in a leisurely charge overnight.
Is 3 mA drain a concern at all?

I believe I'll start a new thread in the 'Charger' department and get further into the weeds with this...thread link to follow.

Link to new thread...HERE.

[AZguy]

3mA is very small in the scheme of things. For 3mA to drain the battery even just 1% of a smallish 11.5Ah battery... 11.5Ah / 3mA = 38 hours - not much to worry about =]

You could probably PM the moderators to move the thread there.


A very funny thing too... I'm very familiar with diodes in series. I've got several batteries so I like to keep them about 80% charged and I can either run them from 80% or throw them on a charger for an hour to bring them to 100% before using. I put two in series and then when the charger shows complete I unplug it and end up with about 80% charge. To bring it to 100% and balance I jump the diodes out with an alligator clip jumper. If I forget and leave them on the charger no harm, it will just charge it more than the 80% but very slowly. I guess eventually it would bring them all the way to 100% but I've never forgot for more than a day or two and it's not gotten more than about 90%. Poor man's approach to 80% charging... not perfect but plenty good enough for me!

[AZguy]

That was odd - I must have been posting when this got moved and your post and my comment went bye bye...

Anyway, here's what I wrote:


3mA is very small in the scheme of things. For 3mA to drain the battery even just 1% of a smallish 11.5Ah battery... 11.5Ah / 3mA = 38 hours - not much to worry about =]


A very funny thing too... I'm very familiar with diodes in series. I've got several batteries so I like to keep them about 80% charged and I can either run them from 80% or throw them on a charger for an hour to bring them to 100% before using. I put two 5A Si diodes in series and then when the charger shows complete I unplug it and end up with about 80% charge. To bring it to 100% and balance I jump the diodes out with an alligator clip jumper. If I forget and leave them on the charger with the diodes in series, no harm, it will just charge it more than the 80% but very slowly. I guess eventually it would bring them all the way to 100% but I've never forgot for more than a day or two and it's not gotten more than about 90%. Poor man's approach to 80% charging... not perfect but plenty good enough for me!