Which schottky diodes for parallel charging?

[kilou]

Hi,

I'd like to parallel charge two Li-ion batteries, one is the original 36V 11Ah of my ebike and the other will be a 36V 13Ah extender battery pack. For this I'll use the original charger of my ebike battery and add a Y cable with XT60 connectors to charge the extender at the same time. To prevent any cross current flow between the batteries in case of unequal voltage when they're hooked up, I'm going to use schottky diodes. I have two questions regarding that:

1) The charger output is 36V 4A so when charging only one battery the diode should handle the 4A and when charging both batteries in parallel each diode will see roughly 2A. I could find schottky diodes rated for 60V and either 5A or 8A. Should I go for 8A or is it preferable to go with the 5A (if the diode has to handle 4A)? In other words should the diode work close to its nominal voltage (in that case the 5A should be preferable) or is it better if the diode can handle a larger current such as 8A?

2) Is it OK to solder the diode onto a wire and cover everything with heatshrink tubing?

Thanks for your input!

 

[dnmun]

if they are identical batteries you will not need a diode. they will always be at the same voltage.

 

[kilou]

They are not exactly identical batteries (use different cells and possibly different chemistry though that's not yet sure) and one is older while the second is new. Most importantly I may not always use them in parallel. This means that they could possibly have different voltages when hooked up for charging and in these situations diodes are needed.

 

[ddk]

... In other words should the diode work close to its nominal voltage (in that case the 5A should be preferable) or is it better if the diode can handle a larger current such as 8A?

other than the current ratings, there is no difference between each diodes circuit effect... IOW -use 8A-rated diodes

2) Is it OK to solder the diode onto a wire and cover everything with heatshrink tubing?

[mr rodgers]"sure it is"[/mr rodgers]
you could also install the diodes inside the connectors for added protection to the wiring (my personal preference)

 

[kilou]

Thanks!

 

[dnmun]

i recognize it is useless to argue with you since you got your degree in EE long before me and since it is not the same chemistry because one pack is older than the other which must change the chemistry and since you need to charge the other battery through the diode also then you can use whatever size diode you wanna use and it will make no difference at all.

but just for education of the masses who read this why not calculate the current through the diode when you plug the two batteries together, with the diode present. use 2 volts difference in voltage, .4V as the forward bias and assume .0000005mR for the wire that connects the two packs. lotsa people will learn something from you here.

 

[kilou]

@dnmun: I feel sorry if I disturbed you with my question... As you might have guessed I have no degree in EE and this is precisely why I ask such basic question here. However I'm following what other users of my system (bosch ebike) have done to build their extender battery and they have used ideal diodes to run the two packs in parallel on their bikes, and schottky diodes to charge in parallel too. I just ask this here because their forum is in german and I'm not doing well in german. I also asked the manufacturer of the ideal diodes that are specifically made for ebikes, and he also said that diodes MUST be installed for parallel charging. Of course people with EE like you can do without because they know what they are doing and check the voltage of the packs. I don't want to do this, I use my ebike to commute and want a simple system that I can charge without bothering the voltage of either pack.

For your information one pack use samsung 22P cells (old) and the other pack will use either konion V3 or VTC4 (new). That's why I wrote the two packs are not exactly identical. I mentionned that I need the ability to also use one pack at a time. Now suppose one of the pack is fully charged (41V) and the second is half used and maybe at 35V. What will happen if you hook them both to the charger for parallel charging without diodes in between the packs? Suppose also that for some reason power is interrupted for whatever reason during the charging process (power failure etc). According to many educated people I may have issues, according to you I will have no problem. Since you mentionned that diodes would not be required I assume that they wouldn't harm anything either. So I prefer staying on the safe side and use diode since I'll need to parallel charge at home and office, not in a fireproof underground bunker.

The question was which diode to use not whether diodes must be used or not. In that sense ddk answered my question, not you. I'm fine if you can do it without diode and I'm also fine if you want to educate others. Just remember that other people suggested using diodes and these are probably as well educated as you are, especially considering a manufacturer of ideal diodes recommends using simple schottky diodes (not their product) when parallel charging... I can send you their email if you wish to educate them as well Regarding your last point and the calculations, as I have no degree in EE I cannot answer but I'd be more than happy if you could enlight me, explain the process and provide the answer instead of waiting for me to educate others, something I humbly recognize I cannot do.

Cheers

 

[dnmun]

ok, for the example. if there is 2V difference and .4V forward bias on the diode, then there is 1.6V left to push current through a resistance of .00005mR

from ohm's law I= 1.6/.00005 = 32,000 Amps

so the diode will have to be like the ones they use for the power transmission lines from Hoover dam which is why i said it did not matter what diode you used. if you plug a battery of higher voltage into a lower voltage pack then it will burn up the diode.

if you charge them with the diode present then the diode will prevent charging current from entering the pack with the diode on top so it will never charge up.

if you are using two different chemistries such as using SLA and lithium then you would use a diode to prevent the current from the li pack from charging the SLA pack. if you are using two li packs of identical cell count and voltage in parallel then you would not have to worry that the one li pack would charge the other one up since they perform and have the same leak down rates so that the diode is not an issue.

i use diodes, 100A total, to tie my lipo pack and the lifepo4 pack together in parallel to protect them from overcharging each other during the charging cycle and can run some power through them initially when the batteries begin with slightly different voltages but i have to eventually connect a jumper to bypass the diodes when the current becomes too high for the diodes to handle as the batteries become identical in voltage. i also have an identical 100A of diodes on top of the SLA battery that is tied in parallel with the two lithium packs to prevent the li packs from losing charge into the SLA pack but the SLA pack is a lower voltage and delivers power at the very end of the combined pack's discharge. imo that is how the didoes should be used.

i do not follow that blogspot so not sure what they are doing, but if you have two batteries with the same series cell count and the same chemistry, linmnco is what it looks like, then you do not have to isolate the batteries from each other, even if one is from a different reseller or a different brand. same chemistry and same cell count, they can be wired up in parallel and you can even tie the packs together at the cell level with a balancing wire so the BMS can balance both packs at the same time.

 

[kilou]

Thanks! That's helpful.

ok, for the example. if there is 2V difference and .4V forward bias on the diode, then there is 1.6V left to push current through a resistance of .00005mR

from ohm's law I= 1.6/.00005 = 32,000 Amps

so the diode will have to be like the ones they use for the power transmission lines from Hoover dam which is why i said it did not matter what diode you used. if you plug a battery of higher voltage into a lower voltage pack then it will burn up the diode.

I'm not sure to follow here: as far as I understood a diode allows the current only in one direction (from the charger to the battery pack). In the opposite direction the resistance should be almost infinte so that no current can flow. Is this not correct? I do not get why the larger voltage pack would drain into the lower voltage pack and blow the diode in that scenario. I mean it would defeat the whole purpose of building diodes, not only in that specific application but any other, no?

if you charge them with the diode present then the diode will prevent charging current from entering the pack with the diode on top so it will never charge up.

Doesn't it depend on which way around you install the diode? I mean the diode should allow current from the charger to the pack and block the reverse path. In that case, it shouldn't it work?

if you are using two different chemistries such as using SLA and lithium then you would use a diode to prevent the current from the li pack from charging the SLA pack. if you are using two li packs of identical cell count and voltage in parallel then you would not have to worry that the one li pack would charge the other one up since they perform and have the same leak down rates so that the diode is not an issue.

One pack is 11Ah and the new extender pack is supposed to be 13Ah. So not exactly the same cell count.

 

[dnmun]

the capacity of the pack is not the same as the voltage or series count of the pack.

i am really sorry, i tied to explain, just like i tried to explain why the current does not flow through the bolt in your other thread about the ring terminals.

when you get to college you can take physics classes and they will cover electricity and magnetism and after you get some experience with ohm's law it may make more sense.

a diode does block current flow in one direction but when you short the two batteries together then the diode will not be able to handle that amount of current. it will blow up.

 

[kilou]

I don't have a EE degree but I do at least know about Ohm's law The thing that I don't get is that what you're saying would mean that using ideal diodes when using the 2 batteries in parallel (not while charging but while riding the bike) would cause the same thing: diodes should blow up as well. The ideal diodes used by others have a max current rating of 30A. So simply putting the 2 batteries in parallel should blow up these diodes as well. However this is not what happens since several guys have done that and the diodes do work. I thought that the ampere rating for diodes (schottky) relates to forward current only (direction in which the diode allows current to flow). In the opposite direction, the resistance should be almost infinite so unless the reverse voltage exceed the nominal voltage of the diode, I do not see why it should blow up, and I also do not understand why some people use this design without issue while their diodes should blow up. Maybe you could keep on explaining but you might already be fed up with me :lol:

 

[Jonathan in Hiram]

My understanding of the way this is done is that the two diodes are back to back across the like polarity terminals of the two batteries, the two other like terminals being connected directly, and no current flows in either direction because any voltage imbalance between the batteries will always reverse bias one of the diodes. You then feed the charger output into the common connection between the back to back diodes and current flows into either battery in proportion to its voltage and internal resistance, the battery with the lower voltage will accept more of the charger current until both batteries are at equal potential (neglecting any slight mismatch between the forward drops of the diodes).

If one battery is really low and the other near full when you connect them then all or almost all of the current will flow through only the low battery diode, they should be sized to handle at least twice the charger current particularly if you plan on wrapping them up in something that will keep them from shedding heat easily.

 

[dnmun]

I don't have a EE degree but I do at least know about Ohm's law The thing that I don't get is that what you're saying would mean that using ideal diodes when using the 2 batteries in parallel (not while charging but while riding the bike) would cause the same thing: diodes should blow up as well. The ideal diodes used by others have a max current rating of 30A. So simply putting the 2 batteries in parallel should blow up these diodes as well. However this is not what happens since several guys have done that and the diodes do work. I thought that the ampere rating for diodes (schottky) relates to forward current only (direction in which the diode allows current to flow). In the opposite direction, the resistance should be almost infinite so unless the reverse voltage exceed the nominal voltage of the diode, I do not see why it should blow up, and I also do not understand why some people use this design without issue while their diodes should blow up. Maybe you could keep on explaining but you might already be fed up with me :lol:

you were talking about taking a charged battery with a diode on top and plugging into a discharged battery as a second battery to backup the discharged first battery. the current that flows out of the new battery going through the diode into the first battery will burn up the diode.

what i am trying to tell you is that if you have two identical batteries then they do not have to be connected through diodes. i don't know why those other people told you otherwise since it is not needed but i know that everyone here also thinks that is needed. but they don't know why either.

 

[kilou]

you were talking about taking a charged battery with a diode on top and plugging into a discharged battery as a second battery to backup the discharged first battery. the current that flows out of the new battery going through the diode into the first battery will burn up the diode.

No, in my first post I mentionned I want to parallel charge two batteries with possibly different voltages. I never mentionned I want to use one battery to charge the other one... Once again, the idea is to use schottky diodes to allow current from the charger to both battery packs and preventing current to flow from the battery with the highest charge to that with the lowest charge, such as on the attached picture.

If one battery is really low and the other near full when you connect them then all or almost all of the current will flow through only the low battery diode, they should be sized to handle at least twice the charger current particularly if you plan on wrapping them up in something that will keep them from shedding heat easily.

Yes that's correct. A single diode should be able to handle all the charge current, also because I also want to be able to charge each battery pack separately if needed. Since the charger has a max output of 4A, I figure that a diode with 50V and 8A should be able to handle that. Do you think that the diode would heat-up badly if 4A goes through it and it is encapsulated below a heat shrink tubing? The voltage drop od my diode is about 0.7V. At 4A this makes 2.8W. Is that ok or does it requires a heatsink?

 

[Jonathan in Hiram]

I'd parallel two of those on each battery if I were going to wrap them up, they are small and are going to build up heat badly I think if wrapped up in something but I'm just going on my own experience letting the magic smoke out of a bunch of semiconductors.

I found some 60V 8A with a 0.53V forward drop on Amazon, the shipping on four is only a little more than for two.

http://www.amazon.com/Schottky-Diodes-Rectifiers-60V-ULTRA-LOW/dp/B00MEI5V2I/ref=sr_1_5?ie=UTF8&qid=1411136212&sr=8-5&keywords=8a+schottky+diode

 

[chvidgov.bc.ca]

I like diodes. Dmun...can't you understand that the two packs may not be at the same voltage when they are put together? It's all well and good to not use the diodes if the packs are connected together all the time. In this case they won't be.

 

[kilou]

I'd parallel two of those on each battery if I were going to wrap them up, they are small and are going to build up heat badly I think if wrapped up in something but I'm just going on my own experience letting the magic smoke out of a bunch of semiconductors.

I found some 60V 8A with a 0.53V forward drop on Amazon, the shipping on four is only a little more than for two.

http://www.amazon.com/Schottky-Diodes-Rectifiers-60V-ULTRA-LOW/dp/B00MEI5V2I/ref=sr_1_5?ie=UTF8&qid=1411136212&sr=8-5&keywords=8a+schottky+diode

I heard that paralleling two diodes may not work as if there is a slight difference in voltage drop between the two, the one with the highest voltage drop will take more current... Don't know whether that makes sense or not but at least you seem to suggest that I'll build too much heat with a single diode for each pack so I should consider that. Maybe a P-mosfet would be better here, what do you think?

Actually I may not need to wrap the diodes in heatshrink tubing. I could enclose them in a plastic case. Not sure if that would be any better but after all, if the diode is supposed to work up to 8A, it should handle 4A without too much problem no? I agree the heatshrink tubing might be too much for them but what about the plastic case?

 

[Jonathan in Hiram]

The plastic case should be fine I think, you just need a little air to circulate around the diode, the area of a case would be a lot more than a bit of shrink tube and should shed heat a lot better.

Yeah, diodes can vary a bit in forward voltage drop but at a couple of amps each I don't think it's going to make maybe a few mA of difference in the current each diode is carrying, diodes of the same part number from the same batch should be fairly well matched.

If you already have the diodes you want to use, try it out and see if they get warm in the open air, then you can better decide how to proceed.

 

[kilou]

The plastic case should be fine I think, you just need a little air to circulate around the diode, the area of a case would be a lot more than a bit of shrink tube and should shed heat a lot better.

Yeah, diodes can vary a bit in forward voltage drop but at a couple of amps each I don't think it's going to make maybe a few mA of difference in the current each diode is carrying, diodes of the same part number from the same batch should be fairly well matched.

If you already have the diodes you want to use, try it out and see if they get warm in the open air, then you can better decide how to proceed.

Great! yes I've already ordered the diodes and will try to install them in the plastic case and check the heat build-up.

Thanks a lot for your help!

 

[kilou]

One additional question regarding schottky diodes in this application:

The diode I'm considering has a voltage drop of 0.7V. Since the charger charges the battery up to 42V (I guess although the charger output is said to be 36V 4A), if I use a diode in-between it will charge up to 42-0.7=41.3V. Is that correct? If yes then will the charging process end at some point with the battery charged "only" up to 41.3V or will the charger enter a loop and never stop charging? I mean if the charging process stops at 41.3V instead of 42V that's actually good for the battery pack but the charger shouldn't start/stop charging all the time... What do you think?

 

[Jonathan in Hiram]

One additional question regarding schottky diodes in this application:

The diode I'm considering has a voltage drop of 0.7V. Since the charger charges the battery up to 42V (I guess although the charger output is said to be 36V 4A), if I use a diode in-between it will charge up to 42-0.7=41.3V. Is that correct? If yes then will the charging process end at some point with the battery charged "only" up to 41.3V or will the charger enter a loop and never stop charging? I mean if the charging process stops at 41.3V instead of 42V that's actually good for the battery pack but the charger shouldn't start/stop charging all the time... What do you think?

I think the charger will simply see it as the battery packs being a little higher charged than it "thinks" they are just looking at the voltage across the terminals.

Watch it the first time and I think it will terminate normally.

For what it's worth my 8A 8 cell RC charger takes a lot longer to charge a battery if the connecting leads have much resistance, short fat wires on the connecting leads are noticeably quicker to charge than longer skinny ones, high resistance wires fool the charger into thinking the battery is higher internal resistance than it actually is and it goes into constant voltage mode a lot sooner. Eventually the same charge point is reached and the same Ah but the end of the charge is prolonged.

 

[dnmun]

the forward bias on the junction will decrease as the current passing through it decreases.

you can isolate each battery from the other and charge them in parallel if there is a separate charging mosfet C- from the power output P- on each battery. you would disconnect the two battery packs from each other at P- so they do not connect through the drains of the output mosfets.

then you can charge with the same charger connected to both batteries at C- and with both batteries connected to each other and the charger at B+.

that allows you to charge both and maintain control over the charge so you do not overcharge either battery. then reconnect at P- of each BMS. that allows both batteries to deliver power in parallel without pushing current through a diode.

 

[ddk]

One additional question regarding schottky diodes in this application:

The diode I'm considering has a voltage drop of 0.7V. Since the charger charges the battery up to 42V (I guess although the charger output is said to be 36V 4A), if I use a diode in-between it will charge up to 42-0.7=41.3V. Is that correct? If yes then will the charging process end at some point with the battery charged "only" up to 41.3V or will the charger enter a loop and never stop charging? I mean if the charging process stops at 41.3V instead of 42V that's actually good for the battery pack but the charger shouldn't start/stop charging all the time... What do you think?

simply put, those diodes are not Shottky diodes, which have a voltage drop on the order of 0.3V.
Possibly mislabeled silicon diodes is what you're looking at.-or two shottkys in series...
ref: https://en.wikipedia.org/wiki/Schottky_diode
...which also references active rectification as previously mentioned by dnmun

+ we're only talking about a 4 amp source and minimal diode leakage.

 

[dnmun]

yep, doping of the junction in schottky diodes is different. the speed is also controlled and you pay a ton extra for high speed schottky diodes, but for some applications you gotta have the speed to prevent other problems in the circuitry.

for batteries, there is no need for diodes if you charge the batteries separately with two chargers into the two BMSs or if you separate the drains of the BMSs at P- when charging with one charger.