Dual Battery Connection Adapter

Az.

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Sorry to be a party pooper, but I must ask...

Why on earth would you want two batteries on your bike???
 

Sturmey

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I use the schottky diodes (since 2018) all the time to parallel up to 3 batteries as I do long runs. I find them more or less foolproof and simple in practice once they are properly installed. The batteries can be left on the bike when charging if you want and the batteries can be charged together with separate chargers or one at a time or in any order. Batteries can be popped in or out if necessary and there is no need to worry about their state of charge or if they balance. There is a little under half a volt loss in the diodes but there is well over a volt gained by the reduction in sag so there is an overall voltage gain in my case.I find it useful with older batteries that have increased their internal resistance and sag badly when used on their own. These older batteries can still give good service when paralleled as the load is shared. So far, so good.....
Diodes are not used with direct drive motors especially with regenerative braking.
 
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saneagle

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Unless the sag characteristics are identical, every time load is applied and then released there will be a degree of current flow between the batteries.

The common parameter with directly connected parallel batteries is the voltage. Current under load is shared based on pack internal resistance, and if one gives much more than the other there will be a voltage difference when the load is removed, and so a current flow.

You could do some modelling in a spreadsheet to see how much of a problem it is likely to be. In the middle flattish part of the discharge curve probably no big deal, but there might be funny effects at close to empty.
That's not right. The current comes out of the batteries. The only way it could go into one is if the other had a higher voltage, but it can't because they're wired together.

It doesn't matter how much any of them sag. Whatever any sag is, it'll be less than the battery on its own because less current will be flowing from each. One cannot sag more than the other when the wires are tied to each other.
 

matthewslack

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That's not right. The current comes out of the batteries. The only way it could go into one is if the other had a higher voltage, but it can't because they're wired together.

It doesn't matter how much any of them sag. Whatever any sag is, it'll be less than the battery on its own because less current will be flowing from each. One cannot sag more than the other when the wires are tied to each other.
Yes, the voltage under load stays the same. But a higher proportion of the energy extracted comes from the pack with lower internal resistance, so its position on the discharge curve changes more than the other one.

When the load is removed, the packs therefore want to be at different voltages, and so a current must flow to equalise voltages. Size of current depends on internal resistance of the battery packs.
 
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Ghost1951

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Yes, the voltage under load stays the same. But a higher proportion of the energy extracted comes from the pack with lower internal resistance, so its position on the discharge curve changes more than the other one.

When the load is removed, the packs therefore want to be at different voltages, and so a current must flow to equalise voltages. Size of current depends on internal resistance of the battery packs.
I suppose this must happen inside any battery which has paralleled cells. One of my batteries has a weak group of cells which always loses voltage more than the others. Over time the pack becomes unbalanced even when the bike is out of use. It isn't just that one group is powering the BMS, because even when I get the whole pack properly in balance, it will lose voltage inside ten miles by a couple of tenths of a volt. I reckon that one cell in that group has less capacity, so it is constantly being charged by the others, more or less depending on whether the battery is being loaded or not.
 

sjpt

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Sorry to be a party pooper, but I must ask...

Why on earth would you want two batteries on your bike???
More range is one reason, though it seems simpler just to switch two batteries for that.

Another is if the main battery has proprietary handshaking and replacement/extras are very expensive. That relies on being able to connect the second (non-proprietary) battery in such a way that the handshaking with the main battery still works but the second battery can feed in power.
 

matthewslack

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In numbers:

Take two fully charged 36V packs connected in parallel, battery one with 0.05 ohm internal resistance and battery two with 0.1 ohm. Apply 30A load.

By ohms and kirchoffs laws, for voltage to stay equal where the cables join, the voltage drop across the internal resistances must be equal, and so twice as much current flows from battery one as battery two: 20A vs 10A.

Suppose that load is maintained for half an hour. Battery one will have provided 10Ah, and battery two only 5Ah.

Suppose they are 15Ah packs. When the load is removed, battery one is at one third capacity and battery at two thirds. If they were not connected, their voltages would be different: roughly 36V and 38V. But they are connected, and by ohms law 2 volts across 0.15 ohms results in a current of 13.3A.

In practice, it is a very dynamic situation, with the 'cross current' varying according to load. Is it a problem? I don't know, but it might be.
 
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matthewslack

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I suppose this must happen inside any battery which has paralleled cells. One of my batteries has a weak group of cells which always loses voltage more than the others. Over time the pack becomes unbalanced even when the bike is out of use. It isn't just that one group is powering the BMS, because even when I get the whole pack properly in balance, it will lose voltage inside ten miles by a couple of tenths of a volt. I reckon that one cell in that group has less capacity, so it is constantly being charged by the others, more or less depending on whether the battery is being loaded or not.
Indeed. The consistency of cells keeps the effect extremely small for a properly manufactured battery in good order.
 
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Benjahmin

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I have been using two batteries in parallel for a few years now. Living in a very hilly area I find that voltage sag happens much later in a ride than if I use just one battery at a time.
I use a simple homemade Y splitter with Anderson connectors all round. No diodes. I used good quality silicone cable.
However, one has to be very mindful when charging and connecting.
1/ The interconnection between batteries MUST always be removed before charging.
2/ I use two chargers. I'm aware that one of them outputs around 0.1v more than the other. So I always use this one to top up both batteries.
3/ You MUST always check the individual battery voltages before interconnecting them. Maximum voltage difference I go with is 0.1v. If more than this I wait for voltages to settle after charging, or top up the lower.

The batteries are of slightly different ages though roughly the same capacity. They serve me well and I'm not aware of any preoblems.
I cannot overstress the need for the need to be mindful and extremely aware/present at the time of inter-connection. So long as this is practiced, diligently, then there isn't any problem here.
 

saneagle

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When the load is removed, the packs therefore want to be at different voltages, and so a current must flow to equalise voltages. Size of current depends on internal resistance of the battery packs.
The packs can't be at different voltages because they're tied together. It's no different to having cells in parallel inside the pack. There are no diodes between them.
 

Ghost1951

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In numbers:

Take two fully charged 36V packs connected in parallel, battery one with 0.05 ohm internal resistance and battery two with 0.1 ohm. Apply 30A load.

By ohms and kirchoffs laws, for voltage to stay equal where the cables join, the voltage drop across the internal resistances must be equal, and so twice as much current flows from battery one as battery two: 20A vs 10A.

Suppose that load is maintained for half an hour. Battery one will have provided 10Ah, and battery two only 5Ah.

Suppose they are 15Ah packs. When the load is removed, battery one is at one third capacity and battery at two thirds. If they were not connected, their voltages would be different: roughly 36V and 38V. But they are connected, and by ohms law 2 volts across 0.15 ohms results in a current of 13.3A.

In practice, it is a very dynamic situation, with the 'cross current' varying according to load. Is it a problem? I don't know, but it might be.
So - taking the OP's first post about two batteries of different capacity, imagine that one is 25Ahr and one is 15Ahr like he mentioned, and imagine that they are connected to the controller via diodes so one battery can not charge the other, what happens as a long journey is undertaken? Does the smaller battery become depleted and eventually stop putting out current, or does the smaller battery just produce less current all through the trip so it depletes at the same rate as the bigger one? I am thinking the latter is probably more likely.
 

Sturmey

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In practice, it is a very dynamic situation, with the 'cross current' varying according to load. Is it a problem? I don't know, but it might be.
I done a rough measurement of that 'cross current' once. The 36v batteries were an 11.6 Ah (samsung 2.9) and a 7ah (samsung 3.5) . I put an ammeter in series with one of the batteries. I allowed the batteries to equalize. Using my 5.5 ohm dummy load, I subjected the parallel batteries to loads of 6.6 amps for different duration's and observed the cross current when load was stopped. I got cross current readings in the order of 300mA that taper down with a minute or two to about 100mA and to about 50mA after 10 minutes.
The batteries cells above were of similar age and manufacturer. I suspect the readings may be greater if say one battery is new and the other is well worn/high internal resistance/saggy.
Its an easy experiment to do if you have some type of load. I suppose it could be done on the bike with a bit of imagination.
 
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saneagle

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I suppose this must happen inside any battery which has paralleled cells. One of my batteries has a weak group of cells which always loses voltage more than the others. Over time the pack becomes unbalanced even when the bike is out of use. It isn't just that one group is powering the BMS, because even when I get the whole pack properly in balance, it will lose voltage inside ten miles by a couple of tenths of a volt. I reckon that one cell in that group has less capacity, so it is constantly being charged by the others, more or less depending on whether the battery is being loaded or not.
The cells are in series. There is no way a low cell can be charged by the others, otherwise batteries would never go out of balance. The BMS does the balancing by draining off the high cells at the top of the charge. The reason your battery voltage reduces a bit from maximum after you take it off charge is because the bleed resistors are still open.
 

saneagle

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In numbers:

Take two fully charged 36V packs connected in parallel, battery one with 0.05 ohm internal resistance and battery two with 0.1 ohm. Apply 30A load.

By ohms and kirchoffs laws, for voltage to stay equal where the cables join, the voltage drop across the internal resistances must be equal, and so twice as much current flows from battery one as battery two: 20A vs 10A.

Suppose that load is maintained for half an hour. Battery one will have provided 10Ah, and battery two only 5Ah.

Suppose they are 15Ah packs. When the load is removed, battery one is at one third capacity and battery at two thirds. If they were not connected, their voltages would be different: roughly 36V and 38V. But they are connected, and by ohms law 2 volts across 0.15 ohms results in a current of 13.3A.

In practice, it is a very dynamic situation, with the 'cross current' varying according to load. Is it a problem? I don't know, but it might be.
You're not thinking it through. Let's say they were both 10 Ah packs. If one battery had given 5Ah, it would be at around 38v, the other, having given 10 Ah, would be 31v. That can't happen when they're tied together. Your theory is therefore flawed. Have a think about cause and effect instead of jumping on random theories.
 

matthewslack

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You're not thinking it through. Let's say they were both 10 Ah packs. If one battery had given 5Ah, it would be at around 38v, the other, having given 10 Ah, would be 31v. That can't happen when they're tied together. Your theory is therefore flawed. Have a think about cause and effect instead of jumping on random theories.
I give up!

I thought you understood internal resistance and ohms law!
 

matthewslack

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So - taking the OP's first post about two batteries of different capacity, imagine that one is 25Ahr and one is 15Ahr like he mentioned, and imagine that they are connected to the controller via diodes so one battery can not charge the other, what happens as a long journey is undertaken? Does the smaller battery become depleted and eventually stop putting out current, or does the smaller battery just produce less current all through the trip so it depletes at the same rate as the bigger one? I am thinking the latter is probably more likely.
A lot of the time, under load, both diodes are conducting, but with currents that may be very different. Whenever the load is light enough for the voltage at the connection point to be higher than that of one of the batteries, the other battery will take all the load, until it falls yo the level of the first.

Think of two water tanks linked with low loss non return valves (diodes), but having differently sized output pipes (internal resistance). The water levels mimic the battery voltages. Imagine one pipe is 15mm and the other is 28mm. At low flows they stay in step, but at high load/flow, much more comes out of the big pipe. Go back to low flow and the small pipe tank is now at higher pressure/voltage, so it will 'catch up'.
 

Ghost1951

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The cells are in series. There is no way a low cell can be charged by the others, otherwise batteries would never go out of balance. The BMS does the balancing by draining off the high cells at the top of the charge. The reason your battery voltage reduces a bit from maximum after you take it off charge is because the bleed resistors are still open.
We are miscommunicating.

I was talking about my weak cell group which i am convinced is caused by a single dodgy cell inside a group of six that are in parallel. My point was about how that parallel connected group must be flowing current into the weak cell.

I know how series charging and balancing works.
 

Ghost1951

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A lot of the time, under load, both diodes are conducting, but with currents that may be very different. Whenever the load is light enough for the voltage at the connection point to be higher than that of one of the batteries, the other battery will take all the load, until it falls yo the level of the first.

Think of two water tanks linked with low loss non return valves (diodes), but having differently sized output pipes (internal resistance). The water levels mimic the battery voltages. Imagine one pipe is 15mm and the other is 28mm. At low flows they stay in step, but at high load/flow, much more comes out of the big pipe. Go back to low flow and the small pipe tank is now at higher pressure/voltage, so it will 'catch up'.
Yes. Thanks Matthew. Nice illustration. Confirms what i was thinking.

Just been out for a 12 mile ride in the wind and colder weather on the wee Argos, £245 folder. What a bargain that was. Mind, now that it is colder, that little 8.5Ahr battery is showing some loss of range. I might think about fitting another battery sledge for my other battery to be fitted on top of the rack in case i want to go far. In that case, this discussion would be important. That set up would have an 8.5Ahr and a nominal 13.5 Ahr working together.
 

saneagle

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I give up!

I thought you understood internal resistance and ohms law!
I do, but you don't seem to understand the other factors in play, like how a battery's internal resistance changes during operation and how you need a voltage difference to make current flow.