Schottky Diode Parallel Pack Adaptor

WheezyRider

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People often ask, “can I put two packs in parallel to increase capacity and reduce strain on each battery pack?” Yes you can, but you need to make sure that the two packs are of a similar voltage before connecting them. Also, if you disconnect one pack, the socket will still be live from the other battery, which could be hazardous. A way of overcoming this issue is to use blocking diodes, so that current cannot flow from one pack to the other. Unfortunately, normal diodes induce a voltage drop of more than 0.5V, which means that significant power is lost and waste heat generated. However, we can utilise Schottky diodes (widely used in the solar power industry) which only induce a voltage drop of 0.1 to 0.2 V to reduce this power loss to an acceptable level.

This is a concept that occurred to me a long time ago, but I didn’t try it out until quite recently, when I suggested it in a thread a couple of months back and forum member “Sturmey” said he had already tried it and it worked well for him. So I do want to acknowledge Sturmey for that, I don't claim to be the first to do this.

I don’t want to disparage anyone’s technical ability, but I will say that although this is a fairly simple thing to do, if you are not sure what you are doing, get some help. The power packs used on e-bikes can deliver a big punch if you do not respect them and, here we are dealing with two packs together. Power packs are some of the most expensive bits of e-bike kit and you could destroy two in one go…not to mention starting a fire you can’t put out easily and the fact that you could burn yourself quite badly if you do not do things carefully and pay attention to what you are doing.


To make a Schottky Diode parallel pack adaptor, you need suitable wire (eg silicone coated 12AWG) 2 suitable Schottky diodes, a heatsink, two male XT60 connectors and one female XT60 connector.

The conceptual diagram is as follows:

37598


37599


The diodes I used were 60A 100V (MBR60100CT), which is probably overkill, but I prefer to over engineer a bit, as you don’t want this to be the weak link in the chain and you might want to use it for a higher power project later. These diodes are actually dual diodes in one package. We will use two of them in parallel. As each diode in the 60A package is 30A, in principle you could get away with just using one of these components and using the two diodes inside it for this application, ie, running a 15 A controller (using one half of the dual diode component for each battery pack). However, for me that seemed a bit marginal as the current trip in my pack is 30A and the BMS is rated to 40A. So, I used two components to give me 60A per pack, which would give plenty of margin for safety.

A modest heatsink is needed as the voltage drop from these diodes means some power is consumed. I measured this to be 0.18V for these diodes, so if drawing 15A, it means 2.7W is dissipated in the diodes, which needs to go somewhere.

37600

The centre connection to the dual diode is the common cathode, and this also is connected to the metal tab on the diode package. As I was soldering the dual anode pins together to make a 60A device, I decided to cut off the middle common cathode connection and solder onto the tab for the cathode connection (my soldering was a bit messy, but you will be able to do a better job ;)). You need to take care to isolate the cathode from the heatsink with something like this polyimide tape, and ensure that the solder does not come into contact with the heatsink screws. It’s a bit fiddly, but it can be done (there is also a plastic insulator under the screws you can't see in the picture). The reason for this is that you do not want your heatsink to be at battery voltage, in case it shorts against anything.

37601


Now you can solder on your connections, tidy it up with some shrinkwrap and put on the XT60 connectors. For safety, use male connectors on the battery side and a female connector to go to the controller.

37602

Now you are ready to connect. Here I have two battery packs. One is at 40.6 V and the other at 38.8V.

37603

Here they are connected to the Schottky Diode Adaptor:

37604

When both packs are connected and switched on, the voltage read by the voltmeter is approx. 0.2V lower than the most charged pack (in this case the pack at 40.6V, giving 40.4 V). When only one pack is switched on, or connected, the voltage is approx. 0.2 V lower than the pack voltage (ie in this case 40.4 or 38.6V).

37605

So there you have it, an adaptor cable that allows you to have two packs connected together, without having to worry about matching their state of charge, or having a live connector present when one pack is removed. Not the most elegant or efficient solution, but simple and effective. I think the electronic experts out there will be able to make a little voltage comparator circuit that switches on or off a couple of MOSFETS or similar, which would be a lot more efficient, but I think this is a nice simple fix that most people could easily put together.
 

Woosh

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excellent write-up!
 

Benjahmin

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Thanks Wheezy.
Can I ask, doesn't the electrical insulation, between the diode tab and the heat sink, also act as a thermal insulator, so somewhat defeating the purpose of the heatsink?
I'm wondering if youcould mount the diodes back to back, each side of the heatsink, and use the heatsink as the controller connection, then heatshrink the lot? Or would that hold too much heat?
 

Woosh

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for a minimum installation, you only need one diode to connect the second battery to the controller + primary battery when the output voltage of the primary battery falls below that of the secondary battery.
However, there are two issues to address: anti spark and hysteresis. The first needs to limit the Amps (eg 5A max) so not to blow the diode in the first place and reduce heat and the second to debounce the switching, you want the switch to stay stable and not oscillate. It's the improvements that make the better designs more complicated.
 

Bikes4two

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@Woosh - the hysteresis - now that is an interesting point and I'm struggling to remember my electronics, but I think the issue is something like this:

Scenario:
> no/low load current and V1 greater than V2 so D1 conducts (D2 doesn't so Pack 2 isololated)
> load is applied (you switch the motor on) and power is drawn from Pack 1 - due to voltage sag, V1 drops at some point to below V2 so D2 conducts and D1 switches off and the controller/motor is powered from Pack 2
> Pack 1, now 'isolated' sees it's V1 voltage rise, probably higher than V2 and the diodes D1/D2 switch state

and umpteen combinations of the above that see the two diodes switching backwards and forwards - or at least you do if there is no in-built hysteresis in the circuitry.

So is the switching a problem or in practice, not? (which @WheezyRider hasn't found it to be?).

----------------------------------------------------------
Anyway, that's my numpty view of things - happy to see a more educated explanation of course as my electronics is a bit ropey these days (sold my soul to IT 25 years ago - noughts and ones can be easier to understand!).
37611
 

Woosh

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for simplicity, consider using BAT2 to recharge BAT1 instead (battery charging regulator).
That simplifies the installation because connecting BAT2 output to BAT2 input is a lot simpler than a Y connector and the current is smaller too.
 
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WheezyRider

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Thanks Wheezy.
Can I ask, doesn't the electrical insulation, between the diode tab and the heat sink, also act as a thermal insulator, so somewhat defeating the purpose of the heatsink?
I'm wondering if youcould mount the diodes back to back, each side of the heatsink, and use the heatsink as the controller connection, then heatshrink the lot? Or would that hold too much heat?
The polyimide tape used for electrical isolation has quite good thermal conductivity and it is very thin. Years ago, they used to sell very thin sheets of the mineral mica for this purpose. The heatsink and polyimide film I used actually came from a defunct controller. In controllers, the tabs of the MOSFETS are often electrically connected in the devices, so they need to be insulated from the heatsink.

You could connect the diodes back to back on a heatsink and insulate the whole heatsink, but you have to make sure enough heat is able to escape through the heatshrink. It's less than 3 Watts, for a 15A system so less heat than an average LED front lamp, so it's not a huge amount of heat.
 
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WheezyRider

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@Woosh - the hysteresis - now that is an interesting point and I'm struggling to remember my electronics, but I think the issue is something like this:

Scenario:
> no/low load current and V1 greater than V2 so D1 conducts (D2 doesn't so Pack 2 isololated)
> load is applied (you switch the motor on) and power is drawn from Pack 1 - due to voltage sag, V1 drops at some point to below V2 so D2 conducts and D1 switches off and the controller/motor is powered from Pack 2
> Pack 1, now 'isolated' sees it's V1 voltage rise, probably higher than V2 and the diodes D1/D2 switch state

and umpteen combinations of the above that see the two diodes switching backwards and forwards - or at least you do if there is no in-built hysteresis in the circuitry.

So is the switching a problem or in practice, not? (which @WheezyRider hasn't found it to be?).

----------------------------------------------------------
Anyway, that's my numpty view of things - happy to see a more educated explanation of course as my electronics is a bit ropey these days (sold my soul to IT 25 years ago - noughts and ones can be easier to understand!).
View attachment 37611
I don't think there is an issue with hysteresis. The diodes do not switch on and off. They just act like valves, stopping current flowing the wrong way, and they have the potential at the cathode side which is the pack voltage minus the voltage drop of the diode.

If the packs are at different states of charge, current only flows to the controller from the pack with the highest potential.

As you say, when the pack with a higher level of charge experiences voltage sag due to load, it's potential will drop until it reaches the same potential (minus the diode voltage drop) of the other battery pack and then current flows from both batteries.

Think of water pressure in pipes with check valves, it's a similar thing.
 
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Woosh

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I mean you should implement a hysteresis circuit to prevent oscillations. The analogy with water pipes is good. You can sometime hear the water pipes going into resonance.
 
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vfr400

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Great write up, thanks.

I have a few comments:

You mentioned 12g battery wire. Personally, I'd use 14g, which is good up to around 30 amps. The wire in your photo looks thinner than 12g - more like 14g!

I would have used a bit of flattened copper pipe as the heatsink and I don't see the need for an insulator. Just screw the two diodes to it keeping the legs clear, then you can solder the output wire directly to the copper where you want.

The thing that worries me is the inrush current to the controller's capacitor/s. It's massive, and if it blows a diode, you wouldn't know, which then becomes dangerous. I think you need to add pre-charge resistors as insurance.
 
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WheezyRider

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Great write up, thanks.

I have a few comments:

You mentioned 12g battery wire. Personally, I'd use 14g, which is good up to around 30 amps. The wire in your photo looks thinner than 12g - more like 14g!

I would have used a bit of flattened copper pipe as the heatsink and I don't see the need for an insulator. Just screw the two diodes to it keeping the legs clear, then you can solder the output wire directly to the copper where you want.

The thing that worries me is the inrush current to the controller's capacitor/s. It's massive, and if it blows a diode, you wouldn't know, which then becomes dangerous. I think you need to add pre-charge resistors as insurance.
Glad you liked it. It's not perfect, lots of adaptations and improvements are possible and I hope this will stimulate discussion.

Someday I hope a proper electrical engineer will come up with a MOSFET based system that is more efficient, but for the moment it does the job.

I expect the copper pipe would be fine if it's big enough and you can stop it making contact with things it shouldn't. Might be a nice solution to insulate the outside, but still be able to get airflow up the middle of the pipe.

Maybe it's my photos, it does say "12 AWG" on the wire! Probably overkill to use such thick wire, but better too thick than too thin :). Some of my projects involve currents well in excess of 30A, so where possible I like to have cables that can cope with that.

Not had any problems with inrush current so far, they are quite heavy duty diodes and it is designed so that even one leg can more than handle the operation of the system on its own, but probably a good idea to implement something that prevents inrush issues in any setup.
 
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awol

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Thanks for the great writeup WheezyRider, I've raided my box of bits and come up with this so far.
IMAG1209.jpg
 
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Sturmey

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Below is picture of my version. The common cathode dual diodes are rated at 20 amp each side, so its adequate for the 15 amp controllers. Its fitted into a 'puncture repair kit' plastic box. The wire need to be clamped well as any movement could brake the legs off the diodes.

PS. Diode spec below. Will take surge of 400 Amps
37630
 

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WheezyRider

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Here is the datasheet for the 60A version I used:


The surge current is 350A per diode, so if you use both diodes in the component, you get 700A.

It's also interesting that the voltage drop increases with increasing current. So you might want to put several diodes in parallel to keep the individual current per diode as low as possible and keep the voltage drop as low as possible.
 
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WheezyRider

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Did a bit of digging today. This sort of thing has been looked at before for UPS systems etc, where Schottky Diodes have been replaced with MOSFETs and dedicated ICs to improve efficiency and there are various modes of operation:


The ideal IC seems to be this one, the LTC4370, for current balancing applications:


The downside is that the IC only goes up to 18V max. I'm not an electronics expert, but could something like two voltage dividers feed the two battery pack voltages into the chip at the correct level, but have the full pack voltages across the output MOSFETs?
 

WheezyRider

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Below is picture of my version. The common cathode dual diodes are rated at 20 amp each side, so its adequate for the 15 amp controllers. Its fitted into a 'puncture repair kit' plastic box. The wire need to be clamped well as any movement could brake the legs off the diodes.

PS. Diode spec below. Will take surge of 400 Amps
View attachment 37630
That's nice Sturmey. How long have you been running this set up? Ever had any connection power surge issues?
 

WheezyRider

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for simplicity, consider using BAT2 to recharge BAT1 instead (battery charging regulator).
That simplifies the installation because connecting BAT2 output to BAT2 input is a lot simpler than a Y connector and the current is smaller too.
It sounds like a great idea, but I think that might be more complicated to implement than it sounds.

You'd probably have to have a 48V pack charging a 36V pack and a regulated Buck converter to ensure constant current, then constant voltage...you would probably lose at least 5 to 10% efficiency doing that.
 

Sturmey

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That's nice Sturmey. How long have you been running this set up? Ever had any connection power surge issues?
I am running this setup for about 2 years/20,000 km and on 3 different bikes. The batteries are charged off the bike so there is a slight spark at plug in. But I never had any trouble with power surges. It simple and just works. The loss is about (point) .4 volts but the reduction in sag more than makes up for this.
 
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WheezyRider

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These look interesting, "smart" diodes:


They have a voltage drop of only 26mV!

They only go up to 30V at 15A, so you would probably have to put several in a series parallel arrangement for an e-bike, but they have the potential to be highly efficient for this sort of application. It will be interesting to see what they are like at coping with inrush currents.
 
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