# How to make a voltage multiplier of 1.075

I have an 80v power supply that's rated to put out 18 amps, and I'm planning on using it to charge an 80v battery pack. Unfortunately the pack's charge voltage is 86v, so what I'm wandering about is how to make a voltage multiplier that can multiply the 80v by 1.075, I was looking at making a Charge Pump but I don't know how to make anything other than a doubler, http://en.m.wikipedia.org/wiki/Charge_pump any advise welcome. Thanks

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A boost converter is probably the general term you're looking for. Although at 86V and 18A it will probably be a non-trivial design. – PeterJ Feb 3 '14 at 10:15

You already have an 80V power pack and all you need is another 6V at approaching 18A. This 6V could be produced by an isolating flyback or forward converter powered from the 80V supply. Because the output will be isolated, it can be wired in series with the 80V to make 86V.

So, 80V feeds an isolating forward/flyback converter that generates 6V dc. This output is then wired in series with the 80V to make 86V to feed your battery pack.

The problem has simplified to trying to find a 6V near-18A isolating converter.

You could of course buy a 6Vdc power supply that is fed from AC - providing this has an isolated output, it can also wire in series with the current 80Vdc output.

I had a quick look in Farnell and they have a small 6V 20A isolated brick but it runs from 36V to 75V so it's a few volts short but maybe use a buck converter feeding it that lowers the 80V to nominally 48V: -

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That's a neat idea! – PeterJ Feb 3 '14 at 10:55
Wow didn't even think of that. It certainly simplifies the issue. – Szymon Morawski Feb 3 '14 at 10:58
actually now that I think about it, wouldn't the output of the dc-dc get burned out if it puts out 6v but you run essentially 86v through it since you're running it in series? – Szymon Morawski Feb 3 '14 at 13:18
You might want to put an inverse diode across the output to keep it from going negative in case of an overload or short, but it should work just fine. – Spehro Pefhany Feb 3 '14 at 13:29
@SzymonMorawski if you had an 80V battery and a 6V battery and put them in series you get 86V - the 6V supply doesn't actually know it has been raised to 80V above ground. – Andy aka Feb 3 '14 at 15:13

It is not mentioned in your question any awareness of the safety issues associated with charging batteries. You might be able to get away with "brute force" 86V for a lead-acid battery. However, if it's NiMH, LiIon or other more sophisticated battery technology you must be very careful to follow the required charging profile and monitor temperature, voltage and current. An 80V battery pack is going to have many cells in series, and that adds another level of risk if you're not monitoring each individual cell temperature and/or voltage.

If you've already taken that into consideration, that's great. Otherwise, there is a non-trivial risk of burning down the lab... If nothing else think what that would do to your project schedule.

It is conceivable that you could use the external reference to increase the output voltage. It depends on the capability (absolute maximum rating) of the supply. But you cannot simply set a lower voltage on the feedback -- think of how an op-amp behaves open-loop -- so you will need to design and tune a control loop. There may not be enough detailed information about the power supply to do this except by trial and error -- and there again is the risk of letting the smoke out.

If you have the experience to know what you're doing, I wish you the best of success. Otherwise, please reconsider, or at least find some local hands-on help from someone who does have the experience. Good luck!

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+1 for note of caution. The description suggests a non-trivial application with serious energy levels, not worth damaging expensive kit or people to save a few $. – John U Feb 3 '14 at 23:44 I appreciate the concern, already talked with manufacture about modding supply to get 86v so that's taken care of. As for the batteries they are lithium nano phosphate batteries and have a charge voltage of about 86v but are able to charge at up to 90v. They also have a battery management system built into the pack so that during storage, charge, and use the cells are kept balanced – Szymon Morawski Mar 12 '14 at 5:08 You can try to tweak the PSU to get needed voltage. It is often easier than crafting converters, though it voids warranty. Normally, there is some tolerance in ratings of the components in power supply units. The tolerance is about 10-20%, however it might be zero or negative for cheap PSUs :)). 8% should be okay for a good PSU. Check the ratings and do it on your own risk, though. If it is a transformer power supply you can try adding few more turns to the secondary winding of the transformer and connect them in a series with the existing one. Just use thick enough wire (>2 mm2 in your case) and make sure it is connected to the right winding. The polarity matters, wrong polarity would decrease the voltage. If it is a switching power supply, you can try adjust the voltage. Normally it has a voltage divider that is used for a feedback loop. The voltage that comes from the divider is then compared to a reference (normally few volts) and used to adjust the power delivered to the transformer. If you increase the resistor in the high part of the splitter by 8-10 %, the output voltage will be about needed 86V. In switching PSUs there might be a potentiometer used for final tuning of the PSU at production. In this case, the adjustment can be made with a screw driver. If you have no clue about transformers, voltage dividers etc, then the safest way would be to get a proper PSU. Your 1.5kW PSU can be quite dangerous if you do something wrong with it... UPD. With external voltage feedback, i guess, one could do the trick. 6v Zener diode inserted into the feedback line would increase the voltage. Of course careful testing is required. PS I fully agree that if you are not 100% sure what you are doing, it is better to get a proper PSU.. Fireworks for 4k$ might be too expensive...

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Honestly I wouldn't want to mess with the inside of the PSU since A. it cost over 4000$, and B. The calibration certificate would be voided – Szymon Morawski Feb 3 '14 at 13:20 Though it does have input for external voltage reference so if I told it that it was getting 6v less than its actually putting out would it give me the extra 6v to make 86v? – Szymon Morawski Feb 3 '14 at 13:22 If the power supply cost$4000 and presumably the battery pack isn't for a child's toy either, would you be better off seeking out a "proper" battery charger for this application? I'm all for hacking stuff up, but when there's a bigger / more expensive job to be done and things have to work right, it isn't always worth the time/effort/money to bodge it. – John U Feb 3 '14 at 23:42