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During its first outing, I managed to burn out the (control circuitry) of two of the batteries connected to my LED jacket (see this, this and this question for context.) I thought I should have been well within spec for the total current I was drawing, so I'm now wondering if it wasn't the total, but just how quickly I was trying to draw it (the draw is quite "peaky".)

In order to mitigate this in the next iteration of my design, I was thinking I could put a few hefty (10,000µF?) capacitors across my power bus. But I understand that they will draw a large amount of current themselves while charging. Can I prevent this by putting a resistor in series with each capacitor?

What resistor values would be appropriate if the voltage across the bus is 5V and I am using four 10,000µF capacitors? Or is there a better way to limit the current drawn by the capacitors on startup?

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Robert Atkins
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  • I'll take advice on any other ways to accomplish not being so hard on the batteries. I intend to cap the power draw at 40W total and split the bus in two with one battery powering each half this time—my other theory was that the batteries' control circuitry didn't like being wired together in parallel for some reason. – Robert Atkins Oct 12 '14 at 18:25
  • batteries of mismatched voltages can certainly have problems. But a series charge resistor in series to make a linear (rather than exponential) charge on the caps is an okay idea. You can also get constant-current supercap charger ICs which can do a known limit, like, 2 amps or 4 amps or whatever you want. Texas instruments has lots of these. – KyranF Oct 12 '14 at 18:33
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    @RobertAtkins See Majenko's answer: Are you connecting things in full-parallel? Stuff with "seemingly similar" voltages may compete violently at 20A+ output capacities. In all the previous discussions we aimed our advice on one battery+one protector+one DCDC to one circuit. They are allowed to share grounds, but not positives. – Asmyldof Oct 12 '14 at 18:53

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My money is the fact that the "batteries" are in parallel.

In actual fact, what you have there aren't just batteries, but batteries with boost converters.

If you open them up you will find a LiPo battery at (around) 3.7V. That feeds a boost converter to raise the voltage to 5V.

Putting voltage regulators (of any sort) in parallel is never a good idea as they will basically fight each other for control over the target voltage, resulting in nasty things happening.

You would be better off splitting your system into separate power zones with different groups of LEDs powered off different batteries. That way they only share a common ground, not common power.

Majenko
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  • The first post is quite outdated, the system is now powered with s higher voltage pack and a buck high-power DCDC converter. However, your answer is not entirely without value. – Asmyldof Oct 12 '14 at 18:52
  • @Asmyldof It's not *yet*—the question that resulted in your awesome answer was asked as part of the planning stages for V2. This question about adding caps to smooth out the load off the batteries is for V1.1 :-) – Robert Atkins Oct 12 '14 at 18:54
  • @RobertAtkins Ah, well. Maybe add this to your new question to avoid confusion? And also: I have shutted-upped now. – Asmyldof Oct 12 '14 at 18:55
  • @majenko Given the above, what do you think of connecting both outputs from one of those Limefuels in parallel? It's doubtful they'd have separate boost circuits on each port isn't it? – Robert Atkins Oct 12 '14 at 18:57
  • @RobertAtkins That would probably be fine, yes. Note the maximum amperage (on the 4 port it's considerably less than the total of the 4 ports together), so the current limit on each port is most likely the limit of the connector, not the supply. – Majenko Oct 12 '14 at 18:59
  • Yeah, it's 4.2A per pack, so I'll set the max draw accordingly (which I can do via software). – Robert Atkins Oct 12 '14 at 19:01
  • A thought: could I put a diode in series with each positive output of the battery packs to prevent the boost circuits from "seeing" each other? – Robert Atkins Oct 14 '14 at 20:20
  • Turns out that even connecting both ports of an individual battery pack together in parallel is a no-no. I've wound the current draw back to under what a single port is spec'd to supply and it's working fine now. Getting at least 4 hours out of two battery packs split across two halves of the power bus, with grounds connected. – Robert Atkins Oct 21 '14 at 22:24
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As another small answer to the question with many potential answers, I think it might not be an altogether bad idea to add some capacitance.

Guess I hadn't really shutted-upped after all ;-).

But, that's more because the strips will be "transient-ing" like mad. Try a smaller cap, with a slight series resistance at each end of each strip.

Which values are a bit dependant on the strip length / expected peak current per strip.

The capacitor is only useful if it can win from the resistance and reluctance of the wiring in it's low series resistance.

For example, with a 10Ohm resistor you will not be using the caps much, because the wiring is quite sturdy and short. On gut feeling I'd probably go with 470uF ~ 1000uF with a normal (non-low) ESR rating on all ends and leave it at that.

Asmyldof
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  • Aha, that's the clue I was looking for—a series resistor will limit the current _out_ as well as the current _in_ won't it? So better to put smaller caps (that won't be a problem vis inrush current) at the end of the strips with no resistor. Gotcha. – Robert Atkins Oct 12 '14 at 19:05
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    You can keep the small resistor, but not have it arranged quite as you think. The resistor would be in series with the incoming power before the capacitor, then the output power would come direct from the capacitor. – Majenko Oct 12 '14 at 19:07
  • @RobertAtkins Yes, in this case the Low-ESR types might cause significant inrush currents if you get to 20 ~ 30 caps and you'll be spending extra money on a negative feature. – Asmyldof Oct 12 '14 at 19:07
  • @Majenko true, but then for sustained currents of 1A, you don't want a drop of larger than 0.2V, so that'd be 200mOhm, which still gives a 25A cold current, so a small extra resistor may not help, but adds extra parts. I'm not expecting the final system to have problems with a total load of 40mF. This'll get more complicated if it needs to be a cover-all for both now and at burning man with full gear. – Asmyldof Oct 12 '14 at 19:10
  • Sounds like is really wanted is a "soft start" circuit that would initially limit the current until the capacitors are charged, then allows full current from then on. More tricky than just a resistor, but a resistor with a large MOSFET to bypass it might be doable. – Majenko Oct 12 '14 at 19:13
  • Yeah, I was afraid of that kind of complexity @Majenkoj :-(. Dead-bugging a resistor onto some caps across the power bus is something I have the knowledge/facilities/capabilities to do easily, design and production of a soft-start circuit not so much. The original reason I went for the Limefuel batteries is I thought they'd have this current limiting built in, but it's turned out they have other problems, vis connecting in parallel. – Robert Atkins Oct 12 '14 at 19:51
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For inrush current limiting, varistors are available. At high currents, they increase the resistance limiting the current. Once capacitors are charged and current demand drops, the resistance drops. Digikey should have these. Seetharam

Seetharam
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  • I think you are confusing "varistors" (voltage-dependent resistors) with "thermistors" (temperature/current-dependent resistors). Normally, one would use a PTC (positive temperature coefficient) thermistor in this application. (But not the "latching" kind that are frequently used as "resettable fuses".) – Dave Tweed Aug 27 '15 at 13:50