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Add wall-wart in parallel with batteries?

  1. Jul 2, 2009 #1
    We have several Roombas in the house and we use a lot (12) of those little "invisible walls" to control where they can and can't go. They're all programmed to daily schedules, but even so, the invisible walls eat up a pair of C-cells every 60 days or so. 24 C-cells every couple of months gets expensive fast. I mean, it isn't going to leave me homeless, but it's pretty annoying to have a device sitting directly under a power outlet, chewing through $20 in batteries each year (and as luck would have it, every one of them is within a foot or so of an outlet).

    Awhile back we opened all of them up and converted them to run off 3V wall-warts. I had planned to just leave them turned on all the time -- a constant IR beam isn't going to bother anybody. Unfortunately, I didn't realize the invisible walls automatically turn themselves off after about 90 minutes if they're manually activated (instead of being programmed). Worse yet, it isn't feasible to simply program all of them using main power, since the programs would be frequently lost anyway: JEA, our electric provider, is extremely flaky. It isn't unusual to get a brief flicker once a day.

    I've been wondering -- would it be safe and feasible to wire the battery tray in parallel with the wall-wart jack? (Obviously the two batteries would be wired together in series, and that would then be wired in parallel with the AC adapter to the circuit board.) Would the wall-wart feed the circuit and more or less indefinitely extend the battery life, which would only see a large drain when the main power went off?

    Or would the lower amperage of the two-battery "cell" be the limiting factor, and the AC adapater would simply act as a second cell and match that amperage, and all I'd do is double the battery life? (Which is still nice, but not really the goal.)

    When the power fails, is it reasonable to expect that the batteries could pick up the load of supply the memory with power after losing the wall-wart "cell"?

    Thanks!
     
  2. jcsd
  3. Jul 2, 2009 #2

    vk6kro

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    Roombas are robotic vacuum cleaners?

    If you are using alkaline C cells, the makers usually say you shouldn't try to recharge them and that is what would happen if you put a wall wart across them.
    You could put a diode in series with them so they can only discharge, but that would use up some of the battery voltage. Schottky diodes use less voltage than normal silicon ones, so maybe that could be worth a try.

    Or, you could convert them to use NiMH cells and recharge them with the wall wart. These are fussy about overcharging though, so you would have to limit the charging to a trickle charge if they are to be left on all the time.
     
  4. Jul 2, 2009 #3
    Thanks. Yes, Roombas are the robotic vacs. Sorry, with several bumping around every day, I forget that not everyone may recognize the name. (And the situation is slightly worse, I had forgotten, but these are actually chewing up a pair of D cells per unit, not C cells...)

    It looks like Schottky diodes tend to drop by 0.3v to 0.5v which I would worry might be too much -- but it's easy enough to try. The programming only turns on the virtual wall LEDs for about 90 minutes per day, so that's only a 6% chance that we'd lose power during the maximum power drain...

    I also wondered about ways to keep the batteries "out of the loop" during normal powered operation, but switch to the batteries when main power was lost. (And in this case the batteries wouldn't necessarily have to be giant D-cells, but just something that could supply 3V for just a few minutes or hours, instead of months.) I'm familiar with relays at a large scale (I wired up a 12VDC-120VAC relay so our 25kW whole-house generator will automatically open exhaust vents and a giant exhaust fan when the generator auto-starts), but at this much smaller scale I'm not really familiar with how the same relay-like switching is accomplished. I would think a big old hardware relay would be too slow to keep power flowing to the memory. Is that a reasonably valid assumption or do physical relays switch much faster than I'm thinking?
     
  5. Jul 2, 2009 #4

    negitron

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    I wouldn't worry about it. Most electronics can handle that much of a voltage loss. Switching from alkalines to NiCds or NiMH will lose approximately .3 V per cell, since NiCD is 1.2 and NiMH is 1.25 V per cell, nominally, while alkaline is 1.5 V. Since most stuff that runs on alkalines can also run on rechargeables, they should also work fine with alkalines through a low-drop diode like a Schottky as the final voltage is about the same.
     
  6. Jul 2, 2009 #5
    How long a period are your electrical flickers? If only a few cycles, I'd simply put a little capacitance across the wall wart outputs. Try 100 uF 10V electrolytics, and see how it goes.
     
  7. Jul 2, 2009 #6

    negitron

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    Pfff. If you're gonna go for capacitance backup, go with one of these babies.
     
  8. Jul 3, 2009 #7
    More complicated method:
    microproccesor controlling a mosfet with the batteries.
     
  9. Jul 3, 2009 #8

    vk6kro

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    Depending on the current, maybe you could just run them all off one small lead acid battery or bank of NiMHs (or even a very large capacitor) which were on permanent trickle charge.

    You could have a series of small regulators all giving 3 volts and if the mains did vanish for a few minutes it wouldn't matter because the battery could supply charge for several hours, probably.

    It would be good for the batteries if the power was deliberately turned off for a while to let the batteries discharge a little.
     
  10. Jul 3, 2009 #9

    MATLABdude

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    With a little bit of work on perfboard and some electrical tape (or small enclosure) you could wire in a battery backup switcher IC (some also fall under the aegis of battery supervisor, or voltage supervisor), which switches to your battery backup whenever your main power source fails. Unfortunately, you now have to check your batteries every so often (however, you can probably drop down from Cs or Ds to coin cells or AAs, depending on how often you expect these to have to kick into action).

    For instance, this chip from Intersil came up when I Googled for battery backup switcher (I've never used it before and haven't looked at the datasheet):
    http://www.intersil.com/data/fn/fn3183.pdf

    Maxim has another, the MAX 690 series, but they only have something like 20 or 50 mA of supply current:
    http://www.maxim-ic.com/quick_view2.cfm/qv_pk/1334
     
  11. Jul 7, 2009 #10
    Thanks for all the suggestions! We have 12 of these, so I'm looking for the simplest possible angle, which makes the diode a great idea if it works. Not looking forward to building 12 copies of anything that involves soldering IC, or making breadboards, etc... They're distributed all over the house, so a centralized power source wouldn't work.

    The leakage through a Schottky diode wouldn't be too much juice flowing into the battery?

    Would I actually want two diodes? One to keep the power from flowing back through the transformer when the AC power goes down? Does my little Visio schematic look right?

    roombavirtualwall.jpg

    It looks like low-voltage low-amperage Schottky diodes no more than a buck from Digikey -- although I need to do some reading to figure out exactly what some of the terms mean so I can choose the right one. (I'm not 100% sure which is the forward voltage and which is the reverse... although I guess I'd want them to be the same, right?)
     
  12. Jul 7, 2009 #11

    negitron

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    That looks correct except for the reversed polarity on the Roomba unit. You shouldn't need the diode on the wall-wart supply, however, since it already incorporates one of more diodes, either in series or as a bridge; these convert the AC from the transformer into DC.
     
  13. Jul 7, 2009 #12
    Oh yeah -- sorry about that, I hit the wrong button in Visio (looking for h-flip to put the text on the right and accidentally hit v-flip then forgot to flip it back).

    Am I right in thinking that as long as the forward and reverse voltages are somewhere above my 3V requirement (there are quite a few out there in the 4V, 7V, 8V ranges), practically any Schottky diode should work?

    I'm not really sure how to estimate the amperage they need to handle -- it was cheap and easy to find transformers that put out 1.5A and I knew the unit drew less than that, so I didn't worry about guessing what it actually needs. But now I suppose I have to know. I've seen "RF" diodes that seem to have very, very tiny ratings (like 50mA) and I would assume I don't want that, but those are mostly what I'm seeing in the lower voltage ranges.

    Any suggestions?
     
  14. Jul 7, 2009 #13

    negitron

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    Do you have any sort of multimeter, particularly one with a low-range current measurement?
     
  15. Jul 7, 2009 #14
    Actually I did just buy a meter recently which measures mA. (Previously I just had one of those cheapo Radio Shack pocket deals which only did resistance and voltage.) I guess I have to put it inline if I remember correctly, yes? (Sorry for the basic questions...)
     
  16. Jul 7, 2009 #15

    vk6kro

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    You would also need to make sure the wall supply delivered more voltage than the batteries.
    This is necessary to reverse bias the diode in series with the batteries when power is available.
    Otherwise, the batteries would be delivering power all the time.

    One way to do this would be to use NiCd or NiMH batteries and put a suitable resistor (maybe 470 ohms) across the diode to trickle charge the batteries when the power is available.

    If you don't have a multimeter, they are available cheaply. I found this one:
    http://www.harborfreight.com/cpi/ctaf/Displayitem.taf?itemnumber=92020
    but there must be lots of them around.

    Just put the meter on the 10 A range initially and measure the current into your LED device. (You put it in series with one power lead to the device). If you are sure the current is a lot less than this, move to a lower range to get better accuracy.
     
    Last edited: Jul 7, 2009
  17. Jul 7, 2009 #16

    negitron

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    Right. Break the circuit at any convenient point in series with the battery and insert the meter. Remember that the reading you'll get only represents the load at the time (obviously) and that the demand can go up or down depending on what the device is doing. You'll probably want to monitor the current over a time period to get an idea of the peak draw.
     
  18. Jul 7, 2009 #17
    vk6kro, please define "put a resistor across the diode"... are you saying basically "loop around" the diode with a connection through the resistor? The diode blocks the full current, but the resistor circuit allows a small amount to bypass it? How does one determine what size resistor is appropriate for this (out of curiosity)?

    I suppose I need to see peak amperage, and I assume that would be with the virtual wall "on" -- spewing it's IR signals through its two LEDs. I can just power it up and watch it for awhile, the meter I have can hold a peak value.
     
  19. Jul 7, 2009 #18

    negitron

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    That's it. The diode allows full current to flow from the battery to the circuit under power, but the resistor allows a small charging current to flow through the battery when the AC power is available. Obviously, for this to work the AC supply voltage needs to be a few volts above the battery voltage. The resistor value will be chosen to limit the current through the batteries to a safe charging level, typically C (the battery capacity in mAH) / 20 for a trickle charge.
     
  20. Jul 7, 2009 #19
    Awesome. I think I have everything I need then.

    Hmm -- any tips on how to spec out the correct diode? Digikey had a rather vast number of search parameters for diodes...
     
  21. Jul 7, 2009 #20
    Also -- why mAH/20? Where does the 20 come from? (Again, just curious.)
     
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