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Convert device from Alkaline batteries to DC

  1. Mar 22, 2008 #1
    I have a device that is currently running on 6 1.5V D batteries which I have to replace every 2 months as they drain very quickly. In theory would it be possible to take (say) a phone charger that puts out 9v and connect and make a circuit drawing from current from a wall socket? Obviously some rewiring would be involved -- but does it have a chance of working?
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  3. Mar 23, 2008 #2


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    Yes, just be sure to get the polarity right. You can buy the little connector bits at Radio Shack (or equivalent store) and solder up an adapter.
  4. Mar 27, 2008 #3
    How do I know how much current I need. The current application drives a small motor for a couple of seconds at a time. I am wondering if the 9V cell phone charge will provide enough current.
  5. Mar 27, 2008 #4


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    Staff: Mentor

    You should measure it with a DVM. You'll need to figure out how to make a series connection with the batteries in order to make the measurement. There's a trick I can tell you about if you aren't able to figure out a way yourself (I have to dash to a meeting).
  6. Mar 29, 2008 #5


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    Welcome to Physics Forums!! You will find there are lots of interesting topics and discussions on this site. And as you have already found, there knowledgeable and helpful people too.

    What is it that you are trying to power? (If you have brand and model, that would be useful)

    Do you know if your 6 cells are in series? Another possible configuration for your batteries is, 1st set of three cells in series = 4.5V, and 2nd set of three cells in series = 4.5V, and both sets in parallel = 4.5V. With this configuration you can draw twice the current (mA-hr) from your supply, compared with a single set.
    Last edited: Mar 29, 2008
  7. Mar 29, 2008 #6


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    "D" batteries are capable of sourcing several amperes of current, so just make sure your DC adapter is also capable of several amperes of current, and you'll be totally fine. You should be able to solder your DC adapter wires to those leaving the (empty) battery holder and the device should operate normally. Drill a little hole in the battery cover and add a little stress relief to make it look professional.

    edit: I'm assuming all six 1.5V cells are in series to deliver 9V at a maximum of ~1-2 amperes.

    - Warren
  8. Apr 7, 2008 #7
    The batteries are in series. I have a digitial voltmeter if that is what a DVM is. But it only measures voltage not current. Is it also necessary to measure the amperage under load? I have a rather crude knowledge of electrical stuff ... but that seems to be the most sensible way to go. Then I put the measuring device in series with the batteries and then put the circuit under load.

    The device that I am powering is a garbage can operated a photo-sensor. The motor raises and lowers the lid (which is only two or three ounces. The battery compartment has no specifications in it -- which is not surprising.
  9. Apr 15, 2008 #8


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    Phone charger wouldnt supply enough current and is usually onlt about 4.5V. It really depends on how much current your device uses and how the batteries are arranged. ie parallel/series. At a guess Id say series, giving you a voltage of 9V. Buy yourself an AC adaptor capable of supplying 9V. You will also need to look at the current rating on it. Usually around 100mA, 300mA 500mA so you know what youre looking for. Sometimes they're labelled in Watts or VoltageAmps, eg 10W, 12W or 10VA 12VA etc. Guessing again and from electronic experience, I'd say you need about 500mA. so thats gonna be about labelled something like 5W or 5VA. Anything higher is better. Since its your device is only operated occasionaly you will probably get away with 250mA.. Hope this helps.
  10. Apr 16, 2008 #9


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    Knowing the current under load is important to selecting a power supply.
    Since your DVM does not have a current range, you can fix this by putting a small value resistor in series with the battery pack and then measuring the voltage across the resistor.
    A good resistor value might be 0.1 ohms. You can buy one or make one out of some magnet wire.
    Warning: The resistor can get very hot.
    The current can be found by the equation E=IR.
    Or you can buy a better DVM with a 10 or 20 amp range:smile:

    For best results the power unit you select should be able to provide twice the current that you measure.
    Unless you understand you units sensor circuit the power supply needs to be regulated.
    A cheap unregulated 9v DC wall wart actually puts out a peak of about 12v which might damage the sensor (or it could work just fine).
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