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Replicating exact battery voltage and current for devices

Discussion in 'Power Electronics' started by NuLED, Jan 29, 2012.

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  1. NuLED


    Jan 7, 2012
    Hello everyone,

    I want to be able to power up electronic devices even when I don't have the right size batteries.

    Using a DMM, if I test the battery voltage (just the batteries themselves in series, such as 3 x 1.5v watch batteries) and then test the current when the device is turned on (DMM in series with the batteries and device connectors) then it is OK to duplicate a power source (using resistors) that provides the same voltage/current to turn on the device, right? (Such as AA batteries in series, or a solar panel, or even 9v battery, etc. and jumper wires clipped onto the battery connectors of the device)

    Anything I should be aware of?
    Last edited: Jan 29, 2012
  2. jackorocko


    Apr 4, 2010
    Best way to work is with a voltage regulator, cheap easy to use and few external components. Common ones work up to around 1A which should suffice for anything running on batteries. A device will only sink as much current as it needs, having more is always better. :)
  3. Resqueline


    Jul 31, 2009
    Well, generally yes, but in the case of chinese-made lasers & flashlights with watch batteries they rely only on the internal resistance in the batteries for current limiting.
    In those cases you'll have to measure the voltage under load as well as the current to get an idea about the load characteristics.
  4. BobK


    Jan 5, 2010
    Electronic devices do not generally draw a constant current. Think of an audio amplifier. If you turn up the volume, it is going to draw more current. In fact, it is going to draw a variable amount of current over each cycle of the output signal. So you cannot reduce the voltage to it by just using a resitor in series with a higher voltage.

  5. cjdelphi


    Oct 26, 2011

    You seem to be a little confused, I've seen TV programs where they hook up clamps from a 12v car battery to the nipples of a guy as torture and shows him being electrocuted lol, now think about a 9v battery, it's not far off 12volts, yet a 9v battery is 100x less heavy 20x smaller so what's going on?...

    Bigger the battery (generally) means that it can supply more current (not voltage) you could have a battery the size of a small truck but only deliver 1.5volts or you could have a battery the size of your hand and deliver 50 volts....

    a 9v battery if you rip apart one, you'll find a series of 1.5v batteries which makes that 9v, and to answer your question.,.....

    1. What's the voltage required of device.
    2. The polarity (this can be important if there's no protective diode)
    3. The battery Capacity

    how much you want to spend? Linear Regulator like the LM317 does a good job, but it will dissipate a lot of heat if the battery you use is significantly higher than what the device requires..... eg 12v > 5v if you have no choice, you could use a switching DC - DC regulator but it's a lot more expensive..

    The other choice would be to use a battery of 6volts or 8volts and use a linear regulator, this will work... ir all depends on how much current the device needs and what you have to work with.... eg, LM317T will do a good job of converting 12v down to 5v providing the devices uses less than 100ma of current... etc etc, or you could buy a nice heatsink and don't worry about the power wastage...
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