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Dropping voltage (variable) of a sealed lead acid battery

  1. Jul 15, 2010 #1

    rwt

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    Hi,

    firstly apologies for what might seem like a 'basic' question, but I've been trying to work this out for a while.

    I have a sealed lead acid (SLA) battery - 6V 4AH. I would like to use this battery to power a 4V coil (for a display) - that is capable of (potentially) drawing the full 4A from the battery.

    I would like to know what value potentiometer (or more likely rheostat) I need to use in the circuit. The output of the pot should be betweeen around 3V and 4V (2V - 4V would be ideal), at the full 4A.

    The coil is for display, so I'd like to use a 'manual' device such as a rheostat / variable resistor, rather than a step-down transformer or voltage regulator circuit.

    Thank you in advance for any hints / suggestions.

    Russell.
     
  2. jcsd
  3. Jul 15, 2010 #2

    uart

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    Science Advisor

    You cant realistically achieve a constant 4V in that scenario using a voltage divider (potentiometer), it would just need to be too lower impedance.

    So realistically your options are :

    1. Voltage regulator if you really need constant voltage.

    2. Series resistance (reostat) if you want adjustable (but not load independent) voltage.

    To design this you'd need to tell us more about the load.
     
  4. Jul 15, 2010 #3

    rwt

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    Hi,

    Thank you for your reply. The load is basically an induction coil (just for demo) - throwing sparks only 2-3cms. The issue is that I have several 6V and 12V SLA's - good for the "big" coils, but some of the smaller coils only need 3V or 4V.

    The constant voltage is not critical - hence the need to use something variable. And the battery is capable of delivering 4A - and these coils can easily 'consume' all of that.

    The coils will only ever be run for a minute or less at any time, so the 4A would not be drawn for very long.

    The rheostat sounds like the best option - but I'm just not sure what resistance and wattage I should look for (although I know it has to be a wire-wound model).

    Thank you again for your reply.
     
  5. Jul 16, 2010 #4

    uart

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    Ok well you just need to measure the resistance of the coil primary and that will give you a good idea of what resistance you need. An external resistance equal to the coil resistance will give the equivalent of about 3 volts.

    BTW. A 4 AH battery doesn't mean that it can only deliver 4 Amps. For a short time it can deliver very much more than that.
     
  6. Jul 16, 2010 #5
    A couple or three of 5 to 10 amp diodes in series with the coil would do the job nicely. At 4 amps they are likely to drop more than .7 volts perhaps as much a 1.2 volts each. This is why I say 2 or three.

    You could also use diode connected power transistors from the junkbox for this.

    For short periods the semiconductors won't need heatsinking.
     
  7. Jul 16, 2010 #6
    Ball-parking with Ohms law.... If your coil draws 4A at 4V it has a resistance of 1 Ohm --it's so easy with these simple round numbers-- Drawing 4A from a 6V source, you have a resistance of 1.5 Ohms. Therefore a series resistor of .5 ohms would be indicated. It will have to dump 2V*4A == 8 watts of power into thin air.

    However the previous comments about the battery being able to supply _way_ more than 4A for a short period of time should be headed. Unless you have measured the coil current at the correct voltage I would proceed with caution as there may be smoke in your future.
     
  8. Jul 22, 2010 #7

    rwt

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    Thanks everyone for your helpful advice. I'm going to give the diode method a go over the weekend - if I can get my hands on the diodes easily enough.

    the alternative was to buy 2 x 2V SLA batteries - but at $A31 each it seems like an expensive option - especially since I've already got 6V and 12V SLA's lying around.

    Having said that, a quick search for a "suitable" rheostat, shows that I'm looking at $A60+. I didn't realise they would be that expensive.

    Thanks again,

    russell.
     
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