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Limiting current draw from Lead Acid Battery

Discussion in 'Electronic Design' started by Mr Bob, Oct 23, 2005.

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  1. Mr Bob

    Mr Bob Guest

    Need a little help.

    Say I have a device that draws 90 amps to operate. It will function
    with 25 amps as well. I want to limit the current draw from a lead
    acid battery. I've read about rheostats but with this many amps I
    understand that quite a bit of heat is generated. I also read abit
    about poteniomters but am not quite sure of the difference between the
    two. Can anyone point me to a source where I can purchase what I need
    to do this?

    Thanks,
    Bob
     
  2. Kryten

    Kryten Guest

    They are both variable resistors.

    Not wishing to sound rude but if you don't know basic electrical info you
    are best off not touching this with a barge pole.

    12V batteries are not likely to kill you but when they push 90A then that is
    just over a kilowatt of power going somewhere. Enough to heat wires hot
    enough to melt insulation and flesh.
    Maybe get an engineer to rip a motor speed controller from one of those
    invalid scooters? Though make sure you ask the invalid for it first :)
     
  3. kell

    kell Guest

    Need to know nature of device...
     
  4. Fred Bloggs

    Fred Bloggs Guest

    Need a little help.
    Ho can someone do that when no one, including yourself, knows what you
    need? The only way to reduce the current draw from the battery is by
    developing a voltage to counter the battery voltage. Your device
    develops some voltage VD with 25 Amps through it, and VD may even be
    zero, so you will need to develop a counter voltage VC that makes
    BATT=VC + VD. This can be done with a resistance R so that 25Amps x R
    equals the required VC counter voltage. R has to sized to withstand a
    power dissipation of 25Amps x VC Volts which may be quite hefty. Going
    back to what you have said, the device and the battery resistance
    develop all the counter voltage at 90Amps, so that it must be
    12/90=0.133 ohms. Then at 25Amps they will develop VD=25x0.133=3.33V. So
    the counter voltage from an added series resistance must be
    BATT-3.33=12-3.33=8.66V. The required resistance is R=8.66V/25Amps=0.346
    ohms, and this must dissipate 25Amps x 8.66V=220Watts. Where you go from
    here depends on other information like is this battery being charged by
    an alternator so that the output voltage rises, how important is it to
    maintain 25Amps exactly or is this a ballpark figure, how is the device
    connected into the circuit, operating environment temperature range, etc...
     
  5. Mac

    Mac Guest

    What is the device? If it is a commercial product with a detailed
    Datasheet, maybe someone here can help you. If it is a simple device, then
    just tell us what it is.

    --Mac
     
  6. Jasen Betts

    Jasen Betts Guest

    It's real hard to say, some devices can be damaged if underpowered.

    Bye.
    Jasen
     
  7. Fred Bloggs

    Fred Bloggs Guest

    The "device" is a fictional entity in the troll's homework assignment.
    All of these losers will be washed out by December and we won't have to
    put up with their deceit then.
     
  8. Mr Bob

    Mr Bob Guest

    Well I guess this is the first time that someone has referred to me a
    Troll! lol........

    Simple question. DC circuit, cold crank amps max to 300. Small DC
    motor will draw full rated amps if I let it. If I limit amps according
    to product line sheet I can slow the revolutions of the motor down. So
    what would I need to limit volt/amps to operate the motor at various
    speeds?

    Thanks,
    Bob
     
  9. Mr Bob

    Mr Bob Guest

    Oh I almost forgot....I no longer am required to complete assignments
    by the semester's end since graduating some time ago with my MBA. I am
    very good connecting people in various serial or parallel circuits to
    achieve business objectives. However, when it comes to completing such
    circuits with electricity, I am rather at a loss to do so..... :)

    Bob
     
  10. Fred Bloggs

    Fred Bloggs Guest

    I don't know what you can use- either something incredibly primitive
    like a lump of coal with a wedge wiper, or something modern and expensive.
     
  11. Rich Grise

    Rich Grise Guest

    There is insufficient information provided with your original question
    to suggest a meaningful answer.

    Please be more specific - what are you trying to accomplish?

    Thanks,
    Rich
     
  12. Mr Bob

    Mr Bob Guest

    I have my answer. Thanks for all the help.

    The speed of a DC motor is directly proportional to the supply voltage,
    so if we reduce the supply voltage from 12 Volts to 6 Volts, the motor
    will run at half the speed. How can this be achieved when the battery
    is fixed at 12 Volts?

    The speed controller works by varying the average voltage sent to the
    motor. It could do this by simply adjusting the voltage sent to the
    motor, but this is quite inefficient to do. A better way is to switch
    the motor's supply on and off very quickly. If the switching is fast
    enough, the motor doesn't notice it, it only notices the average
    effect.
     
  13. kell

    kell Guest

    Think about the motor, not the battery. You will get more constructive
    responses if you ask how to control the motor. Like, how do you keep
    it from drawing too much current when it stalls from a heavy load, or
    maybe you just want to learn about motor control in general, whatever.
    You could drive the motor from a battery or any other very low
    impedance voltage source and you would get pretty much the same answers
    to questions about the motor, I would think.
     
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