current drop

Discussion in 'The Projects Forum' started by jetmldt, Dec 15, 2011.

  1. jetmldt

    Thread Starter New Member

    Dec 15, 2011
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    Please help me...... . I want to obtain from a battery charger rating 48v/9a to 48v/3a. How am i drop output current from 9amps to 3 amps ? U MAY MAIL ME IN <SNIP>
     
    Last edited by a moderator: Dec 15, 2011
  2. MrChips

    Moderator

    Oct 2, 2009
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    The current rating is the maximum current that the charger can supply. The actual current drawn will vary depending on the load applied, i.e. the current is automatically reduced or increased depending on the load. You can limit the current yourself by putting a resistor in series with the load. Use Ohm's Law to calculate the resistance for a max current.

    For example, if V = 48V and you want to limit the current to 3A,
    the series R = V/I = 48/3 = 16 ohms.

    You then subtract the load resistance from 16 ohms to give you your desired series resistor.

    You also have to determine the wattage that the resistor has to handle. P = V*I = 48*3 = 144W
    This would be physically a VERY LARGE resistor.
     
    Last edited: Dec 15, 2011
  3. crutschow

    Expert

    Mar 14, 2008
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    If you are charging a battery then a small resistor in series will reduce the maximum charging current. The needed resistor value depends upon the open circuit voltage of the charger and its internal resistance. You may have to experimentally determine the required value. I would guess you need in the neighborhood of a few tenths of an ohm. You could try varying the number of 1 ohm resistors in parallel to find the right value.
     
  4. MrChips

    Moderator

    Oct 2, 2009
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    The op did not say that he was using the battery charger to actually charge a battery. We really need to know what he is trying to do. Again, the importance of proper communication.
     
  5. crutschow

    Expert

    Mar 14, 2008
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    That's why I said "If".;)
     
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