need circuit change 6 to 4 volts

Discussion in 'The Projects Forum' started by mrel, Jun 29, 2011.

  1. mrel

    Thread Starter Active Member

    Jan 20, 2009
    Have a 6 volts 1.2 ah gel cell rechangeable battery also got 4 volts led lamp (cree xlamp) only need 4 volts and 350 ma.
    So howi do remove 2 volts from the gel cell to make only 4 volts?.
    Is there circuit?

  2. wayneh


    Sep 9, 2010
    Ohms law is your friend. You want to drop 2 volts across a resistor while it is conducting 0.35 amps. V = IR 2 = 0.35R R = 2/0.35 = 5.7Ω The resistor needs to be rated for 1W or more of power dissipation.

    So, putting a resistor in series with your LED will limit the current and protect the LED. I'd start with a larger value, say 10Ω, and see if it's bright enough for you. Your LED will last much longer if you don't supply the full rated current.

    You might even want to try putting two of your lights in series. If they will operate on 3v each, you'll get a lot of light more efficiently than burning off power in a resistor.
  3. SgtWookie


    Jul 17, 2007
    With high-power LEDs like that, you really need a switching current limiter using an inductor. Resistors are cheap, but you'll wind up wasting quite a bit of power as heat in the resistor. Besides, as the battery discharges, the current through the LED will decrease.

    I don't happen to have a decent buck converter design handy for you; and such a thing could be intimidating for an electronics newbie.

    Be aware that a resistor will get quite hot.

    The basic formula for the resistance is:
    Rlimit >= (Vsupply - Vf_LED_typical) / Desired_Current
    Then you also need to calculate the resistor wattage:
    RWatts >= (Vsupply - Vf_LED_typical) / Rlimit * 1.6
    The * 1.6 is for reliability. Less than that, and the resistor will get very hot.

    Your LED could get pretty warm, too. 4v * 350mA = 1.4 Watts. Most of the higher power LEDs need to be mounted on a heat sink of some sort. If you don't use a heat sink, they will overheat and burn up.
  4. NLightNMe

    New Member

    Apr 20, 2011
  5. castley


    Jul 17, 2011
    To get rid of 2 volts you need a resistor in series with the lamp. The resistor value can be calculated: R= V/I ( 2 divided by .35= 5.7 ohms ) The wattage value of the resistor should be .7 watts minimum.(use a 1 watt value for safety) A 5.7 ohm resistor would be hard to find, but a 6 ohm resistor should do the job.
  6. iONic

    AAC Fanatic!

    Nov 16, 2007
    Probably your fastest and safest path to success is a device like the BuckToot LED Driver.


    This will efficiently give you the fixed 350mA the LED needs and deal with the extra 2V.
    It operated from 5V - 24V.