Hello
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?

mrel

 

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.

 

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.

 

Using a device like this National's LM3407 (http://www.national.com/mpf/LM/LM3407.html#Overview) should be pretty straight forward and give you an efficient LED driver.

 

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.

 

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

BuckToot

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