Resistor choice for cell increase

[degarb]

Simple electronic question for anyone that knows the basics. Not I, since this isn't my job. And this knowledge just doesn't stick in my brain, no matter how I study it.

I wish to modify a 3 cell (3x1.2 rechargeable) to a 4 cell. The emitter is an xr-c cree, which has a maximum of 500 milliamp max. So what ohm resistor should I choose so as not to exceed 500 ma? I assume is is being run now at 300 to 350 ma by the manufacturer.

So far, my best guess is based on formula: R = (Vbattery - Vled) / (Iled),

where R is the resistance in ohms, Vbattery is the voltage of your battery or battery pack, Vled is the LED's "forward voltage", and Iled is the "forward current" of the LED.

so, R = 1.2 volt change/ desired current drop, which I don't know.
So if I wish a .6 amp drop, R= 1.2/.6 = 2 ohm ?

Is any of this correct, and without testing the led with no resistor, how do I find out how much drop I need? (I could try progressively smaller resistors, 4 ohm then 3 then 2 then 1.)

 

[Mr Happy]

Just plug in your numbers:

Vbattery = 4 x 1.2 = 4.8 V.
Vled = 3.5 V.
Iled = 350 mA.

R = (4.8 - 3.5) / 0.350 = 3.7 ohms

Your power rating on the resistor would be:

P = 0.35^2 * 3.7 = 0.45 W,

so you would need a half watt resistor at least.

 

[degarb]

Just plug in your numbers:

Vbattery = 4 x 1.2 = 4.8 V.
Vled = 3.5 V.
Iled = 350 mA.

R = (4.8 - 3.5) / 0.350 = 3.7 ohms

Your power rating on the resistor would be:

P = 0.35^2 * 3.7 = 0.45 W,

so you would need a half watt resistor at least.

Thanks. Just to be sure: are you considering the resistor already built in to the flashlight onto which i am adding a 1.2 volt cell, or are your numbers for building a light from scratch?

 

[Mr Happy]

No, that's as if building a light from scratch. If there is already a resistor in the circuit you would have to subtract its value from the calculated value above.

 

[Mr Happy]

Now if you are simply adding one more cell to a light that is already designed and working with three cells, then you can calculate the additional resistance like so:

Additional battery voltage = 1 x 1.2 V = 1.2 V.
LED current = 0.350 A.

We want to drop the extra 1.2 V over our additional resistor.

So V = IR, or 1.2 V = 0.350 A x R

Therefore R = 1.2 / 0.350 = 3.4 ohms.

[SIZE=-2]Note that you wouldn't get much benefit from this modification except about 30% more run time from the extra battery.

Edit: Oops, Alan_P is right of course. You won't get more runtime because the current remains the same. About the only advantage is more voltage available to drive the LED at higher currents.
[/SIZE]

 

[degarb]

Just to check one other assumption: xr-c can be ran at .5 amp? Is this true? And then instead of .350, we should use .5?


And to check, I believe XR-E and Rebel 100 can be done to 1000 ma if heat sinked, or for short periods. (though I run my rebel 100 at .4 amp for hours on end.)

 

[baterija]

Here's a useful calculator that might help. You just fill in Vf, Voltage supplied, and desired current and it calculates the proper resistance. Still useful to get the theory but this is simple.

 

[Alan_P]

Adding a third battery will not increase runtime. You will still be discharging each battery at 500 ma. The additional power will be dissapated in the resistor and not in the LED. Four battery flashlights have higher runtimes than three battery lights because they have different regulators to convert voltage to current.

Alan