Help picking a resistor?

[Tekno_Cowboy]

My DC electronics isn't what it used to be, so I'm wondering if someone could suggest a resistor to limit the current through a Nichia 183 LED (max Vf of 3.6V) to about 500-600mA when driven by an 18500.

Any suggestions?

 

[lctorana]

1.0 ohm half watt

 

[Norm]

For future reference LED Pro
Normhttp://www.jtice.com/led_pro/led_pro.htm

 

[Tekno_Cowboy]

Thanks for the link, but I don't use Windows.

 

[Justin Case]

When you say that the max Vf is 3.6V, at what forward current is that? Is it at the 500mA-600mA max drive current that you desire, or at some other If?

All you need to know to calculate the resistor value is V=IR. You know the desired If (500mA to 600mA). Presumably you know Vf at the current of interest. And you know Vbatt, or at least an estimate of it (e.g., 4.2V). Vbatt-Vf gives you the voltage drop that you need to achieve the max current desired. Divide the voltage drop by the current to get the required resistance. Calculate the power from I*I*R.

lctorana's post that you need a 1 ohm, 1/2 watt resistor is correct if you assume that Vbatt=4.2V, Vf=3.6V@600mA, and If=600mA. Thus,

Vbatt-Vf = 4.2-3.6 = 0.6 = 0.6R and R = 1 ohm. Power is 0.6*0.6*1 = 0.36W, so 1/2 watt works out fine.

 

[Billy Berue]

Power is 0.6*0.6*1 = 0.36W, so 1/2 watt works out fine.

Isn't that going to get really hot really quickly? Why not use a 5- or 10-watt, 1 ohm resistor instead?

 

[Tekno_Cowboy]

I'll have to check the spec sheet. IIRC the If was listed for several Vf's.

 

[Justin Case]

Isn't that going to get really hot really quickly? Why not use a 5- or 10-watt, 1 ohm resistor instead?

You can use any resistor whose power dissipation capability is equal to or greater than the min requirement. Of course, it is always wise to leave some headroom so you don't operate right at the edge. 1/2 watt is already ~1.4X greater than the calculated 0.36W power dissipation. Sure you can use a big power resistor like a 10W wire wound, assuming it will fit into the host flashlight.

 

[Illum]

Vin: 4.2V
Vled: 3.6V
Qty of LEDs: 1
Iout: 550ma [500+600/2, you didn't specify...but resistors are only good for +/5% anyway]

R = V/I
R = [Vin - (Vled x Qty)]/Iout [This is used for all series wired LEDs.]
For parallel its R = (Vin - Vled)/(Iout x Qty)

Solving for resistance:
R = [Vin - Vled]/Iout
R = (4.2 - 3.6)/0.55
R = 1.09 ohms, Closest resistance is 1 ohm, although one can retrieve a 1.1, 1.2 ohm half watt resistors it isn't the most common resistor to look for.

Solving for resistor thermal dissipation:
Pd = I^2 x R
Pd = 0.55^2 x (1.09)
Pd = 0.329725, meaning the resistor will dissipate ~330mW of heat

this should compliment wattage, if calculated would yield
W = Vdrop x Iout
W = (4.2 - 3.6) x 0.55
W = 0.33, or about 1/3 of a watt, so use a resistor at at minimum can dissipate half a watt, 1/4 watt resistors might burn out in this case. It may be a good idea to heatsink the resistor as well as the LED.

I hope your cell is at least protected because a resistor regulated LEDs can be battery vampires

 

[Tekno_Cowboy]

Thanks for all the input. It's all starting to come back to me now. (Surprising the amount of knowledge you can lose when you don't use it for 5 years)

I'm using an AW protected cell, which I don't plan to let drop to the protection level.

I think if I do a similar design in the future, I'm going to use a driver to simplify things.