using a resistor

If all I want is some alternative dimming position and the led is directly driven, then would it be efficient to use resistor to achieve the dimmed light? i.e. would it waste some battery power in heating the resistor and that a driver board will be more efficient? What more is it really that much to worry about?

Hi, what you are describing works fine. The ideal way to do this, is to try to draw enough current to be happy with the brightness, while keeping the current low enough that the battery discharge voltage remains reasonably flat.

As an example of a light I did (the CR2 based light in my sig line)
- 2 x CR2 or RCR2 in series
- Driving a Lux V
- Approx 700ma in "hight"
- Approx 100ma in med
- Approx 20 - 30 ma in low

I used an S bin Vf LED (quite low for a 2 die in series LED), a 2 ohm, 10 ohm, and 30 ohm resistor from a rotary switch. All are mounted on a custom, pro made PCB.

The good - simple, it works.

The challenging - The resistors I used were 1206 size, and the 2 ohm resistor power rating requirement exceeded the wattage rating of the off the shelf resistors. I ended up having to have some special high power ones made, which wasn't that bad per resistor, but it was not that easy to find a supplier and the minimum number kicks in.

If you use a larger physical size resistor, it is a non issue.

The same math works for other configurations like 123 or 18650 size cells.

I still have parts if you want to buy some.

Good luck

Harry

Thanks HarryN for your reply, I am sure that it works fine but is it efficient? Let me put it another way, if I adjusted it so that I have the same brightness (say on half maximum) on 2 identical torches but one using a driver board and one using a resistor, which one will draw more mA? Furthermore how much different?
For example if the torch using the driver board draws 700mA and the resistor version draws 720mA then I'd say that it is not important, technically the driver board is more efficient but the difference is not great for, after all, in real life it come to burn time and the difference there could amount to just say an extra 5 minutes over a period of say 2 hours

The answer is slightly more complicated, but let's run through an example to get close to your numbers.

- 1 LED, Lumileds K2 or Cree P4 type LED
- Assume a nominal Vf of 3.3 volts at 700ma for either (actually not exactly dim)
- Probably 100 ma is more of a dim mode and then the Vf would be more like 3.0 volts.

Power to run the LED
= 3.3 volts x 0.7 amps = 2 watts (approx) or
= 3.0 volts at 100ma = 0.3 watts

To run for 2 hours, you will need a cell that can deliver 2 watts for 2 hours

2 watts x 2 hours = 2 watt hours.

For a R 123, they have about 3.7volts @ 0.7 amp hours

3.7 volts x 0.7 amp hours = about 2 watt hours, so a pretty reasonable choice.

The "gain" from using a driver is that it can take in "power", which is amps x volts, and output it as "power" - effectively transforming any "excess voltage" into current. The "excess voltage" is the difference between V battery and the LED Vf.

The loss from a driver, is that it is not perfectly efficient, so there is routinely 15 % lost in many constant current drivers. In addition, with single cell setups, the battery voltage will often swing from being "above" the LED Vf to being "below" the LED Vf. This is a real pain, as it means the driver must be a "buck / boost" type to deal with it, and they are harder to design efficiently.

Interestingly, most resistored flashlights will turn on when the cell is too low to run a driver. Why - because the driver is trying to run at full spec, and the resistored LED just drops to a lower current draw naturally.

How much is an extra 10 % of run time worth to you? More than carrying an extra cell? If so, then use a driver (buck boost).

I hope that helps.

Harry

PS - Feel free to post an example configuration and use the above info to run your own numbers. We will help you with the corrections / details.

Thanks for your comprehensive explanation, I do not have such detailed idea, even the 2h runtime is a simple quotation, but from your answer I gathered the following matters:

1 - A driver will keep a more consistent brightness output, but may stop working abruptly at the end (as it buck and boost until it can no longer do so)
2 - Resistor will make the light dimmer and dimmer as the power sag (although this is not that great given the discharge characteristic of LiON battery being more linear).

But overall, even if the driver circuit has some losses, it is still a bit (10%?) more efficient than resistor, but resistor is simpler and therefore work to the last amount of power (bad for the battery though).

The issue now is that 3x high power resistor is just as big as a driver circuit.

Give and take and none is really that much better than the other.

I add one more advantage though, through a driver board we could have a situation where a board failure can cause the whole light failure whereas resistor are more reliable (theorectical), is that true?

Power resistors can be large, but they do not have to be.

I had some custom made for my CR2 light (in my sig). 2 ohms and 10 ohms, but in 1206 size. (surface mount). Way more watts rating than you will ever need. Normal 1206 resistors are rated for about 0.12 watts, these can take MUCH more.

They will fit on the leg of most any power LED.

Here is an example of their size.

http://www.curiousinventor.com/guides/Surface_Mount_Soldering/Resistor

I probably can spare a few if you want some.

Harry

Wow, interesting these 1206 resistors that can take the power, let me crystalyse my battery and led config and will ask you for the correct size.
Thanks