<BLOCKQUOTE><font size="1" face="Verdana, Arial">quote:</font><HR>Originally posted by Someguy:

With four cells you can run down to the nominal 0.9V per cell.

But thats irrelevent if you're only using a dropping resistor. With 4 cells, you'll have a nominal voltage 6 volts and you have to design the circuit so the dropping resistor will have voltage of 2.4 volts.

Irrespective of the voltage of the cells, the resistor will always drop the voltage proportional to the resistance of the LEDs. If the voltage of each cell dropped to 0.9v, the LED will definately not get the whole 3.6v

<HR></BLOCKQUOTE>

Actually, the current to voltage relation of the LEDs is not nice and linear, so you can't think of it as a resistance.

A better approximation of an LED is to think of it as a constant voltage drop in series with a low value resistor. The LED is then placed in series with the external dropping resistor. As the battery voltage drops, the constant voltage drop of the LED remains constant, and the voltage drop in the various resistances must fall. This happens naturally as current drops.

So when the battery voltage falls to 3.6V, the LED will see all of the 3.6V, but the current through the resistor will be zero (no light).

With a 3 cell pack, you go from full current with fresh cells at 4.5V to zero current at 1.2V per cell or 3.6V for the battery pack. With a 4 cell pack, you go from full current with fresh cells at 6V, to about half current at 1.2V per cell with a pack voltage of 4.8V to zero current 0.9V per cell.

At a cell voltage of 1.2V, with the 3 cell pack you are finished, but with the 4 cell pack you are still running at 50% current, which is probably >50% light output.

If you can switch to a lower resistance, as some folk do here, then you can get full brightness as the battery discharges.