Thank you for the words of encouragement:) That was a long tirade against the US...
If it comes to the discharge of the batteries, how much dropout voltage do I need to take into account? If only one of the batteries discharges to the level of 1,1V and the other two not, then I have 3,5V on the input, which is exactly the shutdown level of the LE33CZ (dropout is 0,2V). I mean, maybe the one "discharged: battery still can work with the others two still "intact"? Maybe I should choose an LDO with lower output voltage (3,2V) or another one, 3,3V output and ultra low dropout (0,1V)?
I have attached a small image to show, how I wanted to manage to decrease the current from 100mA to 80 mA. My leds have a forward voltage of 3,2-3,5V and input current 20mA each. There are four of them. I have found really a simple way to reduce the current without affecting the voltage, with a FET transistor. There is a resistor built in, do I really need i? If yes, how should I calcultate the resistance? According to:
http://discuss.biohack.me/discussion/250/new-tdcs-regulator-designs/p1it is called a bias resistor:
"Basically, the current through the resistor creates a negative potential at the gate (law of ohm, V = I * R, so a current of 2 mA through a 100 ohm resistor biases the gate by -0.2 V), and if it gets lower than a certain threshold value, the gate slams shut.
I'm using BF245C myself, but any depletion-mode JFET with an appropriate gate-source threshold voltage will probably work."