I'd like to add one more LED (1->2) but don't know how to keep the same current
For a supply voltage of 4.5V, what you're asking is impossible. Apparently, if you run 86uA through your LED, it has a forward voltage of 2.3V. (Forward voltage increases as the current through the LED increases.) So for two LEDs in series you'd need at least 2*2.3V=4.6V

On a side note: Applying excactly 4.6V to your two LEDs in series without a resistor is not a good idea. LEDs need a series resistor, or something else to regulate the current through the LED. LEDs work on constant current, not on constant voltage.

So two LEDs in series need 2*2.3V=4.6V. At 6V supply, that leaves 6V-4.6V=1.4V across the resistor.

U=I*R so R=U/I &gt; R=1.4V/86uA=16.279kOhm


You can't get a resistor with that excact value, so get 16kOhm. That will give you a current of 1.4V/16kOhm=87.5uA

Now looking at the datasheet of your LED, you should realise that you're not even close to using the LEDs at the intended current and brightness. The LED is intended for a current somewhere around 350mA, where their light intensity is quite high. They are called Xlamp for a reason... :-) You might get better results with an LED intended for lower current and brightness. When operated near the recommended or intended current, you will probably get more consistent light output.

Also, as others have commented already, brightness might be different for the two LEDs, even though the current is the same. Especially at these low brightness levels, a small difference in LEDs will be noticeable very quickly.







simulate this circuit &ndash; Schematic created using CircuitLab