hazes

I overlooked that completely! Thanks very much! The right calculation is: (5-3.5)/0,06=25 Thanks again!

Thanks for replying Wilba. Just to make sure I soldered a pot between B+ and the anode pin for the backlight on the lcd. When powered on, I slowly decreased te resistance on the pot until my volt-meter indicated 4,2V, which is the voltage needed to power the backlight according to the data-sheet (measured between Vss and LCD anode pin). I then measured the resistance on the pot and it was about 12 Ohm. That means that a 13 Ohm resistor (which I calculated) should be okay, shouldn't it?

I think I just answered my own question :) (5-4,2)/0,06=13,3 I'll just buy a 13 Ohm resistor and solder it in the right place. OT: I just saw the backlight on for the first time. Awesome!

one little question :) The data-sheet of the LCD tells me the led needs 4,2V on it's anonde. The Core-module delivers +5V on the B+ connection. Can I just connect the two of them, or will I damage the LCD because of the higher voltage? If so, what can I do to drop the voltage by 0.8V? Put in a series-resistor of ?-Ohm?

thanks! I think I'll go for the blue ones :)

Hi, I searched the forum and the wiki, but since i'm a complete beginner at this, i'd really appreciate some expert-advice on my choice of display :). I found the following display, but i'm not quite certain wether it's ok. http://www2.produktinfo.conrad.com/datenblaetter/175000-199999/181659-da-01-en-LCD-MOD_STN_BLAU_NEG_LED_WEISS_40X2.pdf Thanks for your reply and i'm really sorry if this question is asked before.