I have an HD44780 compatible character LCD display (2 rows 16 characters a row) powered with 5v through a regulator (4.97V according to voltmeter). If I connect a 2.2K resistor between the LCD contrast pin and ground and test the LCD, I get expected results of one row of boxes and the other row blank.

The problem with the simple approach is that when the battery voltage drops too much, the LCD screen fades accordingly. This is what I don't want.

I now replaced that circuit with the following circuit because I wanted the same contrast over a larger span of voltages. so that people can see an indication on the screen when the battery is low (instead of a fading screen)

So I tried this circuit with a 1N5224 zener (I think thats the model), but anyways, I measured the voltage across the diode and it outputted 3V but then again I wonder if I chose the wrong diode.

I chose a 680 ohm resistor after the diode because the voltage is lowered and I feel choosing a higher resistor will make the display blank, but somwhow my math is messed up because with the parts chosen I still get a blank display even when using good batteries (to supply power to the regulator).

How do I adjust this circuit so that I get stable contrast on the LCD even when the power coming in is down to about 3 or 4V?

LCD contrast

  • \$\begingroup\$ Have you performed experiments to determine the desired contrast for each voltage? \$\endgroup\$ – Ignacio Vazquez-Abrams May 12 '18 at 4:05
  • \$\begingroup\$ No but I would think an equation exists? I know with a simple led and resistor its brightness (aka a weird form of contrast) is determined with ohms law i=v/r \$\endgroup\$ – Mike May 12 '18 at 16:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.