I am following this Adafruit tutorial on 16x4 LCD, and I managed to make it work. https://learn.adafruit.com/drive-a-16x2-lcd-directly-with-a-raspberry-pi/wiring

From this tutorial, the potentiometer for the LCD contrast (3rd pin) is also connected to the 5V? Why should it be like that? it still worked the same when I disconnect it. I also tried directly connecting it to GND with 10K resistor, and it worked well.

It's really confusing for me as I just starting out playing with electronics.

Edit: I followed the exact same wiring as this enter image description here

2 Answers 2


Both way you described (either using a potentiometer as a voltage divider, or connecting a resistor between the contrast control pin (Vo) and ground) are valid configuration. The contrast of the LCD is determined by the voltage (VLCD) between the Vcc and Vo. When terminating the Vo with a resistor to the ground, the internal impedance between Vcc and Vo formed a voltage divider with the external resistor.

VLCD for contract control of 1602 LCD dispaly


Your question is not really answerable without details of the internal circuitry of the display.

Assuming you are using the actual device you linked just wire it as described.

Connecting the potentiometer as described is applying a variable voltage, between 5V and 0V to the contrast pin.

The display almost certainly has some active circuitry controlling the contrast, and presumably would work at full contrast without an input.

PS I haven't used the device listed, but those I use have a "piggy back" board with a I²C interface (which only requires 2 pins on the Pi) and includes the contrast potentiometer on the board.

  • Thanks, I just got that LCD, much more convenient with less cables.
    – Gooch
    Mar 5, 2018 at 9:43

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