I'm quite new to electronics and am doing it as a hobby. I recently bought a 16x2 LCD display online to use it with an Arduino. When I received it, it noticed that there were a few components missing on the back. Is the display still functional or do i need to buy new components? If i do, which components do I need?

The back of the display looks like the display on the top in this photo. enter image description here

  • \$\begingroup\$ it's fine mate. They can be for alternative designs using the same PCB, for different products/features. You obviously got one with "less features" or a different display that doesn't need that particular IC \$\endgroup\$
    – KyranF
    Mar 27, 2015 at 2:48
  • \$\begingroup\$ In fact it looks like it's meant to be a regulator or a logic-level converter, for the "3.3V version" shown in the picture. \$\endgroup\$
    – KyranF
    Mar 27, 2015 at 2:49
  • \$\begingroup\$ Looks like a voltage doubler on the 3V3 part. Could be needed to generate a voltage above 3V3, where on the 5V part that's not needed because 5V is enough. Also note that jumper J1 is open on the 3V3 part and connected on the 5V device. This is probably used to connect either Vcc (5V) or the generated voltage (e.g. 2x 3V3) to the appropriate sub-circuits. \$\endgroup\$
    – JimmyB
    Mar 27, 2015 at 11:16

1 Answer 1


Typical bias for the glass on those displays is ~4V below VCC, otherwise the contrast will be too low to see anything. To make the display visible at 3.3V VCC you need bias below ground. The parts present in a second picture are there to achieve that.

The contrast is also temperature-dependent. You will need negative bias for ambient temperatures below 10 deg C; "wide-temperature" displays have a bias circuit populated as well.

  • \$\begingroup\$ If possible, can you get the numbers from the IC package that is missing on one of the boards? I am curious which part they are using to generate the bias voltage. \$\endgroup\$
    – hwengmgr
    Mar 27, 2015 at 10:44
  • \$\begingroup\$ It will probably be some form of switched capacitor inverter. Fairly common - they basically use a h-bridge to charge one capacitor and then flip it round and dump the charge into the second capacitor. Looking at the traces on the PCB in the image, probably an LM2662 or similar - note that pin 1 is connected to VDD which would be fine for the LM2662. \$\endgroup\$ Mar 29, 2015 at 19:32

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