Query regarding 16x2 blue lcd display.

Other Hardware Displays
1

I am using a 16x2 white on blue lcd display.Initially I could not get the contrast right as I connected pin number 3 to ground.Later I learned that this pin needs to be provided a voltage between 0.4 to. 0.9 V.I made a voltage divider circuit but by mistake provided 4.4 V instead of 0.6 V as I interchanged the 5V and ground pins on the voltage divider circuit.Will this harm my lcd display?It seems fine though.

2

"Will this harm my lcd display?"

No. If you hook up the recommended potentiometer and then turn it all the way to one end you will be supplying the full 5V to the contrast pin.

Don

3

As - of course :grinning: - floresta says.

In fact, a voltage divider is completely unnecessary. The correct component is a single resistor from pin 3 to ground - try values between 220 ohms and 1k, pick the best and use that.

Alternately, do not connect the potentiometer end to 5V, just adjust it to the best point and measure the resistance with your multimeter and substitute the nearest standard resistor value. It is not that critical.

4

I was merely answering his question and supporting my answer with an example based upon information available in the datasheet.

That's a lot different than giving him an alternative method to connect his LCD.

Don

5

floresta:
No. If you hook up the recommended potentiometer and then turn it all the way to one end you will be supplying the full 5V to the contrast pin.

Don

Thank you.

6

Paul__B:
As - of course :grinning: - floresta says.

In fact, a voltage divider is completely unnecessary. The correct component is a single resistor from pin 3 to ground - try values between 220 ohms and 1k, pick the best and use that.

Alternately, do not connect the potentiometer end to 5V, just adjust it to the best point and measure the resistance with your multimeter and substitute the nearest standard resistor value. It is not that critical.

Thank you.
Can you please explain how this method works(resistor from pin 3 to ground)?
How will this supply the required voltage?

7

The trick is that there is a string or "ladder" or resistors R1 to R5 on the LCD board, each 2k2 ("222") which run from Vo - pin 3 - to Vcc and total 11k, so when you connect a resistor from Vo to ground, this actually forms a voltage divider which puts Vo at about 0.5V so that the actual contrast voltage used for the LCD itself is then about 0.5V.