1
\$\begingroup\$

I have a LCD i want to use for a project but still can't get it working on a breadboard.

I have used LCD's before so I'm familiar with them, this one however has something i haven't seen before.

From my experience pin 15 is usually LED+, in the datasheet of this LCD it says that pin 15 is Vee and is a negative voltage output. enter image description here

I read somewhere that Vee controls the contrast of the LCD, but that would be strange because according to this datasheet the contrast is controlled with pin 3.

If i connect everything and apply power i see this only: enter image description here Anyone any ideas?

\$\endgroup\$
2

1 Answer 1

Reset to default
2
\$\begingroup\$

in the datasheet of this LCD it says that pin 15 is Vee and is a negative voltage output.

On this particular LCD module, pin 15 (Vee) is indeed a negative output voltage, which is needed in order to produce the necessary voltage on pin 3 (Vo). This is shown on datasheet page 6 - here is an extract:

extract from datasheet page 6

and again on datasheet page 10, where it shows that you should aim for 6.5V at room temperature on pin 3 (Vo) (measured with respect to Vss). Since Vdd is only 5V, you need the negative voltage from pin 15 (Vee) on the external contrast potentiometer, to achieve Vo (which they also call "VOP" and "\$\small V_\textrm{OP}\$") of 6.5V:

extract from datasheet page 10

That is the reason for the "Note" at the bottom of that table, indicating that the circuit shown (similar to that on page 6 of the datasheet) is required to provide the correct voltage on pin 3 (Vo).

Try measuring the voltage on pin 15 (Vee), with respect to pin 1 (Vss), and you should find it is a negative voltage.

Some other LCD modules require a negative voltage on the contrast potentiometer (so that Vo can be greater than Vdd) but don't provide that negative voltage themselves. Then we need to provide one e.g. from an oscillator, capacitor and diode arrangement. This LCD module removes the need for those external components, by supplying that negative voltage for use with the contrast potentiometer.

From my experience pin 15 is usually LED+

You could analyse some of the circuit on the back of the LCD module's PCB, to confirm that pin 15 does not go to the LED backlight.

\$\endgroup\$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.