3
\$\begingroup\$

I have a circuit that runs on 3.3V and want to use an HD44780-based LCD with it (which normally requires 5V). The LCD logic and backlight works fine with 3.3V, but the contrast doesn't, so nothing can be seen on it.

As the circuit runs on batteries, using a 5V voltage regulator won't work well. I read somewhere that few people got the LCD to work by supplying a negative voltage (mostly 1V) to the contrast pin.

Unfortunately I didn't find any explanation why this works, and how my circuit could benefit from it. The only solution I found made use of a PWM to generate the voltage, but I don't have any micro controller pin available.

My idea would be to use an ICL7660 charge pump to generate -3.3V from my 3.3V input, and supply this directly to the contrast pin. The IC seems to be efficient enough to work on a battery-powered device.

Can you tell me if this will work or if there is a better way?

\$\endgroup\$
  • 1
    \$\begingroup\$ If you are going to use a charge pump inverter IC, you may was well just use a charge pump boost IC. \$\endgroup\$ – Passerby Jan 31 '16 at 20:31
  • 1
    \$\begingroup\$ @Passerby thank you, of course, you are right. Nevertheless, I would be interested in whether it would be possible or not, and why and what negative voltage can be supplied to the LCD. \$\endgroup\$ – muffel Jan 31 '16 at 21:17
5
\$\begingroup\$

Yup, it will definitely work. You could even throw together two quick (preferably Schottky) diodes and say... Two 100nF ceramic caps, (like the setup I made as a proof of concept for this answer) instead of an IC charge pump. Anyway, before I get too out of scope...


An LCD's Bias Voltage affects the contrast of the backlight (background) to characters (foreground), as well as the available viewing angle. Normally, a potentiometer's wiper is connected to Vo; this is the LCD's V Bias Input. The potentiometer -normally- acts as a voltage divider, between Vdd and Vss. Like I mentioned, -normally- at +5V Vdd and 0V Vss; this is important because of your datasheet!

For example, if you can find a Shenzhen datasheet floating around the web (as it's likely you have a ubiquitous Chinese LCD), you' ll find something like this:

Operating Voltage for LCD: Vdd-Vo Typical 5.0V

Now, this means that with our hypothetical 'normal' situation, we have Vdd at +5V, Vss at 0V, and our potentiometer divides that into something in between - thus providing Vo a voltage which affects our contrast and view angle!

So, how does running one of these puppies at +3V3 work? Well, I'm sure glad you asked. Remember that datasheet? It says we ideally want our operating range to equal about 5 volts, but of course there's a bit of lee-way there. Different applications require different contrast and view angles because... get this.. People's eyes are different!


Now, if we powered the LCD from +3V3 and kept our Vo referenced to a 0V Vss (Ground), we wouldn't have enough voltage bias on Vo to allow us to see. (At least not very well...)

Hypothetical:

Vdd = +3v3

Vss = 0v

Potentiometer connects Vdd on one leg to Vss on the other leg. Vo is connected to the wiper.

Our LCD's potential range is now only 3v3... Not enough to see much, if anything!


What if we use a negative voltage bias?

Vdd = +3v3

Vss = 0V (still)

Potentiometer connects Vdd on one leg to our Negative bias on the other leg. Vo is connected to the wiper.

Negative bias = -1.6V

3v3 minus -1.6 = 4.9v!

Wowee! Suddenly Vo has nearly a full 5v potential range! Now our potentiometer can properly divide this ~5v range into something our eyes can see! With light!

\$\endgroup\$

Your Answer

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

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