-4.2v

[diono]

Hi,

Can someone shed light on what is meant by -4.2v. I have an LCD I want to connect to a pic18. Pin 15 and 16 on the LCD need to be connected to 4.2v and -4.2v source. I understand that the 5v source can be reduced to 4.2v via resistors, but how do you get -4.2v.

Thanks,
Dion

 

[OLIVE2222]

Modern LCD don't need negative supply anymore (and they are deadly cheap). However you can use a pump charge like the generic 7660 or the negative output of a RS232 translator like the max232 or 3232
Olivier

 

[Harald Kapp]

Welcome here, diono.

-4.2V means that the voltage is negative with respect to ground (0V). You can use e.g. a charge pump inverter (see e.g. here) to create a negative volateg from a positive voltage. These are fairly easy to handle.
Another method involves switch mode inverters, which I don't recommend for your level of experience as tehy can be nasty to get running smoothly.

You may be able to find such a voltage inverter in kit form.

Alternatively you can use a second 5V supply, which is isolated from the first 5V supply. Connect 5V from the second supply to gnd of the first supply (that's why they need to be isolated) and the gnd from the second supply is then at -5V. A voltage divider can reduce this to -4.2V in the same way as you would do for a reduction from +5V to +4.2V.

 

[diono]

Welcome here, diono.

-4.2V means that the voltage is negative with respect to ground (0V). You can use e.g. a charge pump inverter (see e.g. here) to create a negative volateg from a positive voltage. These are fairly easy to handle.
Another method involves switch mode inverters, which I don't recommend for your level of experience as tehy can be nasty to get running smoothly.

You may be able to find such a voltage inverter in kit form.

Alternatively you can use a second 5V supply, which is isolated from the first 5V supply. Connect 5V from the second supply to gnd of the first supply (that's why they need to be isolated) and the gnd from the second supply is then at -5V. A voltage divider can reduce this to -4.2V in the same way as you would do for a reduction from +5V to +4.2V.

Thanks for all the replies. I forgot to mention that pin 15 1nd 16 are for backlighting.

 

[Harald Kapp]

I forgot to mention that pin 15 1nd 16 are for backlighting.

Backlighting or contrast?
It is unusual to require a negative voltage for the backlight, this is typical for the contrast adjustment.

However, you need to know what current is required to drive these pins. A resistive divider is not a good choice if you need to drive any significant current into (or out from) these pins.

Maybe we can come up with some more help if you tell us the type of the LCD, better yet, upload a datasheet.