Hello and thank you always.

I have a question about a very basic topic I think. So far in the different projects I have made, I have worked with ICs that work with 5V. (Actuators and other components sometimes used 12V or 24V). I have never used 3.3V parts. So I have the following questions:

1) Are there PICs that use 3.3 V instead of 5V?

2) Take a look at the following link of a LCD:
http://www.wvshare.com/product/3.2inch-320x240-Touch-LCD-C.htm
In it , it says: Note : this product should work with MCUs with 2.2V~3.6V IO voltage ONLY. What does this exactly mean? Surely not that this LCD pins should all be 3.3V based, since I see pin 1 goes to 5V! Perhaps that the data pins should take 3.3V as logical one??

3) If the above is true, then how can I connect it to a 5V based PIC?? I am guessing using something like the 74HC4050 as in the following (unrelated) picture:

As you can see there a 5V pic is connected to a 3.3V component. Is this correct??

Any opinion, pointer or advice greatly appreciated. I dont want to burn anything :~

 

hi KR,
Many 3.3V modules have an onboard 3.3v regulator,powered by an external 5v source, check the TFT documentation for this regulator.
As you may know, the 4050 and 4049 CMOS buffer can accept input voltages higher than their supply rail voltage, used as voltage level shifters.
E

 

use following IC you will get both 5 volt as well as 3.3 volt

 

hi KR,
Many 3.3V modules have an onboard 3.3v regulator,powered by an external 5v source, check the TFT documentation for this regulator.
As you may know, the 4050 and 4049 CMOS buffer can accept input voltages higher than their supply rail voltage, used as voltage level shifters.
E

Thanks for the reply eric.
As far as I am understanding this TFT can accept either a 5V source (pins 1 ans 2) or a 3.3V source (pins 33 34), but if you use the 5V source then pins 33 34 outputs 3.3V. SO I was thinking to connect this output to the 4050 to connect the 5V PIC to this TFT... What do you think?




On the other hand, I have never done it but... I think I can use a PIC18F4553 with 3.3V, can't I???

 

hi KR,
I would use the 3.3v from the TFT to power the 4050 level shifter.

If you power the PIC from 3.3v, you will still have to power the TFT with 5v, but the 4050 will not be required between the PIC and TFT

E

The datasheet states that the 16F876A will work at 3.3v.

 

TI was thinking to connect this output to the 4050 to connect the 5V PIC to this TFT... What do you think?

You could also use series resistors to limit the current through the input protection on the TFT.

CD4049 and 4050 are manufactured without the input clamp to the positive supply, so don't be tempted to level shift from high to low voltage with a normal CMOS inverter/buffer.

 

Just one-way you can use red leds.

And upto 20ma to drop voltage.

There are regulators of course. A 3v Lcd isn't compatible with 5v, simply put.

Most 5v pics can work from 3v.

 

Most 5v pics can work from 3v.

This is true but at lower power supply voltages the PIC will not operate at maximum oscillator/clock frequency. Take a look at the graphs in the data sheet for details on the limitations.

 

You could also use series resistors to limit the current through the input protection on the TFT.

CD4049 and 4050 are manufactured without the input clamp to the positive supply, so don't be tempted to level shift from high to low voltage with a normal CMOS inverter/buffer.

Thank you for your reply. I am sorry I didnt understand the last line and it seems really important. Could you clarify please?

 

I am sorry I didnt understand the last line and it seems really important. Could you clarify please?

The input of CD4049 only has the clamp to ground:

This is the input protection for a typical CMOS input:

If you attempted to interface logic operating at 5V with CMOS operating at 3.3V, 5V on a 3.3V input would forward biase the VDD clamp diode. Sufficiently high current can cause latch up and/or electromigration of the metal. A series resistor can be used to limit the maximum current through the diode in overdrive situations; but you want the resistor small enough to minimize the RC delay.