Calculating current limit resistor value

Hi
I have a 33R resistor and a 39pF cap to 0V. Note: First it is the 39pF connected the PIC I/O then to 0V. The 33R is then from the capacitor to a 3 pin header.
The 33R is connected to the TX line of a PIC. I think the PIC can sink or source upto 25mA.
If the output of the TX line was VIH ie. Typically 0.8VDD, then at 3V power to the PIC this would be 0.8x3=2.4V
To restrict the current to say 2mA
R=V/I = 2.4V / 2mA = 1.2K. Does that seem correct.
Also is it advisable to connect an internal weak pull up to one side of the 33R to VDD internally.
Thanks in advance

 

Hi
Please could you elaborate with the example above? Do you mean it will be different when the I/O lineis either teceivjng or transmittibg.

 

Ok. I will draw out the schematic for you today. Thanks for your help.

 

Hi
Here is a schematic showing only the TX line. The RX line is the same. They both go to the connector on the right hand side which is used to read data from a unit.
Do i need an internal pull up? The PIC can source or sink 25mA, i want to adjust the value of R1 to limit well below this current.
Look forward to hearing from you.
Thanks in advance.
Raj

 

I cant seem to upload the file but this is what i have:


PIC ---TX-----------------------33R resistor-----------
|
|
-------
------- 39pF
|
|
----- 0V

 

Hi Bob
Thanks. Is there away to calculate this properly? I thought it would be R= 0.8VDD/ 2mA = 1.2K. Is that wrong? Also do i need an internal weak pull up on the PIC? I do not know what the point is by having it.
Thanks in advance.

 

Hi Bob
Connecting the unit to a PC via FTDI lead, so we can download information from the units EEPROM.
Thanks.