PIN_A4 "Open-drain" Problems...

kd5uzz · Joined: 28 May 2006 Posts: 56

I'm sorry if this is a little too low level for you guys...

I've seen mention of this problem elsewhere on the forum, and I thought I understood how to fix it. It seems I was wrong. I'm using a 16F876 (not an 'a') and I'd like to use PIN_A4 as an output. As I understand it A4 is an 'open-drain' output...what that means exactly I'm not sure. Basicly if I pull that output high I can then have the PIC drive it low. Ok, gotcha. Except it doesn't work. I'm pulling it high with a 4.7k resister and watching the output of the pin w/ my multi meter. When I run:

kender · Joined: 09 Aug 2004 Posts: 768 Location: Silicon Valley

kd5uzz · Joined: 28 May 2006 Posts: 56

The actual code does include another delay_ms(75) line at the bottom of the while, my mistake. Is there no way to turn off the LED by driving the output low (i.e. simulate the function of all other i/o pins)?
-Daniel

Ttelmah · Guest

You are hitting a 'classic' problem, known as the 'RMW' problem. This stands for 'read modify write'. When you set/clear a pin on the PIC, the current value on the port is read, the value is modified with the change you want, and then the result is written back to the port. Now your code, then does the following:

Read current value on Port A
Set the bit for A4.
*)Output this to the port.
At this point Pin A4, _starts_ rising, driven by the resistor. The time it takes to get 'high', will depend on the loads present (both resistive and capacitive).
You then carry on.
*)Read current value on Port A
Set the bit for A3.
Output this to the port.

Now the 'key' is whether pin A4, has got high enough in voltage between the two asterisks, to be _seen_ as high. If not, at the second read, the value will be read as a 'low', and it is this that will be written back to the port, and the pin will no longer be driving high!...

Now the 'read, modify, write', is all performed in one instruction of the PIC, with the read at the start of the instruction, and the write at the end. Hence there is very little time indeed, between the value being 'output' in the 'output_high(pin_A4)' instruction, and needing to be read again, for the next output instruction.
So, if you want to perform successive output operations like this, _you_ have to ensure that enough time is allowed for the pin to reach the required level, between one instruction and the next (in fact you also have to ensure that the drive is enough, so that it _will_get to the required level). Hence in some cases when driving devices that limit the voltage the pin can reach, the problem can appear at any time.
Provided 4K7, is a low enough resistor, to actually pull the output up to _4v_ (pin A4, has a 'schmitt' input buffer, so the pin must reach 0.8*Vdd, before it'll be seen as a 'high'), then all you need to do, is add a short delay between the A4 output instruction, and the A3 output. Otherwise you need to reconsider the resistor value needed...
The problem will also occur on pins wih normal drives, when a significant load is present. This is a hardware limitation of the way the PIC works.

Best Wishes