Electrical – Understanding PIC32 digital output with open-drain

open-drainpic

I'm trying to understand how to operate digital pins configured as open-drain outputs on a pic32. As I've read on other forums, setting TRISx to 0 and ODCx to 1 configures the pin as open-drain output, and writing 1 to LATx will drive the output high-Z (w/pullup resistor) and writing 0 will drive it low.

Source: http://www.microchip.com/forums/FindPost/789946

If you have an actual open drain output, you can keep TRIS low and control the transistor with LAT. LAT=0 => transistor closed to ground and the output will be low, LAT=1 => transistor open and output will be floating or clamped high with a resistor (or pulled low by another signal).

What I'm caught up on is this part of the diagram for the pins:

From the PIC32 IO Port Reference Manual

What does ODCx do if TRISx is set 0? The output of the multiplexer controlled by ODCx will be low no matter what, as its either TRISx or (TRISx AND 'LATx), which will always be 0 when TRISx is 0.

Best Answer

p12-3 from I/O Ports under 12.2.4 Registers for Open-Drain Configuration (ODCx):

n. If the ODCx bit for an I/O pin is a ‘1’, the pin acts as an open-drain output.

and:

The ODCx register setting takes effect in all the I/O modes, allowing the output to behave as an open-drain even if a peripheral is controlling the pin. Although the user could achieve the same effect by manipulating the corresponding LATx and TRISx bits, this procedure will not allow the peripheral to operate in Open-Drain mode

To enable Open-Drain mode, you have to use ODCx.

p12-3 under 12.2:1 Registers for Configuring Tri-state Functions (TRISx)

If data direction bit is ‘0’, the corresponding I/O port pin is an output

and under 12.2.3 Registers for Configuring Latch Functions (LATx)

A write to a LATx register latches data to corresponding port I/O pins. Those I/O port pin(s) configured as outputs are updated.

So ODCx = 1, TRISx = 0. Then output a 0 to LATx and output is driven low. A 1 and output is open-drain with pull-up to a max of \$V_{IH}\$.

Which agrees with your thoughts, web. But it is always better to go to the data sheet.

Related Solutions

Electronic – PIC18F2550 I2C Open Drain

You can do open drain on any digital pin on the PIC yourself, it's easy! I'll show through a quick chunk of example code using RB0.

_TRISB0 = 1; // Set the pin to high impedance
_LATB0 = 0;  // Set the output low (this would be _PORTB0 on some pics)
// As long as you're using pin RB0 as a open-drain, don't touch _LATB0

// To output a "low" (drain) do this:
_TRISB0 = 0; // Set the pin to output, it's already low, so it will "drain"

// To output a "high" (open) do this:
_TRISB0 = 1; // Set the pin to high impedance, the pullup resistor will pull high

The I2C module will do something similar to this internally, without you doing anything. In other words: yes! You'll need a pullup resistor!

EDIT: To answer your second question which I did not directly address, the pin does not need a pullup resistor during normal digital output operation. My example was to show you how a normal, fully functioning digital I/O pin could operate as a open-drain without any additional behavior or behaving unexpectedly. I should also note that I have done this exactly what I showed here to do open-drain in real programs, so this isn't a toy example, it's really how you do it!

Electronic – Using Open drain to get +5V

An open drain pin can only sink current, so you are right about the pullup resistor, but a transistor is not necessary. Also Vss is the ground pin which should be at 0V:

4052 Pinout

Vdd is the positive supply pin, and Vee is the negative supply pin (as mentioned in the 4502 datasheet Vee can be connected to Vss/circuit ground if you don't have a negative rail)

Internally, the open drain pin looks like the part in the dotted rectangle:

Open Drain Example

Simulation - notice how the MOSFET is sinking the current through the resistor to pull the pin to 0V intially, then when the MOSFET shuts off at 100ms, the resistor pulls the line back up to 5V (in this case - could be +3.3V, or another supply voltage within specs of the ICs involved)

Open Drain Example Simulation

So all you need is the pullup resistor to the 4052 positive supply rail. This can be anywhere between, say, 2.2kΩ and 100kΩ. Generally for something that only switches slowly and speed of switching is not important, something like 10kΩ-22kΩ would do fine.
The higher the resistance, the more susceptible to the line being affected by noise, and the slower the line switches, so if the PCB is in a noisy environment (or there is a lot of high current/high speed switching on the PCB itself) you may want to choose a lower value (the downside is increased power consumption when the line is pulled low)

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Related Solutions

Electronic – PIC18F2550 I2C Open Drain

Electronic – Using Open drain to get +5V

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