[PATCH] pinctrl: document the "GPIO mode" pitfall

[laurent.pinchart@ideasonboard.com (Laurent Pinchart)]

Hi Linus,

On Thursday 25 April 2013 23:39:18 Linus Walleij wrote:
> On Tue, Apr 23, 2013 at 3:33 PM, Laurent Pinchart wrote:
> >> +And your machine configuration may look like this:
> >> +--------------------------------------------------
> >> +
> >> +static unsigned long uart_default_mode[] = {
> >> +    PIN_CONF_PACKED(PIN_CONFIG_DRIVE_PUSH_PULL, 0),
> >> +};
> >> +
> >> +static unsigned long uart_sleep_mode[] = {
> >> +    PIN_CONF_PACKED(PIN_CONFIG_OUTPUT, 0),
> >> +};
> > 
> > I'm having a bit of trouble with PIN_CONFIG_DRIVE_PUSH_PULL and
> > PIN_CONFIG_OUTPUT. Strictly speaking, when configured in output mode, the
> > pin will be in a push-pull configuration.
> 
> For your system or for any system? Open drain, open source are also
> output modes, and none of them are push-pull.

Indeed. I was actually thinking about the opposite, push-pull is output.

> > Could you clarify the exact scope of the two configuration parameters ?
> 
> PIN_CONFIG_OUTPUT is left a bit unspecified, but here the idea was a passive
> drive, like just connecting the pin to VDD or GND without any driver stage
> at all.

Isn't that a driver stage ? :-)

> Maybe I should patch the documentation since we seem to be the only user?
> 
> In the above case (which is derived from the ABx500) I think what is
> happening is that the pin is connected to ground during sleep, without any
> enabled driver stages, which saves a lot of power, since you do not need to
> bias the totempole during sleep in that way.

Right. What is unclear to me is the interaction between OUTPUT and DRIVE_*. 
That's the part I would like to see clarified. Does DRIVE_* imply that the pin 
is driven by the selected function, and OUTPUT imply that the pin is driven to 
a fixed level ? If so, how do you configure the drive type of a pin that will 
be used through the GPIO API ? What about cases where I want to drive the pin 
to a fixed level in a non low-power output mode (for instance because I need 
more current that what the low-power output mode provides) ?

-- 
Regards,

Laurent Pinchart