I think you have to change it in uboot, but that is beyond my Linux
expertise. Look at your original question's other replies. I think
somebody explained how to do that. The register programming info you
need is all in the am335x technical reference manual, but I have only
manipulated the I/O pins in user space.
On 5/8/2014 1:29 AM, r van dam wrote:
@Guy I have been away for a bit thus my late reaction.
Can you give me a hint how to change the pullup to pulldown at bootup?
Op maandag 21 april 2014 21:02:52 UTC+2 schreef Guy Grotke:
I would not fight the enabled pullup with my own pulldown: Either
change
your control program and circuit to take high as inactive, or
change the
boot software to program that GPIO with no pull resistor (so you
can add
your own external pulldown) or program that GPIO with the internal
pulldown
enabled.
Fighting the internal pullup with a higher-current pulldown is
just asking
for trouble.
-----Original Message-----
From: [email protected] <javascript:>
Sent: Friday, April 18, 2014 12:11 PM
To: [email protected] <javascript:>
Subject: [beagleboard] Re: Change default state of GPIO pin
If there is a pullup then your pulldown will have to be several times
stronger to make sure that the floating value becomes a logic low.
You now
have an effective voltage divider with a pullup / pulldown
configuration.
Fighting against the configured on-chip pullup is going to mean
that to
output a high you're going to need many times the drive current
you would
normally need as you sink current into that low-value pulldown
resistor.
Not sure what your threshold on the buzzer is but if the pullup is
say 30 to
50K then to get a solid 10% default low on the pin you'd need a 3
to 5K
resistor on the pulldown. That would be a 1.1 to 0.6mA load on
the pin
when it swings high. You're also burning 0.1mA when the pin
floats since
the voltage divider will always be present. That may or may not
impact
your design.
Assuming I'm thinking of this correctly.
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