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]
Sent: Friday, April 18, 2014 12:11 PM
To: [email protected]
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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