I found this thread on http://electronics.stackexchange.com/questions/94297/transistor-which-opens-circuit-reverse-transistor I think I will go for the solution from alexan_e, making a reversed circuit with two transistors. It only costs one transistor more.
I have tried changing pull-up to pull-down with uboot, but that didn't work for me. Thanks for all the help! Op vrijdag 9 mei 2014 01:34:01 UTC+2 schreef Guy Grotke: > > 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] >> 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. >> >> -- >> For more options, visit http://beagleboard.org/discuss >> --- >> You received this message because you are subscribed to the Google Groups >> "BeagleBoard" group. >> To unsubscribe from this group and stop receiving emails from it, send an >> email to [email protected]. >> For more options, visit https://groups.google.com/d/optout. >> >> -- > For more options, visit http://beagleboard.org/discuss > --- > You received this message because you are subscribed to the Google Groups > "BeagleBoard" group. > To unsubscribe from this group and stop receiving emails from it, send an > email to [email protected] <javascript:>. > For more options, visit https://groups.google.com/d/optout. > > > -- For more options, visit http://beagleboard.org/discuss --- You received this message because you are subscribed to the Google Groups "BeagleBoard" group. To unsubscribe from this group and stop receiving emails from it, send an email to [email protected]. For more options, visit https://groups.google.com/d/optout.
