John Syn <[email protected]> wrote: > > On 2/24/14, 2:14 PM, "[email protected]" <[email protected]> wrote: > > >John Syn <[email protected]> wrote: > >> >"Steady state max. voltage at all I/O pins" > >> >"-0.5 volts to IO supply voltage +0.3 volts" > >> > > >> >So, even with power off, some voltage *is* allowed and in fact it > >> >should be fairly easy to keep the voltage within these limits using > >> >Schottky diodes for clamping. > >> Schottky diodes aren¹t going to clamp the voltage to this range. Simply > >> use the 3V3 output from the BBB to enable the supply to your board. > >> > >??? > > > >How would using the 3.3v IO output from the BBB be any different from > >clamping the input voltages to the 3.3v IO output from the BBB? It's > >the same thing surely? > I didn't say 3V3 IO, I said 3V3 output or 3V3 supply output (P9-3 or > P9-4). If you have a regulator on your 12V side, you probably have an > enable pin. Connect P9.3 to the enable pin. > > We're not talking about things with power supplies, I'm measuring battery voltage. Strangely enough most batteries don't have an 'enable pin'. :-)
I'm after the simplest way I can think of to do this. A simple resistor divider to drop the battery voltage I'm measuring down to something in the range 0 to 1.8 volts and then some way of protecting the BBB input from spikes and when it's turned off. Adding buffers (op amps presumably) with their own power supplies etc. is (to my mind) unnecessary complexity. > >One Schottky diode prevents the voltage going below 0.3 volts (it will > >conduct such that the voltage doesn't go below 0.2 volts), another can > >clamp the input to prevent it going above the 1.8 volt ADC supply > >voltage. > I'm not sure you understand how a schottky diode works. Voltage drop at > forward biases of around 1 mA is in the range 0.15 V to 0.46 V, but if you > conduct more that 1mA, your volt drop will be higher. Also, the reverse > voltage will not clamp at all. > The ones I am looking at have a forward voltage drop of around 0.2 volts at 1mA, since the series resistors on the input (after dividing the voltage down to 1.8 volts) will be 10k or so the current won't even get to 1mA - so less than 0.2 volts. There is one diode clamping to the 0v rail, that prevents the input voltage going below about -0.2 volts. There will be another diode clamping to the ADC supply voltage which will prevent the input voltage going above ADC supply voltage plus about 0.2 volts again. I started my electronics career in the 1960s, I'm not entirely inexperienced or unqualified! :-) -- Chris Green · -- 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/groups/opt_out.
