17:20 <zmatt> Siegurd: -5V on the input of the circuit doesn't result in a negative voltage on the input of the opamp, the R1+R2 voltage divider will make it a positive voltage

17:21 <zmatt> specifically, the opamp drives its output to whatever voltage ensures that its IN- reaches the level of its IN+, which is the fixed voltage R3/(R3+R4)*1.8V

17:31 <zmatt> which, with the resistor values given, will be about 0.755V

17:35 <zmatt> interestingly he gives gives two different values for the output range of this circuit with these resistor values, and both sets of values are wrong, even though he has the correct formula printed right in between

17:36 <zmatt> actually maybe he included the influence of the 100 kΩ resistor in the first values, he didn't include that in his formula

17:37 <zmatt> "at VIN = 5V, VOUT = -0.05V" is obviously definitely wrong, the opamp is not going to output a negative voltage since it has no negative supply

17:37 <zmatt> actually no that 100kΩ should have no effect

17:43 <Siegurd> I built a model in a simulation and it seems to work. And in a few days I will assemble this circuit on a breadboard to make sure the solution works for +-5V input

17:43 <zmatt> this input circuit looks potentially dangerous though... what happens if the beaglebone (and therefore the opamp) is powered off?

17:43 <zmatt> while the input is still driven at -5 or +5V

17:44 <Siegurd> it will damage the board pin or CPU

17:45 <zmatt> datasheet shows the opamp output-current (short-circuit) drops to zero when the power supply voltage drops below 0.7V or so

17:47 <zmatt> no output current means both beaglebone and opamp will see the full ±5V on their input (that 100 kΩ pulldown probably isn't going to do much)

17:56 <Siegurd> I think the easiest way to get around this is to control the +-5V sensor power like in USB via the EN signal from the Beaglebone