It is much more likely that there is a digital signal i.e. If 0V = "door closed", there is no advantage to them choosing such a small change of 0.3V to represent "door open", when a higher voltage (with its greater noise immunity) could be chosen instead. IMHO it is very unlikely that this is the full or the correct story, as there are much better options for a signal which must travel, reliably, perhaps over long wires, back to the control panel.Ī door being open vs. a door is open, a motion detector detects motion) the corresponding pin goes from 0V to about 0.3V. I have snooped on the control box with a voltmeter and it looks like when a sensor is tripped (i.e. Another idea I've considered is to implement a voltage comparator with Vref at 0.15V) Simulate this circuit – Schematic created using CircuitLabĪny ideas on why my circuit is behaving like this? (Or if you have another way of solving the problem, that would help too. (Another data point: when I tested with V1 set to 3.3V but everything else the same, the Output stayed at a constant 2.1V)
However, when I implement the circuit on a breadboard the output stays at a constant 3.9V, regardless of if I implement the circuit as designed or disconnect the V0 input. When I simulate the circuit, it works exactly as I intend. Using the op amp as a non-inverting amplifier, and choosing my resistors as shown in the schematic below, I should see a gain of 11, which should give Output = V0 * 11. My idea is to use an op amp (I have a couple of LM2902N's lying around) to multiply the voltage. What I would like to do is monitor the pins with my RPi, so I figure that the input voltages to the RPi will need to range from 0V to 3.3V in order to work with the GPIO pins. Much like this question, I am trying to interface a Raspberry Pi 3 Model B to an alarm system, but this time is is a GE NetworX NX-8V2.
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