dc motorgroundvoltage-regulator
I have a circuit involving a DC motor and voltage regulator as following:
simulate this circuit – Schematic created using CircuitLab
What would I realistically wire the ground to? Would it be the negative side of the DC motor or some random hunk of metal? (Side note: am I using the voltage regulator correctly here?)
Ground is not 0V.
Ground is merely a point of reference from which all other voltages can be measured.
Note: A voltage is a potential difference between two points. When you measure 5V you are measuring a 5V difference between two points.
It is entirely up to you what point in your circuit you choose as ground. Normally it is the point of "least potential" (such as the - terminal of your battery), but not necessarily.
So your "ground" is always "ground" because that is what you have chosen it to be. The voltage at ground, with respect to ground will always be 0V, since nothing can have any difference to itself.
When you press one of your buttons (with those two resistors removed, please) the lower side of the internal resistor (I assume that is what you mean in your question) is connected to ground, thus making that point measure 0V. When the button is released there is no connection to ground, so the voltage at that point is the supply voltage minus whatever is dropped by the resistor caused by any current flowing through it (typically a very small amount for an input pin).
You're saying the ABS logic has been reversed in that one system pulls low to indicate 'ABS on' and the other pulls low to indicate 'ABS off'. Your schematic is correct.
Good practice when switching a relay from an electronic device is to add a flyback diode to catch the negative inductive kick that the relay gives on turn-off.
simulate this circuit – Schematic created using CircuitLab
Figure 1. The complete circuit with lamp and guess of likely ABS unit internal transistor switch.
Best Answer
Edit:I misread your question for the 2nd time...I was answering it in a rush. I didn't realize that the motor was being used as a generator, and I thought it was the output of the circuit. Here is a diagram with the capacitors and the correct connections. The diode should be chosen, so that the max motor output voltage minus the diode's forward voltage drop is less than the maximum input voltage of the regulator's input:
simulate this circuit – Schematic created using CircuitLab