Gary_Arduino:
The last time I ran a powered up test and had my meter hooked in-line ... seems like the magnet was pulling maybe .5 amps.What I do know is that it was like .1amp over what the USB output of my computer was rated for so that was the point I decided that the magnet needed it's own power supply.
Well, you did say:
Gary_Arduino:
I am powering the magnet with 5volts stepped up to 12v through a small (eBay special) amplification circuit.
So if you multiply the voltage by 2½ times, then the input current to your boost regulator will be multiplied by at least three times, so a magnet current of .5 Amp will require 1.5 Amps at 5 V.
Just measure the magnet resistance with the Ohms range of your digital multimeter. Subtract the resistance of the multimeter probes measured when you hold them (firmly) together.
Gary_Arduino:
I will also add that, while using that same power supply to provide the reverse power spike does simplify things somewhat (one circuit to rule them all) ... 12v is WAY overkill for what is required to eliminate the Remanence.
Clearly. So you will need at least one resistor and for the circuit with the DPST relay that will be the total of the two "crossover" resistors. If you control the "shutdown" input of the boost converter, you can avoid it consuming power once the "release" pulse is delivered. Now that I think of it, an "H-bridge" may actually be arranged very simply in this situation as the boost converter itself would form half of the H-bridge and avoid using a relay.
Gary_Arduino:
I really am curious as to how little it will take. We already know that .7volts from a AA battery was more than sufficient. I may spend my next bit of bench time, setting up and testing that.
Well, an AA (Alkaline) battery gives 1.5 V, not 0.7 V. But if 0.7 V is sufficient (and you do need to double check that measurement), then that is also the voltage across a power diode if you specify the "crossover" resistors accordingly.