Using diodes to protect from reverse polarity

Gary_Arduino:
When you say "2*AC" what does this refer to? Not AC as in AC vs. DC ?

Referring to the terminals of a bridge rectifier.
Leo..

Could have a resistor in the 12volt line to the "reverse curent relay", for a lower reverse current.
Leo..

Wawa:
Could have a resistor in the 12volt line to the "reverse current relay", for a lower reverse current.

Could have? Must have to fulfil the objective.

But MarkT's circuit with a DPST (or DPDT) relay should actually work just fine, except that it continues to drive the reverse current.

Paul__B:
But MarkT's circuit with a DPST (or DPDT) relay should actually work just fine...

The diode limits reverse voltage across the coil to 0.7volt.
Leo..

I don't like drawing higher currents (did the OP ever say how much?) thru NC contacts, most relays I've ever replaced that were switching motor current had the NC contacts burned or welded.
Also, I used a FWB (1 part, 4 leads) instead of 4 diodes (4 parts, 8 leads), me lazy :slight_smile:

Wawa:
The diode limits reverse voltage across the coil to 0.7volt.

Which as I did point out out in reply #13, the OP had stated in reply #6 was quite sufficient to provide the counter-magnetism so should be quite effective.

But if more voltage was required, this could be arranged by putting two or more diodes in series which would equally (or arguably even more) effectively manage the "kickback" since allowing a higher voltage quenches the current more rapidly. :grinning:

FredScuttle:
I don't like drawing higher currents (did the OP ever say how much?) thru NC contacts, most relays I've ever replaced that were switching motor current had the NC contacts burned or welded.

More support to using NO contacts only, though I suspect this actually relates to breaking current through NC contacts, which does not happen in this case. 12 V was mentioned, but not the actual current.

Sorry about not posting the current. I actually don't know.
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.

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 Remenance.
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.

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.

Wawa:
The diode limits reverse voltage across the coil to 0.7volt.
Leo..

Who cares, the bulk of the voltage is across the current limiting resistors.

Paul__B:
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.

That reading of .7v was the voltage of the AA passed through a diode during an early test to make sure I understood how the diode worked and if it would pass enough current/volts through it to clear the Remenance.

One of the underlying purposes of this project is to give me hands on experience as I teach myself electronics Sadly, I learn better by doing (and frequently failing) then from JUST reading something in a book.

BTW - the resistance of the magnet is .064ohms. So subtracting the probe resistance gives me a reading of .056ohms

The completely random other approach is to stick tape on the electromagnet pole till there's enough
magnetic gap to defeat the remanence (assuming this doesn't reduce the active force too much for
the application).

Gary_Arduino:
BTW - the resistance of the magnet is .064ohms. So subtracting the probe resistance gives me a reading of .056ohms

And what current would flow if you applied 12 V to .056ohms?

Paul__B:
And what current would flow if you applied 12 V to .056ohms?

Wait wait, I think I remember THIS one!
It's I = V/R
So 12/.056 =
... that can't be right.
214 amps?!?

If it is pulling that much shouldn't all of this be on fire?

I am clearly confused.

You did the calculation correctly. 214 amps is the correct answer.

But I suspect the measurement was wrong. Might that be kilo-ohms? Decimal point moved? Most multimeters won't read micro-ohms.

More data -

I have a Drok USB Tester and hooking that in line between my power supply (A USB battery supply) and the power amplifier I see that the supply is putting out 5.08v and the amplifier is drawing .51amps.

Putting that tester between the amplifier output and the electro-magnet, I see that the amplifier is putting out 10.5V It is adjustable and apparently that was the setting I decided I needed when I first set it up a few months ago.
The magnet is drawing .21amps

I am now I am wondering if one of the Arduino's PWM pins (with a circuit to dial down the 3volt power supply) might be able to put out the small pulse I'd need to remove Remanence.

Gary_Arduino:
I am now I am wondering if one of the Arduino's PWM pins (with a circuit to dial down the 3volt power supply) might be able to put out the small pulse I'd need to remove Remanence.

And that is exactly what I was thinking.

If you use an Arduino pin to switch the boost converter on or off by its enable pin, connect it to one end of the magnet. The other end of the magnet is grounded by a diode and connected through a resistor to a second Arduino pin. You have a logic level FET, also controlled by the same (that is, second) pin which grounds the output of the boost converter.

You now have an H-bridge. When the boost converter is enabled, it powers up the magnet, losing only the voltage drop of the diode. When you shut off the boost converter, waiting for its output capacitor to discharge, you then switch on the second Arduino pin for a short time which applies 0.7 V to the diode and grounds the first end of the magnet.

PWM has absolutely nothing to do with this.

Paul__B ... Is this diagram what you were suggesting?

Gary_Arduino:
Paul__B ... Is this diagram what you were suggesting?

Errr, no! That would not do anything.
Magnet01.png

You could indeed use a relay in this configuration to switch the "hot" end of the magnet between your 12 V and ground if it were not practical to find the enable pin on the boost converter, but I figure what I have suggested is substantially more compact and - avoids the relay (current draw).

OK, you just put the diode in the wrong place. :astonished:

Ok.
I'm going to need to study this a bit.

Wow - so now I have many questions. It's embarrassing really as this has gone well beyond asking for advice and has now moved into the area of tutoring. I apologize.

Starting with a clarification. My power booster is VERY simplistic. Unlike the booster in your diagram mine only has + and - Vin and Vout. No other connections.

  1. As it is wired now, my 5 volt power supply and the power booster are entirely separated from the rest of the Arduino circuitry. They go directly to the magnet with just an in-line relay to interrupt the power flow. That relay IS controlled by the Arduino but I believe the "power" side of that relay is opto-isolated(?) from the Arduino.
    Your circuit drawing seems to have everything sharing the same ground. Would that provide a possible path for some errant power surge from the magnet to get to my Arduino?

2)What is "Logic Level" in your circuit? Is that a transistor or maybe a relay?

  1. It appears that the "release pin" is wired directly to the "ground" side of the magnet. What keeps the 12v power from traveling up that wire to the release "output pin" and frying the Arduino? I thought that is where a diode would need to go because it would only allow power to travel from the pin and not up into it (like a one way valve).