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Topic: Using diodes to protect from reverse polarity (Read 2551 times) previous topic - next topic

Wawa

#15
Jul 22, 2018, 04:39 am Last Edit: Jul 22, 2018, 04:49 am by Wawa
SPDT relays are more common in the Arduino world.
Both NC conatcts to ground, both NO contacts to supply, solenoid between the two CO contacts.
Maybe easier to understand when drawn as an H-bridge.

Four diodes can be a bridge rectifier.
2*AC to solenoid, + to supply, - to ground.
Leo..

Gary_Arduino

#16
Jul 22, 2018, 07:11 pm Last Edit: Jul 22, 2018, 07:12 pm by Gary_Arduino Reason: adding image
Wawa's idea seems do-able. Although it still has me trying to squeeze yet another relay into my project box. When I initially designed this project I didn't know about Remenance. If designing a circuit to deal with it seems like such an after thought ... that's because it is.  :smiley-confuse:

Four diodes can be a bridge rectifier.
2*AC to solenoid, + to supply, - to ground.
Leo..
When you say "2*AC" what does this refer to? Not AC as in AC vs. DC ?

BTW - here is a picture of the type of Relay I have been using.

Software - Arduino 1.8.5 on a Mac 10.7.5
Hardware - Arduino Uno (ATMEGA328P processor)

"Have patience. Sometimes it's hard to believe someone could know less than you do ... but those that know more than you, are struggling with the same thought."

Wawa

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..

Wawa

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

Paul__B

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.

Wawa

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..

FredScuttle

#21
Jul 23, 2018, 04:49 am Last Edit: Jul 23, 2018, 04:53 am by 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.
Also, I used a FWB (1 part, 4 leads) instead of 4 diodes (4 parts, 8 leads), me lazy :)
Awww! Who needs an instruction manual to use a simple chain sa......

Paul__B

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.  :smiley-lol:

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.

Gary_Arduino

#23
Jul 23, 2018, 06:03 am Last Edit: Jul 23, 2018, 06:05 am by Gary_Arduino
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.
Software - Arduino 1.8.5 on a Mac 10.7.5
Hardware - Arduino Uno (ATMEGA328P processor)

"Have patience. Sometimes it's hard to believe someone could know less than you do ... but those that know more than you, are struggling with the same thought."

Paul__B

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:
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.

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.

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.

MarkT

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.
[ I will NOT respond to personal messages, I WILL delete them, use the forum please ]

Gary_Arduino

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
Software - Arduino 1.8.5 on a Mac 10.7.5
Hardware - Arduino Uno (ATMEGA328P processor)

"Have patience. Sometimes it's hard to believe someone could know less than you do ... but those that know more than you, are struggling with the same thought."

MarkT

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).
[ I will NOT respond to personal messages, I WILL delete them, use the forum please ]

Paul__B

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?

Gary_Arduino

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.
Software - Arduino 1.8.5 on a Mac 10.7.5
Hardware - Arduino Uno (ATMEGA328P processor)

"Have patience. Sometimes it's hard to believe someone could know less than you do ... but those that know more than you, are struggling with the same thought."

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