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Newbie question about step down voltage

  1. Dec 24, 2014 #1
    I need to step down 12vac to 9vac.
    The details.
    I am making colloidal silver.
    I have a 12vac 2amp timer that switches polarity between the two silver rods every 2 mins.
    12v to the silver rods is a bit much. 9v would make a better product.
    So what is the simple way to do this, and still allow for the polarity to switch?
    I know how to solder, and can tell a hot wire from a ground. That's about as far as my electrical knowledge goes, so please keep it simple for me.
    Sorry for my ignorance on this subject.
    Thanks in advance for any advice.
     
  2. jcsd
  3. Dec 24, 2014 #2

    mfb

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    If you don't care about the energy cost and if the current through your material is constant, you can put a resistor in series. Not very elegant, but very easy. Multiple diodes would also work (use them in pairs to allow both current directions) and they are more stable in terms of voltage drop.
     
  4. Dec 24, 2014 #3
    Would you recommend a particular resistor? As well, since it reverses polarity, should I put them (it) on both wires going to the rods? Keeping in mind that the rods are suspended in water and not physically connected. I have a local Radio Shack that I can go to. Thanks.
     
  5. Dec 24, 2014 #4

    mfb

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    The resistor should be rated for the expected power value and have the right resistance value of course, everything else does not matter. It also does not matter on which side you put it into the circuit.
    The rods are connected to your power supply. And the water connection is like a resistor as well.
     
  6. Dec 24, 2014 #5
    So when the polarity switches to the side without the resistor the rod will only get 9v from the 12v source?
     
  7. Dec 24, 2014 #6

    mfb

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    The resistor is always in series with the cell, so it will always reduce the voltage applied to the cell. Resistors do not care about the direction of current.

    It does not make sense to say "the polarity switches to this side" - the voltage is always applied between the two sides, so everything that changes is the direction of current and voltage.
     
  8. Dec 24, 2014 #7
    I think I get it. So even if I only have one resistor on one side it will only allow a certain amount of voltage to complete the circuit no matter what direction the flow is?
    I have tried this before on my setup and it seems that one rod without the resistor will produce lots of silver into the water and the other one with the resistor does not. Without any resistor they produce the same amount as the polarity flips. I mean to say the rods take "turns" producing silver into the water as the polarity switches. Curious. That's why i'm here. lol.
    I really appreciate your time on this btw.
     
    Last edited: Dec 24, 2014
  9. Dec 24, 2014 #8

    mfb

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    It won't act as a fixed voltage limit, but it will always reduce the voltage applied to your cell, independent of the current direction.
    Then something else is wrong.
     
  10. Dec 24, 2014 #9
    I think I've got it.
    You said the water acts like a resistor. That makes sense.
    Taking into account that the silver rods take "turns" producing silver as the polarity switches, this means that the rod that is getting the full or "hot" flow of current is dumping it's energy into the water and the other rod is not fully completing the circuit. An "anode, cathode" thing here. One sending and one receiving voltage. One will have a buildup of silver particles on the rod while the other does not. I want to avoid that. This is why I make the polarity switch every couple of mins. That solved that problem.
    So by putting a resistor on both legs of the circuit I should achieve a balance of power to both rods. Hence, a step down from 12v to 9v.
    So my assumption is that I need a 3v resistor on both legs or wires going to the rods?
    I will try this and report back with my results, unless others have some input as well.
     
    Last edited: Dec 24, 2014
  11. Dec 24, 2014 #10
    So what do you think?
     
  12. Dec 24, 2014 #11

    NascentOxygen

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    Dividing your resistance into 2 separate resistors will make not a jot of difference in a series circuit like this.

    I suspect the timer switching every 2 minutes may have been implemented with DC in mind? It will achieve nothing different to were it not present when you are using AC. What is your source of 12V AC?

    What do you dissolve in the water to make it conductive before hanging the rods in it?

    Do you have a web link to point to where you are following some instructions?
     
  13. Dec 24, 2014 #12

    dlgoff

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    I think there may be a misunderstanding with what you mean by "switches polarity". Do you mean switch power from one rod to the other every 2 minutes like this not so good paint picture where only one resistor is required?

    ?temp_hash=93dff923df427e3493bc5de629165aa2.jpg
     

    Attached Files:

  14. Dec 25, 2014 #13

    mfb

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    That does not make sense.

    Please show a sketch of your setup.
    There is no "3V resistor". The voltage drop at the resistor will depend on the current. Which current do you have in your setup?
     
  15. Dec 25, 2014 #14
    My setup is very basic.
    I have a polarity switch with variable timing plugged into my wall outlet at my house. The adapter for the timer DOES convert AC to DC as the timer is a DC unit. Sorry for leaving that out. I imagine that point might be important.
    This polarity switch will run a standard motor in forward then reverse depending on the time I have set for it to do so. I tested this.
    There are 2 leads from the switch that go to each silver rod suspended in water.
    I used to hook up a standard 9v battery to make the silver solution. I noticed that only one rod would produce unless I turned the plug on the battery around. At that point the other rod would produce.
    I use distilled water, so the process takes some time to get going. I usually spike the water with silver solution to get the ball rolling faster.
     
  16. Dec 25, 2014 #15
  17. Dec 25, 2014 #16
    This is the polarity timer/switch. The output is 12.3vdc. I need to get that output to the rods down to 9v. 101_2318.JPG
     
    Last edited: Dec 25, 2014
  18. Dec 25, 2014 #17

    NascentOxygen

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    And the outcome being that after some hours you end up with a tiny pile of silver "dust" under one rod? Is that it? While your reversing timer may well evenly share the effort between the rods, you may just end up getting half as much silver from each, so no net gain? Was this something you devised, or is it standard recommended practice? Does one rod liberate a white fog of miniscule bubbles into the water---if so, then indeed periodic reversal is a good technique.

    The best way to drop the voltage from 12.3V to 9V here is not with a resistor, but with a couple of rectifier diodes. You'll need 4 or 5 diodes, the type you would buy to make a low voltage power supply. Rated at 50V or 100V and 1 amp will be good and robust. Connect them in series, and place them in one of the leads where polarity does not reverse. Measure the voltage across the silver rods to see how close to 9V it becomes. Each diode loses about 0.7V at low current. These are better than a resistor here.

    Can you explain how you concluded that it will perform better with 9V? What probem are you seeing the 12.3V cause?
     
  19. Dec 27, 2014 #18
    NascentOxygen.. You pretty much hit the nail on the head in your first paragraph. If I don't reverse polarity I end up with globs of unusable silver dust that attach themselves to one rod.
    On the other hand you recommend to place the diodes in one of the leads that does not reverse polarity. They both reverse polarity.
    As for 9v being better than 12v, the more the voltage the bigger the molecules of silver comes off the rods. 9v tends to create smaller molecules that are more useful. 1v would be the best but it would take a year to make a batch. 9v tends to be the standard in the DIY realm.
    Maybe the simplest way to accomplish my goal is to contact the person who made the switch and ask him to make one in 9v. I was hoping not to have to do that.
     
  20. Dec 27, 2014 #19

    NascentOxygen

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    Back in post #2 in this thread, mfb pointed out you can pair each diode with a parallel one that's reversed then it will conduct in both directions, showing 0.7v drop regardless of current reversals. So you need 8 to 10 diodes now. Almost any sort will do, I just recommended power diodes because they have stronger leads and are more likely to survive an accidental touching of the silver rods! Open up the power supply from a junked VCR or desktop computer box and you'll see big diodes you can carefully salvage using a pair of sidecutters. (First short out big capacitors in case they have held a charge, if recently used. Make sure it's unplugged from the mains!!) Don't bother trying to straighten out the leads of diode you salvage, more bending can weaken them. Just solder them into two strings, one going one way, the other in the reverse. Good luck!

    We'll recognize you as the blue-skinned guy who doesn't suffer from winter colds & flu??
     
  21. Dec 27, 2014 #20

    NascentOxygen

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    Have you tried hanging 6 or 7 rods in the one jar and powering all in series? Use something inert to keep them from touching, and I think you'll put 3 times as much silver in solution for no extra time or effort. The 12.3v may be a perfect match for half a dozen rods!

    You didn't answer my Q as to whether in operation a stream of white gas can be seen drifting away from either electrode?
     
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