Dual Silver Cell Setup off One PC Power Supply

[kadriver]

Here is the dual cell setup. I finally got it running last night.

I want to make a larger cell in a 300 series stainless container like the one described by Harold.

But for now I used what I know works to get some of this silver processed. I have gotten to where the metal is coming in faster than I can refine it. Not a bad problem to have, unless the price of silver heads south.

That pile of anode bars is only about half of what I have to run through the cells. I have about another 1600 grams or so of cemented silver that needs to be washed and melted into anode bars.

In the last week, I have been working so much that I neglected to check the cell. The silver bridged the gap between cathode and anode basket and shorted the cell.

It has happened twice. In both cases I took the cell apart and cleaned, added fresh electrolyte and restarted, using any silver crystals to make anode bars.

I was wondering if anyone has come up with a solution (other that a constant watch) to solve the problem of the cell shorting. I may have asked this before - sorry if it is a repeat.

I was thinking of getting a wall plug timmer, and setting it for an hour, so if I get busy and forget to check the cell, then the power will shup off after an hour. I can then just reset the timer and carry on.

Thanks - kadriver

 

[Grassbur]

I have a power supply that has a built in Overload shut off. If the short occurs causing a power surge it will automatically shut off the power source. The Timer sounds like a good idea. Hey there is an idea for an iphone app! lol

How do you have your cells wired? Is your neg feeding into the next cell?

 

[kadriver]

The power supply has about four outputs with 3.3 volts. So I used a seperate output plug from the same power supply.

 

[Grassbur]

Nice. My cell is getting 3.5 at the end of the Cathode wire so I was considering just hooking another cell up right behind it. Have you tried this?

 

[element47]

Kadriver, it would be my belief that while it appears that your supply has four orange "wires" coming out of it rated at 3.3 volts, they are simply four wires and not 4 "independent outputs". In other words, if you were open the box your supply came in, internally, you would see that those 4 wires go to the same place inside. Or, if you were to remove AC power from the supply and measure ohms from one of those four orange wires to any other, you would get zero ohms (you of course never really get zero ohms, you always get like .2 or .3 ohms which is the resistance of your test leads. Some meters allow you to zero that out just like scales allow you to "tare" out a mass you wish to weigh to zero out the weight of the container)

Now the question is, how can those four wires be converted to "independent outputs". This matters in your case because with the type of load a cell presents to a power supply, I strongly suspect you will encounter "load swamping". In practice, it may not matter critically, but if you are asking and if you wish to keep your setup producing predictable results....

You need to supply a current-limiting resistor in series with each "leg" or each cell you are driving. The following diagrams should explain this. Basically, the power will not share "equitably" without such current-limiting resistors. The cell with the lower resistance will hog the great majority of the power. Will it matter? Maybe not, maybe the higher resistance/lower voltage cell will just take longer to plate out your silver than the other one. But it terms of correct practice, you should have BOTH a current limiting resistor AND a fuse in series with each cell. (In the diagrams below, your cell is represented by the LED symbol because I have stolen said images) Same "swamping" consideration. The lower resistance item will hog most of the current/power without a current limiting resistor. By the way....there would be NO such limitation if you hooked up the cells in SERIES, though since each cell would be getting "half" of 3.3 volts, 1.65 volts might not be enough to run the reaction. Is there a 5 volt portion of your supply? That would solve the "splitting 3.3 volts in half" issue.

You could also place another (ordinary, non light-emitting) diode in series with each cell. But it would have to be somewhat of a highish current device, I am thinking at LEAST 5 amps and probably 10. It/they would of course have to be correctly polarized or they would not conduct. In other words, the "bar" side of the diode has to go towards the "lower volts" wire of your supply.

 

[niteliteone]

Kadriver
If this is the same power supply as in your "power supply demo" from oct 10, 2011, it has a current rating of 28 amps on the 3.3v side and each one of the orange wires used will take it’s share from the 28 amps.

As a simple solution take 2 orange leads and connect together, then run through a 10 amp fuse or breaker in series with your cell. Also be sure to use 2 black leads for each cell also.

Do the same for the second cell and you should be just fine.

No need for any current limiting or voltage dropping circuits that will change any part of the process you have already developed and use. Your only adding a fuse or circuit breaker

Each cell with its own resistance will only pull the current needed to operate, with the fuse or breaker limiting the maximum draw to the 10 amps.

Two cells will operate in parallel, each receiving 3.3v @10 amp max. Total load on power supply will be limited to 20 amp max. well below the 28 amp maximum for the power supply.

I hope the description is understandable. If drawing is needed, let me know.
Tom C.

 

[goldsilverpro]

Grassbur,

I always wired 2 cells, or more, in series.

The current is what does the work. It makes the silver deposit on the cathode, dissolves the anodes and, therefore, is much more important than the voltage. In a series circuit, the current is constant throughout the components in the circuit and the voltage is additive. Therefore, if the meter says 10A, you know that 10A are being applied to each cell.

In parallel, the current is additive and the voltage is constant (who cares). Therefore, if the meter reads 10A, you only know that the total current in the 2 cells is 10A. Unless the 2 cells are identical (hard to do) in their resistances, the current in each cell is different. You might have 4.5A in the first cell and 5.5 in the 2nd.

When I ganged 3, 30 gal silver cells together, I wired them in series to a 250A, 12V rectifier. Therefore, I could run 250A and up to 4V, average, in each cell. In parallel, one could get 12V in each cell and only an average of 83A. Of course, to get in the ballpark of 250A in each cell, I could have bought a 750A/4V rectifier (if I could find one), and run the tanks in parallel.

I like series.

 

[kadriver]

I understand each post - thanks for the guidance you guys.

Putting a fuse inline sounds like a good idea.

Both cells seem to be working fine after 24 hours of constant operation.

Here is a question; If I measure the amps across each cell and get say 4 amps, can I safely use a 7 amp or 8 amp fuse?

Having a fuse in there would save me lots of work if the crystals were to bridge the cathode and anode again.

I understand that I have these two cells in parallel curcuits right now.

I understand that this splits the amps between the two cells.

If I ran in series, then how would this be accomplished?

After connecting the orange lead from the power supply to the anode on cell one, would I then run the cathode lead from cell one to the anode lead of cell two, then connect the cathode of cell two to the black wire from the power supply? Is it that simple?

This series wiring would allow full current flow through each cell - correct?

Having full current flow will speed the formation of silver crystals, or using a parallel wiring setup, like I seem to have now, will slow down crystal formation - are these statements true?

I am good with mechanical, but I am not good with electricity. While in the Navy an electronics tech said this to me, "Wires have smoke in them, the trick is to not let the smoke out."

This pretty much describes the extent of my electrical training.

I also have another PC power supply that I could use.

For now, all is well and my output should nearly double. But if I can speed the production of silver crystals up, then I will make the necessary changes to make that happen.

Thanks for the valuable input from all of you.

kadriver

 

[niteliteone]

Here is a question; If I measure the amps across each cell and get say 4 amps, can I safely use a 7 amp or 8 amp fuse?
Correct

Having a fuse in there would save me lots of work if the crystals were to bridge the cathode and anode again.
Correct

I understand that I have these two cells in parallel curcuits right now.
Correct

I understand that this splits the amps between the two cells.
No, each cell is independent of the other

If I ran in series, then how would this be accomplished?

After connecting the orange lead from the power supply to the anode on cell one, would I then run the cathode lead from cell one to the anode lead of cell two, then connect the cathode of cell two to the black wire from the power supply? Is it that simple?
Yes,Correct

This series wiring would allow full current flow through each cell - correct?
No, a reduced flow and equal in each cell

Having full current flow will speed the formation of silver crystals, or using a parallel wiring setup, like I seem to have now, will slow down crystal formation - are these statements true?
See statement

kadriver

Having maximum current flow will give the fastest crystal growth.

Using the parallel wiring setup will allow the fastest crystal growth in each individual cell while isolating each cell from the other.

Using the parallel wiring setup will allow maximum current flow to each cell.

The current flow is regulated by the resistance of the electrolyte. Even though the power supply will max out at 28 amps, the cell can only use what the resistance will allow.

Lower resistance = higher current

Assume this to be your parallel cell
If cell 1 has 10 ohms resistance at 3.3 volt the current will be 0.33 amp
If cell 2 has 15 ohms resistance at 3.3 volt the current will be 0.22 amp
Load on power supply is 6 ohms resistance at 3.3 volt and 0.55 amp

Now connect in a series circuit
Cell 1 + 2 has 25 ohms resistance at 3.3 volt the current will be 0.132 amp
Cell one has 0.132 amp x 10 ohms and the voltage will be 1.32 volt.
Cell two has 0.132 amp x 15 ohms and the voltage will be 1.98 volt.
Load on power supply is 25 ohms resistance at 3.3 volts and 0.132 amp.

Theoretically the parallel circuit will give more current flow. So if it is the current flow that controls the speed of growth the parallel circuit is the way to go.
However if lower voltage and lower current will cause slower growing but more dense crystals then that is the way to go.

As long as you add the fuses the smoke will stay in the wires. :shock: :lol:

Ok guys, I'm ready for it. :shock:
I can take it.
Please point out any mistakes in my math or logic or share with the forum your practical experience. ( agree's or disagree's)
Tom C.

edit spelling

 

[butcher]

One thing to consider if you are trying to run two cells in parallel,
In parallel circuits the voltage is the same (voltage drop on each cell) the current divides.

The resistance in these cells is constantly changing, and two resistances in parallel circuit would not divide evenly, but split according to each cells resistance, this can make a situation where one cell starts getting more current than the other (which will again change its resistance and the other cell stops working cascading the problem.

In the series cells the same current flows through both cells and the voltage divides determined by the cells resistance.

Since these are current driven cells my thinking is series would perform better, parallel cells would create work trying to keep the juggling act balanced.

Notice in the LED picture above how trying to run the LED's in parallel will burn out the lights as one LED will start hogging the current from the power supply and burn itself up.

 

[kadriver]

I am beginning to think that I may just convert another PC power supply and stack it onto the other and use two individual power supplies. I have several laying around in my shop.

I want to leave it setup as it is right now and record the results of using the power supply in parallel.

When it comes time to harvest the silver I will configure the power differently and see if I get a different quantity of silver over the same period of time.

The current flow is regulated by the resistance of the electrolyte. Even though the power supply will max out at 28 amps, the cell can only use what the resistance will allow.

If the resistance to flow of current is low (as determined by the concentration of the electrolyte), then the amps will be higher. If my electrolyte is weak in concentration, then the resistance to current flow will be higher, and current flow will be reduced - right?

niteliteon, I like you examples. I can see how the math works and derive some knowlege from these equations.

Thanks to all for your help.

kadriver

 

[goldsilverpro]

niteliteone & kadriver,

If you had a PS that would only provide 10A, plus enough voltage to overcome the total resistance, the total amps applied in 2 cells in series would be 20A, if the full 10A were applied. In a parallel circuit, you would only get 10A total of the 2 cells. Thus, from the same PS, you would dissolve and deposit twice as much silver in the series circuit. If you applied 10A, you would get about 80g of silver per hour in a series arrangement and only 40g/hr in parallel. To get the same maximum silver production from a parallel circuit as you would from a series circuit, it would take a PS twice as large.

If the connections are all tight and the right size wiring, etc., is used, there will be little resistance. In series, if the electrode surface areas are not the same in the 2 cells, you could get varying crystal characteristics, due to the different current densities. It is best to try to make the 2 cells as identical as possible.

 

[niteliteone]

niteliteone & kadriver,

If you had a PS that would only provide 10A, plus enough voltage to overcome the total resistance, the total amps applied in 2 cells in series would be 20A, if the full 10A were applied. In a parallel circuit, you would only get 10A total of the 2 cells. Thus, from the same PS, you would dissolve and deposit twice as much silver in the series circuit. If you applied 10A, you would get about 80g of silver per hour in a series arrangement and only 40g/hr in parallel. To get the same maximum silver production from a parallel circuit as you would from a series circuit, it would take a PS twice as large.

quote]

With the bold words in mind please explaim how you would get any crystal growth. :shock:
In the real world as I understand it if the pwer supply ONLY has 10A then it will stop working if you try to take 20 A from it.
In my example I was using real world equipment that the builder has on hand. The laws of physics do not change no matter how we change the formula E = I x R that cannot be disputed for this purpose.

edit, Sorry I hit the submit button to soon, I ment to save.

 

[niteliteone]

In the example above: Posted by kadriver November 11th, 2011, 9:25 pm
I was using equipment already in use. The voltage is already defined (3.3V), the cell current draw is already defined (4A), this means the system resistance is (0.825 ohm) and the process already produces a crystal growth.

I am assuming that this is per cell in a parallel system.
Cell 1 has 3.3V and draws 4A this means the total resistance is 0.825 ohm
Cell 2 has 3.3V and draws 4A this means the total resistance is 0.825 ohm
So this system is consuming 13.2W of power

Now let’s rearrange this system to a series system.
Cell 1+2 is now going to read 3.3V divided between 2 cells, the total resistance is going to be 1.65 ohm, so the current will only be 2A

Cell 1 has 1.65V and resistance of 0.825 ohm this means the current is 2A
Cell 2 has 1.65V and resistance of 0.825 ohm this means the current is 2A
So this system will consume 6.6W of power :shock:

In this example of actual equipment being used by this member,
You can clearly see that the parallel circuit produces twice the work as the series circuit. :mrgreen:
W = watts = actual work done. A real and measurable number.
The parallel circuit clearly has twice the current flow to each cell.
The power supply is fully capable to handle the load safely. :idea: :shock: :?:

If this system were to be connected in series the resistance of each cell is additive.
So only half the current will flow.
Only half the work will be done.
Only half the growth will happen.

Again open to feedback. I can handle good or not. Always open to new knowledge.

Tom C

 

[HAuCl4]

Hi GSP. On a tangent topic re. silver cells.
Do you know of any closed circuit process for re-cycling and cleaning the electrolyte continuously in a silver cell or is it better done at intervals/in batches?. :?:
Someone asked me, and I told him my opinion was it was simpler and cheaper to replace a portion of the electrolyte every so often, but I'm not certain it is the best answer.

 

[kadriver]

I added a jumper wire and wired the cells in series earlier this afternoon.

When I checked it after about 2 hours, I thought the crystals had stopped growing. But after looking closer, I saw that they were actually growing much thicker and lower down in the container.

The crystals (when it was parallel) were thin and would built up quickly.

This slow thick crystal growth is much more desireable as I won't have to keep tending the cell to knock the crystals down as often.

I checked the voltage and it was 3.34 volts DC. I have a good meter, but I think the fuse for the amps is blown so I can't check that.

I am very pleased with this new series wiring configuration. I think this is the way to go.

I clipped both orange leads on the anode of cell one, and both black leads to the cathode of cell two.

I then put a nice fat jumper wire that runs from the cathode of cell one to the the anode of cell 2.

I think someone earlier in the post recommended doing this, I guess to allow for a higher flow of current.

I should be able to get about 50 troy ounces in 4 or 5 days.

Thanks you guys!

kadriver

 

[niteliteone]

I think someone earlier in the post recommended doing this, I guess to allow for a higher flow of current.

I should be able to get about 50 troy ounces in 4 or 5 days.

Thanks you guys!

kadriver

Actually for a lower flow of current. :shock:
The current should now be 1/2 as much as it was in the parallel circuit and the voltage in each cell will be 1/2 of what it was also.

I mentioned this in my earlier post, but the part I don't know the answer to is "will you get the weight per time ratio you were getting with the parallel circuit"
I guess you can run a time trial to figure out which wiring method will produce more weight per time period. Series vs Parallel.

Thicker crystals are easier to work with, but if you are needing quantity it would be good to know which method would produce more ounces per hour of usable crystals.

Also note that what happens in each cell will affect the other cell in series. If one cell shorts it then applies full voltage and current to the other cell which will produce as fast as a single cell in parallel. :shock: This can cause it to quickly short out as well if not caught in time.

I'll let you get back to eork for since I saw the weekend load you have to start working on. Some guys get all the luck. :mrgreen:
Wish I had someone to go out and find treasures for me. 8) Tell your wife what a good job she is doing . 8)

Tom C.

 

[kadriver]

She found another 2800 grams of nice sterling flatware in a wooden case earlier this past week.

Thankfully she wanted to keep it (I am almost getting tired of digesting silver). She went to her stash (in a bank safety deposit box) and gave me 60 grams of 14k gold to work with and she kept the flatware.

She keeps a box full of gold (and I don't even get to look at the key). When she buys something she "reimburses" me to cover the cost of what she has spent.

This woman is amazing. She also found over 380 grams of sterling jewelry and about 12 grams of 14k and 6 grams of 10k scrap this morning - all at local yardsales.

She is bringing in the metal faster that I can refine it.

But this probably won't last for long. We are entering out slack period - Thanksgiving to about the middle of Feb.

I don't know why, but the metals just seem to dry up during this period.

Thanks - kadriver

 

[niteliteone]

She is bringing in the metal faster that I can refine it.

But this probably won't last for long. We are entering out slack period - Thanksgiving to about the middle of Feb.

I don't know why, but the metals just seem to dry up during this period.

Thanks - kadriver

Pace yourself and Stockpile. :mrgreen:

Tom C.

 

[Harold_V]

If you had a PS that would only provide 10A, plus enough voltage to overcome the total resistance, the total amps applied in 2 cells in series would be 20A,

Not true. Voltage and cell resistance dictate current flow, which, combined, will be only as great as the meter indicates, with the cells combined.

Ohms law dictates that it takes one volt to push one amp through one ohm of resistance. Cells in series require greater voltage to operate, so if you produce ten amps, silver production by the combined cells would be no greater than if you ran a single cell at the same amperage. You do NOT gain amperage by running in series. If that was the case, you could use a ten amp power supply to operate any number of cells, multiplying the silver production by the number of cells involved. It doesn't work that way.

Series cells require greater voltage to produce greater amperage. Parallel cells require constant (and low) voltage, but a huge amperage capability. The resistance of multiple cells in parallel will be lower than that of a single cell.

Fusing is to eliminate problems from shorting. If cells are allowed to operate without constant supervision, especially if copper content rises, crystal production changes to fine silver wire production, which grows rapidly. That creates a perfect opportunity for the cell to short. You risk burning out the power supply (if it's not fused).

Be aware---if you short your cell, pretty good chance you will have burned a hole in the filter basket (don't ask). That will require that all of the silver produced to be reprocessed to eliminate the contamination that is released.

Harold