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My PSU's keep dying -_-


xeluc
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I'm very interested in electrochemistry and I believe that I can learn a lot from it, but I can't even get it to work for more than a few hours. So far, I just have been using wall outlet power supplies designed to supply power to phones or what ever. These are rated 9-12 volts and 300 mA. I understand a higher amperage would be more efficient, but the power supplies heat up quickly and eventually die. This is apparently because the power supply is almost shorting out or something. Maybe... Anyhow, I'm ready to invest some money into this hobby as it has become to me, so what can I do about getting something that would stand up to what I am doing? Of course, my Dairy Queen paychecks don't supply me with huge amounts of capital, so something of moderat cost would be nice. I just don't know where to look really. What have you guys done?

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Then you don't need to worry about the current either as the current is directly proportional to the voltage and resistance. I personally think the results in a 2V cell are pretty ridicules and really only useful in theoretical chemistry demonstrations. Whatever floats your boat I guess.

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PC power supply is best you can get if you are in low budget and can not wire your own. Best for lab use are old AT ( 286 type PC ) power supplyes that were in wide use 10 ... 12 years ago. Most of these will survive any short circuit (they just switch itself of) and you have choice of 12V and 5V rails (some even have 3.3V rail). ATX power supplies and other modern devices are generally less foolproof and usually require some extra connections to work at all.

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If you are tight on budget, then indeed I would go for a cheap computer power supply. With the 12 Volts output you can do many things. However, you also need to be able to regulate the voltage output (between say 1 volts and 10 volts). Even better would be to make a current source of say 0.5 A to 2 A, which automatically adjusts its output voltage in order to keep current constant. These are really great for electrolysis.

 

With an LM317 you can make an adjustable voltage source, ranging from 1.2 V and upwards to appr. the raw voltage minus 3 volts. These are perfect in combination with a computer power supply.

 

If you have a little more budget, then you can buy a lab power supply in an electronics store. I do not know American prices, but over here in The Netherlands I can buy one for $150 to $200 with the following specification:

 

Adjustable max. output current 0 .. 5 A

Ajustable output voltage 3 .. 30 V

 

With these you can implement a current source and a voltage source. If the current limiting is set to 3A and your circuit draws less than 3 A then the selected voltage will be output. If you want a current source, then set the voltage to 30 V and adjust the max current at e.g. 1 A. Now, the power supply adjusts its voltage, such that the current drawn is 1 A. The latter works perfectly for more concentrated liquids when they are electrolysed.

 

When you want a device with a lower adjustable output voltage, then you need to spend more money, but such a shortcoming can easily be adjusted for. Buy a few power diodes and place these in series with the voltage output. Each power diode takes around 0.6 V, so with three of these your Vmin will be 1.2 Volts.

 

the specs I have given are just indicative. With some searching you might be able to find better devices for that price. Especially the lowest voltage setting is important. Electrolysis can be done at voltages as low as 1.5 Volts.

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Xeluc, in the mean time, if you still have a working mains adaptor like you mentioned, if you use a 9v 3W bulb in line, no matter what, you`ll not blow the thing, even a dead short, as the bulb won`t allow you to draw any more power than the 3 watts (well within the adaptors limits). :)

 

I`m with Woelen though, I use a homemade PSU with an lm317 and a 2n3055 output driver, they work great and I`ve had this thing now for a good 15 years!

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Xeluc' date=' in the mean time, if you still have a working mains adaptor like you mentioned, if you use a 9v 3W bulb in line, no matter what, you`ll not blow the thing, even a dead short, as the bulb won`t allow you to draw any more power than the 3 watts (well within the adaptors limits). :)

 

I`m with Woelen though, I use a homemade PSU with an lm317 and a 2n3055 output driver, they work great and I`ve had this thing now for a good 15 years![/quote']

 

Goodv idea. I actually already killed mine..v Nexttiem though :) I really don't mean to sound stupid or like I'm unable to do anything myself, but could you elaborate as to how a home made device like this would be made?

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If you want to make your own power supply you have to get suitable transformer at first. This must be rather powerfull not like those in telephone chargers but much bigger. Output voltage of the transformer must also be suitable. Something form 15 to 25V is good. You need 4 diodes and big electrolyte capacitor to convert voltage from transformer output to DC. When converting to DC voltage will go higher - aproximately 1.4 times compared to AC output. This must still remain less than 40V otherwise you can not use simple circuits with common LM317. If this all is done then connecting LM317 as voltage or as current regulator is very easy.

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there a little bit more specific info here also: http://www.scienceforums.net/forums/showthread.php?t=16469&highlight=lm317

just look around the bits in RED.

 

as for the transformer, I used an old 13.8v 3-5 amp CB power supply transformer, it was center tapped so I only needed 2 diodes for the bridge, but you could use an old car battery charger and convert that just as easily :)

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I as thinking of using a battery charger but I think the one I have now has some kind of swithc off or something becasue I quickly threw steel wool on both clamps and put it in salt water whiel plugged in and nothing happened. Shoud lI have it on a specific setting?

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Many battery chargers are not simply voltage sources. They contain quite a lot of intelligent sensing electronics, which determines what voltage the battery to be charged has and how much current has to be emitted.

 

When no battery is present, then probably no current is applied at all.

 

Have a look at this: http://www.hw.cz/data_ic/lm317.pdf

 

It is the LM317 datasheet, which also contains some circuits you can build around this chip.

 

If you are not confident enough on building your own electronics, then another fairly inexpensive alternative is the simple PC power supply, with some wiring.

 

http://web2.murraystate.edu/andy.batts/ps/powersupply.htm

 

You can adjust voltage by buying a set of power diodes. Each diode takes 0.6 volts. If you put them in series, then you can make many voltages in steps of 0.6 volts. If you also can get two germanium diodes, then you can also adjust voltage in steps of 0.2 volts. That should be perfectly OK and provide you good flexibility in your electrolysis experiments.

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I like the PC Power supply route :) So jsut using the wires I want with diodes.. will it last or do I need to add a resister or something.

It will last, provided your diodes are sufficiently large. So, you need to take good power diodes. Suppose your electrolysis takes 5A maximum, then take some headroom, make 10 A of that for headroom, so buy your diodes, such that they can stand 10A of current.

 

I have been thinking about this and an even better approach would be to buy a bunch of 10 Ohm power resistors and just 2 silicon power diodes. With that setup you have very high flexibility at low cost.

 

Buy 10 resistors, each 10 Ohm and each allowing between 5 and 10 Watts of power dissipation.

 

Now you can implement a reasonable approximation of a current source with your resistors. Electrolysis voltages are around 3 volts. Some cells require 4 volts, others less than 3 volts, but 3 volts is a fairly average voltage. Fluctuation around that is small.

 

Now some theory. Because the voltage of your voltage source is 12 Volts, and the voltage of your cell is around 3 Volts +/- 1 volt, the voltage accross your resistor network is 9 volts +/- 1 volt. So, the current remains constant within let's say 11 to 12% of your selected current.

 

Selecting currents can be done with resistor networks. Remember: Never put a single 10 Ohm resistor in series with an electrolysis cell. That may destroy your resistor with the rating I have provided you!

 

The current you select for electrolysis is 9 / Reff.

 

You select Reff with your resistors by means of series and parrallel circuiting of resistors. I'll try to explain to you how this can be done.

I introduce a mathematical operator // with the property x//y = (x*y)/(x+y). Examples: 2//3 = 6/5 = 1.2; 3//3 = 9/6 = 1.5

 

 

Now, I also need to introduce a notation for networks.

 

I use notation RRR for a series of three resistors (30 Ohm), RRRRR for a series of 5 resistors (50 Ohm).

 

I use notation RRR//RRRR for a series of three resistors, and a series of four resistors, with both series put in parallel. The resistance of such a network equals 30 // 40, which is appr. 17 Ohms.

 

You never may have a single R in one branch of your parallel network. That can lead to destruction of that resistor. So don't use networks like R or R//R or RR//R.

 

Now, some useful Reff values:

 

RRRRR --> 50 Ohm

RRR --> 30 Ohm

RR --> 20 Ohm

RRR//RRR --> 15 Ohm

RR//RRR --> 12 Ohm

RR//RR --> 10 Ohm

RR//RR//RRRR --> 8 Ohm

RR//RR//RRR --> 7.5 Ohm

RR//RR//RR --> 6.7 Ohm

RR//RR//RR//RR --> 5 Ohm

RR//RR//RR//RR//RR --> 4 Ohm

 

With electrolysis experiments, you'll most frequently use between 10 Ohm and 20 Ohm networks. For lower resistance you need a good and thick graphite anode, otherwise it will be pulverized in minutes, so going with networks, lower than 10 Ohm should only be done with high quality heavy electrodes.

 

Now, what can you do with the diodes? These can be used for fine-adjustments. Suppose you want to electrolyse bromide, to make bromate and you see some bubbling at the anode. Then oxygen is formed and you know that the voltage is too high. You can decrease your current in that way by selecting a network with higher resistance, but you can also put a single or even two diodes in series with your resistor network. In fact you make a new network ReffD or ReffDD, with Reff your original resistor network.

 

Never ever put a diode in parallal to your network, e.g. RR//RR//D or RR//RR//DD WILL destroy your diodes and most likely also your power supply.

 

What is perfectly safe is a network like this: RRD//RR or RRDD//RR, as long as in a single branch there are at least two resistors, you are safe.

 

It may seem a little awkward when you start with this, but when you have some experience, then you'll get a quick feeling for this.

 

When your solutions are very dilute, then the current source approximation does not work anymore, then the voltage accross the electrolysis electrodes will rise to much higher values than 3 Volts and in that case you have a low current and high voltage and you mostly produce oxygen at the anode.

 

For easy working, make nice plastic boxes, one for each resistor, with heavy duty connectors for both sides of the resistor and thick short wires (e.g. loudspeaker cables) with contra-connectors at each end, such that you easily can plug your networks together. I'm going to make such a thing as well, that is why I showed you the page for the PC power supply project. I have an old power supply left and with $20 I can buy all resistors, connectors and plastic boxes to make it robust and easy to use. I already have an LM317 circuit, but for electrolysis I have discovered that current control is much better, and then the resistor networks are superior.

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Hm. I don't understand most of what you said Woelen. However, I remembered last summer i bought a Windows 3.1 computer at a garage sale for $5. So I gutted the power supply. I wentto radio shack and bought alligator clips (I've needed those FOREVER) and a 10 Watt 10 Ohm resistor. I hooked it up to the 5 Volt 3.8 Ampere? Wire. This power supply has lasted 48 hours of continuous operation and that's a record. I will regard my problem with psu's dieing solved. Now for some manipulation. My father has a volt meter so I will be using that to see evactly what im pulling form it. SO let me see if I got this straight. Ignoring losses from the wires and the actualy operation or electrolysis. With a 10 Ohm resistor, is my theoretical 3.8 amps turned into .38? If so, that SUCKS. Thats no different form my little wall plug in Power supply. Does this mean that I would need to make a transformer to raise the amperage? I know how to do it and have actually wanted to, but... this power supply is already like a freaking foot long (not exagerating). Ahem... Anyhow.... Thanks for all your help so far. Any New Ideas?

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Hm. I don't understand most of what you said Woelen. However, I remembered last summer i bought a Windows 3.1 computer at a garage sale for $5. So I gutted the power supply. I wentto radio shack and bought alligator clips (I've needed those FOREVER) and a 10 Watt 10 Ohm resistor. I hooked it up to the 5 Volt 3.8 Ampere? Wire. This power supply has lasted 48 hours of continuous operation and that's a record. I will regard my problem with psu's dieing solved. Now for some manipulation. My father has a volt meter so I will be using that to see evactly what im pulling form it. SO let me see if I got this straight. Ignoring losses from the wires and the actualy operation or electrolysis. With a 10 Ohm resistor, is my theoretical 3.8 amps turned into .38? If so, that SUCKS. Thats no different form my little wall plug in Power supply. Does this mean that I would need to make a transformer to raise the amperage? I know how to do it and have actually wanted to, but... this power supply is already like a freaking foot long (not exagerating). Ahem... Anyhow.... Thanks for all your help so far. Any New Ideas?

 

A ten ohm resistor in series with a 5V source will limit the current to 0.5A. If you wanted a higher current you could just use a lower resistance. Resistors are a very crude way of limiting current however. You would be much better off using a voltage regulator to regulate the current.

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Easy way for higher current in your case is to put some of those 10W 10 ohm resitors in parallel. Each one will give additional 0.5A. You can step to 3.5A such way - no more as 3.8A is limit of your power supply.

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The solution, I have proposed, gives you the opportunity to use any current between let's say 100 mA and 2.5 A. It also has the advantage of current control, instead of voltage control. This is a great added advantage. Besides that, it also is a VERY cheap and very simple solution if you already have a working PSU from an old PC.

 

The basic idea behind my R-networks is to allow a safe and reliable way of controlling current from a 12 V power supply (which every PSU from a PC has). The reason that you have current control is that 12 V is much more than 3 V, which is common in electrolysis.

 

Of course, you can buy a regulated and adjustable power supply, but that is more expensive. My idea is the idea of Raivo in his last post, but worked out in more detail with a lot of different preset currents and the most important difference being the fact that I use 12 V instead of 5 V. With 5 V the current regulation is much worse, because the electrolysis voltage is quite large, compared to the 5V of the power supply.

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So if I understand you guys correctly... 10 Ohm resistors in parallel would give me amperages in steps of .5 amps, up to 3.5.? If I had those resistors in my 12 volt, 2.8 amp line, does that mean 10 ohm resistors in parallel would give me up to 2.4 amps? This whole thing with resistors confuses me. The equation V=IR would say adding more resistors would increase voltage, not current. Maybe this has to do with them being parallel instead of in series.

 

Should I be using my 5 volt 3.8 amp wire or my 12 volt 2.8 amp wire? Also, If I connect multiple 5 volt 3.8 amp wires together with sufficient neg and pos wires to equal out, will my amperage or voltage increase. Is this safe??

 

Remember: Never put a single 10 Ohm resistor in series with an electrolysis cell. That may destroy your resistor with the rating I have provided you!

 

I have one resistor in the line right now.. The persons powersupply on the webpage you gave me also only has one? So... Is this wrong?

 

Also Woelen. I finally understand what your talking to with your resistor thing, to a degree. when you have like RR//RR//RR I don't understand how you calculate THAT. Also, when you take RRR//RRR and do the (x*y)/(x+y), you get 1.5 . However the resistance is 1.5 Ohms. I did the same with RR//RRR. I got 1.2 but the ohms are 12. So do I just multiply by 10? SO anyohw... yes.. Thanks to whoever helps me

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Putting outputs in parallel will not increase the available current because all outputs are coming from the same transformer and all outputs of the same voltage are coming from the same winding.

 

It depends. Computer power supplyes are not simple transformer circuits. Switchmode converters are used that relay on inductors and contain quite complex circuits. I have seen computer power supplyes that have multiple 5V modules inside that CAN be paralleled for higher amperage. There are others that can supply almost all output power to any output rail if other rails draw no current. It depends on circuits and is impossible to tell just by description. 3.8A is a bit low for normal PC. So there may well be other 5V rails that can be paralleled.

 

Nevertheless 3.5 A is not too low for electrolysis experiments. In my opinion its better to stick with that for some time.

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Just summarizing it all:

 

Try to approach a current source as much as possible. That gives much more consistent results with your electrolysis.

 

In order to approach a current source, it is MUCH better to use the 2.8A 12 volts line than the 3.8A 5 volts line. Take for granted the lower maximum amperage. In fact, 2.8 A is quite a lot already for electrolysis and if you want to use that current, then you need a GOOD anode, otherwise it will pulverise in minutes.

 

Why is 12 V better than 5 V? Suppose you do an electrolysis. The current as function of voltage, applied over the electrolysis cell is highly non-linear. Suppose for a certain electrolyte, you have the following function I(V), current as function of voltage:

 

I(0.5) = 0.0 A

I(1.0) = 0.0 A

I(1.5) = 0.0 A

I(2.0) = 0.0 A

I(2.5) = 0.1 A

I(2.8) = 0.4 A

I(2.9) = 1.0 A

I(3.0) = 2.0 A

I(3.1) = 3.5 A

I(3.2) = <almost like short circuit>

 

A characteristic like I sketched above is very common. Now you also understand, why having a voltage source has such inconsistent results. If you adjust your source at 3 V, then the current depends very sensitively on the applied voltage. A little fluctuation in your output voltage, due to imperfections in your circuitry gives a huge fluctuation in your current. This makes it very hard to obtain consistent results.

 

If you use a current source of 1 A, then the voltage will be 2.9 V. You'll also see that in the range from 0.1A to 3.5A there only is around 0.5 V voltage change.

 

Now, suppose you have a 5V power supply and you want to select a certain current as function of resistance, then you need to determine the voltage over the cell. Over the entire range, it may fluctuate between 2.5 V and 3.1 V. The voltage over the resistor is 5 - 3.1 to 5 - 2.5 volts, hence it is between 1.9 and 2.5 volts in this example. You see that the variation is quite strong in relative sense (over 25%). So, selecting a certain current is not that easy.

 

Now, if you use a 12 V power supply, then the voltage over the resistor is between 8.9 and 9.5 volts. Now, this voltage is constant withing 6%. The absolute variation is the same, but the relative variation is smaller.

 

If you use a resistor to control the current, then you have quite an accurate (more than enough of accuracy) for electrolysis. If the characteristics of your electrolyte change somewhat, due to heating, change of chemical composition, etc, then with a voltage source you'll observe a huge change of current, while with a current source, you'll only notice a small change of voltage and if you use a 12 V power supply, then your current hardly will change, because the total voltage drop over the resistor network hardly changes.

 

----------------------------------------------------------

 

Now the question, why not take a single 10 Ohm resistor in series with 12 Volts. In the power supply scheme, I posted, a 10 Watt resistor is used, which is connected to the case and this is sufficiently cooled. For the other ten resistors I suggest to buy lower rated 5 Watt resistors (these are a lot cheaper). If you use a single 10 Ohm resistor and you have a short circuit, then the power dissipation will be V*V/R = 12*12/10 = 14.4 Watt. This may burn out a resistor, rated for 5 Watt.

 

------------------------------------------------------------

 

Parallel circuits. Why 10 pieces of 10 Ohm resistors? With suitable combinations of series and parallel circuits all resistances, mentioned in my previous long post can be reached.

 

As an example, I show you how to compute the resistance of the network RR//RRR//RR.

 

RR is a series of 2 resistors, having a total resistance of 20 Ohms (series --> addition of resistances).

RRR is a series of 3 resistors, having a total resistance of 30 Ohms.

 

The total resistance of this network is

 

20//30//20

 

How do you evaluate this?

 

20//30//20 = ((20*30)/50)//20 = (600/50)//20 = 12//20 = (12*20)/32 = 240/32 = 7.5 Ohm

 

Like with addition and multiplication, the order of evaluation of // operators does not matter: (a//b)//c = a//(b//c) = (a//c)//b.

 

How do you build a network RR//RRR//RR?

 

You take 7 resistors. Two of them you connect in series, three of them you connect in series and the final two are connected in series. This yields 3 series resistors. These three series resistors in turn are placed in parallel.

 

-------------------------------------------------------------------------

 

Finally, this theory only holds when the used power supply allows the selected current to be delivered. From a flimsy cell-phone charger you'll never obtain 2.5 A or so, but a PC power supply certainly can deliver that kind of currents.

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OK Woelen. I understand everything you've said. Thanks a lot everyone for your help. So basicly, I can't get my 12 volt wire to work. There was a 12 + wire, 2 "com" wires and soemthing else... I don't know... regardless. I reconnected everything to the 5 volt output. Just for experimentation i made a resistance of 5 ohms. I should have 1 amp running through although i dont have a meter to check for sure. I did notice however that even using zinc electrodes, I get Cl2 at the Anode. Ususally this would only happen with graphite electrodes. Also I beleive that O2 was produced because it was bubbling and i doubt it was all Cl2 or it's stink horribly. Now I just need to control my Voltage also. Did I need to buy one of those lm whatchamacallits to do so? Also, Could I jsut use a potentiameter instead of the resistors? I think that'd be more flexible although if someone hasnt mentioned it, theres prolly something wrong with it.

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If you can find potentiometer that can be used at 10W (better still 30W) then you can use it. Otherwise it does not last long.

 

Those lm317 's are cheap (just looked in ebay - $2.49 for 10pcs), buy some, there is a lot to do with them.

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Wow guys, check this out. I went to a website and they are GIVING me 5 LM317's. :eek::D I'm not even paying shiping! I requested a free sample and they replied saying they would be more than happy to ship 5 to my... business... :rolleyes: :rolleyes: Ha, anyhow, they said that the parts were being shipped from singapore (I hope I'm not benefitting from sweat shops over there :-( ), so it will take "3-15 days to ship". So ha! Anyhow, when I get them, and play with them, I'll tell you guys how everything went.

 

Using a current source has the advantage' date=' that it automatically adjusts at the halogen to be electrolysed. For an iodide solution a lower voltage is 'chosen' by the current source than for a chloride solution. This eliminates the problem of using too high a voltage, resulting in unwanted formation of oxygen and associated pulverization of the anode.

[/quote']

 

So, do I have a "current source Woelen? so will my voltage adjust itself or do I need something else in order for this to happen. I wouldn't think it'd do it by itself.. who knows.

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