My PSU's keep dying -_-

[woelen]

Making a current source with an LM317 can be done, but you are limited to 1A. An LM 317 is intended as a voltage controller. Without external power transistor you can go up to 1A. With a powerful power transistor and a good power supply you can get much more. But, then you can get voltage control.

 

You can also have ccurent control. You do that by placing a resistor between the VR pin and ground. The current then is 1.2/R and can be draw between the VR and VOUT pins (current flowing from VOUT to VR, the cathode hence being the VR-pin).

 

For electrolysis experiments I prefer the simple setup with the resistors and a 12 V DC PSU source with a large elco for ripple voltage reduction. It is more robust and virtually fool-proof. If you really want to use an LM317 for current control, then use a power transistor, otherwise you are severely limited. A 2n3055 is OK. Also use a huge cooling block and good heat-conducting paste if you use a 2n3055. Remember, the 2n3055 is not shortcircuit-protected and can be blown out fairly easily. The LM317 is shortcircuit protected and is not blown up immediately.

 

If you use the LM317 in the standard way, then you have a voltage source and not a current source. Voltage sources are much less suitable for electrolysis than current sources, as explained before already.

[raivo]

woelen, i do not know for sure but i suppose that several LM317 based current sources can be paralleled for higher current. I do not see any reason why this should not work.

 

Proper cooling is required of course otherwise even LM317 will die.

[YT2095]

Remember' date=' the 2n3055 is not shortcircuit-protected and can be blown out fairly easily. The LM317 is shortcircuit protected and is not blown up immediately.

[/quote']

 

it might be worth mentioning at this point that the 2n3055 output on my PSU has a 5 amp fuse inline at the output side.

and that in the case of my 3 amp constant 5 amp surge transformer with a voltahe range of 0 to 30v, unless I`m at the extreme top end of the voltage range (I`ve never had to power anything above 24v anyway) a 2n3055 can handle 115 watts, so it`s unlikely to "Blow" even without a fuse on a momentary short.

something to bear in mind when designing your power supply

[woelen]

it might be worth mentioning at this point that the 2n3055 output on my PSU has a 5 amp fuse inline at the output side.

and that in the case of my 3 amp constant 5 amp surge transformer with a voltahe range of 0 to 30v' date=' unless I`m at the extreme top end of the voltage range (I`ve never had to power anything above 24v anyway) a 2n3055 can handle 115 watts, so it`s unlikely to "Blow" even without a fuse on a momentary short.

something to bear in mind when designing your power supply [/quote']

You can also add the 5 amp fuse in front of the feedback circuit, such that the real output voltage is not affected by voltage drop accross the fuse. By incorporating it in the feedback circuit, the LM317 compensates for voltage loss accross the fuse.

 

@Raivo, why whould you want to use multiple LM317's for high current? You can modify YT's circuit with the 2n3055 easily for making a very high-current source.

 

An example:

 

Apply a resistor of 0.24 Ohm between the 1.2 V ref pin and GND. This means that 5 A must flow through this resistor in order to maintain voltage of 1.2 V at the ref pin.

 

Now, you take a 2n3055, with the base at the VOUT output of the LM317, the collector at the power rail and the emitter is the current output. The current source now is between the emitter of the 2n3055 and the ref pin of the LM317. Suppose that you short circuit the current source, then the LM317 will apply 1.8 ... 1.9 volts at its output. Approximately 0.6 volts is across the base-emitter junction, the LM317 will adjust the precise voltage, such that the voltage at the vref pin equals 1.2 volts. Because you have a resistor of 0.24 Ohm at that point, going to GND, you'll have 5 A. Now, supposer a 1 Ohm resistor is between the emitter of the 2n3055 and the Vref pin of the LM317, then the VOUT will be adjusted to 6.8 .. 6.9 volts. This causes the voltage accross the 1 Ohm load to be 5 V, the voltage accross the GND and the Vref pin again 1.2 V.

 

The only point is that the current source is not between a certain point and GND, but it is between the emitter and the Vref pin. For certain applications, where other circuits are driven by the same power supply this may introduce problems, but if the current source is used on its own, without any other circuits driven from the same power supply, then thus is no problem at all.

 

With this setup, the output current, delivered by the 2n3055 is 5A, the collector current is 5A - x, with x being the base current, delivered by the LM317. This setup is very accurate, because all non-linearities of the transistor are inside the control-feedback loop.

 

I hope you understand what I mean. Unfortunately I cannot draw in this software, otherwise I would make a nice drawing over here .

[YT2095]

actualy the best way is use a 24v relay in Latch configuration with a front mounted momentary reset switch, as soon as the power gets shorted, current to the relay coil drops below threshold and breaks the power

as soon as the short is corrected, you simply press the switch at the front and you have power again, saves alot of time messing with fuses and as a failsafe, it`s unbeatable

 

constant current can be similarly acheived using a transistor to power the relay, set to measure the PB across a fixed inline resistor, if the voltage dropped across it reaches a set threshold, the power cuts out, using a series of different value resistors on a rotary switch you can select your upper current limit to whatever you like (transformer permiting).

I`ve employed Both of these designs to great effect in the past also, not only do I like my work over engineered, but Idiot-Proof is always nice too

[woelen]

Keep in mind, I'm not talking about a current limiting thing, I'm talking about a true current source in my previous post.

 

Current limiting is very useful, when you have configured your device as a voltage source. What I described in my previous post is a current source, where the current drawn from the supply is independent from the load.

[YT2095]

well if you`re using series regulation, then whereas you use an emiter follower for Voltage regulation, you simply use a Collector follower for the Current regulation

[woelen]

In both cases, you use an emitter follower. For the voltage regulation, you use a fixed resistor (e.g. 120 Ohm) between vref and GND of the LM317 and a variable resistor between the emitter and the vref pin for the LM317. The load them is placed between the emitter and GND. The base is connected to the output pin of the LM317 and the collector to the power supply rail.

 

For the current regulation, you use the same setup, but now without a resistor between the emitter and the vref pin. The load itself must be placed between the emitter and the vref pin, not between the emitter and GND. For a decent current, now you have to use a quite small resistor between vref and GND. For 1 A you need 1.2 Ohm, for 5 A you need 0.24 Ohm.

[YT2095]

Interesting Method

 

I guess mine`s a little "Old Fashioned", the current reg is all transistor with my method, basicly comes down to Voltage ampifiers and Current amplifiers at the end of the day

[woelen]

Finally I have found the time to make a good power supply for electrolysis experiments. The power supply itself can be used for very high currents (mine goes up to 15A, according to documentation, but in practice it may be a little less). I designed a set of resistors, for current control. This allows currents up to over 3A. Of course, adding more resistors in parallel allows increasing the current, but in practice one hardly needs more than 2A, because the current density should not be too high. The setup is quite versatile and it is cheap.

 

http://woelen.scheikunde.net/science/chem/misc/psu.html

 

The page is a recipe for how to build such a thing yourself. Emphasis is put on safety and ease of use. No working with wires, which have to be wound around each other and so on. The concept uses nice connectors and configurations can simply be changed by plugging/unplugging.

[xeluc]

Thanks a lot Woelen. I can tell you have put a lot of effort into it. It is very informative. The problem with my PSU is that if I try to use a salt bridge, it wont turn on. It's the stupid thing where it shuts off if theres nothing being drawn or something. I'll work on it...

[woelen]

Yes, I've put a lot of effort in it, but it was worth it. Now I can do electrolysis experiments very easily and safely. Simply plugin the resistors and switch on the power supply. My next project is to make Ba(BrO3)2 electrolytically from potteries BaCO3 and NaBr. That would be a nice oxidizer.

 

The only problem I have left now is that the size of my graphite anodes is limited. They are 8 mm diameter and can be immersed in the liquid for 5 cm. That only makes an area of approximately 13 cm². I read that graphite anodes do not allow more than 40 to 50 mA current density per cm², so in order to keep erosion low I must limit the total current to 0.5A through the cell, which is quite meagre . I can use multiple cells in parallel, but that is quite cumbersome. I'm looking for thicker/longer anodes, but they are hard to find.

 

If I were you, I would buy two cheap ceramic 10 Ohm/10 Watt resistors and put these in series, making a 20 Ohm resistor. Use this as load. That should provide sufficient load for your PSU to keep it working always. Do you use an older PSU? For me, a load of just 20 mA is sufficient to allow the PSU to switch on.

[xeluc]

I already have the reisistors. With electrodes in a weakish solution of CuCl2, if the electrodes are more than a few inches from each other, the power is cut. FORGET about using a salt bridge with the setup I have here...

[woelen]

That setup of yours must be quite frustrating . What resistors do you have and what model of PSU do you have, is it is very old one? As I stated, my PSU always powers up, even without load. The builtin 20 mA load of the LED and 470 Ohm resistor are sufficient. Can you determine an approximate value of the minimum current drain, needed by your PSU, before it shuts down? You can measure this by taking an ampere meter and a very concentrated NaCl solution and then slowly diluting this and looking at the measured current.

[xeluc]

yeah I could find the minimum current drain. I'll need an ampmeter first. I think my failure has something to do wiht me only using two wires.. I dont have those other wires connectd that you said. They're all clipped. This was a VERY old PSU. Went to a 3.1 computer. It looks like I may be getting another. They aren't exactly hard to find. This time I'll use your site as a reference

[woelen]

Oh yes, it is important to have these other wires connected, especially the green, orange and brown wire. I personally experienced that if these wires are not connected properly, then the PSU works very unreliably and then it seems to have a will of its own, which is not good at all for a PSU . I did not cut away the wires, before I was absolutely sure that the thing powers up in a stable way. That is why I set up the page on my site. I did the trial and error already, so you do not need to go through all that again.

 

The PSU contains a circuit with some feedback electronics and it determines from the signals it senses, whether the raw power supply has come up properly. Only if this is the case, it releases the power for the CPU and other hardware on the mainboard. This is absolutely necessary for PC's, because otherwise the CPU, memory and so on may be initialized improperly or even be permanently damaged. These devices do not tolerate a large deviation from their rated voltage and especially immediately after switch-on, voltages may vary wildly. The PSU simply isolates al these "wild" voltages from its output wires and only when the sense signal becomes stable it releases the outputs for the CPU, mainboard and so on.

 

The normal sequence of operation for a PSU is that it

1) senses the signals, present at the mainboard, through the sense wire(s).

2) as soon as the signals are within tolerated levels, it connects the regulated voltages (3.3V, 5V, 12V) to the output wires.

3) activates the power OK line (grey wire), which is connected to the CPU and other components on the main board and adapters. This line is not released before step (2) is taken.

[xeluc]

I did not cut away the wires, before I was absolutely sure that the thing powers up in a stable way.

And that right there is what is different about us. I'll pick up another sometime this week and redo everything the right way.