electric circuit for chemical reaction

Discussion in 'Electrical Forum discussion & Blog' started by aeacfm, Jul 12, 2010.

1. aeacfmNew Member

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dear nukeman i hope you have a beautiful day

i want to clear some thing to you , the part i pointing to i didnot built it in the circuit because i cant understand it clearly and due to the lack in my electric knowledge .
so can i neglect it in future or the circuit must include it

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2. nukemanNuclear Engineer

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If you remove that, you will need to still have R2 in there. Where R2 connects to Q1 in the diagram, that point will now connect to ground (the -5v power supply will not be used). On the + side of the power supply, use the +12v instead of the +5v. Otherwise you will not have much adjustment range.

Basically, you are building what is on the 1st page of this link:

http://www.national.com/ds/LM/LM117.pdf

You can use the values of R1, R2, C1, C2 in the link or you can use the ones that you already have (assuming that they are still okay).

If you are going to rebuild the circuit with new parts, this is probably what I would use for parts:

- ATX power supply (+12v, Ground)
- C1 = 0.1 uF
- C2 = 1 uF or 10 uF (doesn't really matter)
- R1 = 620 Ohm

Without the stuff in the diagram that you circled, the minimum output voltage will be about 1.2v-1.25v. The maximum voltage depends on the power supply voltage and what you select for R1 and R2.

Iadj is small (50 - 100 microamps), so you can simplify the equation and say Vout is about = 1.25(1+R2/R1).

If R2 is adjusted to 5k and R1 is 620, the max voltage is about 10v. However, the regulator likes to have the input voltage 3v or higher than the output. Since you would be using 12v on the input voltage, the max output voltage would probably be more like 9v.

The minimum output voltage would be about 1.2v.

So using these values, you could adjust between 1.2v and about 9v.

The purpose of the parts that you circled in the diagram is so you could build a circuit that could go down to 0v instead of having the 1.2v minimum. It just depends on what you need. If you need to go below about 1.2v, you will need to build the more complex circuit.

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4. nukemanNuclear Engineer

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Correct, but you would remove R3 also. Take a look at the link that I gave..that should help. You will also remove the -5v that is shown in the diagram and use ground instead.

For the current, you don't have to worry about it. Basically use can either have a voltage source (what this circuit is) or a current source. With a voltage source, you set the voltage and the current will adjust to what it needs to be based on the resistance between the electrodes (Ohm's Law). The only issue you could run into is if this current was more than the regulator could handle. The regulator can handle a few amps (depending on which one you are actually using). If the current you need is in the mA range, you will have no issues at all.

With a current source (not what this circuit is), you send a fixed current through the system and the voltage will change depending on the resistance. Current sources are good for some things, but voltage sources are usually a bit easier to work with.

Are you at a univeristy or perhaps a lab? In either case, I would think that you could find someone with basic electronics knowledge to help you out. I certainly don't mind helping, but there is only so much help that I can provide without being there to see how you actually built the circuit.

Yes, you will be limited to about 1.2-1.25v as a minimum voltage with this circuit. However, if you get this working, then you can add the parts that were removed and then you will be able to adjust all the way down to 0v. Circuits are often designed/drawn/tested in blocks. Each part of the circuit does one simple job. You can build/test each of these blocks and then combine them as the circuit becomes more complicated. So, there is nothing wrong with starting off simple (1.2v minimum circuit) and then adding in the other parts later to be able to get down to 0v once you know the simple version works.

5. aeacfmNew Member

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hi, nuke
i hope you still remember me

i find two instruments in the electrical department in my company one variable current , and the other is variable voltage.

so is there any way to make connection between the two instruments so that they can work as variable Amp, V as one device? i hope !!!!!

6. ThatguyHomeowner

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Do you mean one is a current source and one is a voltage source? Wikipedia will explain the difference.

7. aeacfmNew Member

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yea , one is current source and the other is voltage source can i make them in one device to be one instrument of DC power supply with variable Amps , Voltage

8. ThatguyHomeowner

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That depends on if each real world instrument can both source and sink current like their ideal, mathematical, on-paper counterparts. This info may be hard to get.
Some setups may work as long as there is a load but will go crazy when there is no load. Also, some power supplies may not start up properly if they are looking at weird loads.

Why not just use a voltage supply with adjustable current limiting? Or, if you can come up with an equivalent circuit for your load it should be easy to find some way to power that load. If there is a negative resistance region, like in an arc, this could get messy.

You can use interpolation on a graph of the voltages and currents to come up with whatever precision you need.

BTW, two "D" cells in series with a diode will give you 2 to 2.3 vdc and you can put a resistor in series with this voltage supply. One D cell will give you ~1.5 vdc.
Make some V and I measurements on the load with different resistor values and we'll go from there. Start with a 2 ohm, 2 W resistor. It may get hot to the touch.

Last edited: Oct 10, 2010
9. nukemanNuclear Engineer

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I would go for the voltage supply. What is the range of adjustment on it?

From the data that you gave earlier, it looks like the previous experiment used fixed voltages (1.8, 1.9, 2.0v, etc.) and the current/current density was measured. If for some reason the current was too high and you needed to limit it, you could use a voltage supply thta had a built-in current limiting or you could add a resistor to limit the current.

The way that I see it, the experiment would go like this:

1. Set the voltage (maybe 10 steps changing 0.1v at a time)
2. Measure current
3. Divide current by the area of the electrodes to determine current density
4. Measure reaction efficiency

This is what I understand that you are trying to do. If this is not correct, you should give us detailed information on exactly what you want to measure and how you plan to do it. The more information that you give us, the better we can help.

Good luck.

10. aeacfmNew Member

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ok
i will give you it

11. aeacfmNew Member

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here is the whole conditions i will work at it , i took it as prtscn to you
some symbols like
- C/M is molar concentration and it equals the no of moles of substances per liter of solution .
- C carbon (graphite electrode).
- pt platintium electrode.
i think you know the rest better than me

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13. aeacfmNew Member

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i forgot to say that the key condition in this test are :

- electrode type
- current density
- temperature

14. nukemanNuclear Engineer

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That helps a bit. Now the question is: What are you looking to do?

1. Repeat this experiment and see if you get the same results (1.7, 1.8, 1.9v, etc.)?

2. Are you looking to improve these results (finer adjustments between 1.8 and 2.0v and find the best conditions, perhaps using 0.02v steps)?

3. Have you determined how you will measure the current efficiency?

Ideally, you would want the voltage to bounce around less that say 10% of your resolution. So if you were using 0.1v steps (like in the table above), you would want the voltage stable to 0.01v or better (meaning if you wanted 1.8v, the actual voltage might bounce between 1.79v and 1.81v... less variation would be better). Now if you wanted 0.01v steps to improve on that table, the voltage should be stable to 0.001v or better. The other thing to consider is voltage drift. If these experiments take some time, the voltage might change while doing the experiment (depends on your voltage supply). So if the voltage supply isn't properly designed to compensate for this, you might start the experiment with the correct voltage, but the voltage might be different towards the end of the experiment.

So, this is why I ask what you are trying to exactly do (what are your goals?). Finer adjustments require a better voltage source as well as better equipment to measure the fine voltage/current changes.

Do you have the information on that voltage source from your electrical department? You could post the brand and model number and I could see if I could find the specs online for it.

15. aeacfmNew Member

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reduction of oxysulfur acid containig water to sulfide
in other words convert HSOx to S (hydrogen is the most interfering spcs. here

Last edited: Oct 13, 2010
16. aeacfmNew Member

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- exactly i want to improve the results , also the conditions may be different (but not too much)
- i also want to try with these data to see if i will reach the same results or not.
- i dont know to measure the current effiency but , also i want to control it .

17. aeacfmNew Member

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what happened nukeman
is that mean that i must work on the previous circuit

18. nukemanNuclear Engineer

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Sorry. I didn't see that you had responded to this before.

Let us know what the electronics department has for a voltage source. You can get the name/model of it if you don't know any more info than that. We can then look up the specs online and see if it will work.

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ok
many thanks

20. ballvalveGeneral Engineering Contractor

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I think he is really from Iran and is finalizing the nuclear bomb detonator details. Call Homeland security!

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