electric circuit for chemical reaction

[aeacfm]

I don't think that the actual output needs to be all that precise. ]

you can see how i need the precision from the above data

[nukeman]

Okay. That helps some. It looks like you need something with precision to 0.01V or better. In the table that you listed, I assume this is taken from another experiment. To me, it looks like they controlled the voltage (0.1V steps) and the current density and reaction are just measured. Normally, you control one thing (volts, amps, or resistance). Ohm's Law relates everything to what you did change. For the current density, it was likely calculated be measuring the current (amps) and dividing that by the electrode area.

I do have a bit of a chemistry background. It is not my specialty, but physics/engineering undergraduate and graduate work had me involved to some extent in all of the physical sciences.

[aeacfm]

. In the table that you listed, I assume this is taken from another experiment.
I do have a bit of a chemistry background. It is not my specialty, but physics/engineering undergraduate and graduate work had me involved to some extent in all of the physical sciences.

you looks Professional!!!! aha In my opinion, at least
this is for acid i want to reach them then i will try them for different acids ok!!!

so in your opinion what you suggest to do but Taking into account that iam not electricity man iam only chemistry man and you make me like electricity
thanks again

[aeacfm]

hey
no body here i really need what i posted it is my last chance in my work !!!!!!

[nukeman]

Try the 1st page of this pdf. See if that helps.

http://www.edn.com/contents/images/6598370.pdf

This is a 0-3v adjustable voltage source built around the LM317 (very common device). If the voltages you were working with were higher, the circuit could be made more simple (especially if you needed say 3v and larger). The only trick with this circuit is it calls for (5-10v) on the positive side and (-5 - -10v) on the negative side. So, you couldn't use the +/- 12v on a computer power supply. However, there is also a +/- 5v on a computer power supply that would work. Take a look at this:

http://www.wikihow.com/Convert-a-Com...b-Power-Supply

So, the combination of these should be able to do what you need for a small cost. On the output, you would add a current meter (inlne between the circuit output aand the electrode) and a voltmeter between the output and ground.

This circuit should be able to provide 1 amp or so. I don't know the exact electrode size becuase when you say "A battery", I don't if you mean "A" in terms of cell size (like AA, AAA, etc.) or if you mean (a battery) like saying it is a battery but not saying what size. Anyway, if you know the electrode size (in cm^2), mulitply that number with the maximum current density that you need and that will tell you the maximum current that is needed. If you need more than 1 amp or so, then we'll have to go for a different design.

[aeacfm]

ok friend let me read this carefully and if there is some thing that is not clear i will notice

what i mean by A battery ia the size A (torch battery)

[jimbo]

Some basic things you need to specify in this setup are:
►input voltage
►desired output voltage
►expected range of current ( all the components such as transformer, diode, etc. can fit in a spoon, or in the back of a pickup, or anywhere in between)
►do you need controlled voltage, or controlled current, or both
►with what degree of accuracy do you need to measure the volts and current? Basic test equipment would give you better than 5%. Reasonably priced equipment will give 1%. If you need better accuracy than that, add $$$$$$

[aeacfm]

i will use the above described circuit in electrolysing or electrochemical reaction .
1-if you have chemical back ground i will electrolyse aqueous acidic solution in which i dont need the hydrogen to evolve with me ok!!!! thats first
2-this would achieved using carbon electrode (internal graphite in A battery )
3- this electrode would be of high hydrogen over potential to prevent the evoultion of hydrogen
4- the best conditions to aciece the abone conditions are as follows
voltage v current density Amp/Cm2 efficiency of reaction (%)
1.5 0.017 60
1.6 0.024 61
1.7 0.055 63
1.8 0.076 67
1.9 0.147 80
2.0 0.247 61
2.1 0.846 23
2.2 1.456 15

from the above data you can see how i really need it precise as the efficiency decrease from 80 to 61 by increasing the voltage by 0.1 v !!!!!!!!!
but if it went like what you noticified 0.05 increments so it may work i think !!!!!

thats all what i need

[aeacfm]

hello gent. men
i was so busy trying to prepare for the chemical components for my work , In spite of this i took a quick look about
http://www.edn.com/contents/images/6598370.pdf

but iam a little confused about is there a need for the diode system or not because he told it is required to give little voltage than the refrence voltage (even i dont know what is the refrence voltage ), also it make temp. drifting so he used the Fairchild Semiconductor LM185 or an Analog Devices to solve this problem .

in any case i was trying to attach these questions in diagram of the circuit in that file but its size was too large !!!!!! lol!

at the end i has imagination for the circuit and i need who correct to me if that was possible arranged in numbered sequence :

1- mains power (110-220 AC V)

2- (12 V DC) transformer.

3- Voltage regulator (0-3 V DC)

4- Computer power supply

5- A current meter (inlne between the circuit output aand the electrode) and a voltmeter between the output and ground.

finally i cant get the trick of adjusting the transformer

many thanks

[aeacfm]

no body here at once

[nukeman]

It is a little confusing in what he talks about. You just need to do it like he shows in the circuit without any extra diodes.

Here is what he is talking about:
- normally, with the LM317 , the lowest output it can do is about 1.2v. However, we want to be able to get down to 0v. One way to do this would be to use the LM317 and add a pair of diodes (and have a negative voltage supply) to shift this 1.2v down to 0v. See, each diode drops about 0.6v across it, so 2 diodes would be 1.2v. Basically, you are just shifting everything down by 1.2v (so if you had a design that went from 1.2v-5v, now would go 0v-3.8v). However, he mentions that using the 2 diodes, there is a problem with drifting. What happens is that the actual voltage drop across the diode changes with temperature. This would not be very good, as your reference voltage would change and then your output voltage would change. His circuit removes this problem.

- for power, all you need is the computer power supply. The computer power supply will convert your mains (110v-220v AC) to DC power. There are several DC voltages inside the power supply available (+12v, -12v, +5v, -5v, etc.). Take a look at the 2nd link that I gave you. You will use the +5v and -5v from the computer power supply to power the circuit.

The circuit has two adjustments. One adjusts the reference voltage (to set where 0v is) the other adjusts the output voltage (0v-3v).

You would then use a current meter and voltmeter to measure what is being applied to your electrode. I'll try to draw something up for you to make it clear.

[nukeman]

Try this:



This should work for what you need. If you can give me the exact electrode sizes, then I can verify that this circuit will be able to handle the current load. If it can't we'll have to try something else.

Good luck.

[aeacfm]

ok thats better
but other question just to understand the (D1 RED,R4 510, BCW33, R2, R6, R5, ) all in one component or what i cant get it ???

other question if i will work for volts in the range from 1.2 V to 3 V you said that it would be easier , so how it will look like ????
i am sorry i know i am wasting your time

[aeacfm]

This should work for what you need. If you can give me the exact electrode sizes, then I can verify that this circuit will be able to handle the current load.

Good luck.

for one electrode the surface area approximately = 14.326 cm2

[nukeman]

Each of the parts listed are individual components (resistors, capacitors, etc.). On the resistors (R1, R2, etc.), the number listed is the resistance in Ohms. R2 and R6 are adjustable resistors. D1 in an LED. Q1 is a transistor.

The most simplified version of this circuit (1.2v minimum) can be found in this data sheet (1st page). You would not the the negative side of the computer power supply in this case. Also, you could use either +12v or +5v for power (it would be better to use the 12v to make sure you can easily get above 3v (but using the 5v would work fine too).

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

If your electrodes ar 14cm^2, we might need to go with a different design. Say you want a current density of 0.2 A/cm^2:

0.2 * 14.326 = 2.87A

This circuit will only handle 1.5A. (current density of about 0.1A/cm^2 using those electrodes).

Would it be possible to use smaller electrodes for this experiment, or do you have to use this size? If you need to use this size of electrode, we will need something that can handle more current.