electric circuit for chemical reaction

Users who are viewing this thread

aeacfm

New Member
Messages
42
Reaction score
0
Points
0
Location
egypt
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

Plumber
Messages
8,918
Reaction score
18
Points
0
Location
San Diego, CA
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

New Member
Messages
42
Reaction score
0
Points
0
Location
egypt
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

New Member
Messages
42
Reaction score
0
Points
0
Location
egypt
sorry

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
 

Nukeman

Nuclear Engineer
Messages
707
Reaction score
1
Points
18
Location
VA
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

Nuclear Engineer
Messages
707
Reaction score
1
Points
18
Location
VA
Try this:

vreg.jpg

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

New Member
Messages
42
Reaction score
0
Points
0
Location
egypt
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

New Member
Messages
42
Reaction score
0
Points
0
Location
egypt
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

Nuclear Engineer
Messages
707
Reaction score
1
Points
18
Location
VA
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.
 

aeacfm

New Member
Messages
42
Reaction score
0
Points
0
Location
egypt
to me ; as the surface area of the electrode increase it gives more efficiency

Unfortunately i need this high surface area electrode
 

aeacfm

New Member
Messages
42
Reaction score
0
Points
0
Location
egypt
ok sir
but if i dont need voltage below 1.2 V what i will eliminate
 

Nukeman

Nuclear Engineer
Messages
707
Reaction score
1
Points
18
Location
VA
ok sir
but if i dont need voltage below 1.2 V what i will eliminate


Just use the link that I gave you in post #31. There is a diagram in there for a regulator that goes from 1.2v-25v. In order for it to reach 25v, the input voltage has be be over 28v. You don't need this, so you can simply use the +12v from a computer supply to power it. This will allow you to do 1.2v-9v and supply up to 5A. The LM138 (or LM338) will also need a heatsink to prevent it from getting too hot
 

aeacfm

New Member
Messages
42
Reaction score
0
Points
0
Location
egypt
ok i will try
thanks for every thing

i will show you the results as soon as possible

see you soon
 

aeacfm

New Member
Messages
42
Reaction score
0
Points
0
Location
egypt
dear nukeman

how are you friend

i Almost assembled the circuit you figured to me (iam sorry for this long time but that because i dont have such experience in electricity) but my problem here the Bottom of the figure as i dont know how to connect them together

many thanks
 

Nukeman

Nuclear Engineer
Messages
707
Reaction score
1
Points
18
Location
VA
I am doing well.

I assume that you are talking about the ground (downward pointing arrow). C1, C2, and one side of R2 are connected together. This will also connect to the ground on your computer power supply.

One other thing: R2 is an adjustable resistor. These have three terminals (left, center, right). You will connect the center terminal to "ADJ" of the regulator. Then one of the other terminals (either right or left) will connect to ground. One of the termianls (either right or left) will not be connected to anything.

Let me know if you have any questions.

Good luck.
 

aeacfm

New Member
Messages
42
Reaction score
0
Points
0
Location
egypt
hello nuke man
i am not sure if still remember me

i had abig problem because i am a stubid man because the current was in milliampere not in ampere so what u give me was very high so the board burned
iam sorry i have lost your effort with me.
so i it would be of pleasure of you if the last values was in milliampere by my calculations it would be from 1 mAmp to 3 or 4 mAmp

many thanks for your help and time
 

Nukeman

Nuclear Engineer
Messages
707
Reaction score
1
Points
18
Location
VA
You must have had a problem in the circuit (possibly a short). This is a voltage souce, so it will put out whatever current is needed (up to the maximum value). If the current is in mA, then you could build the circuit with the LM317 (probably easier to find than the LM 138/338).

Can you post a picture of the circuit that you built? Maybe we can see where the problem is.
 
Top
Hey, wait a minute.

This is awkward, but...

It looks like you're using an ad blocker. We get it, but (1) terrylove.com can't live without ads, and (2) ad blockers can cause issues with videos and comments. If you'd like to support the site, please allow ads.

If any particular ad is your REASON for blocking ads, please let us know. We might be able to do something about it. Thanks.
I've Disabled AdBlock    No Thanks