hazardous setup without the hydrogen and oxygen being separated. Re-design of the electrolyser should be a first priority ahead of power supply considerations.

Not only the things I talked about in that post, but his electrial connections are in the mixed gas cloud! A loose wire causing a spark?? The normal way of doing this is having each pole of the plates incased in it's own output chamber, to keep it from mixing.

There is a reason industry doesn't make their gasses this way. With all the other things being done as shown he is probably using tap water and using table salt, also making chlorine gas with the hydrogen and oxygen.

 

All my gasses are vented to the atmosphere during these stages so there is no risk, even if I had hypothetical sparks flying around.

Now back to my PSU issues.

I have made two new power supplies to replace the cheap off the shelf option that could deliver 4A but the max voltage across the electrolyser could only reach about 2.5V. This I am assuming is due to the voltage drop occurring as current passes through the electrolyser and the somewhat exaggerated performance of the PSU.

The first additional PSU I have built is based on an old Pentium 4 AXT supply (see Pics 1&2) but this still demonstrated a voltage drop across the output/top of electrolyser (see Pic 3). [The reason why I have to use a separate ammeter and voltmeter with this setup is that, due to the voltage drop, the inbuilt meter supply falls below the 6V it needs to operate and so the display shuts off.]

So thinking it might be due to the same issue I went a step further and built a beefier PSU based on converting an HV transformer from an old microwave to a step-down unit with rectification and smoothing (Pic 4) and based on a design shown earlier in this thread. Yet again, when delivering about 7A, the voltage across the electrolyzer reads a bit under 3V.

I then tried the theoretically ideal constant voltage source for these requirements, namely a car battery, which delivered over 10A but with a voltage drop to 8V from its resting 12.8V.

So to my query, which I suppose is actually more theoretical:

Electrolysis requires a minimum of 1.2V to start and, for good operation, I would like around 2V for each set of plates. So in my device, which has 2 groups of 4 pairs of plates, I require 8V for the most efficient operation of these cleaning and conditioning stages for the electrolyser. [The main gas generation stage after plate preparation will be using an auto-tuning pulsed square wave system that is nearing completion]

So if I have, say 8V from my supply with no load, and I am then losing volts when it's under load and the current is flowing through the electrolyser, does this mean that I am no longer supplying about 2V for each pair of plates or does the voltage drop only reflect the electrical work that has already been done and so I needn't worry about it? In other words, if I have an 8V supply and then I connect it up, is it really relevant what the resulting voltage across the electrolyser reads?

If the answer is that the resultant voltage across the electrolyser is not important then I can use one or more of my available PSUs. If the resultant voltage does matter then I will probably be better off using the car battery since that has the minimum voltage drop and the value it drops to is close to what I need anyway.

Thanks

Jules

 

The answer is no. If the voltage at the electrodes fall then that is what the cell is seeing.
Ohm's Law applies.

How old is your car battery?
A good battery can deliver 500A CCA = Cold Cranking Amps, for 30 seconds at -18°C without dropping below 7.2V.

All power sources, power supplies, batteries have internal resistance.
If the voltage at the load falls then you are witnessing the effect of the internal resistance of the power source at that load current.

 

Study Ohm's law, until you get it. It is the most important concept in electronics, and it is enlightening.

 

The answer is no. If the voltage at the electrodes fall then that is what the cell is seeing.
Ohm's Law applies.

How old is your car battery?
A good battery can deliver 500A CCA = Cold Cranking Amps, for 30 seconds at -18°C without dropping below 7.2V.

All power sources, power supplies, batteries have internal resistance.
If the voltage at the load falls then you are witnessing the effect of the internal resistance of the power source at that load current.

Battery is new 40Ah one. Yes I’m aware of the ‘Ir’ losses but hoped that in theory at least the loss might be realised after the electrons had done the work of releasing the Hydrogen.

The funny thing is that despite the 3V reading all the plates were gassing in a small way. Or at least there were small bubbles forming on all the plates which were rising up to the top.

Also the resistance across the cell terminals reads quite high, tens of kilohms, I will check again tomorrow and yet with the current flowing of around 7A it implies around 1ohm. The resistance must drop considerably when ions start to migrate.

 

I have just measured the resistance across the electrolyser terminals with no volts applied to it. It's 1 MOhm. When voltage is applied, based on the V and I, then it's less than 1 Ohm.

Regarding using a car battery, with 5-10A being drawn I wouldn't expect to lose 4.6V. That implies that the internal resistance is around 0.5 Ohm. I thought a Lead Acid battery was quite a bit lower than that. Anyway I think the car battery is my best bet to do what needs to be done.

 

As I understand post #22, your car battery is bad. Any decent car battery should be able to provide 12 A (i.e., 12 V through a 1 Ω resistor) without showing a 4 V drop due to internal resistance.

Are you sure electrolysis cells operating at 8V to 12 V and plenty of current are adequately represented by as an ohmic resistance measured at a much lower voltage and current. Where does the work to split water come from?

 

As I understand post #22, your car battery is bad. Any decent car battery should be able to provide 12 A (i.e., 12 V through a 1 Ω resistor) without showing a 4 V drop due to internal resistance.

Are you sure electrolysis cells operating at 8V to 12 V and plenty of current are adequately represented by as an ohmic resistance measured at a much lower voltage and current. Where does the work to split water come from?

My battery may well be bad but it shouldn’t be as it’s new (but relatively cheap).

Electrolysis requires a minimum of 1.2V to start but 2V is usually provided to ensure it and offset any resistances. That’s where the energy to split the water comes from.

 

I understand a little about electrolysis. My point was, when you measured the resting cell's ohmic resistance, was electrolysis going on too? Was the rate the same as you see with your 12 V battery attached. That extra energy needs to come from somewhere. right? And of course, it will show as more current, so your V/I will be lower.

 

I understand a little about electrolysis. My point was, when you measured the resting cell's ohmic resistance, was electrolysis going on too? Was the rate the same as you see with your 12 V battery attached. That extra energy needs to come from somewhere. right? And of course, it will show as more current, so your V/I will be lower.

The 1MOhm Reading was with no battery or power supply attached. I had based the 1 Ohm value on the V and I readings with a supply attached and I suppose I could measure resistance across the cell when it’s drawing power but the value on the resistance meter would be affected by the power supply being in parallel with it.

 

The 1MOhm Reading was with no battery or power supply attached. I had based the 1 Ohm value on the V and I readings with a supply attached and I suppose I could measure resistance across the cell when it’s drawing power but the value on the resistance meter would be affected by the power supply being in parallel with it.

No. You cannot measure resistance with an ohmmeter when the circuit is being powered. Measure voltage and current and apply Ohm’s Law.

 

It was pretty clear what the 1 MΩ reading represented. Was it valid? How do you know? What was your control? As for the rest, if you don't explain what your actual test conditions are, how do you expect answers.

 

No. You cannot measure resistance with an ohmmeter when the circuit is being powered. Measure voltage and current and apply Ohm’s Law.

Which is what I did.

 

Most of ATX have rather high 12 V channel with voltage loop what is easy to convert toward the current loop. Its zillions of publicats at web for this theme.
RE:""Study Ohm's law, until you get it"" Electrolyte is non-linear resistor. At small voltage the resistance is very large, at large voltage it may be near zero.

 

Most of ATX have rather high 12 V channel with voltage loop what is easy to convert toward the current loop. Its zillions of publicats at web for this theme.
RE:""Study Ohm's law, until you get it"" Electrolyte is non-linear resistor. At small voltage the resistance is very large, at large voltage it may be near zero.

Yes that fits The mobile ions change everything once they get moving