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I've built myself a electrolysis tank, using a plastic barrel as the container, rebar as the sacrificial cathode, a 12v 15A car charger (DC), and a Sodium Carbonate solutions as the eletrolyte. The anode is a rusty piece of metal that I need cleaned.

This works wonders, but I noticed that when I charge a 12v battery with the charger, the analog voltmeter jumps to around 12v as soon as I connect the terminals, but it doesn't when I connect it to the electrolysis tank?

I've seen it jump once, in the electrolysis tank - All the other times it's been near zero (Maybe slightly over?)

How come this happens? Shouldn't I measure 12v since that is what I output, or am I missing something? I am thinking that could be a huge voltage drop in the electrolytic solution, and that I'm actually only seeing a few volts, could be hard to read and display on the analog voltmeter (However, I'm not sure this is how it works...)

The charger/powersupply's negative wire is connected to the anode and the charger/powersupply's positive wire is connected to the cathode. The voltmeter is build into the charger/powersupply.

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  • \$\begingroup\$ "Shouldn't I measure 12v since that is what I output", what? \$\endgroup\$ – Harry Svensson Jul 27 '18 at 10:10
  • \$\begingroup\$ Yes, that is a very noobish question. I haven't done electronics in years and I'm a software developer by trade... So electrics is basically black magic for me :D \$\endgroup\$ – MadsRC Jul 27 '18 at 10:12
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    \$\begingroup\$ Show a picture how you connect your charger, meter and bath. \$\endgroup\$ – Andy aka Jul 27 '18 at 10:49
  • \$\begingroup\$ I'm unable to post a picture right now, but I've updated the question with an explanation of how it's connected. \$\endgroup\$ – MadsRC Jul 27 '18 at 10:55
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You have already identified impedance to be the key word.

In this case, the (output) impedance of the battery charger is relatively high, while the resistance of your 'circuit' (the water tank) is very low.

Basically, you have a voltage divider made from the impedance of the charger and the resistance of the tank in series. What you are measuring is the voltage across the small resistance of the tank, because you probably cannot measure the voltage across the internal resistance of the charger.

schematic

simulate this circuit – Schematic created using CircuitLab

In more detail, the charger is able to provide, say, 14V maximum or 15A maximum. When the current draw tends towards zero, the voltage will max out at about 14V; however, when the current drawn approaches 15A, it will be limited there and the voltage will drop in response to Ohm's law (you cannot push 15A through a 0.1 Ohm resistance and at the same time have 14V drop across that resistance.)

The charger may either employ some 'intelligent' charging controlling both voltage and current, or it may just limit the voltage to 14V and provide as much current as it can, which is implicitly limited by the impedance of some parts of the charger like e.g. its mains transformer.

I assume the "analog voltmeter" you mention is the one integral to the charger. This voltmeter is made to display voltage when charging batteries, so will probably be most 'accurate' (or responsive) for voltages between 10 and 14V; nobody usually needs the very-low-end scale below, say 8V, and nobody would expect the meter to be precise anyway; it's just some indicator to coarsely estimate the state-of-charge of the battery. These often don't even have a linear scale so that you may have a very hard time distinguishing 1V vs. 2V; otoh, 11V vs. 12V will probably be much more visible.

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Sounds like the voltage is dropping in your bath because the current it is taking is more than the charger can put out. [Forget the battery charging stuff - batteries have a voltage "pushback"]

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  • \$\begingroup\$ Sounds like I have some reading up to done on the fundamentals of electricity. Would it be correct to state that the voltage drop can be attributed to the impedance of the circuit (With the large, 60+L body of water being the main culprit)? \$\endgroup\$ – MadsRC Jul 27 '18 at 10:43
  • \$\begingroup\$ @MadsRC Yes. The impedance, or in this case resistance is probably too low for the charger to properly supply. Either the voltage is severely drooping under the load or a safety feature has cut in and killed the output \$\endgroup\$ – Dirk Bruere Jul 27 '18 at 12:32

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