[SOLVED] PSU current limiting on 12volt rail (Electrolysis)

[Earendil86]

Hello,

Wondering if someone might be able to help.

I'm using an 850watt power supply that can handle 70 amps on the 12v rail for rust removal (electrolysis) of larger than average items. In order to do this, I opened it up and combined ALL of the 12v wires and GND wires so that I have enough to handle the amperage without starting a fire. I also added a 60 amp inline fuse right out of the case to ensure I don't hit the 70 amp maximum draw.

My issue is that the current draw is based on a few factors but size of object and number of electrodes is one. The best way for large electroysis that I've seen is using a manual battery charge or DC welding supply, however, the first is harder to come by since most are automatic these days and the second is expensive.

My question is this, is it possible to limit the maximum current that the power supply will deliver rather than let the load pull as much as it wants and then trip the unit when it goes above? I ask this because I am hoping to do a larger item than I currently am working on, however, based on what I've seen, I imagine it'll pull more than 70 amps. With how electrolysis works, I don't need that high amperage anyways, but I need a way to limit it.

Thoughts?

 

[jay32267]

You could use a DC circuit breaker....but the trip point isn't very exact.

What if you used less electrodes on a larger item?

 

[kanewolf]

You need a 1000 watt .2 ohm resistor or something to limit current -- https://www.mouser.com/Passive-Components/Resistors/_/N-5g9n?P=1yz8ij3

 

[Earendil86]

Could definitely use less electrodes but then it's still guesswork in terms of what that upper bound is going to be until the switch is pulled. Also, having less electrodes can affect how well the item is derusted.

As for the 100watt .2ohn resistor. What's the math on that. Can you explain it a little? It's been a while since my electrical classes. An example would be awesome.

Thanks

 

[kanewolf]

Could definitely use less electrodes but then it's still guesswork in terms of what that upper bound is going to be until the switch is pulled. Also, having less electrodes can affect how well the item is derusted.

As for the 100watt .2ohn resistor. What's the math on that. Can you explain it a little? It's been a while since my electrical classes. An example would be awesome.

Thanks

Not 100 but 1000 W resistor. 12V * 70A max = 840W total -- To limit to 60A (720W) use ohm's law calc -- About .2 ohms will limit to around 60A at 12VDC.

 

[jay32267]

Not 100 but 1000 W resistor. 12V * 70A max = 840W total -- To limit to 60A (720W) use ohm's law calc -- About .2 ohms will limit to around 60A at 12VDC.

The issue you would have with this is...and it may not be a big deal...but it might be...is....much of the energy will be dissipated in the resistor.

At 60 amps...ALL of the energy will be dissipated in the resistor.

 

[kanewolf]

The issue you would have with this is...and it may not be a big deal...but it might be...is....much of the energy will be dissipated in the resistor.

At 60 amps...ALL of the energy will be dissipated in the resistor.

It is quite possible that the copper wire being used to connect the leads is the limiting factor.
Have you calculated (or measured) the resistance of the lead wire you are using?
This table can give you an estimate -- http://hyperphysics.phy-astr.gsu.edu/hbase/Tables/wirega.html

 

[Earendil86]

I'm using 0 gauge wire from the PSU leads to the electrolysis tank. Much thicker than needed and that chart shows 0 guage as being pretty low resistance. That chart shows solid wire and mine is stranded 0 Gauge from my car audio days but I imagine similar.

Now an interesting item to note, I've tested the PSU over current protection and that works and I ended up blowing my second 60 amp AGU fuse that I have inline on the power lead 2 times. So I ended up tossing in an 80Amp, knowing that I am then relying on the PSU's internal overprotection to make sure it doesn't blow.

Now, recently, it blew up the 80amp fuse but the PSU stayed running... ? Is it possible that the fuse block is seeing higher current that the PSU? I'm curious as to why the PSU wouldn't have tripped if the 80 amp fuse did. Any thoughts appreciated.

And to the resistor point, isn't Jay right in that at 60 amps the tank itself would not see any amperage? And does that actually prevent the circuit from going higher than 60 amps? or just that any amps over 60 would be consumed by the part?

 

[Earendil86]

I'll also add this.

I found this during some of my online searching:

Another way to control current and voltage is to use a 12Volt light bulb in series with the cathode lead. The bulb will limit the current (it's equivalent to adding a fixed value resistor in series) and will drop voltage across it. This is great for small parts where you don't want the current too high.
Tip: If you need say 4 amps to de-rust a larger piece, use 4 bulbs wired in parallel. The effective resistance is 4 times less than a single bulb. 4 times less resistance, allows 4 times the current flow. If each bulb alone would allow 1 amp to flow, then 4 in parallel will allow 4 amps to flow.

Is this saying that adding a 12v light bulb will prevent the tank from pulling large current amounts? Does it act like a straw in a glass of water? ie: will this prevent the circuit from pulling 60 or 70 amps of power?

Thanks,

 

[jay32267]

"I'm curious as to why the PSU wouldn't have tripped if the 80 amp fuse did. "

There are MANY variables that determine when a device trips for overcurrent.

For instance....and 80 amp breaker or fuse....generally doesn't trip at 80 amps.....at least not right away.

There are time vs current curves for fuses and breakers as well that describe this behavior.

"And does that actually prevent the circuit from going higher than 60 amps? or just that any amps over 60 would be consumed by the part? "

The .2 ohm resistor actually prevents the current from going higher than 60 amps.

 

[jay32267]

I'll also add this.

I found this during some of my online searching:

Another way to control current and voltage is to use a 12Volt light bulb in series with the cathode lead. The bulb will limit the current (it's equivalent to adding a fixed value resistor in series) and will drop voltage across it. This is great for small parts where you don't want the current too high.
Tip: If you need say 4 amps to de-rust a larger piece, use 4 bulbs wired in parallel. The effective resistance is 4 times less than a single bulb. 4 times less resistance, allows 4 times the current flow. If each bulb alone would allow 1 amp to flow, then 4 in parallel will allow 4 amps to flow.

Is this saying that adding a 12v light bulb will prevent the tank from pulling large current amounts? Does it act like a straw in a glass of water? ie: will this prevent the circuit from pulling 60 or 70 amps of power?

Thanks,

"Is this saying that adding a 12v light bulb will prevent the tank from pulling large current amounts? "

A 12V bulb for all practical purposes is the exact same as a resistor.

 

[Earendil86]

thank's for those updates. I'll add this here as I found it an interesting read on using cheap ceramic resistors to do exactly what I'm trying to accomplish. I'm now thinking of limiting the circuit to 56 amps and then building some extra combinations to give myself options (see chart at bottom). For smaller pieces I want to be able to run less amperage. My calculations come from the following site which also brings some questions and I'd like your thoughts:

https://woelen.homescience.net/science/chem/misc/psu.html

So basically, the author states that electrolysis uses roughly 5 volts and they use 7 volts for their resistor and amperage calculations. From that they calculate amperage using a combination of cheap power resistors. Unfortunately, I only have 5 - 43 Ohm, 10W resistors on hand, so with that logic as presented that puts the current flow at 0.81 amps with 5 of those in parallel so I need to buy others.

That being said, using that logic, your suggested 0.2 Ohm resistor (at 7 volts) only provides 35 amps.

So my updated question is...

1) why 7 volts instead of 12 volts?

Here is my calculated chart assuming 7 volts. I ensured that my total wattage is less than the Max wattage understanding that this ends up with some pretty large bundles of power resistors which might bring their own challenges. Thoughts on that would be appreciated.

QTY (in Parallel)	Resistor Value (Ohms)	Resistor Wattage	Total Resistance	Amperage	Wattage	Max Wattage
20	2.5	20	0.125	56	392	400
20	3	20	0.15	46.67	326	400
15	3	20	0.2	35	245	300
10	3	20	0.3	23.33	163	200
5	3	20	0.6	11.67	82	100
4	5.6	20	1.4	5	35	80
2	5.6	20	2.8	2.5	17.5	40

 

[kanewolf]

Since I don't do electrolysis rust cleaning, I don't know why 7v rather than 12. That almost 1/2 voltage change is what changes the amperage. I did my calculations with 12VDC...

 

[Earendil86]

Just to complete this thread for anyone who happens to stumble upon it. I ended up trying this with some H6054 headlights that I had with the following specs:

12.8V / 65W (Hi Beam)

I put 2 in parallel and measured some parameters then I put 3 in parallel and repeated. Here is a chart of what I found.

	2x H6054	3xH6054
Total Voltage (V)	12.2	12.2
Vdrop Lights (V)	8.33	7.88
Vdrop Electrolysis (V)	3.9	4.3
Current flow (A)	7.2	10.5

So the website that I found was pretty good with it's estimation of how the resistance would affect the voltages across the electrolysis tub.

I have a few more headlight bulbs that I'm going to add into the circuit tomorrow to bring up the amperage to save time with such a large piece, however, even now I have it bubbling away producing hydrogen and turning the water brown.

Thanks for all the reply's Kanewolf and Jay. Very much appreciated.