Galvanic Isolators

[lymmranger]

Interesting. Chris - can I ask why? Given the thrust of the thread (before it deviated into how to electrocute passing swimmers on the cut) was as I understand it that no GI could be guaranteed to avoid the risk of AC conduction in the circumstances previously discussed, why a GI rather than a IT? Purely cost or do you have a reason for believing this GI is more reliable?

 

And more to the point why, given your knowledge are you not building your own to incorporate previously discussed "add ons"

[chris w]

1. It can handle a continuous current that's 2.5 times the value at which my circuit breaker should have already cut the current.

 

2. The chances of the RCD not activating anyway under fault conditions in a properly bonded hull are a very remote possibility

 

3. It contains a capacitor to bypass any stray low voltage AC around the diodes so allowing the GI to do its job 100% of the time

 

4. After any power outage caused by the RCD and/or breaker, I would always manually check the viability of the GI. (I'll probably rig up a test circuit permanently)

 

5. The IT option is heavy, bulky, expensive and still needs monitoring as it, too, does have failure modes.

 

6. I can't mitigate against every failure possibility but one has to take a view and, on balance, IMHO this is a low risk solution to galvanic corrosion.

 

7. At only £129 (less a discount) it's not worth faffing around making my own aluminium heatsinks and epoxy potting stuff etc etc

 

Chris

Edited November 16, 2006 by chris w

[Yoda]

Actually, with all of this dicussion about galvanic isolaters and having high voltage circuits bonded to the hull, why don't we ask ourselves another pertinent question.

 

As we are dicussing steel hulled boats, can someone tell me why we still fit anodes to the hull?

[chris w]

Yoda

 

Because, while a GI will stop galvanic corrosion caused by dissimilar metals being connected together via a shore earth line (even two steel boats will have slightly different steel compositions) , it won't stop cathodic corrosion caused by dissimilar metals which are connected together on the same boat. eg: props, hull fittings, the hull etc.

 

The anode is designed (by its composition) to be the most negative metal and therefore sacrifices itself (ie becomes the donor metal) rather than some other part of your boat.

 

Anodes are usually magnesium in fresh(ish) water because zinc forms a coating of zinc hydroxide after a few weeks in fresh water which severely reduces its anode effect.

 

Chris

Edited November 16, 2006 by chris w

[Yoda]

Yes, thank you, I know this. The only problem is sometimes they don't work. Can't find the thread I was looking for.

 

Are well back to ........

[Grasshopper]

It contains a capacitor to bypass any stray low voltage AC around the diodes so allowing the GI to do its job 100% of the time

 

 

A long, long time ago in a galaxy far, far away..........or possibly just page 4 of this thread....... I asked:

 

Earlier there was mention of GI's to ABYC standards that have a capacitor wired in parallel to the diodes. Would one of these reduce the risk?

 

 

And Allan replied:

 

Gibbo ended up in an extremely long and very technical debate on uk.rec.waterways about the merits or otherwise of putting a capacitor across a GI. It wouldn't significantly reduce the risk of a failure, but there is an outside chance it may reduce the level of RF current passing through the GI, although theory says it's pointless unless you can experiment on your own particular setup until (and if) you find something that works.

 

Now I read 'outside chance' and 'although theory says it's pointless' as meaning a capacitor wouldn't reduce the risk by much if anything at all.

 

I take it, you disagree?

[chris w]

At the (high) risk of spawning another "War & Peace" , AC theory shows that the reactance ("resistance" in simple terms) of a capacitor is 1/[2 x (pi) x frequency x capacitance] so for a 25000uF (microfarad) capacitor its reactance works out to be 0.13 ohms at 50Hz mains frequency. At higher frequencies (like the 30KHz switching frequency of a power supply/charger) it will be infinitesimal (about 200 micro ohms).

 

The capacitor is wired across the GI diodes, so stray AC current can either pass through the diodes or through the capacitor. The resistance of the diodes will be typically much greater than the reactance of the capacitor so the AC will bypass the diodes and pass straight into the shore earth lead. (For the purists, combining resistance and reactance is not as simple as just combining pure resistances but for this analysis it will suffice).

 

Since the AC bypasses the capacitor rather than passing through the diodes, the AC can't bias the diodes into forward conduction and so the diodes present a full 1.2 volt "wall" to galvanic currents. The maximum galvanic voltage is around 0.8v so the diodes act as an effective block once more even in the presence of stray AC voltages.

 

Having weighed everything up, this is my own personal decision on the way to go

 

Chris

Edited November 16, 2006 by chris w

[Gibbo]

At the (high) risk of spawning another "War & Peace" , AC theory shows that the reactance ("resistance" in simple terms) of a capacitor is 1/[2 x (pi) x frequency x capacitance] so for a 25000uF (microfarad) capacitor its reactance works out to be 0.13 ohms at 50Hz mains frequency. At higher frequencies (like the 30KHz switching frequency of a power supply/charger) it will be infinitesimal (about 200 micro ohms).

 

The capacitor is wired across the GI diodes, so stray AC current can either pass through the diodes or through the capacitor. The resistance of the diodes will be typically much greater than the reactance of the capacitor so the AC will bypass the diodes and pass straight into the shore earth lead. (For the purists, combining resistance and reactance is not as simple as just combining pure resistances but for this analysis it will suffice).

 

Since the AC bypasses the capacitor rather than passing through the diodes, the AC can't bias the diodes into forward conduction and so the diodes present a full 1.2 volt "wall" to galvanic currents. The maximum galvanic voltage is around 0.8v so the diodes act as an effective block once more even in the presence of stray AC voltages.

 

Having weighed everything up, this is my own personal decision on the way to go

 

Chris

 

I'm going to say this once and once only. Then I will not comment on it again. I refuse to do the same thread again as on urw. You can always look up the thread via google groups.

 

In *very* basic theory what you say is, of course, perfectly correct. However this assumes ideal components, which simply do not exist. In practice when you add in ESR, ESL, inductance of the wiring etc you find that the presence of the cap in the GI can actually make matters *worse* at RF not better. That is the last comment I will make on the subject.

 

You will need some *serious* theory to underdstand it. You may possess that theory, you may not, I have no way of knowing.

 

Gibbo

Edited November 16, 2006 by Gibbo

[chris w]

I do have the knowledge [bSc (Hons) in Electronics inter alia] and, as you know, one can't do a complete AC theory treatise when answering these posts. I am fully aware of the second and third order effects of adding a capacitor but, having spoken to the manufacturer's technical people, I am convinced that it will solve the problem as they understand them too.

 

If, in practice, it doesn't I may return it for a full refund so I have nothing to lose and lots to gain.

 

Chris

Edited November 16, 2006 by chris w

[smileypete]

OK, OK, I'll repeat what I have previously asked, to try and make things crystal clear

 

The results I would like to see is some real world measurement including:

 

The type and size of boat.

The type or design of the GI.

The equipment used which generates RFI.

The amount of AC current and most importantly DC current conducted by the GI.

 

It's not the presence of voltage across the GI but any DC current flowing through it that's important.

 

It's DC current flowing through a GI and the earth connection that contributes to galvanic corrosion.

 

After a lot of detailed discussion, no one has offered any useful real world measurement of this.

 

If the voltage across the GI is resulting in a DC current of 10uA, is that really bad enough to spend a few hundred on an isolation transformer?

 

Is the DC current 10uA? 1mA? 100mA 10A? Why not measure it and find out???

 

cheers,

Pete.

[chris w]

Pete

 

The GI voltage is very important. If it's below about 1.2v there is NO current flowing though it. That's what we need to achieve.

 

Even if this voltage is exceeded and a current flows, I have no idea and probably no-one knows exactly what corrosion, in quantitive terms, is caused by10pA or 10uA or 10mA of current.

 

But no current = no galvanic corrosion

 

Chris

Edited November 16, 2006 by chris w

[Gibbo]

Pete

 

The GI voltage is very important. If it's below about 1.2v there is NO current flowing though it. That's what we need to achieve.

 

Even if this voltage is exceeded and a current flows, I have no idea and probably no-one knows exactly what corrosion, in quantitive terms, is caused by10pA or 10uA or 10mA of current.

 

But no current = no galvanic corrosion

 

Chris

 

A chemist on urw calculated how much steel is lost per mA of galvanic current. Apparently it's simple if you kinow what you're doing.

 

But essentially I agree, no current = no galvanic erosion. Surely that must be the aim?

 

Gibbo

[Amicus]

If I might take advantage of the lull just to say thanks team, this has been very edumacenshuul. Even this ol’ fart half-wit has been enlightened. Good stuff.

[Amicus]

If I can buy a 5KVA one of these

 

http://cgi.ebay.co.uk/BRAND-NEW-240V-110V-...1QQcmdZViewItem

 

for 75 quids, and its fully isolated, how come a boaty type transformer cost hundreds, and, need to be isolated in a case?

Can I use two of the above, back to back, or front to front IYSWIM

[Gibbo]

If I can buy a 5KVA one of these

 

http://cgi.ebay.co.uk/BRAND-NEW-240V-110V-...1QQcmdZViewItem

 

for 75 quids, and its fully isolated, how come a boaty type transformer cost hundreds, and, need to be isolated in a case?

Can I use two of the above, back to back, or front to front IYSWIM

 

Yes you can, and as usual there's a but.......

 

You'll get a bit of voltage variation, but not enough to cause a problem.

 

You'll have to take the lids off and play with the earthing arrangements.

 

The biggest problem is the rating. They "tool rated" which means intermittent rating. If you run them at full power for a long time they will burn out, or at the very least keep cutting out on thermal overload.

 

Gibbo

[Osprey Sprinkler]

If the voltage across the GI is resulting in a DC current of 10uA, is that really bad enough to spend a few hundred on an isolation transformer?

 

Is the DC current 10uA? 1mA? 100mA 10A? Why not measure it and find out???

 

cheers,

Pete.

 

Hi Pete and all

 

Here are some examples of how long it would take to erode away 1Kg of magnesium anode for various boat to boat resistances therefore various currents (that is 1Kg total shared among the anodes, not 1Kg per anode):

 

1000 Ohms – 1Kg loss in 300 years (0.8mA)

100 Ohms - 1Kg loss in 30 years (8mA)

10 Ohms - 1Kg loss in 3 years (80mA)

1 Ohm - 1Kg loss in 4 months (800mA, and I really don't believe this one)

 

I've given a sample calculation below if anybody cares to check. These are absolute maximum corrosion rates for each current - there is no way a given current can free more than this much anode but there are many ways you could lose less. Iron would be lost a little faster for a given current because of its higher relative atomic mass but, as all of this is driven by a very small voltage, I think it is fair to assume that the magnesium would take the majority of the loss.

 

Of course you will still lose plenty of metal in the usual way through chemical corrosion and galvanic effects between different metals on your own boat; this is just a measure of how much extra you will lose through being connected to shore power with no isolation.

 

I’ve heard salt water types mention keeping currents under 20mA, which is around 1Kg loss every 12 years if you are connected 24 hours a day, 365 days a year – longer if you ever unplug and go cruising. I would guess that currents are lower in fresh water. If this is achievable then I for one would rather put up with replacing my anodes at most one year sooner and have the reassurance of an unimpeded non-isolated shore power earth. (But I’d still unplug it when not in use. )

 

 

If somebody could measure an actual current earth to hull we'd know where we really stand. All this would take is someone brave enough to disconnect their shore power earth for a few seconds and stich the ammeter in.

 

Can I assume that the reason why it is hard to measure the resistance between adjacent boats is because the galvanic PD confuses the meter?

 

Ashley (physicist not chemist )

 

 

 

 

 

Calculation:

Time to corrode 1Kg of Magnesium (sorry, notation is total carp in text mode) =

(oxidation state of Mg x Avogadro's number x grams per Kg x charge on an electron x resistance hull to hull) /

(Galvanic voltage x atomic mass of Mg x seconds per year)

 

Substituting

oxidation state of Mg = 2

Avogadro's number (atoms per gram mole) = 6e23

grams per Kg = 1000

charge on an electron = 1.6e-19

resistance hull to hull = 1000

Galvanic voltage = 0.8

relative atomic mass of Mg = 24

seconds per year = 60x60x24x365

 

gives 317 years - say 300 years to one sig. fig. given the finger-in-the-air nature of all this.

Can justify this in detail if anybody can stay awake long enough.

[chris w]

Thanks Ashley, very interesting

 

You and I should definitely start a maths thread on here.

[PeterF]

Do it NOW Peter. Remember Gibbo is never wrong!

 

Chris,

 

Oh you meant do the FE cal now, I obviously thought "it" meant something else.

 

Anyway I put together a model and the result is below. I know this has lost its place as the discussion has moved on quite interestingly.

 

This is to scale and shows quite wide current lines. The model was of 2 boats sat on the surface of the water to simplify model. Base plates were 2m wide. The model was done as 2D ignoring end effects. The boats were spaced 2m apart, hence mid points of the base plates were 4m apart. The water was 4m deep. I could be accused of being disengenious here (if only I could spell it) as water is not normally that deep. I have also assume the water is conatined in an insulated container so there are no grounding effects. The water is 200µS/cm, the boat on the left is at +1V, the boat on the right at 0V. The coloured areas are contours of constant voltage, the bright red from 0.9 to 1.0V, the dark blue from 0 to 0.1V and each one representing a 0.1V range. The black arrows represent current stream lines, the area below the first line has 10% of the current flow, the area below the uppermost line has 100% of the current flow with the space between pairs of lines containing 10% of the current flow. The current streamlines are orthogonal to the contours of constant potential as they should be.

 

The model predicts a total current flow of 0.185A for a 15m long boat giving a resistance of 5.4 Ohms. I ran the model with 2 flat plates 4m apart 2m wide and 15m long and this gave 6.7 Ohms as had already been established and hence 0.15 Amps.

 

I agree if I had restricted the water depth to say 0.5m below the hull then the resistance would have been higher and the contours shallower. The effect would also have been different had the surroundings benen modelled as earthed at 0V.

 

However, to say the surrent only flows in a direct thin strip is wrongbecause if no current was flowing from the bottom of the left hand boat downwards this would mean that there was no voltage differential, hence the water under the left hand boat would be at 1V. The same goes for the ater under the right hand boat which would have been at 0V. therefore you would have 2 volumes of water from boat to bottom of container at different voltages, which would not be sustainable, thus the current can flow in arcs between point sources in a large volume.

 

Peter.

 

Edited November 17, 2006 by PeterF

[Osprey Sprinkler]

This is really interesting Peter. Certainly gives a good picture of the potential gradient. Are you using your own home-brew modeling tool?

 

I'm trying to envisage how things would change for two hulls with conducting bases 0.5m deep and insulating sides in a marina 1m deep with a grounded floor. I'm sure you're right in saying that the boat to boat resistance will be larger. My guess is that the hull base to marina floor path would be dominant, being 2m wide and only half a metre long. My chemistry isn't up to knowing what the charge transport mechanism might be between steel or magnesium and mud so I couldn't guess if it involves loss of metal.

 

Wouldn't it be tha case that the steel base plate would be only half a volt or so above ground though, and an even smaller fraction of a volt different from one base plate to another? If there is a magnesium anode at each end of the hulls (or maybe just one of the hulls) which is the better part of a volt higher than the base, won't most of the current flow preferentially through this? I expect this is getting a bit complex to model.

 

Good food for thought, though.

 

Ashley

[PeterF]

This is really interesting Peter. Certainly gives a good picture of the potential gradient. Are you using your own home-brew modeling tool?

 

I'm trying to envisage how things would change for two hulls with conducting bases 0.5m deep and insulating sides in a marina 1m deep with a grounded floor. I'm sure you're right in saying that the boat to boat resistance will be larger. My guess is that the hull base to marina floor path would be dominant, being 2m wide and only half a metre long. My chemistry isn't up to knowing what the charge transport mechanism might be between steel or magnesium and mud so I couldn't guess if it involves loss of metal.

 

Wouldn't it be tha case that the steel base plate would be only half a volt or so above ground though, and an even smaller fraction of a volt different from one base plate to another? If there is a magnesium anode at each end of the hulls (or maybe just one of the hulls) which is the better part of a volt higher than the base, won't most of the current flow preferentially through this? I expect this is getting a bit complex to model.

 

Good food for thought, though.

 

Ashley

 

I knocked that one up on Excel, a lot of my job involves mathematical modelling of various chemical, thermal and mechanical processes and Excel can be used for rough and ready models. I am not volunteering to do any more stuff though. I only started this as a question around conductance in the water in a marina. What it has shown is as per GIbbo's earthing comments that potential differences can build up in the water between hulls with a raised voltage and local earthed surfaces be they other boats or the bed of the canal. I also have to state that I assumed no resistance between the hull and the water - the hull will be in some unkown oxide form rather than bright steel. Certainly in a shallow marina the most direct path is to earth - but what if the marina was lined with a membrane during construction, can the earth be taken as true ground in this case. In terms of current flowing then depending upon which way it is flowing and to / from what your boat could either be the the giving up its metal or your boat could be OK (or even being electro plated). In terms of adding anodes to this type of model, yes it can be done but it will be more complex and I am not going to do it.

 

Cheers,

 

Peter.

[Gibbo]

Chris,

 

Oh you meant do the FE cal now, I obviously thought "it" meant something else.

 

Anyway I put together a model and the result is below. I know this has lost its place as the discussion has moved on quite interestingly.

 

This is to scale and shows quite wide current lines. The model was of 2 boats sat on the surface of the water to simplify model. Base plates were 2m wide. The model was done as 2D ignoring end effects. The boats were spaced 2m apart, hence mid points of the base plates were 4m apart. The water was 4m deep. I could be accused of being disengenious here (if only I could spell it) as water is not normally that deep. I have also assume the water is conatined in an insulated container so there are no grounding effects. The water is 200µS/cm, the boat on the left is at +1V, the boat on the right at 0V. The coloured areas are contours of constant voltage, the bright red from 0.9 to 1.0V, the dark blue from 0 to 0.1V and each one representing a 0.1V range. The black arrows represent current stream lines, the area below the first line has 10% of the current flow, the area below the uppermost line has 100% of the current flow with the space between pairs of lines containing 10% of the current flow. The current streamlines are orthogonal to the contours of constant potential as they should be.

 

The model predicts a total current flow of 0.185A for a 15m long boat giving a resistance of 5.4 Ohms. I ran the model with 2 flat plates 4m apart 2m wide and 15m long and this gave 6.7 Ohms as had already been established and hence 0.15 Amps.

 

I agree if I had restricted the water depth to say 0.5m below the hull then the resistance would have been higher and the contours shallower. The effect would also have been different had the surroundings benen modelled as earthed at 0V.

 

However, to say the surrent only flows in a direct thin strip is wrongbecause if no current was flowing from the bottom of the left hand boat downwards this would mean that there was no voltage differential, hence the water under the left hand boat would be at 1V. The same goes for the ater under the right hand boat which would have been at 0V. therefore you would have 2 volumes of water from boat to bottom of container at different voltages, which would not be sustainable, thus the current can flow in arcs between point sources in a large volume.

 

Peter.

 

 

And this is now *way* off topic!

 

Nice simulation. I assume that's based on a matrix as you mentioned earlier?

 

This will work for a metallic type conductor, i.e. where the majority carriers are electrons. As you will know the majority carriers in an electrolyte are ions. The speed of these is *ridicluously* slow compared to electron current due to their much higher mass and physical size (Stokes' law - fluid dynamics - the mathematics of which leaves me completely stumped!). Even the speed of (individual) electron flow is *much* slower than many people imagine.

 

I'll make some assumptions that you accept that a moving charged particle (of whatever nature) creates a magnetic field? You accept that charged particles of the same polarity moving in the same direction create a magnetic field that tends to attract the charged particles?

 

This isn't some microscopic effect that only becomes apparent at the atomic scale or over microscopic distances. It can be demonstrated simply enough by passing a current, in the same direction, down two parallel non magnetic conductors (say copper wire). The conductors will pull together with a force proportional to the current. Due entirely to the charged particles being attracted together by their respective magnetic fields.

 

In an electrolyte this effect is massively increased due to the extremely low speed of the charge carriers (if there are any boaters left reading this [which I doubt] think of the effect of a side wind on a narrowboat at 2mph compared to the effect of the same wind on a speed boat at 60mph). The magnetic field created by any ions of opposite polarity travelling in the opposite direction will be the same. The result is that the current is forced to flow in a narrower channel than would be the case in a metallic conductor.

 

I did subtly allude to this in an earlier post when I specifically referred to current in a conductive fluid (I should *really* have said in an electrolyte) as opposed to in a solid.

 

You state that the current flow will be at 90deg to the gradients of electrical potential. This is, of course, correct. But it could also be argued that the gradients of electrical potential will be at 90deg to the current flow. Put a low resistance path somewhere in your simlation. This will force the current to flow mainly through that low resistance path. This, of course, will move the potential gradients. Cause and effect can be determined either way round in so far as either one can affect the other. In this case the magnetic field will force the current into a narrower channel than a simple Ohms law calculation will show, this will have the effect of moving the potential gradients thus the resultant simulation would still obey Ohms law.

 

There used to be a demonstration of this somewhere on the net but I cannot for the life of me find it. It was a video clip of some electrodes in an electrolyte passing a current. The electrolyte was made of water in which was dissolved some chemical (which completely evades me at present) that was highly visible (there may have been ultraviolet illumination but I can't remember). The movement of ions could clearly be seen and therefore the current flow could clearly be seen. It was much narrower than would be expected by Ohms law.

 

It has to still be out there somewhere but as I say I just cannot find it anywhere.

 

I suspect there maybe something out there related to electroplating that may show this effect.

 

I don't expect you to write a simulator for this but if you really feel ike it the results would be interesting.......

 

Gibbo

[chris w]

Even the speed of (individual) electron flow is *much* slower than many people imagine.

This is because (and I appreciate Gibbo will know this already but I thought others might be interested) an electron doesn't per se flow from one end of the wire to the other. It leaves its atom and fills a vacant hole in the next atom, thus leaving behind a hole in its original atom which is filled by an electron from a neighboring atom and so on.

 

The usual analogy given is to imagine a row of people seated at the theatre with an empty seat at one end. If they all move up one place, they have only moved a short distance at slow speed but the "hole" (the vacant seat) shot down to the other end at fast speed in the opposite direction.

 

Originally, current was arbitrarily said to flow from positive to negative. Then electrons were discovered and so it became that "conventional" current flowed from pos to neg but that "actual" current (consisting of electrons) flowed from neg to pos. However, conventional current and actual current (the movement of "holes") are once more in line with each other as holes move in the opposite direction to electrons just like the empty seat moves in the opposite direction to the people. The "majority charge carriers" are nowadays accepted to be these "holes" and not the electrons themselves.

 

Chris

[Gibbo]

Originally, current was arbitrarily said to flow from positive to negative. Then electrons were discovered and so it became that "conventional" current flowed from pos to neg but that "actual" current (consisting of electrons) flowed from neg to pos. However, conventional current and actual current (the movement of "holes") are once more in line with each other as holes move in the opposite direction to electrons just like the empty seat moves in the opposite direction to the people. The "majority charge carriers" are nowadays accepted to be these "holes" and not the electrons themselves.

 

Chris

 

That majority carriers depend upon the material. Whichever it has the most of. For instance metallic conductors have an abundance of free electrons, so the majority carrier in metals are the electrons.

 

Semiconductors come in both types, p-type and n-type. P-type having an abundance of holes (the holes are the majority carriers) and n-type an abundance of electrons.

 

I like the analogy of a tube full of ball bearings. Put a new ball in one end, and *instantly* one comes out of the other end. Even though each ball has only moved one ball diameter the overall effect is that the "information" moved the entire length of the tube in the time it took to put one ball in.

 

Thinking about this, if you make the tube long enough, you can exceed the speed of light

 

Gibbo

[ChrisPy]

Thinking about this, if you make the tube long enough, you can exceed the speed of light

 

 

................ er, no .......................

[chris w]

Interesting thought but even two light beams approaching each other, each at the speed of light ©, have an approach speed which is still c.

 

One can exceed the speed of light, in a given medium. As when particles leave a nuclear reactor at the speed of light and hit cooling water. For an instant they are travelling faster than light normally does in water. This gives a blue glow called Cherenkov Radiation.