# How Do Power Inverters Work?

Discussion in 'Physics & Math' started by cb767, Aug 21, 2005.

1. ### MetaKronRegistered Senior Member

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I think the device is more commonly called a dynamotor. That's dynamo plus motor, and dynamo is generally accepted as the term for a DC generator. It is true that this setup would have trouble responding to a changing load, but a good design for a radio has high immunity to swings in B+ voltage. Tube or transistor, in my hobby work I avoid designs that have any great dependence on either the exact supply voltage or the exact characteristics of the components.

Someone mentioned the "ground return." I don't know that anyone, for any large scale power generation, ever used the ground as a return for the power, one wire as one leg, the ground as the other. The "ground return" is used for unbalanced consumption because the current is generated as three-phase and generally consumed as single phase. The three legs of the supplied power from the generating station will never, ever carry anywhere near the same wattage. This is why the center tap on the secondary side of a "pole pig" is grounded and it is also the neutral line. The house ground is for grounding electrical boxes, switches, and outlets. The neutral is not a good substitute for this. The house ground's function is to prevent electrical shock. If a wire comes loose in your appliance the electricity goes to ground, not your fingers.

One problem that I think that people will find in a DC line is that each line acts as a heavy-duty condenser. If you have a wire in the first place that is insulated from the ground its mass will charge up to hundreds or thousands of volts and when you have runs kilometers long that's a lot of stored energy. It's a really good idea to have drains here and there.

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It's correct that a dynamotor produces DC - but we were talking about AC output and that device is rightly termed a motor-alternator.

HVDC transmission systems are well desigined and a part of that design calls for it to operate at very high voltages to overcome I-square-R losses just as it is with AC transmission facilities. There's no need for "drains" which would also just create additional losses in the system.

5. ### Billy TUse Sugar Cane Alcohol car FuelValued Senior Member

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Until one of us gets some references to back our opinions up, we will just just disagree on much of this. You are correct in that you do not need to have feromagnetic materials to have a loss. There is the eddy current loss in any conductor subjected to an AC (voltage) field, but as noted earlier this loss and the hysterisis loss in the feromagnetic materials does not exist for the DC transmission line which got us speaking about transmission line loses.

I believe the energy loss in high voltage AC line are, in order of importance:
(1) I^2R loss due to current flowing in the line
(2) Corona loss due small wires have sufficient electric field at their surface to exceed the dielectric strength of the air. BTW this is why all high voltage lines do not use one single conductor, but usually 4 with at least 10cm separations so the the field at the wire surface is much less.
(3) Magnetic hysterious Loss in the iron of the towers. (Beats 4 and 5 as parts of tower are much closer to wires.)
(4) Eddy currents in the ground, when wet.
(5) Dielectric losses in the ground when dry
(6) Birds, their nests mainly, kite strings and tree limbs etc. until removed.
(7) Rain drops carring charge to ground after contact with the line.
(8) Shorts, including intentional ones with lightening induced "circuit breakers" briefly operating.

I will estimate the relative importance, annually, using ">" to be "order of magnitue" and >- to be factor of 5.
E.g. >>- is a factor of 50 difference and>>>>- is factor of 5,000. (All just my guesses.)

For AC line:
1 >>>> 2* >>> 3 >- 4 >- 5 > 6 >> 7 ~ 8
For DC line:
1>>>>- 2* >>>>>>> 7 ~8 and 3,4,5,6 are zero loss.

Post you guesses, or better, researched results with source link. If wildly wrong, I will be happy to learn what is correct. I included the conduction losses on surface of the insulators supporting the wires and dielectric loses in them with items 4 & 5 although a small part of these loses might be consider to be more like items 6 & 8. I also note that the monetary losses probably move item 8 ahead of item 6 for both AC and DC lines as repairman may be needed, especailly if the tower is knocked down by wind or some anti- government nut etc.

I still think that it is for "power factor correction" that capacitors are added along the way. I do not even understand how they can reduce the "inductive loses," which are unchanged if the current is unchanged.

However, perhaps we are not in disagreement much as if the power factor is lowered, then the current is increased to deliver the same power to users. That said, it still seems (to me) silly to call the increased losses without capacitors "inductive losses" as the increase in I^2R (item 1) is a 1000 times or more greater than the increase in "inductive loses." If you had said that the capacitors are added to reduce the "resistive loses," then I would agree. So perhaps it is more the names, than the physics, we are discussing on this point.
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*I am assuming each "wire" is actually 4, spaced properly.

Last edited by a moderator: May 11, 2007

7. ### kevinalmRegistered Senior Member

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Insert # 2.5, magnetic hysteresis losses in the steel wire strands braided together with the aluminum strands of the cable for strength. This may be 1.5, not sure.

8. ### PronatalistRegistered Senior Member

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750
Is earth-grounding a good idea still?

I would have thought that earth-return-grounding was popular, because the earth is so big, that its resistance would be lower than that of the wire, and so costs of wiring can be reduced. Obviously, at least 1 wire is needed, to avoid obvious short-circuiting and deliver power to where it needs to go. Isn't it a similar concept in grounding in cars?

But, are there strange effects, glows or cows being affected, by electrical currents running through the earth, or are these widely scattered enough to have neglible effects? I have heard of weird stuff like that, but it's probably in that unsubstanciated category like the ridiculous stories about UFOs.

Shouldn't most everybody at least have an inverter?

I have a 500 watt invertor. I figure it to be sort of like "a poor man's generator." I thought about all the stuff in my home that needs electrical power, that no longer works during a power outage. So should I buy a generator? Generators are big, heavy, bulky, expensive, and may not even start when needed. The better, more long-lasting generators aren't portable, but run on natural gas. The cheaper ones often have stale oil and gasoline in them, that may sit neglected for years. In comparison, while an invertor won't adequately power a refrigerator to guard against food spoilage, it will run most anything else, that isn't "power hungry." And for no more cost, than a "toy" battery-powered TV, I could theoretically power up my full size 27" fullscreen HD tube TV. Using my long yard power tool extension cords, and hooking up my invertor to my car battery. But should I? Those invertors in the affordability price-range, I wonder if their power is "pure" enough to power such expensive electronics safely? I saw some demo at an electronics store, with Monster surge suppressors, comparing how "noisy" electrical power is from a typical power strip, as compared to one of their pricy surge suppressors. Lots of noisy hum from the power strip, and very quiet from the speaker when the knob is turned to the Monster surge suppressor. Presumably, they are filtering out the 60 Hertz AC hum, since they both would still have it, leaving the customer to hear all the other garbage "noise" on the power line?

So far, I haven't really powered anything other than a test 60-watt lamp, and an electric cordless/corded shaver during a power outage, that had long since lost its ability to hold a charge.

9. ### Billy TUse Sugar Cane Alcohol car FuelValued Senior Member

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Good point - insert accepted as 1.5 as it may well beat out the corona if simple steel wire, but only as 2.5 if some steel with small hysterisis loop is used.

10. ### Billy TUse Sugar Cane Alcohol car FuelValued Senior Member

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If there is a lethal votage inside a metal box (for example a microwave) I think grounding is a very good idea. Not much corrsion damage from ground currents with AC anyway.

Your inverter would let you watch TV during storm that took down the wires etc, but I would not do it with car battery supply for more than a few minutes at a time. What you need is an electric boat battery which looks just the same, but is structurally very different inside.
Car batteries have many thin plates for high surface area so you can draw several hundred amps for 10 or even 15 seconds before the liquid adjacent to the plates is too depleted of the ions doing the conduction in the liquid (ie before significant "polarization set in") A battery deside for continuous current (but lower) for an hour or so is with relatively few heavy plates than will not warp and short out as they get hot (well warm, as they are in liquid) If you take 500 watts out from you inverter then you are taking from "12V" car battery at least 45 amp from it. (nothing is 100% efficient, perhpas even 50 amps). I bet that will cause an internal short in around 10 minutes or less.
Thus, better to just not open your refrigerator door until the power returns.

What you say about the large cross section, low resistance, of the Earth path is 90% true, but even with a long metal stake driven in the ground that first few feet is a high in-series resistance. In the old days, in rural areas, sometimes a one wire telephone was used, but if you wanted it to work well, you had to pour some salt water on the ground around the "grounding stake," at least in the dry season.

11. ### Billy TUse Sugar Cane Alcohol car FuelValued Senior Member

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If there is a lethal votage inside a metal box (for example a microwave) I think grounding is a very good idea. Not much corrsion damage from ground currents with AC anyway.

I tend to keep things a long time. Several transformers inside various devices I have owned have developed a connetion between their case and the primary winding. I am not sure why, but it has alway been relatively near the one of the two AC line wires. (Pehaps one wire goes straight to the iron core and begins wrapping around and is covered by many turns so it gets hotter as the turns make thermal insulation also and the insulation on the inner most layer of turns fails?) Anyway, I learned to paint with red finger nail polish one side of the AC plug and never plug it in with that side was up. - Sort of primative three wire system with only two wires back when there were no three wire plugs. Then the mild "leakage current" shock, which I had been getting from touching the case, ceased.

Your inverter would let you watch TV during storm that took down the wires etc, but I would not do it with car battery supply for more than a few (<5 ) minutes at a time. What you need is an electric boat battery which looks just the same, but is structurally very different inside.

Car batteries have many thin plates for high surface area so you can draw several hundred amps for 10 or even 15 seconds before the liquid adjacent to the plates is too depleted of the ions doing the conduction in the liquid (ie before significant "polarization" set in). A battery designed for continuous current (but much lower) for an hour or so is with relatively few heavy plates than will not warp and short out as they get hot (well, very warm as they are in liquid).

If you take 500 watts out from you inverter then you are taking from the "12V" car battery at least 45 amp from it. (Nothing is 100% efficient, perhpas even taking 50 amps). I bet that will cause an internal short in around 10 minutes or less.

Thus, better to just not open your refrigerator door until the power returns or at least have the frig load on and then disconnect it for TV and lights on. I doubt, once it is running, that the frig takes 500 W but do not know.

What you say about the large cross section, low resistance, of the Earth path is 90% true, but even with a long metal stake driven in the ground that first few feet is a high, in-series, resistance. In the old days, in rural areas, sometimes a one-wire, party-line*, telephone was used, but if you wanted it to work well, you had to pour some salt water on the ground around the "grounding stake," at least in the dry season.
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*Every one had a different ring pattern (and you though different "ring tones" for you cell phome were new!) It was fun to listen in, but not wise to do so as the current divided and sound got weaker.

Last edited by a moderator: May 12, 2007

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10,296
Actually, I don't think we're in disagreement at all. Yes, corona losses are by far the largest. Right at the moment I've no references readily available regarding the others but I do know they play contributing roles.

As to the phase-correcting networks, yes they are capacitors and it's merely a choice of wording. Some engineers call them "phase-correcting" while others say "power factor correcting" or "power factor balancing" networks. But no, they cannot do a thing about resistive losses - they correct the phase angle that has been distorted by induction. Exactly the same as where PF correcting capacitors are used in customer installations that have a lot of inductive loading (large electric motors) in their plant.

Incidentally, for the benefit of the few members that may not be aware of it, the skin effect is quite large at these very elevated voltage levels. For that reason, there's been some work done using hollow (tubular) conductors rather than the conventional solid stranded ones. Again, I don't have any references handy but I've read that it seems to be resulting in some degree of success. It reduces the weight loading on the support towers (and stresses on the insulators) and could result in cheaper construction costs. But so far I've not seen where the jury has returned a verdict.

13. ### Billy TUse Sugar Cane Alcohol car FuelValued Senior Member

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Jeez! We should have stopped when we agreed.

I hope you only mean that higher voltage drops along a wire correspond to / make higher currents and thus the skin effect loses are higher.

For any given metal, the "skin depth" and curent distribution within the metal volume is a function of frequency alone. Voltage has nothing to do with it. (Actual it is very very slighly more complex as the resistance is a function of temperature, so it is not really "the same metal" when carrying more current. Also the skin may thus has higher resistivity than the interior, if rod is short. If rod is very long - not losing heat to the ends - then interior is at same temperature as surface. No sane person worries about this.)

I do not remember exactly* but roughly, much more than half (90%?) of the AC current will flow in only the first wavelength deep into the conductor; but at 60 hz the current is essentially uniform thru out any reasonable thickness of metal conductor because the wavelength is so much larger than the conductor.
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*Skin effect falls out of Maxwell's equations with parallel E field incident on the surface. - A somewhat messy boundary value calculation with real resistivity, even constant, on one side of a plane boundary (infinitely deep) and vaccum on the other, incident E field side. I did it years ago, but can not remember what fraction of the energy lost (not reflected) dies in the first wavelength, but it is surprisingly high.

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Nope, sorry, Billy, but you've slipped another cog.

I've very much aware of the skin effect as a result of frequency since I worked with rectangular and circular waveguides in the 4 - 6 MHz range for a good number of years. But the skin effect is also very real at extremely high DC voltages as well. Basically, the effect is due to the large number of electrons present which are naturally trying to repel each other. And since the conductor is "full" of them, they are pushed toward the surface since the surroundings contain almost no charge to repel them.

I see the misunderstanding, though - I failed to mention that I was still talking about HVDC transmission.

15. ### MetaKronRegistered Senior Member

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5,502
Resistive losses are fought by running as high a voltage as is practical through the lines.

Corona loss is not such an issue for DC because with DC corona simply makes a larger conductor.

Good point about the skin effect of DC, there.

16. ### Billy TUse Sugar Cane Alcohol car FuelValued Senior Member

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OK I guess we do still agree. I had never heard (or even thought about) what you are calling a skin effect. I have done a little work with a Van de Graph machine so know all about the electrons going to the surface, but never described it a due to the interior being "full", but guess there is nothing wrong with saying that. I think of it as the interior keeps net zero charge (number of protons and electrons essentially equal). All the excess electrons on the surface are free to move and keep as far from each other as possible due the repulsion between like charges. (In the interior only the "valence band" electrons are free to move - slide along the "Fermi surface", if that means anything to you.)

If you could put a bunch of electrons in the interior of a metal, they would immediately flow thru it to the surface. - Hey that is exactly how the Van de Graff works. - they ride up the inside belt, get pulled off by the comb of spikes, and out to the surface they go!

What if anything can you tell me about the resistance of these excess surface electrons to movement along the surface by electric field (votage drop along the surface? Is it any different that the "free electrons" that are keeping charge neutrality when they flow (very slowly, compared to thermal velociies, drift) under the field. I bet the "excessones" are all entergetically above the fermi surface. I wonder if they could form Cooper Bardine & S? pairs and be high temperature supercnductors under any conditions? - doubt it but interesting thought. I never though about any of this before and it is now time to go to bed.

In a pure metal there are lattice defects that scatter electrons and phonon waves that do the same, both causing the electrical resistance. At room temperatures, in annealed metals at least, I am almost sure the phonons are by far the more important.

I do not know how the phonons behave when nearing the surface, but bet they reflect back into the interior. There is probably a set of "surface confined" waves also, distinct from the thermal phonons. Lot of interesting questions to think about while falling asleep tonight. If I am still so ignorate about all this tommorow, I may need to open a solid state book again. nite-nite.

Last edited by a moderator: May 12, 2007
17. ### leopoldValued Senior Member

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17,455
110VACRMS is approx 170 VP (peak).
we designate RMS values because a 170 VP gives an equivalent heating value as 110 VDC. this is only accurate for sine wave AC.

edit:
due to read-onlys keen eyeballs i must clarify.
the 170 VP is wrong in both value and nomenclature.
the correct value is app. 155.8 and the nomenclature is peak to peak.

Last edited: May 12, 2007

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Mmmmm... might want to try the math again, Leo.

The peak reading would be about 77.8v and is more commonly expressed as peak-to-peak which is about 155.6v.

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Yes, Billy, I still remember Fermi surface, crystal lattice of metals, semiconductors, and all the rest, But it's been a VERY long time and that was simply studying as opposed to very little application work on my part. So most of it has just eroded away over time.

Interesting thoughts, though, but I'd need to locate some old books and notes that I probably don't even have anymore after moving my household many times in the intervening years. <sigh>

20. ### MetaKronRegistered Senior Member

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5,502
Try it again yourself, RO. The RMS voltage of a sine wave is .707, one over the square root of two, times the peak voltage. Peak to peak is twice that result. About 155.8 is right for the peak, and 311.6 for the peak to peak.

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Incorrect. The original 110V is already peak-to-peak at RMS so the true value of PTP is 155.58698.

22. ### paulfrRegistered Senior Member

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Does not a grounding post need to be deep enough to hit the water table ?
If it does not, I fail to see how you get much conduction thru dry, non homogeneous dirt. But if you hit the water table you have a reliable conductor.
Water itself, pure water that is, is an insulator. Its the impurities that make its resistivity lower. In high voltage engineering, we have to deionize the water to use it as a non conducting coolant.

Skin depth at 60Hz; For both copper and aluminum it is surprisingly smaller than one might imagine. As I recall, around 1/3 inch [not sure about that number, but it is close]. Also if your currents are in the 100's of amps, even 4/0 cable is affected and the design must accomodate this.

Also, note that with a simple diode and capacitor, an alternator becomes equivalent to a generator. Much cheaper than the generator or MG set as some call it.