# transformer winding wiring configuration question

• posted

Well, let's see. You've read the book mentioned in the thread, and still don't understand. You've read, and objected to, the answers posted here, by Bill Shymanski, VWWall, Eric Tappert, John G, Michael Terrill, Daestrom and myself, and still don't understand. You are not willing to experiment. Perhaps a prayer to the patron saint of lost causes would be in order.

Ed

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• posted

|> | Well, the answer goes back to the theory. Remember that when an |> isolated |> | load is connected to any secondary of an energized transformer, the |> | magneto-motive force created by the secondary current is opposing the |> MMF of |> | the primary current. |>

|> This I know. But there is some ambiguity to this. You said force and |> then |> current. The problem I see is that I don't know which it really is that |> determines the orientation. Given 2 identical windings, if the _voltage_ |> ends up being the same on the same ends, then the 2 _currents_ are flowing |> in opposite directions. This is explained by the fact that power is going |> in on the primary and power is going out on the secondary. Understanding |> the "opposing force" would help. But due to the loose usage of terms when |> many people speak about theory, especially "force" vs. "current", I just |> can't be sure what is going on there. | | The 'force' I spoke of is Magneto-Motive Force, not voltage. MMF is | measured in ampere-turns. It is directly proportional to the *current* (not | voltage) through a winding.

Is this a flux density of the magnetic field?

| Rule # 1: The MMF generated by any current flowing in the secondary is | *always* opposite the MMF generated by current flowing in the primary. From | this simple rule, you can figure out all your answers.

So how can that be? How do the electrons ... and whatever it is the magenetic field is made of ... know which is primary and which is secondary?

|> | two). Point the thumb of that hand (for me, it would be the 'left |> hand') in |> | the same direction along the core. Now your fingers of that hand wrap |> | around the core in the direction of current flow in the secondaries. |> | Regardless how many secondaries there are, and which way they are |> wrapped, |> | the current in*all* secondaries will go around in the direction of your |> | second hand's fingers. You're done. |>

|> I'm sure all the secondaries are like any other secondary. I'm sure all |> the |> primaries are like any other primary. It's the relationship between the |> primary and the secondary that I haven't pinned down. |>

|> Looking at the core from one end, I believe the direction of wire wrap, |> e.g. |> clockwise vs. counter-clockwise, is the issue. Whether the wire starts at |> the bottom and ends up at the top, or starts at the top and ends up at the |> bottom, is not. Or a wire could wind CCW going from top to bottom and |> then |> continue winding CCW in a new layer going back to the top, repeating until |> the needed number of windings are done. It would still be CCW from the |> referenced view. | | This is true. Did I say something to make you think otherwise??

I'm paraphrasing it to look at it in a different perspective to be sure my understanding of it is consistent.

|> For consistency, I would bundle all the windings together as a "cable" and |> wind this cable as described. While that may be a lousy way to construct |> a real-life transformer, I think it clearly shows the idea of everything |> in |> the same orientation. |>

|> When I energize the transformer, the primary current is opposed by the |> field. |> Whether that is an actually a current opposition or a voltage opposition |> would not matter (yet) since it affects the same winding. But when we |> look |> at the 2nd winding as a secondary, there would be a voltage potential |> there, |> but not being connected to a load, no current. Now if a load is |> connected, |> is the _current_ on the 2 windings going to be in the same direction? | | No. See Rule # 1 | | The varying magnetic field (created by varying MMF of primary) induces a | voltage in primary that opposes the applied voltage. Since the secondary is | wound right along side the primary (in your 'cable' example), the induced | voltage in the secondary would have the same polarity at any instant as the | applied voltage on the primary. If a load is connected to the secondary,

It would seem to me that the induced voltage on the secondary would have the opposite polarity as the applied voltage. The reason I say this is the magnetic field is going to induce a voltage on one wire, then it will induce a voltage on another wire running along with it in exactly the same polarity. Since that polarity is the opposite of the applied voltage, it would be opposite in the secondary as well.

It gets back to the same ambiguity. How can a magnetic field that is inducing voltage in two wire do it differently on one vs. the other just on the basis that one is a primary and one is a secondary.

| the induced voltage in the secondary creates a current flow to the load. | The MMF *must* of the secondary *must* be opposite the MMF of the primary | (see Rule # 1). Since you have them wound in the same direction, the only | way to get opposite MMF is to have opposite direction of current flow. | | So for some instant in the cycle when the electrons are traveling CW around | the core in the primary winding, the electrons in the secondary winding are | traveling CCW.

Maybe I'm misunderstanding your description of induced voltage. This makes sense about the current because it is consistent with wiring practice. That is, we run 2 wires along side each other to power a load, and the magnetic fields between them cancel out nearly to zero.

| No. The currents would be traveling in opposite directions (see Rule # 1).

I can see that. I just can't see it from Rule #1 yet.

|> If the currents are the same then the |> voltage on the secondary will be opposite because power is being taken out |> instead of being put in. | | The currents are not 'traveling in the same direction', so the voltage on | the secondary is the *same* polarity as the primary. (in your winding | example)

Then I must be misunderstanding because I got a different meaning than this from a previous statement.

|> But is that so, that both primary and secondary currents go in parallel |> when |> power is drawn from the secondary? It seems that can't work because it |> would |> increase the field strength, and something I read suggest the secondary |> has |> to be tapping into the field strength, effectively lowering it, for power |> to |> be taken out. | | Congratulations, you've talked yourself out of your mistake. The currents | do *not* travel in the same direction around the core, so the field does not | increase.

What mistake?

A lot of this posting is paraphrasing what others post. If what I say seems wrong, then one of:

1. What they posted was wrong.
2. What they posted was poorly written.
3. I just didn't understand what they posted.
4. I understood what they posted by paraphrased it poorly.
5. My paraphrasing was misunderstood.

I've always believed, and your post here is confirming, that the currents will be in the opposite direction. Thus if I wire 2 windings together the same way, such as I could do by using a 2-wire cable, with the winding only going from top to bottom (for clarity, not for practical construction) then when at some instant the current is going from top to bottom on the energized winding (primary), it is going from bottom to top on the secondary.

That also tells me if the voltage is - at the top on one where the power is coming in, it will be - on the top of the other where the power is heading out.

| How much clearer do you want me to word this? The *current* in the | secondary travels in the opposite direction around the core as the primary | *current*.

OK, that's clear. But what I read from you about the voltage seems to be the opposite of what I expect. I would expect the voltage polarity to be the same at the same instant, when the current is reversed. But you have it stated the other way.

At least if we agree about the current, I'm half way to understanding this.

| No, you have to mentally 'cut the core' and unwrap it. Once you open up a | toridal core and straighten it out, or 'slide' the secondary winding over to | the same side as the primary, you have a simple solution.

I sort of did that. I just did it the other way around, starting with a straight core and bending it around.

|> Here is a more complex scenario which is one of those things that has |> given |> me the want to find out precisely about this. Suppose I have an "E" core |> transformer, which has 3 vertical bars crossing between a top bar and a |> bottom bar. This is the typical core used for a 3-phase transformer. |> Label |> the three vertical core segments A, B, and C. If I wind a primary around |> core A and energize it, the field will loop around through cores B and C. |> If I wind a secondary around core C, now what happens? | | Well first of all, you don't have a three-phase transformer with just one | primary winding and one secondary winding. Your example is a single-phase | transformer on a three-phase core. Bad, bad, very bad.

Certaing it is not a three phase transformer; it's just the core from one and other windings added for single phase.

| But the answer is still there. When you have current flowing through the | primary around the 'A' leg, suppose that on some particular half-cycle, the | MMF of the primary is such that your 'thumb' points upward. So 'lines of | flux' will rise through the iron from the top of the coil, turn with the | iron and 'flow' downward through the iron in legs B and C. When a load is | connected to your secondary winding, its MMF *must* oppose this 'flow of | lines of flux'. | | Looking down from the top, you can think of the 'lines of flux' rising up | toward you in the 'A' leg, turning and going down away from you in the 'C' | leg. So if the current in the primary is flowing CW at a particular instant | (as viewed from above), then the current in the secondary will also be | flowing CW at the same instant (as viewed from above). The reason it | *seems* to be the same direction this time is because the magnetic field did | a u-turn in the iron. So we still have an opposition (rule #1 still | applies).

I can see that from the reversal of the orientation because the field is turned around in C with respect to its origin in A.

| BTW, with no load on the secondary, the voltage output of the secondary is | *not* going to be what you'd expect from the turns ratio. If the 'split' of | the magnetic flux between the B and C legs were perfect, you would get half | the voltage that the turns ratio alone would predict. Can you see why?

No, I don't see why. In this scenario I see ambiguity. That means I just don't know what would happen. Or why.

|> Will the secondary |> around core C be able to get full power out, despite B being present? Or |> will the B core reduce the available power in some way? | | The 'B' core will greatly reduce the amount of power available on the 'C' | winding. This is because when the secondary current begins to flow through | a connected load, the secondary's MMF (which remember opposes the primary's) | will simply 'divert' much of the magnetic flux of the 'A' leg that was | originally 'flowing' into the C leg of iron, over to the B leg of iron. | This results in much less magnetic flux in the C leg and correspondingly | less induced voltages. (i.e. the voltage output of the C winding will drop | drastically as it is loaded) | |>

|> Suppose I wind a tertiary winding around core B, and monitor the voltage |> by drawing a trivial few microamps. Will this voltage change as more and |> more power is drawn from the winding around core C? |>

| Yes. The exact 'split' of magnetic field from the primary A winding will | vary drastically with load. So a heavy load on the 'C' winding will shift | the 'split' such that the field strength in C is reduced and the field | strength in B is increased. This is why you almost *never* make a | transformer like this. Three-phase transformers are *not* wound as you're | describing.

I would not expect them to be wound that way.

|> What happens if I put a big load on B, or even solidy short it? How will |> that affect the power I can get from C? What if I put a resonant circuit |> on the B core, peaked at a high impedance at the power frequency? | | Now, you can have all sorts of fun. A shorted turn on B would effectively | reduce the amount of primary flux that goes through the B leg, forcing more | through the C leg. Or, you can 'notch' the iron in the B leg so small | amounts of flux pass as before, but 'large' amounts saturate the iron in the | area of the notch. So limiting the amount of power that can be drawn | through that leg. | | A ferro-resonant transformer with a tuned tertiary winding is just such a | 'critter'. With a high-reluctance path (an air-gap) to the winding with a | tuned resonant circuit (a series circuit tuned for *low* impedance at power | frequency) draws little power in the tertiary because of this air-gap, | leaving most of the power for the secondary. But a slight increase in | primary MMF (as caused by a rise in supply voltage) will increase the power | drawn by the tertiary and thus the power supplied to the secondary remains | fairly constant. Such things are also known as constant-voltage | transformers, or by the brand name Solo-Transformers.

OK, this does sound like fun, though I was more interested in playing with harmonics.

|> If I had 3 windings around a single core, I can better visualize what |> might |> be happening, despite a few lingering doubts or ambiguities about how it |> all works. But with the E-core, things are "stranger" here. | | Because E-core are not used for simple, single phase transformers. When an | E core is used for a three-phase transformer, *two* windings are wound | around each leg. A primary, and the same phase's secondary. The three | primaries (one on each leg) are connected similar to how three single-phase | transformers might be connected (either wye or delta). And the three | secondaries are also connected in much the same way (either wye or delta). | | Putting a single phase primary on one leg, and two different secondaries on | the other legs is *not* something I've run across in my many years (except | the ferro-resonant transformer I described earlier).

Well, let me take the normal three phase transformer and vary it just a bit. Instead of varying it by construction (let's leave it at normal), let's vary it by placing a solid short on ONE of the secondary windings. Would this then cause an _increase_ in the MMF of the other two phase cores, thus _raising_ their secondary voltages?

This would be consistent with your statement from above, repeating:

| Yes. The exact 'split' of magnetic field from the primary A winding will | vary drastically with load. So a heavy load on the 'C' winding will shift | the 'split' such that the field strength in C is reduced and the field | strength in B is increased. This is why you almost *never* make a | transformer like this. Three-phase transformers are *not* wound as you're | describing.

I have seen a scenario where, during a storm, I saw the voltage rise quite a bit for a couple seconds, from around 120 to an estimated 140 to 150. A 60 watt light bulb got brighter to about the level somewhere between a

75 watt and a 100 watt bulb. A few more seconds after this I heard from a distance a loud boom like some primary distribution had shorted. My speculation is that a different phase shorted, and the transformer that was supplying power on all these phase raised the voltage on the others due to the high current of the one that was shorted. Plausible?

I'm going to save this whole post, because it must be me who is simply misunderstading an earlier induced voltage reversal description. This I'll have to re-read a few more times to be sure.

• posted

| Well, let's see. | You've read the book mentioned in the thread, and still don't understand. | You've read, and objected to, the answers posted here, by Bill Shymanski, | VWWall, Eric Tappert, John G, Michael Terrill, Daestrom and myself, and | still don't understand. | You are not willing to experiment. | Perhaps a prayer to the patron saint of lost causes would be in order.

I really just dismissed the answers that avoided the theory and focused on how to use some dots or markings that electrons never see.

However, Daestrom has gotten into some magnetic theory in another branch of this thread. Only one aspect still doesn't fit for me, but hopefully it is just a misunderstanding of some terminology or perspective.

• posted

From: snipped-for-privacy@ipal.net On Sat, 09 Apr 2005 03:54:15 GMT snipped-for-privacy@bellatlantic.net wrote: | Well, let's see. | You've read the book mentioned in the thread, and still don't understand. | You've read, and objected to, the answers posted here, by Bill Shymanski, | VWWall, Eric Tappert, John G, Michael Terrill, Daestrom and myself, and | still don't understand. | You are not willing to experiment. | Perhaps a prayer to the patron saint of lost causes would be in order. I really just dismissed the answers that avoided the theory and focused on how to use some dots or markings that electrons never see. However, Daestrom has gotten into some magnetic theory in another branch of this thread. Only one aspect still doesn't fit for me, but hopefully it is just a misunderstanding of some terminology or perspective. ____________________________

I don't know what to tell you Phil I presumed you knew all that well }:-)

My Analysis: The darkened dots are just nodes (or transformer taps) that are neither positive nor negative, (actuallizing when connected or when voltages are applied only) the transformer leads undarkened dots (o) are reversible in the generic diagrams you offered because they are not otheriwse marked.There is no source depicted so they are neither dc nor ac transformer windings. just a group of windings, some in series, some in parallel.

To derive any other item of interest from them there is insufficient data.

Although; Imagination is the limit };-) =AEoy

• posted

Google for: "phil howard"+ka9wgn Phil, (3600 of him!), seems to be confused about a lot of things. :-)

• posted

On Sat, 9 Apr 2005 06:32:24 -0400 Roy Q.T. wrote: | | From: snipped-for-privacy@ipal.net | On Sat, 09 Apr 2005 03:54:15 GMT snipped-for-privacy@bellatlantic.net wrote: | | Well, let's see. | | You've read the book mentioned in the thread, and still don't | understand. | You've read, and objected to, the answers posted here, by | Bill Shymanski, | VWWall, Eric Tappert, John G, Michael Terrill, | Daestrom and myself, and | still don't understand. | | You are not willing to experiment. | | Perhaps a prayer to the patron saint of lost causes would be in order. | I really just dismissed the answers that avoided the theory and focused | on how to use some dots or markings that electrons never see. | However, Daestrom has gotten into some magnetic theory in another branch | of this thread. Only one aspect still doesn't fit for me, but hopefully | it is just a misunderstanding of some terminology or perspective. | ____________________________ | | I don't know what to tell you Phil | I presumed you knew all that well }:-)

I knew some basics, though not at 100% confidence in all of it. Some of what I knew I knew well. Some I wasn't sure but it seemed right. Then I ran into uses of transformers that seemed to contradict some of what I thought I knew.

One thing I have determined partially from what a couple people said is that the schematics themselves do NOT show polarity. I always thought they did, and up until recently never saw any contradictions. OTOH, most uses wouldn't matter if the secondary is reversed in phase. But the relationship of the phase of re-induced voltage or current does matter if you are connecting primary and secondary together, as in a buck-boost transformer.

I did "believe" that transformers worked a certain way, but I did not "know" it to a high degree of confidence. Daestrom's most recent post actually gave me more confidence about the current relationship in just the same way I always believed. But he did say something about voltage that didn't seem to fit. But this could be due to my misunderstand of the terms he used (often I do understand how things physically work, but don't have the terminology just right, leading to confusions).

I'll wait for his reply to my reply before I try to resolve that in any other way. Then at some point I plan to follow up with a closing summary and see if I really got it right.

| My Analysis: The darkened dots are just nodes (or transformer taps) that | are neither positive nor negative, (actuallizing when connected or when | voltages are applied only) the transformer leads undarkened dots (o) are | reversible in the generic diagrams you offered because they are not | otheriwse marked.There is no source depicted so they are neither dc nor | ac transformer windings. just a group of windings, some in series, some | in parallel.

My interest is, and remains, about the "inside" of the transformer beyond any dots, letters, or numbers marked on a manufactureed transformer. What I want to know is how they actually work; e.g. the relationship between the physical construction of the windings with respect to their orientation and the phasing of the voltage and current. Many early replies jumped to the issue of the dots, which is certainly important to know with respect to using a manufactured transformer, but doesn't help if I don't already know the exact relationship between them and the physical construction if I am going to (in theory) build a transformer of my own.

I'd seen dots before, and assumed they were polarity. Right assumption. But that knowledge was established before I realized there is a difference between the polarity of voltage with respect to polarity of current. But my past knowledge of the dots didn't come with a knowledge of whether they apply to voltage or current, so that knowledge became usless to resolve the things I did not yet understand.

FYI: I'm very much oriented to understanding _why_ things work as they do, and not much oriented on understanding things in symbolic ways, which is what people do to a great degree when they use things as component pieces in constructing larger things that use them (usually this works fine, but in some special cases, this can lead to problems). Whatever I think about, I prefer to think in terms of both the fine details as well as how they interact in the big picture.

| To derive any other item of interest from them there is insufficient | data.

I don't know what you mean by this.

• posted

On Sat, 09 Apr 2005 15:28:44 GMT VWWall wrote: | snipped-for-privacy@bellatlantic.net wrote: | |> Well, let's see. You've read the book mentioned in the thread, and still |> don't understand. |> You've read, and objected to, the answers posted here, by Bill Shymanski, |> VWWall, Eric Tappert, John G, Michael Terrill, Daestrom and myself, and |> still don't understand. |> You are not willing to experiment. |> Perhaps a prayer to the patron saint of lost causes would be in order. | | Google for: "phil howard"+ka9wgn Phil, (3600 of him!), seems to be | confused about a lot of things. :-)

Or else he likes to become 100% absolutely certain of every detail, unlike way too many people in way too many so called technical fields.

FYI, I do like to experiment. But I also like to establish a hypothesis first that is as confident as the whole body of knowledge allows. I do not think it appropriate for me to set up a test experiment to discover how a self constructed transformer (so I can relate to how the windings are actually oriented, which would otherwise require the destruction of a manufactured transformer) would be polarized unless, and until, I know as much as is possible to know already. There may be details that I fail to observe because I didn't even know they matter.

• posted

If it takes you this much work to grasp such a simple concept you'll never learn enough to do anything useful. I learned it at 13 by reading old EE college text books picked up at Goodwill and I had no one to ask questions. Either you get it or you don't. If you can't visualize the concept in your mind it can't be hammered in.

• posted

Not quite. Think of MMF as 'voltage' or force that is trying to create a magnetic field. Just how much magnetic field is *actually* created is a function of the MMF and the permeability of the flux path. A given MMF applied to a core with low permeability creates a lower flux density (magnetic field strength) than the same MMF applied to a core with high permiability.

They don't. And they don't have to. The magnetic field acts on the electrons in both coils in exactly the same manner. If the current in the primary is flowing CW, the magnetic field generates forces on the electrons trying to push them CCW (*opposite* the direction they are actually moving). In the primary, this force opposes the externally applied voltage, but is not powerful enough to overcome the external voltage. So the current (electron flow) still flows against the magnetic force. In the secondary, there is no externally applied voltage, so the magnetic force on the electrons has nothing to counter-act its effect.

That is correct, but you're getting tangled up in semantics about the word 'opposite'.

Looking at your simple, 'cable' winding going CW around a core from the 'top' to the 'bottom' in one pass. When the applied voltage to the primary is negative on the wire connected to the top, electrons flow from the top, CW around the core to the bottom and return to the voltage source. As the current increases, the changing magnetic field applies a force on the electrons trying to push them CCW back 'up' the winding. This electro-motive force (i.e. 'Voltage') is trying to push electrons out the top connection, so we would say the EMF is such that the top of the coil is negative with respect to the bottom of the coil. This EMF 'opposes' the applied external voltage and limits the current flow. One *could* say it is the 'opposite' polarity of the applied source, but that can be a bit misleading. Afterall, if you look at the polarity at the 'top' of the winding, it is negative, just like the applied voltage. So is it really 'opposite polarity'??

The increasing magnetic field that is developing a force on the electrons in the primary, trying to push them CCW back 'up' the winding, is doing the exact same thing to the electrons in the secondary winding. They too are being 'pushed' CCW to the 'top' of the winding. But there is no external voltage source applied to the secondary that is stronger than the EMF induced by the magnetic field. If an external load is connected, the EMF created by the increasing magnetic field will push electrons right out the top of the secondary winding and through the load. From a macroscopic standpoint, we say the top of the winding is 'negative' since it is the 'source' terminal of electrons flowing to the load. (while the bottom terminal is 'positive' since it is the 'sink' terminal for electrons returning from the load.

It doesn't. That seems to be your biggest misconception. It works on the two wires *exactly* the same way. But in the case of the primary, it is

*not* strong enough to actually determine the direction of current flow. In the primary, it merely retards the magnitude of current that would otherwise flow due to the applied voltage.

Consider for a moment two batteries, a 12VDC battery and a 11VDC battery. Connect the positive of the 12V to the positive of the 11V using a 1 ohm resistor. Similarly, connect the negative of the 12V to the negative of the

11V. This is similar to what you might have for the primary circuit at one particular instant in the sin wave. The lower voltage of the 11V EMF in the primary winding opposes current flow, but the 12V applied source 'wins' and electrons flow from the negative side of the applied source, *into* the negative terminal of the primary winding (against the magnetically induced EMF).

When the secondary is unloaded, the varying magnetic flux of the core induces the maximum amount into the primary. But this is still less than the applied voltage. But this situation results in very little current flowing in the primary (because the induced EMF is high, and opposes the applied voltage). As load is applied to the secondary, the secondary current creates a MMF that opposes the MMF created by the primary. This results in a slight drop in the EMF induced into the primary winding. Now, with the same applied voltage, a smaller EMF in the primary winding (and other factors the same), the primary current will increase.

In fact, as secondary current creates stronger and stronger MMF that opposes the primary's MMF, the current in the primary will increase so that its MMF counters the secondary's MMF to maintain a balance.

Remember that in a 'load', the electrons travel from the negative terminal of the load, thru the load, to the positive terminal. This is the case in the primary. But in any 'source', internally, the electrons travel from the positive terminal back to the negative terminal. This is the case in the secondary. So while the 'top' terminal of both the primary and secondary may be negative, the current flow at the two terminals is 'opposite'.

With an 'E' core with the primary on A leg and secondary on C leg, the magnetic flux generated by the primary in the 'A' leg, is twice the flux density 'flowing' through the 'B' and 'C' legs (the 'lines of flux' split and return to the 'A' leg by two paths, the 'B' leg and the 'C' leg). With only half the flux density in the C leg as that in the A leg, you will get about half the induced voltage in the C winding that you would expect from the turns ratio alone.

As with so many things, 'it depends'. Are you still energizing the primary of that leg? If so, the magnetic flux in the leg with a shorted secondary will be low and the primary current will be very high (no emf induced in the primary to oppose primary current). If we *assume* that your shorted secondary has zero resistance, then the magnetic flux in that leg will be very low. The applied voltage will have little opposition (just the resistance of the winding) and current flow in the primary for that phase will be extreme.

But the voltage induced in the other secondaries (of a proper three-phase transformer) is a function of the flux in their legs, not the shorted leg. So their voltage will not be significantly affected (the shorted leg has little affect on the flux density in the other legs).

No, because 'my statement from above' was for a different situation. One with a primary on one leg, a secondary on a different leg, and the third leg being the experimental one. In *that* situation the flux density in the leg holding the secondary *is* affected by changes to the experimental leg. Not the same thing at all.

While the voltage surge happened (so it *must* be possible), it most likely wasn't a voltage increase caused by any 'magic' inside a three phase transformer. What does happen often, is a fault on one phase will cause a shift in the neutral point. Instead of the neutral being in the 'center' of the three phases so that the phase voltages of a wye connected system are balanced, the fault 'pulls' the neutral voltage closer to that phase. This 'stretches' the distance from other phases to neutral, increasing the phase-neutral voltage for those phases. (forgive me the 'pull'/'stretch' terms, but that's what it looks like when you see it on a phasor diagram ;-)

daetrom

• posted

|> Is this a flux density of the magnetic field? |>

| | Not quite. Think of MMF as 'voltage' or force that is trying to create a | magnetic field. Just how much magnetic field is *actually* created is a | function of the MMF and the permeability of the flux path. A given MMF | applied to a core with low permeability creates a lower flux density | (magnetic field strength) than the same MMF applied to a core with high | permiability.

OK, so we have a set of measures in the magnetic domain analogous to voltage, current, resistance, and power.

| They don't. And they don't have to. The magnetic field acts on the | electrons in both coils in exactly the same manner. If the current in the | primary is flowing CW, the magnetic field generates forces on the electrons | trying to push them CCW (*opposite* the direction they are actually moving). | In the primary, this force opposes the externally applied voltage, but is | not powerful enough to overcome the external voltage. So the current | (electron flow) still flows against the magnetic force. In the secondary, | there is no externally applied voltage, so the magnetic force on the | electrons has nothing to counter-act its effect.

And with no current in the secondary, it's (almost) as if the secondary is not even there, so the primary is simply behaving like an inductor.

|> It would seem to me that the induced voltage on the secondary would have |> the opposite polarity as the applied voltage. The reason I say this is |> the magnetic field is going to induce a voltage on one wire, then it will |> induce a voltage on another wire running along with it in exactly the same |> polarity. Since that polarity is the opposite of the applied voltage, it |> would be opposite in the secondary as well. |>

| | That is correct, but you're getting tangled up in semantics about the word | 'opposite'. | | Looking at your simple, 'cable' winding going CW around a core from the | 'top' to the 'bottom' in one pass. When the applied voltage to the primary | is negative on the wire connected to the top, electrons flow from the top, | CW around the core to the bottom and return to the voltage source. As the | current increases, the changing magnetic field applies a force on the | electrons trying to push them CCW back 'up' the winding. This | electro-motive force (i.e. 'Voltage') is trying to push electrons out the | top connection, so we would say the EMF is such that the top of the coil is | negative with respect to the bottom of the coil. This EMF 'opposes' the | applied external voltage and limits the current flow. One *could* say it is | the 'opposite' polarity of the applied source, but that can be a bit | misleading. Afterall, if you look at the polarity at the 'top' of the | winding, it is negative, just like the applied voltage. So is it really | 'opposite polarity'??

OK, so the voltage is not opposite. The induced voltage in the primary is in the same polarity as the applied voltage. But since the power direction is opposite (out instead of in) the current will be opposite. The secondary will be induced the same way.

So we have the same _voltage_ polarity between primary and secondary at the same end (assuming wound in the same sense), and opposite _current_ polarity, and opposite power flow (primary in, secondary out).

| The increasing magnetic field that is developing a force on the electrons in | the primary, trying to push them CCW back 'up' the winding, is doing the | exact same thing to the electrons in the secondary winding. They too are | being 'pushed' CCW to the 'top' of the winding. But there is no external | voltage source applied to the secondary that is stronger than the EMF | induced by the magnetic field. If an external load is connected, the EMF | created by the increasing magnetic field will push electrons right out the | top of the secondary winding and through the load. From a macroscopic | standpoint, we say the top of the winding is 'negative' since it is the | 'source' terminal of electrons flowing to the load. (while the bottom | terminal is 'positive' since it is the 'sink' terminal for electrons | returning from the load.

Really, this is the way I thought it was. But the terminology was tripping me up.

Lets see what I have so far:

Once current flows on the secondary as a result of a load attached, now we have this new current, which is in the opposite direction of the primary current, creating a new MMF. This MMF cancels out the primary MMF. Since the primary MMF was opposing its own current, that opposition now gives way and more current flows in the primary. It balances out with the same power level in and out (plus small heating, acoustic, and radiation loses).

|> It gets back to the same ambiguity. How can a magnetic field that is |> inducing voltage in two wire do it differently on one vs. the other just |> on the basis that one is a primary and one is a secondary. |>

| | It doesn't. That seems to be your biggest misconception. It works on the | two wires *exactly* the same way. But in the case of the primary, it is | *not* strong enough to actually determine the direction of current flow. In | the primary, it merely retards the magnitude of current that would otherwise | flow due to the applied voltage.

I got that now.

| Consider for a moment two batteries, a 12VDC battery and a 11VDC battery. | Connect the positive of the 12V to the positive of the 11V using a 1 ohm | resistor. Similarly, connect the negative of the 12V to the negative of the | 11V. This is similar to what you might have for the primary circuit at one | particular instant in the sin wave. The lower voltage of the 11V EMF in the | primary winding opposes current flow, but the 12V applied source 'wins' and | electrons flow from the negative side of the applied source, *into* the | negative terminal of the primary winding (against the magnetically induced | EMF). | | When the secondary is unloaded, the varying magnetic flux of the core | induces the maximum amount into the primary. But this is still less than | the applied voltage. But this situation results in very little current | flowing in the primary (because the induced EMF is high, and opposes the | applied voltage). As load is applied to the secondary, the secondary | current creates a MMF that opposes the MMF created by the primary. This | results in a slight drop in the EMF induced into the primary winding. Now, | with the same applied voltage, a smaller EMF in the primary winding (and | other factors the same), the primary current will increase.

There would be less inductance in the primary?

| In fact, as secondary current creates stronger and stronger MMF that opposes | the primary's MMF, the current in the primary will increase so that its MMF | counters the secondary's MMF to maintain a balance.

Got it.

|> | BTW, with no load on the secondary, the voltage output of the secondary |> is |> | *not* going to be what you'd expect from the turns ratio. If the |> 'split' of |> | the magnetic flux between the B and C legs were perfect, you would get |> half |> | the voltage that the turns ratio alone would predict. Can you see why? |>

|> No, I don't see why. In this scenario I see ambiguity. That means I just |> don't know what would happen. Or why. | | With an 'E' core with the primary on A leg and secondary on C leg, the | magnetic flux generated by the primary in the 'A' leg, is twice the flux | density 'flowing' through the 'B' and 'C' legs (the 'lines of flux' split | and return to the 'A' leg by two paths, the 'B' leg and the 'C' leg). With | only half the flux density in the C leg as that in the A leg, you will get | about half the induced voltage in the C winding that you would expect from | the turns ratio alone.

So I am curious what current the secondary can now have. I believe it will be double. In once sense this is consistent if the power flow is upheld. But another way to look at it is that the double current is creating a double MMF, which goes back towards the primary to cancel out what is there. But only half gets back to core A because half goes to core B.

Even though it would be constructed symmetrically, it would not behave symmetrically since it will step down the voltage. Turning it around and feeding power to core C and loading core A, it's still a step-down transformer. If this is how it really behaves, maybe this is interesting for reflected power situations like a reactive load.

Still, something doesn't seem right about that.

|> Well, let me take the normal three phase transformer and vary it just a |> bit. |> Instead of varying it by construction (let's leave it at normal), let's |> vary it by placing a solid short on ONE of the secondary windings. Would |> this then cause an _increase_ in the MMF of the other two phase cores, |> thus |> _raising_ their secondary voltages? | | As with so many things, 'it depends'. Are you still energizing the primary | of that leg? If so, the magnetic flux in the leg with a shorted secondary

Yes, the short is on the secondary side, and the primary is still energized for the time being.

| will be low and the primary current will be very high (no emf induced in the | primary to oppose primary current). If we *assume* that your shorted | secondary has zero resistance, then the magnetic flux in that leg will be | very low. The applied voltage will have little opposition (just the | resistance of the winding) and current flow in the primary for that phase | will be extreme.

That would be expected ... a high current on the secondary means a high current on the primary, relative to the turns ratio.

| But the voltage induced in the other secondaries (of a proper three-phase | transformer) is a function of the flux in their legs, not the shorted leg. | So their voltage will not be significantly affected (the shorted leg has | little affect on the flux density in the other legs).

The flux density is staying the same despite the short condition?

|> | Yes. The exact 'split' of magnetic field from the primary A winding |> will |> | vary drastically with load. So a heavy load on the 'C' winding will |> shift |> | the 'split' such that the field strength in C is reduced and the field |> | strength in B is increased. This is why you almost *never* make a |> | transformer like this. Three-phase transformers are *not* wound as |> you're |> | describing. |>

|> I have seen a scenario where, during a storm, I saw the voltage rise quite |> a bit for a couple seconds, from around 120 to an estimated 140 to 150. |> A 60 watt light bulb got brighter to about the level somewhere between a |> 75 watt and a 100 watt bulb. A few more seconds after this I heard from |> a distance a loud boom like some primary distribution had shorted. My |> speculation is that a different phase shorted, and the transformer that |> was supplying power on all these phase raised the voltage on the others |> due to the high current of the one that was shorted. Plausible? | | While the voltage surge happened (so it *must* be possible), it most likely | wasn't a voltage increase caused by any 'magic' inside a three phase | transformer. What does happen often, is a fault on one phase will cause a | shift in the neutral point. Instead of the neutral being in the 'center' of | the three phases so that the phase voltages of a wye connected system are | balanced, the fault 'pulls' the neutral voltage closer to that phase. This | 'stretches' the distance from other phases to neutral, increasing the | phase-neutral voltage for those phases. (forgive me the 'pull'/'stretch' | terms, but that's what it looks like when you see it on a phasor diagram ;-)

That actually makes sense now. The neutral thus became closer to the shorted phase, to the extent of it's return voltage drop at that high a current level. Or one can look at it analogous to an open neutral, but instead of being open and subject to an extreme voltage swing, it is simple resistive and has "some" voltage drop on the neutral; like a "partially open neutral".

So in a single phase 3-wire feed, if i short just one phase, the voltage drop in the neutral is going to contribute to being somewhat like an open.

Still, the stretching analogy makes sense just as well, and perhaps easier to calculate with.

My next "transformer brainstorm" probably doesn't exist, yet. Take a square core (with square hole in the middle). Take 5 more just like it for a total of 6. Stand them on end and arrange them in a hexagon so a side of one is adjacent to the side of the next one. Now put 6 windings around each of the side _pairs_ as if the winding wires were now tying the cores together. These will be numbered in sequence 1 through 6. Apply power to windings 1, 3, and 5 from each of three phases at 120 degres phase angle. Attach loads to windings 2, 4, and 6. For simplicity assume the connection is wye-wye. Would there be a 60 degree phase change?

• posted

| snipped-for-privacy@ipal.net wrote: |> |> Or else he likes to become 100% absolutely certain of every detail, unlike |> way too many people in way too many so called technical fields. |> |> FYI, I do like to experiment. But I also like to establish a hypothesis |> first that is as confident as the whole body of knowledge allows. I do |> not think it appropriate for me to set up a test experiment to discover |> how a self constructed transformer (so I can relate to how the windings |> are actually oriented, which would otherwise require the destruction of |> a manufactured transformer) would be polarized unless, and until, I know |> as much as is possible to know already. There may be details that I fail |> to observe because I didn't even know they matter. | | | If it takes you this much work to grasp such a simple concept you'll | never learn enough to do anything useful. I learned it at 13 by reading | old EE college text books picked up at Goodwill and I had no one to ask | questions. Either you get it or you don't. If you can't visualize the | concept in your mind it can't be hammered in.

I learned it. Or at least I thought I did. But I never had a 100% solid confidence because there were vague things going on that were not explained. It's that less than 100% confidence, combined with seeing a buck-boost transformer diagram that looked wrong ... and still looks wrong ... that my understanding was questioned.

I disagree with the concept of "Either you get it or you don't."

• posted

autotransformer

----------- You apparently know the directions that the windings are wound. . Assume a flux direction (It doesn't matter what direction you assume as it is all relative) and marking the "ingoing" points on the individual windings as you follow the assumed flux direction. If a winding is wound in the opposite direction, mark the "outgoing" point.

It is easy to visualise if you mentally cut the core and lay it out flat, so it looks like a multicoil solenoid. -. corresponding points with regard to the flux direction have the same polarity. In this way, the voltages in the windings will all be essentially in phase.

--------------------------------------------------> flux

*coil1. *coil2. *coil3. .coil4* coils 1 to 3 wound in same direction, coil 4 in opposite direction * is the polarity mark All coils will have the marked terminal + at the same time.

However, it is often a lot easier tie one end of a primary winding to one end of a secondary winding. and energise the primary winding at a low voltage and measure the differential voltage across the terminals of the primary and secondary windings. If the voltage is lower than the applied vol tage, the secondary voltage is in phase with the primary voltage. (i.e the non connected terminals are at the same polarity-mark them accordingly).

• posted

voltage.

installation

Ahem. The book says that I can do a "kick test" - with a flashlight battery and a multimeter. Hookup the multimeter to one side, momentarily touch the primary wires to the battery terminals and observe the "kick" of the meter (works best with an analog meter, I expect). When the meter kicks upscale, the primary terminal that touched the + terminal of the battery, and the

• lead of the voltmeter, can be considered the "dot" or H1/X1 terminals. If the meter kicks downscale, then the - lead of the meter was on the "dot".

Having said this, I've never had to do it myself...down to the workshop after breakfast and take the time to try it, I will.

If my powers of visualization were better, I could probably work out from first principles how the winding direction relates to polarity...but it's late, and in a practical case the winding you're interested in is under 10,000 gallons of oil and you *really* don't want to drain it down to feel up the leads.

Bill

• posted

half-cycle)

--------- The polarity markings refer to voltage. If the winding direction is , say CCW for all windings -just assume a flux direction and mark the corresponding ends of the coils. If a coil is in the opposite direction with regard to the flux, mark the opposite end.

-----------

---------- You just about have it. If there is a load on the secondary the current tries to reduce the flux. However, ignoring voltage drops in the winding, the flux is determined by the applied voltage (E=4.44fNABmax) and in order to get this flux there must be a corresponding exciting current (basically the no-load current) so the primary current must increase to produce extra amp turns to counter the secondary amp turns. That is NpIp =NpIe +NsIs where p is primary, e is exciting and s is secondary. The field strength remains the same as determined by Ep. If the primary current is in at the polarity mark (voltage + at mark) then the secondary current is out at the mark (voltage still + at mark).

-----------

--------- You got it.

----------------

-------

In an ideal transformer this would be strictly true. Such a beast doesn't exist but real transformers can be close to ideal.

If the polarity marks are that marked terminals have the same voltage polarity, then there will be a primary current in at the primary terminal. At no load this will be Ie as the exciting ampere turns must be such as to produce the flux called for by a given voltage applied (true for ac- NOT for DC). The core material affects the current but not the voltages. Now there will be a secondary voltage in phase with the primary voltage. If a load is applied, the secondary current (out from the marked terminal which is the source as far as the secondary is concerned) tries to reduce the flux but it can't do this so more primary current must flow to compensate. NpIp-NsIs=NpIe which is generally small enough to neglect.

• posted

| Ahem. The book says that I can do a "kick test" - with a flashlight | battery and a multimeter. Hookup the multimeter to one side, | momentarily touch the primary wires to the battery terminals and observe | the "kick" of the meter (works best with an analog meter, I expect). | When the meter kicks upscale, | the primary terminal that touched the + terminal of the battery, and the | + lead of the voltmeter, can be considered the "dot" or H1/X1 terminals. | If the meter kicks downscale, then the - lead of the meter was on the | "dot". | | Having said this, I've never had to do it myself...down to the workshop | after breakfast and take the time to try it, I will. | | If my powers of visualization were better, I could probably work out | from first principles how the winding direction relates to | polarity...but it's late, and in a practical case the winding you're | interested in is under 10,000 gallons of oil and you *really* don't want | to drain it down to feel up the leads.

No, you definitely don't want to drain all that oil. Do you even have the space in your workshop to drain it to?

Imagine you are the engineer designing the transformer. It's still on the drafting table (or CAD equivalent). Your boss reminds you that the terminal templates are due tomorrow with the all the other manufacturing drawings and that you fly out to the assembly facility in 2 months to supervise the first tests. Did you get them polarized correctly? Of course if you are an old pro at transformer design, you'll have no doubt.

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