wrote:
|> 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?
|> Is this a flux density of the magnetic field? |> | | Not quite. Think of MMF as 'voltage' or force that is trying to create a
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?
--
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| Phil Howard KA9WGN | http://linuxhomepage.com/ http://ham.org/ |
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| Phil Howard KA9WGN | http://linuxhomepage.com/ http://ham.org/ |
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