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The middle one is also two half turns in series. Follow the current path and you'll see the effect in the centre limb, Ampere.turns or volts.seconds, is exactly the same as the first diagram.
I'm trying to understand this. Assuming that the center ferrite cross section is twice that of each outer limb, wouldn't the flux from one side coil split up as 2/3 going through the center and 1/3 going through the other limb?
That would imply a coupling of 1/3 between the outer coils, which means a total inductance of the series circuit to be 8/3 ( =2*(1+k) ) of a single side. For a parallel circuit I get a factor of 2/3 ( =0.5*(1+k) ) compared to a single side loop.
The center loop (top diagram) would have twice the inductance of a side loop due to the doubling of loop area.
Transformation ratios would also be somewhat different from the upper diagram.
Registered Member #72
Joined: Thu Feb 09 2006, 08:29AM
Location: UK St. Albans
Posts: 1659
Uspring wrote ...
Dr. Slack wrote:
The middle one is also two half turns in series. Follow the current path and you'll see the effect in the centre limb, Ampere.turns or volts.seconds, is exactly the same as the first diagram.
I'm trying to understand this. Assuming that the center ferrite cross section is twice that of each outer limb, wouldn't the flux from one side coil split up as 2/3 going through the center and 1/3 going through the other limb?
That would imply a coupling of 1/3 between the outer coils, which means a total inductance of the series circuit to be 8/3 ( =2*(1+k) ) of a single side. For a parallel circuit I get a factor of 2/3 ( =0.5*(1+k) ) compared to a single side loop.
The center loop (top diagram) would have twice the inductance of a side loop due to the doubling of loop area.
Transformation ratios would also be somewhat different from the upper diagram.
No, the top two diagrams have exactly the same inductance, and the lowest one 1/4 the inductance of the others because of the different electrical connection. Starting from the symmetry of the situation is the easiest way to figure it out. In the middle diagram, the two outer limbs have the same current wrapped round them, each producing the same H field at the centre limb. They add together. Same H field for each, same area, same length, so same B field.
In the bottom diagram, the voltages of the two half turns are constrained by the connection to be equal. This means the integrated volt.seconds in the two limbs are equal, again an equal split of the central flux to the outers.
It's an interesting question of why the two half coils are not coupled, as they are wound on bits of the same core. The important point is the symmetry. A magnetic wall exists down the middle of the central leg. You could bisect the central leg into two half areas without changing the performance of the system at all. No lines of flux cross this mid-line, so it wouldn't 'see' the difference. Having got this zero flux magnetic wall, you can of course now physically seperate the two halves, again without any change to the performance, like here, where you can use a pair of C cores instead of E cores You will notice there is no metal between the two sets of C cores, any contact is only incidental. That's because when they are symmetrical, no flux crosses the mid-line.
Now it's possible to understand the difference between using one 'half turn' which wrecks leakage inductance, and two 'half turns' in parallel, which maintain symmetry. Think of it as each outer limb's half turn 'energising' half of the central limb. If you put what you think is a half turn around one outer limb, then what you actually have is a full turn around half of the total transformer core, a very different animal.
A quick way to demonstrate that the first two diagrams give the same inductance is to notice that they are in fact topologically *identical*. Short the input terminals to make a loop of wire. Now continuously deform the loop of the second diagram until it can lie on top of the loop of the first diagram. Note that this is a 2D diagram of a 3D situation, we are not in flatland. The only thing that the Amps and Teslas care about is the topology, how many times a wire threads through what hole in the core, not the big or small, left or right route the wire takes when it's outside the core. You will need to take a wire over the complete core as part of that movement.
A magnetic wall exists down the middle of the central leg. You could bisect the central leg into two half areas without changing the performance of the system at all. No lines of flux cross this mid-line, so it wouldn't 'see' the difference.
The symmetry argument is compelling and thinking of it I had the same idea of bisection before I read your remark.
But think of another experiment: Gradually reduce the center cross section in the middle diagram until it finally disappears. Then you'll have two windings in series but oppositely wound. The resulting inductance would be zero. But also the symmetry argument would hold up to the point the center section gets infintely thin. Any idea about this?
I made an error in the calculation in my previous post. Since the flux is inverted from one side to the other the coupling should be k=-1/3.
Registered Member #72
Joined: Thu Feb 09 2006, 08:29AM
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Hmm, I'm not sure you understand the concept of 'turns' here, or I don't understand what you're proposing. If you thin the core down until it vanishes, then you don't have turns either, just a wiggly bit of wire in fresh air.
If you stop the thinning process just before it vanishes, then the first order approximation of the core flux totally dominating the air flux is broken, and the paths of all the wires matters, not just those threading the core. Whether the thin-core flux is small, infinitessimal or zero compared to the air flux, when you add the fields from either just the 'half turns', or all the wires in the second diagram, I don't see the inductance vanishing in either case. What are you seeing?
You do sometimes see air-cored RF chokes and transformers described as having (for instance) 3.5 turns. Their geometry is a spring with 3.5 turns, and then a straight piece of wire down to complete the circuit. But is it one wiggly turn, or 4 turns of different geometry? Because it's all in air, it doesn't matter. Just do Biot Savart and calculate the inductance, Maxwell and his equations don't care what words you choose to use to describe it. An inductor with that geometry will indeed have more L than a '3 turn' inductor, and less than a '4 turn'. But those are geometric shape turns, not Maxwell topology flux linkage turns.
Did you try yet to see that the first and second diagrams are topologically equivalent, and so have exactly the same inductance?
Registered Member #72
Joined: Thu Feb 09 2006, 08:29AM
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Yes, fractional turns has not been as much a thread hijack as a thread derailment. Sorry for my contribution to that. I wonder if the mods can excise the relevant posts and transplant them to your new thread on the topic.
Thank you Patrick since I want to make another comment to what Dr. Slack said.
Hmm, I'm not sure you understand the concept of 'turns' here, or I don't understand what you're proposing. If you thin the core down until it vanishes, then you don't have turns either, just a wiggly bit of wire in fresh air.
Maybe I misunderstood how your drawing was meant. I thought of 2 E shaped cores put face to face to form an 8. The drawing is in the plane where the cores meet. I was also disregarding the secondary turns for the time being, so there are no wires in air, when you thin out the central leg.
We have a left-right mirror symmetry when looking at your drawings. That implies, that the field also has this symmetry and that suggests, that you can treat the left and right part separately, since no field lines from the left turn go through the right turn and vice versa.
It's an interesting question of why the two half coils are not coupled, as they are wound on bits of the same core. The important point is the symmetry. A magnetic wall exists down the middle of the central leg. You could bisect the central leg into two half areas without changing the performance of the system at all
If you apply voltage to only one limb, the symmetry is broken and there will be field lines coming from the driven limb and going to the other one, inducing a voltage there. So the 2 limbs _are_ coupled and that means, that you cannot just add up the separate limb inductances to get the series inductance. The picture of field lines in the symmetric case is deviously misleading.
Registered Member #72
Joined: Thu Feb 09 2006, 08:29AM
Location: UK St. Albans
Posts: 1659
Hmm. Lots of things, and unfortunately many of them will sound a bit like avoiding the question. I had assumed a model, which is correct in the limit of infinite core permeability, and/or zero vacuum permeability, ie it's just turns counting and topology, and a constant cross section core. Real life will deviate from that, but by percent. The important thing with transformers is first to get something that can work even theoretically in that perfect limit, like the policy of balancing the two half turns, and then optimising it practically, like twisting the lead-in wires together, and striving to make the outer limbs physically the same length and area.
Now many of your thought experiments break that model. So we are left confusing theoretical possibility, and practical optimisation.
I think I now understand that you were thinning the central limb only, essentially leaving a ring core, with the two outer windings phased in opposition. In the limit of zero vacuum permeability, that does indeed have zero inductance. But it's not a transformer to windings on a central limb, so whatever inductance a not-transformer has, it isn't really relevant to the discussion. In real life, a small flux occurs through the air/vacuum between the top and bottom, so there is a very small inductance. This was the non-zero inductance I was referring to when I thought you were removing the entire core. The inductance of the opposed ring-core will be more or less twice than the zero core case, as half of the length of the magnetic circuit is 'shorted out' by the core.
While Patrick still wants to pursue half turns, I think the geometry he has chosen for the cores and formers could make it tricky in practice to achieve a turn on the centre and two balanced outer turns for an effective half turn. Unless of course the former shown is only used to support the secondary, and the primary uses discrete wire wound onto the cores, I don't see room for a primary former.
While I am sympathetic to your looking for the coupling between the two coils, remember that they are *only ever used* in series or parallel, not independently. The two outer limb coils used independently will have finite coupling, but when they are forced, by terminal connection, to operate together, the core central limb can be split by a magnetic wall and the coupling vanishes. Now I can't be bothered to do all the linear superposition stuff, or calculating the inductance in detail as the other coil is shorted and open. There are some things that you have to do in detail, and some you don't. For instance if a complicated system of rotating machinery appears to be producing net power, you don't have to delve into all the detail, you know from conservation of energy that it just can't. Similarly, my first and second diagrams have the same topology, so in the perfect limit of infinite core/vacuum permeability ratio they must have the same inductance. Period.
I thought the original task was 'how can Patrick implement half a turn properly?'. Balancing the outer limb turns achieves that. Leaving them unbalanced causes the horrible leakage inductance problems that burn people who attempt it without maintaining the symmetry. Now you don't really need to figure out what bad happens in detail. If you want to make a 4:1 transformer on a core, you wind 4 turns and one turn. That takes care of the topology, then you try to sort out the insulation, and leakage inductance, and all the other tricky practical optimisations. You don't wind 3 turns and then anguish that it's not 4, or start thinning the central leg down and wondering where your inductance went. Similarly if you want 0.5 turn, you put balanced turns in parallel round the outer limbs, and that takes care of the topology. Then worry about insulation and leakage inductance and heat removal and stray capacitance and all the other non-topology stuff that you have to.
I'm not going to contribute to this thread further, unless I can be bothered to crank up POVRAY, and make an animation showing that my first and second diagrams do indeed have the same topology. Or maybe I'll draw a few intermediate 2D diagrams to show it, but not as convincingly. I do urge you to look at those two diagrams, move one of the lead-in wires over the core (as you are allowed to do in a 3D situation), lengthen and shorten the wires (which doesn't change the topology) and see that they are indeed the same diagram.
I'm reminded of the man who went to the doctor, complaining that whenever he poked himself in the eye, it hurt. The doctor advised him to stop poking himself in the eye. But why does it hurt doctor? The details of why it hurts to self eye-poke don't really matter, stop poking yourself in the eye and all will be OK. An E core transformer with integer turns is balanced with respect to the outer limbs. If you put half a turn round one outer limb, poking its balance in the eye so to speak, resulting in different turns ratios on different parts of the transformer, it hurts. Keep it balanced by putting a turn on the outer limb, connected in parallel. Balanced restored. Eye stops hurting.
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High Voltage Planar Ferrite transformers. ( Intial Experiments )
