But Frank,fdegrove said:
That is also what this fuss is about… It is also covered by AudioNote's patent where it is called a “matching transformer (that) comprises a transformer input resistor”.
That’s interesting.
More interesting news about what can be patented are here (to see drawings, you'll need some TIFF viewer for your browser).
Pedja
DEVIL'S ADVOCATE? NO WAY.
Hi,
The fuss is about SE not understanding some basics.
I couldn't care less about AN's patent here since we're a DIY community and as such this patent can't apply to us.
A patent that, if it was actually granted, shouldn't have IMHO.
But that's another chapter...
As long it's a one off, personal use thing almost any patent doesn't aply to you, moreover nobody can sue for it.
Either way, if you consider going commercial, consult a lawyer first.
Cheers,
Hi,
The fuss is about SE not understanding some basics.
I couldn't care less about AN's patent here since we're a DIY community and as such this patent can't apply to us.
A patent that, if it was actually granted, shouldn't have IMHO.
But that's another chapter...
As long it's a one off, personal use thing almost any patent doesn't aply to you, moreover nobody can sue for it.
Either way, if you consider going commercial, consult a lawyer first.
Cheers,
Konnichiwa,
Please note that under US patent law you can patent pretty much anything not previously published and a material change on "prior art". European Patent law is a little more resrictive but neither require you to provide proof that what you are patenting is NOT "obvious"....
If for example tomorrow, I decided to manufacture a DAC that had a certain arrangement of circuit involving a transformer in the analogue section I'd be talking to some heavyweight IPO Lawyers about if it would be more beneficial to file a "follow on" patent lsiting Audio Note's patent as "prior art" or simply filing for a diamissal of Audio Note's on the VERY OBVIOUS grounds of OBVIOUSNESS.
But as a private individual presenting ones own work on web pages I preferred to leave all that legal mill to keep whirling without me....
Sayonara
PS, I have found a really new way to use Jell-O - Iguess i should patent it and then sue Jell-O for damages....
Pedja said:
Please note that under US patent law you can patent pretty much anything not previously published and a material change on "prior art". European Patent law is a little more resrictive but neither require you to provide proof that what you are patenting is NOT "obvious"....
If for example tomorrow, I decided to manufacture a DAC that had a certain arrangement of circuit involving a transformer in the analogue section I'd be talking to some heavyweight IPO Lawyers about if it would be more beneficial to file a "follow on" patent lsiting Audio Note's patent as "prior art" or simply filing for a diamissal of Audio Note's on the VERY OBVIOUS grounds of OBVIOUSNESS.
But as a private individual presenting ones own work on web pages I preferred to leave all that legal mill to keep whirling without me....
Sayonara
PS, I have found a really new way to use Jell-O - Iguess i should patent it and then sue Jell-O for damages....
This is incorrect. It may well be that you're unlikely to get sued (or even get caught) for doing a personal-use one-off. But it is still infringement. What you are allowed to do is to build a unit for experimental purposes- if it happens to be an I-V for your home CD player and you leave it in your system once you're done diddling with it, you're infringing.
Thorsten, as a theoretical matter (reality is different!), you do have to show some sort of non-obviousness. The classic example is a ball point pen; if you try to patent "writing a screenplay with a ballpoint pen," it is supposed to be rejected for obviousness. After all, that's what the pen is designed to do. But if you find that you can use the tip of the pen to pick and place microchips, that's patentable, even though pens existed before you thought of using them to pick up chips.
Now, the PTOs have not been as good about enforcing this as they should be, but this is how the guidlines for patent examination are set up.
Now, the PTOs have not been as good about enforcing this as they should be, but this is how the guidlines for patent examination are set up.
Konnichiwa,
EXACTLY.
Im quite aware of the origins and aims of the patent system. But reality is different and I tend to concentrate not so much on pretty theories, but on reality.
The current aptent system is a joke and not a funny one either....
Sayonara
PS, I have come up with a few things that are definitly non-obvious and novel and involve transformers.... I suspect I'll be putting them into the public domian with one or two key infos missing just to spoil other people patenting it....
SY said:
EXACTLY.
Im quite aware of the origins and aims of the patent system. But reality is different and I tend to concentrate not so much on pretty theories, but on reality.
The current aptent system is a joke and not a funny one either....
Sayonara
PS, I have come up with a few things that are definitly non-obvious and novel and involve transformers.... I suspect I'll be putting them into the public domian with one or two key infos missing just to spoil other people patenting it....
Patents
Having investigated patents, patent infringements in the technology field, no one going to SUE YOU for using the inventer’s idea in a home project. The lawyers get interested when you infringing and making money at it. That is when they want their cut of your profits “royalties” before the sue, unless you are a serious competitor.
Also, no company is going to spend their money on lawyer and court time to stop you from using most inventions on a hobby, it’s just cost to much money. The next problem is with the burden of proof, that why the patent claim are broad.
There is a large % of patents that are completely worthless. However the engineer get some cash for it.
Having investigated patents, patent infringements in the technology field, no one going to SUE YOU for using the inventer’s idea in a home project. The lawyers get interested when you infringing and making money at it. That is when they want their cut of your profits “royalties” before the sue, unless you are a serious competitor.
Also, no company is going to spend their money on lawyer and court time to stop you from using most inventions on a hobby, it’s just cost to much money. The next problem is with the burden of proof, that why the patent claim are broad.
There is a large % of patents that are completely worthless. However the engineer get some cash for it.
Re: We got nice discussion on “how do the transformers work” topic.
Fred, let me assure I am dead serious here and out to learn something. I hope you can believe me that I know how xformers work, how to convert load impedance from prim to sec and vice versa and all that stuff. BUT the question was: is it current or voltage that makes the xformer work?
The nice things about thought experiments is that it allows you to look at the core of the problem. In my thought experiments I noted a circumstance where you have voltage across the coil without current, and vice versa. IMHO this points out that it really is the current that gives rise to the field that makes it work. That was the original discussion, no? I don't think dragging the loss thing out is relevant here.
Jan Didden
Fred Dieckmann said:
Fred, let me assure I am dead serious here and out to learn something. I hope you can believe me that I know how xformers work, how to convert load impedance from prim to sec and vice versa and all that stuff. BUT the question was: is it current or voltage that makes the xformer work?
The nice things about thought experiments is that it allows you to look at the core of the problem. In my thought experiments I noted a circumstance where you have voltage across the coil without current, and vice versa. IMHO this points out that it really is the current that gives rise to the field that makes it work. That was the original discussion, no? I don't think dragging the loss thing out is relevant here.
Jan Didden
OK, I'll tell you what's wrong. In TE #1, you correctly state that there is no current, but you then incorrectly state that there is a voltage. In TE #2, you correctly state the presence of current, but you incorrectly claim the absence of voltage.
Transformer operation depends on both voltage and current. Arguing which one is the cause, and which is the effect is as futile as the chicken/egg issue. In TE #1, the infinite resistance in the primary wires results in ZERO VOLTAGE impressed upon the primary inductance. The transformer primary consists of leakage inductance in series with winding resistance, in series with magnetizing inductance, where the magnetizing inductance has a parallel resistance across it representing core losses (hysteresis and eddy current), which we can neglect for now. Although a finite (ac) voltage is placed across the primary terminals, the emf impressed upon the transformer core is zero volts, and of course the resultant primary current is also zero. The primary inductance has a finite ac reactance. The resistive component is infinite. As a result, the voltage is all dropped across the infinite resistance, and zero volts appears across the inductive reactance. The voltage across the primary inductance is responsible for setting up the core flux. There is no primary current and no primary voltage. As a consequence, there is no core flux. If the primary resistance was huge, but not infinite, there would be a small primary voltage, and a small primary current, as well as a small core flux.
If you examine the B-H curve of any magnetic core material, you will soon understand the following irrefutable property. AC voltage and AC current are mutually inclusive. They cannot exist independently. If one is present (non-zero), the other is also present (non-zero). If an ac voltage or emf is impressed upon the primary, a flux density "B" is present in the core. Since the B-H curve has a finite inside (non-zero) area, if "B" is non-zero, then "H" is also non-zero. A non-zero "H" means that a current (non-zero) must be present. In order for core flux to exist, there is an emf across the primary producing "B", and an exciting current in the primary producing "H". You can't have one without the other. The only way one could exist independently is to find a core material whose B-H curve is a pure vertical line ("B" will exist without "H"), or a pure horizontal line ("H" exists without "B"). There is NO SUCH material.
In TE #2, the fact that the primary wires have zero resistance (superconducting) does not change anything. There is still a voltage across the primary! Zero resistance simply means that all of the voltage appears as emf with no drops. The primary has a finite inductance value, and an ac reactance is present. The ac current source driving the primary must output a voltage to force current into the inductive reactance of the primary. Remember that the resistance is only one component of impedance. With motors, inductors, and transformers, the resistance is usually small compared to the inductive reactance, except at very low frequencies. Even if the transformer is air-cored, the current source will see an inductance and the voltage it outputs is 2*pi*f*L*I.
As far as induction goes, someone mentioned that induction equals "LIB", and concluded that "I" being current must be the cause, and that the secondary voltage must be the effect. This seems intuitively logical, but our intuituin is wrong whenever we restrict our thinking to one variable. The "B", remember is directly proportional to primary voltage. The secondary voltage is induced according to Vsec = -Nsec * d(phi)/dt, where phi = B * area of core, or magnetic flux in webers. The magnetic flux is created by BOTH the primary voltage and primary current. The same magnetic flux induces BOTH the secondary voltage and secondary current. Even if the secondary is a dead zero resistance short circuit, BOTH secondary voltage and secondary current are present. Even with zero resistance, there is a finite inductance and an associated ac reactance. The secondary voltage is equal to 2*pi*f*Lsec*Isec. This secondary voltage is reflected back to the primary according to the equal volts per turn law. If the secondary is open circuited, BOTH secondary voltage and secondary current exist. The secondary winding has a finite capacitance (electric field) across it. This capacitance produces an ac DISPLACEMENT current. This displacement current is reflected back to the primary and is balanced according to the law of balancing ampere turns (Ampere's Law).
If you don't believe me, please refer to any reference text on E/M fields, transformers, antenna, etc. Induction operates on both voltage and current simultaneously. No one - NO ONE - has ever determined which one is the cause or the effect. Again, ac current, and ac voltage cannot exist independently. If one of them is zero, the other must also be zero. There are NO EXCEPTIONS.
A transformer is not an I/V converter. It reflects both I and V simulataneously. It demands both primary I and primary V to create a core flux. It induces both secondary I and secondary V regardless of terminating impedance (open, short, or anything in between).
Have I made myself clear?
Best regards.
What !!!
I just read this post from last year and my answer to your question is most emphatically NO ! - I'm suprised that no one else responded
If you have a transformer with an open cct S and apply a signal to the P, the P will behave just like an inductor. Current will flow in the P but it will be out of phase with the voltage. The primary will not know that the secondary exists. There will be no actual energy tranfer from P to S. The P will only know that the S is there if the S is terminated. It is only when the S is terminated that normal transformer behavior occurs.
With no S termination there WILL be a EMF attemting to produce a current in the S but there CANNOT be any current because there is no cct for it to flow through.
So the statement
'It induces both secondary I and secondary V regardless of terminating impedance (open, short, or anything in between).'
is not correct.
Perhaps every one else just got tired of talking about this.
It strange that in this thread everyone was so busy auguing about how transformers work no one really explained the benefits of using transformers in I/V conversion cct's ( careful choice of words ).
So what is it that is so good about transformers in this role that makes audionote so keen to patent it ?
Claude Abraham said:
I just read this post from last year and my answer to your question is most emphatically NO ! - I'm suprised that no one else responded
If you have a transformer with an open cct S and apply a signal to the P, the P will behave just like an inductor. Current will flow in the P but it will be out of phase with the voltage. The primary will not know that the secondary exists. There will be no actual energy tranfer from P to S. The P will only know that the S is there if the S is terminated. It is only when the S is terminated that normal transformer behavior occurs.
With no S termination there WILL be a EMF attemting to produce a current in the S but there CANNOT be any current because there is no cct for it to flow through.
So the statement
'It induces both secondary I and secondary V regardless of terminating impedance (open, short, or anything in between).'
is not correct.
Perhaps every one else just got tired of talking about this.
It strange that in this thread everyone was so busy auguing about how transformers work no one really explained the benefits of using transformers in I/V conversion cct's ( careful choice of words ).
So what is it that is so good about transformers in this role that makes audionote so keen to patent it ?
Re: What !!!
There is nothing that is so good about it. It is just another
flavour of I-V. If you compare just R I-V, transformer after R,
GBS I-V they all sound different and different people will all have
their own preference.
I prefer GBS because I personally find it less coloured and
more linear (when done right). However others love the
sound of "transformer I-V" and others prefer just R into a low
noise hi gm tube.
Make one and see what you like.
Cheers,
Terry
mikelm said:
There is nothing that is so good about it. It is just another
flavour of I-V. If you compare just R I-V, transformer after R,
GBS I-V they all sound different and different people will all have
their own preference.
I prefer GBS because I personally find it less coloured and
more linear (when done right). However others love the
sound of "transformer I-V" and others prefer just R into a low
noise hi gm tube.
Make one and see what you like.
Cheers,
Terry
Hi,
Well, the main advantage as far as I can tell is that it provides galvanic isolation...
Other than that, depending on the xformer used, it can keep out digital spuriae due to BW limiting of the device.
I read elesewhere that these AN DACs tend to roll off the mid-highs with a 12dB slope....Yuk.
But some people may like that anyway.
Cheers,
Well, the main advantage as far as I can tell is that it provides galvanic isolation...
Other than that, depending on the xformer used, it can keep out digital spuriae due to BW limiting of the device.
I read elesewhere that these AN DACs tend to roll off the mid-highs with a 12dB slope....Yuk.
But some people may like that anyway.
Cheers,
re : What!!!
Mikelm, please reread my post carefully, as I've explained that an open circuited secondary still has both current and voltage induced into it. Since 1873, it has been well known that ac voltage and ac current cannot exist independently. If one is zero, so is the other. If one is non-zero the other is non-zero. Your countryman James Clerk Maxwell proved this and for 131 years every experiment and measurement has affirmed the same.
The most popular fallacy regarding transformers is that the magnetic core flux exists due to primary current only, and not primary voltage. Also, many believe that an ac flux induces *voltage* into the secondary, and that the secondary current flows only when a load is connected. It seems obvious, since an open secondary has a voltage but "no current", one can falsely conclude that I to V action is taking place. Simply put, many labor under the myth that the primary magnetizing current creates the core magnetic flux, and that the flux induces a voltage into the secondary which in turn causes secondary current only when a load is connected. This is plain wrong and can be demolished with little effort.
Let's start with a *voltage* source, an ac generator. feeding a transformer primary with 120 volts, 60 Hz. The secondary is open. Neglecting losses, 120 volts should appear at the secondary, if the turns ratio is one to one. A 10 ohm load is applied across the secondary, and 12 amps flow. At this point it is easy to believe that the flux induces a voltage into the secondary (S), and that the S current depends on the load. The fallacy here is that it is the ac generator, which is a voltage source, that is responsible for the constant S voltage / variable S current behaviour, not the transformer. In your example, you placed a transformer in parallel with a constant voltage source, and concluded that the constant S voltage, variable S current is a result of transformer action, or I-V. A transformer merely reflects the secondary impedance back to the primary according to the turns ratio squared, or 1:1 in this case. The constant voltage behaviour at the transformer S, is merely a reflection of the constant voltage behaviour of the ac input at the P.
Remember, a current transformer P is inserted in series with the circuit where current is to be measured. The secondary is short-circuited, so that zero volts exists across the S, and the P. A volt-meter or current shunt is connected to the S terminals, and the shunt switch is opened, and a voltage exists across the S terminals. If the terminating impedance is low compared to the external circuit, a constant current behaviour exists. Changing the impedance across the S changes the voltage, but current remains approximately constant. The constant current behaviour is due to the constant current source at the P, not induction action of the transformer. Even if the circuit is voltage source driven, if the terminating impedance is way smaller than the source impedance, constant current results. Both voltage transformers and current transformer operate under the exact same laws. Whether the transformer S outputs constant voltage or constant current depends on the source powering the P.
As far as I-V action goes, the following example demonstrates how absurd the idea is. Back to a voltage source driving a transformer with open circuited S, 120 volts, 60 Hz. If this was an I-V converter, any change in primary I, would produce a corresponding change in secondary V. Let's say that the primary magnetizing current is 0.10 amp. The S voltage is 120 volts. Suppose we decrease the frequency to 50 Hz, but decrease the P voltage to 100 volts, so that the core flux remains the same, and P magnetizing current is still 0.10 amp (neglecting eddy current and hysteresis losses). The S voltage is now 100 volts. The P current remains the same, yet the S voltage has decreased by exactly the same amount as the P voltage. The S voltage follows the P voltage, not the P current.
Now, let's hold the P voltage at 120 volts, while decreasing the frequency from 60 to 50 Hz, increasing the core flux, so that the P magnetizing current is 0.12 amp instead of 0.10 amp. The P current has increased, but the S voltage remains at 120 volts. Again, the S voltage follows the P voltage, not the P current. As long as the core does not saturate, the S voltage value is determined by the P voltage. Differing conditions with regard to frequency and flux density change the value of P current but the S voltage only follows the P voltage without regard to P current.
Finally, the core flux is created by both P voltage and P current. Whether iron or air cored, all materials have a relative permeability greater than zero, and less than infinite. You need both V and I to set up a core flux. The S voltage is the P voltage scaled by the turns ratio, or V-V. The magnetizing current is needed to create the flux, along with P voltage. When a load is placed across the S, the S current is reflected to the P, and a P current flow exists in order to maintain the core flux. THe S current creates an mmf which would change the flux, so an opposing P current flows to cancel this mmf. This is I-I conversion, but it does not include the P magnetizing current. This current is not subject to the law of balancing ampere turns. The I-I conversion is between S load current, and P current due to the S load, but not counting the P magnetizing current necessary to produce the core magnetizing flux. Hence a transformer converts primary V and I into core flux. The core flux induces BOTH V and I into the secondary. This is VI to flux to VI conversion. The S voltage is determined by the P voltage, or V-V conversion. The P current is determined by the S current, or I-I conversion EXCEPT for the P magnetizing current. Whether the S outputs constant voltage or constant current is determined by the source driving the P.
As far as the question, "How can both V and I exist in an open secondary?" goes, Maxwell answered that in 1873 with displacement current. Faraday never quite got it because he only developed his equations with conduction current. Experimental findings never quite matched Faraday's equations. When Maxwell added displacement current to Faraday's law, agreement existed. Remember, each turn of the S has a potential across it and under ac conditions a displacement current will flow due to the ac electric field. Don't forget the turn to turn capacitance present in all transformer windings. Any text on e/m, energy conversion, antennae, etc. will confirm the above. THe bottom line is that ac voltages and ac currents are mutually inclusive, as well as ac electric fields, and ac magnetic fields. No one has ever determined which one, V or I, is the cause and/or effect. Take what I've posted to any PhD, or e/m expert, and they'll concur. Best regards.
Mikelm, please reread my post carefully, as I've explained that an open circuited secondary still has both current and voltage induced into it. Since 1873, it has been well known that ac voltage and ac current cannot exist independently. If one is zero, so is the other. If one is non-zero the other is non-zero. Your countryman James Clerk Maxwell proved this and for 131 years every experiment and measurement has affirmed the same.
The most popular fallacy regarding transformers is that the magnetic core flux exists due to primary current only, and not primary voltage. Also, many believe that an ac flux induces *voltage* into the secondary, and that the secondary current flows only when a load is connected. It seems obvious, since an open secondary has a voltage but "no current", one can falsely conclude that I to V action is taking place. Simply put, many labor under the myth that the primary magnetizing current creates the core magnetic flux, and that the flux induces a voltage into the secondary which in turn causes secondary current only when a load is connected. This is plain wrong and can be demolished with little effort.
Let's start with a *voltage* source, an ac generator. feeding a transformer primary with 120 volts, 60 Hz. The secondary is open. Neglecting losses, 120 volts should appear at the secondary, if the turns ratio is one to one. A 10 ohm load is applied across the secondary, and 12 amps flow. At this point it is easy to believe that the flux induces a voltage into the secondary (S), and that the S current depends on the load. The fallacy here is that it is the ac generator, which is a voltage source, that is responsible for the constant S voltage / variable S current behaviour, not the transformer. In your example, you placed a transformer in parallel with a constant voltage source, and concluded that the constant S voltage, variable S current is a result of transformer action, or I-V. A transformer merely reflects the secondary impedance back to the primary according to the turns ratio squared, or 1:1 in this case. The constant voltage behaviour at the transformer S, is merely a reflection of the constant voltage behaviour of the ac input at the P.
Remember, a current transformer P is inserted in series with the circuit where current is to be measured. The secondary is short-circuited, so that zero volts exists across the S, and the P. A volt-meter or current shunt is connected to the S terminals, and the shunt switch is opened, and a voltage exists across the S terminals. If the terminating impedance is low compared to the external circuit, a constant current behaviour exists. Changing the impedance across the S changes the voltage, but current remains approximately constant. The constant current behaviour is due to the constant current source at the P, not induction action of the transformer. Even if the circuit is voltage source driven, if the terminating impedance is way smaller than the source impedance, constant current results. Both voltage transformers and current transformer operate under the exact same laws. Whether the transformer S outputs constant voltage or constant current depends on the source powering the P.
As far as I-V action goes, the following example demonstrates how absurd the idea is. Back to a voltage source driving a transformer with open circuited S, 120 volts, 60 Hz. If this was an I-V converter, any change in primary I, would produce a corresponding change in secondary V. Let's say that the primary magnetizing current is 0.10 amp. The S voltage is 120 volts. Suppose we decrease the frequency to 50 Hz, but decrease the P voltage to 100 volts, so that the core flux remains the same, and P magnetizing current is still 0.10 amp (neglecting eddy current and hysteresis losses). The S voltage is now 100 volts. The P current remains the same, yet the S voltage has decreased by exactly the same amount as the P voltage. The S voltage follows the P voltage, not the P current.
Now, let's hold the P voltage at 120 volts, while decreasing the frequency from 60 to 50 Hz, increasing the core flux, so that the P magnetizing current is 0.12 amp instead of 0.10 amp. The P current has increased, but the S voltage remains at 120 volts. Again, the S voltage follows the P voltage, not the P current. As long as the core does not saturate, the S voltage value is determined by the P voltage. Differing conditions with regard to frequency and flux density change the value of P current but the S voltage only follows the P voltage without regard to P current.
Finally, the core flux is created by both P voltage and P current. Whether iron or air cored, all materials have a relative permeability greater than zero, and less than infinite. You need both V and I to set up a core flux. The S voltage is the P voltage scaled by the turns ratio, or V-V. The magnetizing current is needed to create the flux, along with P voltage. When a load is placed across the S, the S current is reflected to the P, and a P current flow exists in order to maintain the core flux. THe S current creates an mmf which would change the flux, so an opposing P current flows to cancel this mmf. This is I-I conversion, but it does not include the P magnetizing current. This current is not subject to the law of balancing ampere turns. The I-I conversion is between S load current, and P current due to the S load, but not counting the P magnetizing current necessary to produce the core magnetizing flux. Hence a transformer converts primary V and I into core flux. The core flux induces BOTH V and I into the secondary. This is VI to flux to VI conversion. The S voltage is determined by the P voltage, or V-V conversion. The P current is determined by the S current, or I-I conversion EXCEPT for the P magnetizing current. Whether the S outputs constant voltage or constant current is determined by the source driving the P.
As far as the question, "How can both V and I exist in an open secondary?" goes, Maxwell answered that in 1873 with displacement current. Faraday never quite got it because he only developed his equations with conduction current. Experimental findings never quite matched Faraday's equations. When Maxwell added displacement current to Faraday's law, agreement existed. Remember, each turn of the S has a potential across it and under ac conditions a displacement current will flow due to the ac electric field. Don't forget the turn to turn capacitance present in all transformer windings. Any text on e/m, energy conversion, antennae, etc. will confirm the above. THe bottom line is that ac voltages and ac currents are mutually inclusive, as well as ac electric fields, and ac magnetic fields. No one has ever determined which one, V or I, is the cause and/or effect. Take what I've posted to any PhD, or e/m expert, and they'll concur. Best regards.
Cluade
Well, I'm overwhelmed,
Thanks for such a magnificent reply.
I think that I don't disagree with NEARLY everything that you say.
The crux of the matter may be this displacement current.
You mentioned it but you did not really say what it was or how to measure it.
Can we imagine that we have a T with no interwinding C - perhaps a one turn P and a one turn S - this gives us no real chance for interwinding C. The secondary in unterminated.
If we apply an AC signal to the P there will be some voltage and some current. Out of phase ( I think this may be very significant )
my understanding is:
The change of current in the P creates back EMF that is magnetically transferred in to the S.
this EMF is potential energy - perhaps even theoratical.
Given the even the tiniest opportunity this EMF will produce a current in the S. In our senario we are trying not to give it even that tiniest opportunity.
If there is no interwinding C and there is no termination there is no chance for a current to flow in the S. Is this correct or is this where diplacement current is doing something. ( If it is could you explain it as I have not come across it before )
If you try to measure for voltage you will always find some because a measurement device will always draw some current - this supports your augument. I think this is part of measurement theory.
All I am saying is that before you applied the meter to measure that voltage the EMF was still there as a pure unrealised potential. Just because you are not measuring it does not mean that it is not there.
If you are saying that if you are measuring a voltage, there must be a current, fine I agree. But, if it is just that a particular mathematical model cannot describe a situation where that pure AC potential exists without a current I contend that this does not mean that this is necessarily the case. The model may be flawed, just like an analogy at some point always breaks down.
I have come to the understanding that Voltage/EMF either AC or DC is primary and causal, and current is purely a slave of the applied voltage and the resistive/impedance conditions.
I think there may be a possibility that we are not disagreeing here. Often apparent disagreements occur when terms are not defined properly or understood correctly.
What do you think ? Can you see what I am driving at here ? or are my auguments flawed.
If they are could you explain in laymans terms where and how.
Regards
Mike
ps How would you descibe the difference between a voltage and an EMF would you say that they are one and the same ?
Well, I'm overwhelmed,
Thanks for such a magnificent reply.
I think that I don't disagree with NEARLY everything that you say.
The crux of the matter may be this displacement current.
You mentioned it but you did not really say what it was or how to measure it.
Can we imagine that we have a T with no interwinding C - perhaps a one turn P and a one turn S - this gives us no real chance for interwinding C. The secondary in unterminated.
If we apply an AC signal to the P there will be some voltage and some current. Out of phase ( I think this may be very significant )
my understanding is:
The change of current in the P creates back EMF that is magnetically transferred in to the S.
this EMF is potential energy - perhaps even theoratical.
Given the even the tiniest opportunity this EMF will produce a current in the S. In our senario we are trying not to give it even that tiniest opportunity.
If there is no interwinding C and there is no termination there is no chance for a current to flow in the S. Is this correct or is this where diplacement current is doing something. ( If it is could you explain it as I have not come across it before )
If you try to measure for voltage you will always find some because a measurement device will always draw some current - this supports your augument. I think this is part of measurement theory.
All I am saying is that before you applied the meter to measure that voltage the EMF was still there as a pure unrealised potential. Just because you are not measuring it does not mean that it is not there.
If you are saying that if you are measuring a voltage, there must be a current, fine I agree. But, if it is just that a particular mathematical model cannot describe a situation where that pure AC potential exists without a current I contend that this does not mean that this is necessarily the case. The model may be flawed, just like an analogy at some point always breaks down.
I have come to the understanding that Voltage/EMF either AC or DC is primary and causal, and current is purely a slave of the applied voltage and the resistive/impedance conditions.
I think there may be a possibility that we are not disagreeing here. Often apparent disagreements occur when terms are not defined properly or understood correctly.
What do you think ? Can you see what I am driving at here ? or are my auguments flawed.
If they are could you explain in laymans terms where and how.
Regards
Mike
ps How would you descibe the difference between a voltage and an EMF would you say that they are one and the same ?
Claude,
I just had the thought that what you may be trying to say to me is that for a transformer to do anything useful there has to be some power transfer from the P to the S and for this to take place the has to be both V and I in both windings
If this is the case I think we may agree.
what do you say ?
regards
mike
I just had the thought that what you may be trying to say to me is that for a transformer to do anything useful there has to be some power transfer from the P to the S and for this to take place the has to be both V and I in both windings
If this is the case I think we may agree.
what do you say ?
regards
mike
more on I/V cause/effect xfmr induction etc.
Greetings Mike,
Thank you for your reply.
First, neither I nor anyone alive understands this issue completely. What I state is based on many years of painstaking effort, research, and measurements by many people. I can answer questions to a point, and then I must state that I don't know why it is, but only that it is that way very consistently.
As far as voltage being causal and current being an effect based on impedance, that is incorrect. Here is an expanded view of transformers.
First, let us just start with a 120 volt, 60 Hz, ac generator with a 100 ohm internal source impedance,and open circuited (infinite load resistance). Let us place a 10 kohm resistance across its terminals. The voltage across the 10 kohm load is around 118.8 volts, or 99 percent of the generator's open circuit voltage. The current is around 11.88 mA. If we remove the 10 kohm load and replace it with a 100 kohm load, the terminal voltage is 119.88 volts, and current is 1.1988 mA. The current is almost ten times smaller, but the voltage changed very little. Now, replace the load with 1.0 megohm. The terminal voltage is 119.988 V, and current is 0.1200 mA. With 10 megohms, we get 119.9988 V, and 0.0120 mA. As we can see, each time we up the resistance by a factor of ten, current decreases by very close to ten, while voltage barely changes at all. From 10 kohm, up to 10 megohm, all the way to infinity, this is true. Under these narrow conditions, it is easy to conclude that voltage is independent, or the "cause", while current depends on the load resistance, and is the "effect". But, let's not rush to judgement.
Now, place a 1.0 ohm resistance at the terminals. The terminal voltage is 1.1881 V, and current is 1.1881 A. Lowering the load resistance to 100 milliohms (0.10 ohms) gives 0.11988 V and 1.1988 A. Lowering the resistance by a factor of ten, reduced the voltage by almost ten, but current barely changes. Now apply 10 milliohms and the voltage is 0.011999 V and current is 1.1999 A. With 1.0 milliohm, voltage is 0.001200 V and current is 1.200 A. Each time we decrease resistance by ten, the voltage drops by ten, but the current is very constant. Under these narrow conditions, one can easily conclude that current is the cause, and voltage is the effect, exactly the opposite of the first case. This is as equally invalid as the first conclusion, another rush to judgement.
In the first case the terminating resistance was always much LARGER than the source internal resistance, and constant voltage behaviour was effected. In the second case, the terminating resistance was always much SMALLER than the source resistance, and constant current behaviour was effected. Real world power sources, like batteries or generators, are neither ideal as current or voltage sources. The relationship between the internal resistance and external load determine the actual behavior.
Now, with 100 ohm load, voltage is 60 V, and current is 0.60 A. Varying this load up by 10 (1000 ohms) gives 109.09 V, and 0.10909 A. With a 10 ohm load, we have 10.909 V, and 1.0909 A. Under these narrow conditions, the behavior is neither constant voltage nor constant current.
Now, let's place a voltage transformer across the source, with a 1:1 turns ratio and apply a load to the secondary (S). With a 1.0 ohm load, S voltage is 1.1881 V and S current is 1.1881 A, just like the example without the transformer. With a 0.10 ohm load, we get 0.11988 V and 1.1988 A. With 10 milliohm, we get 0.011999 V and 1.1999 A. The transformer S is outputting a voltage which directly varies with load resistance, while maintaining a constant current. If the S is short-circuited (0 ohms), the S voltage is 0 V, while the S current is 1.20 A. Are we then to conclude that the transformer outputs a constant CURRENT, and that voltage is determined by the load resistance?! Under these narrow conditions, can we rightly conclude that current is the "cause", and VOLTAGE the "effect". Of course not! The transformer is merely reflecting the load impedance on the secondary back to the primary in direct proportion to the turns ratio squared, or 1:1 in this example! That is EXACTLY what is happening. It's that simple folks. The constant current behavior is due to loading the S with a resistance MUCH SMALLER than the generator's internal resistance. Remember the primary (P) voltage and current are identical to those on the S. When the S is shorted (0 ohms), 0 V and 1.200 A also appears on the P. Now, if we load the S with 10 kohm, we get 118.8 V and 1.188 mA. Increasing the resistance upward yields directly varying current, with constant voltage in P and S. Under these narrow conditions, the transformer APPEARS to be outputting a constant VOLTAGE, with current being a result of the load resistance. Thus, many will falsely conclude that the laws of induction describing transformer action, Ampere's and Faraday's, state that ac flux induces VOLTAGE into the S, with current being determined by load resistance. But the previous example suggests just the opposite, doesn't it? Again, what is happening is REFLECTION of the load impedance on S back to P. It is the ratio of resistances ALONE, load vs. source, that account for the constant current or constant voltage behavior, or neither.
The reason for such mass confusion is that almost all persons familiar with transformers, have been exposed to VOLTAGE transformers exclusively, not CURRENT transformers. So, what's the difference? NONE WHATSOEVER. Current transformers are optimized to be placed in SERIES with a circuit in order to measure current. The primary must burden the circuit with the smallest voltage drop possible, for a precise measurement. The secondary has more turns than the P, increasing the voltage, while decreasing the current to a safe value. The S must be short circuited, 0 ohms, and a low valued current sensing resistor is placed across the S. The short is removed from the S terminals, and current now flows in the sense resistor. The voltage across the sense resistor is measured. Before removing the sense resistor, the S must be SHORTED again. A current transformer MUST be terminated in a resistance much smaller than the primary circuit resistance. When S is shorted there is 0 V and a current I in the S. Removing the short results in very near the same I, but a small voltage in S developed across the sense resistor. When shorted, the S is all I and no V. When loaded with small resistance, we get the same I (approx), but a measurable V. So much for the idea that transformers induce voltage. Also, consider the following. A current transformer S is shorted, 0 ohms, with 0 V and 1 A in S. The primary current then doubles. In S we now have 0 V, 0 ohms, and 2 A. In the first case we have 0 ohms and 0 volts, with 1 amp. In the second case we have 0 ohms and 0 volts with 2 amps. If transformers output voltage, with current determined by load resistance, we have a paradox here. If 0 volts is induced in S, and the terminating resistance is 0 ohms, what is the current? Is it 1 A or 2 A? It can be either. The voltage and resistance remained zero, yet the current changed. Why? Because the 0 ohm secondary resistance was reflected back to the primary. With the original primary current, and 0 ohms, in S, there is a primary (and secondary) voltage drop of 0 volts. Doubling the primary current results in a 0 volt drop as well. The S current is determined by P current, which is determined by the load resistance of S reflected back into P.
Along that line, consider the following. If the generator has 100 ohm resistance, and the secondary is loaded with 100 ohms, we get 60 V and 0.60 A in S. What determines the voltage and current in P and S? For those who think that flux produces S voltage (cause), and load determines S current (effect), how do you explain the fact that 60 V and 0.60 A is presented to the P? The generator outputs 120 V, so what determines the 60 V and 0.60 A in the P? The only answer is that the 100 ohm load is reflected back to the primary, 1:1, and it combines in series with the generator's 100 ohms, and the total resistance in the primary circuit is 200 ohms. A voltage divider determines the 60 V across P, and the 120 V and total resistance of 200 ohms determines 0.60 A (Ohm's law). Any other explanation makes no sense. For those who still insist that a transformer is an I/V converter, or that the induced quantity is voltage, with current being an effect of load resistance, please explain this 100 ohm / 100 ohm scenario.
To summarize, a transformer requires BOTH primary voltage and primary magnetizing current in order to produce flux in its core. This ac flux produces induction in the secondary. The induced quantities are BOTH voltage and current. When the secondary load impedance reflected back to the primary by the turns ratio squared is much smaller the the internal resistance of the source driving the primary, an ILLUSION of constant current (cause) is created, with voltage (effect) being determined by the load resistance. Such is the case with current transformers. In the opposite case, with reflected secondary load impedance being much greater than the source resistance, an ILLUSION of constant voltage (cause) and current (effect) being determined by load resistance takes place. This happens with voltage transformers. Since everybody is familiar with voltage xfmrs, and very few are familiar with current xfmrs, it is easy to understand why constant voltage or the "voltage is causal" myth is much more prevalent than the current counterpart. The primary current and voltage are determined by the reflected secondary impedance combined with all impedances in the primary circuit.
This is getting too long. I'll discuss displacement current tomorrow. I hope this helps. Best regards.
Greetings Mike,
Thank you for your reply.
First, neither I nor anyone alive understands this issue completely. What I state is based on many years of painstaking effort, research, and measurements by many people. I can answer questions to a point, and then I must state that I don't know why it is, but only that it is that way very consistently.
As far as voltage being causal and current being an effect based on impedance, that is incorrect. Here is an expanded view of transformers.
First, let us just start with a 120 volt, 60 Hz, ac generator with a 100 ohm internal source impedance,and open circuited (infinite load resistance). Let us place a 10 kohm resistance across its terminals. The voltage across the 10 kohm load is around 118.8 volts, or 99 percent of the generator's open circuit voltage. The current is around 11.88 mA. If we remove the 10 kohm load and replace it with a 100 kohm load, the terminal voltage is 119.88 volts, and current is 1.1988 mA. The current is almost ten times smaller, but the voltage changed very little. Now, replace the load with 1.0 megohm. The terminal voltage is 119.988 V, and current is 0.1200 mA. With 10 megohms, we get 119.9988 V, and 0.0120 mA. As we can see, each time we up the resistance by a factor of ten, current decreases by very close to ten, while voltage barely changes at all. From 10 kohm, up to 10 megohm, all the way to infinity, this is true. Under these narrow conditions, it is easy to conclude that voltage is independent, or the "cause", while current depends on the load resistance, and is the "effect". But, let's not rush to judgement.
Now, place a 1.0 ohm resistance at the terminals. The terminal voltage is 1.1881 V, and current is 1.1881 A. Lowering the load resistance to 100 milliohms (0.10 ohms) gives 0.11988 V and 1.1988 A. Lowering the resistance by a factor of ten, reduced the voltage by almost ten, but current barely changes. Now apply 10 milliohms and the voltage is 0.011999 V and current is 1.1999 A. With 1.0 milliohm, voltage is 0.001200 V and current is 1.200 A. Each time we decrease resistance by ten, the voltage drops by ten, but the current is very constant. Under these narrow conditions, one can easily conclude that current is the cause, and voltage is the effect, exactly the opposite of the first case. This is as equally invalid as the first conclusion, another rush to judgement.
In the first case the terminating resistance was always much LARGER than the source internal resistance, and constant voltage behaviour was effected. In the second case, the terminating resistance was always much SMALLER than the source resistance, and constant current behaviour was effected. Real world power sources, like batteries or generators, are neither ideal as current or voltage sources. The relationship between the internal resistance and external load determine the actual behavior.
Now, with 100 ohm load, voltage is 60 V, and current is 0.60 A. Varying this load up by 10 (1000 ohms) gives 109.09 V, and 0.10909 A. With a 10 ohm load, we have 10.909 V, and 1.0909 A. Under these narrow conditions, the behavior is neither constant voltage nor constant current.
Now, let's place a voltage transformer across the source, with a 1:1 turns ratio and apply a load to the secondary (S). With a 1.0 ohm load, S voltage is 1.1881 V and S current is 1.1881 A, just like the example without the transformer. With a 0.10 ohm load, we get 0.11988 V and 1.1988 A. With 10 milliohm, we get 0.011999 V and 1.1999 A. The transformer S is outputting a voltage which directly varies with load resistance, while maintaining a constant current. If the S is short-circuited (0 ohms), the S voltage is 0 V, while the S current is 1.20 A. Are we then to conclude that the transformer outputs a constant CURRENT, and that voltage is determined by the load resistance?! Under these narrow conditions, can we rightly conclude that current is the "cause", and VOLTAGE the "effect". Of course not! The transformer is merely reflecting the load impedance on the secondary back to the primary in direct proportion to the turns ratio squared, or 1:1 in this example! That is EXACTLY what is happening. It's that simple folks. The constant current behavior is due to loading the S with a resistance MUCH SMALLER than the generator's internal resistance. Remember the primary (P) voltage and current are identical to those on the S. When the S is shorted (0 ohms), 0 V and 1.200 A also appears on the P. Now, if we load the S with 10 kohm, we get 118.8 V and 1.188 mA. Increasing the resistance upward yields directly varying current, with constant voltage in P and S. Under these narrow conditions, the transformer APPEARS to be outputting a constant VOLTAGE, with current being a result of the load resistance. Thus, many will falsely conclude that the laws of induction describing transformer action, Ampere's and Faraday's, state that ac flux induces VOLTAGE into the S, with current being determined by load resistance. But the previous example suggests just the opposite, doesn't it? Again, what is happening is REFLECTION of the load impedance on S back to P. It is the ratio of resistances ALONE, load vs. source, that account for the constant current or constant voltage behavior, or neither.
The reason for such mass confusion is that almost all persons familiar with transformers, have been exposed to VOLTAGE transformers exclusively, not CURRENT transformers. So, what's the difference? NONE WHATSOEVER. Current transformers are optimized to be placed in SERIES with a circuit in order to measure current. The primary must burden the circuit with the smallest voltage drop possible, for a precise measurement. The secondary has more turns than the P, increasing the voltage, while decreasing the current to a safe value. The S must be short circuited, 0 ohms, and a low valued current sensing resistor is placed across the S. The short is removed from the S terminals, and current now flows in the sense resistor. The voltage across the sense resistor is measured. Before removing the sense resistor, the S must be SHORTED again. A current transformer MUST be terminated in a resistance much smaller than the primary circuit resistance. When S is shorted there is 0 V and a current I in the S. Removing the short results in very near the same I, but a small voltage in S developed across the sense resistor. When shorted, the S is all I and no V. When loaded with small resistance, we get the same I (approx), but a measurable V. So much for the idea that transformers induce voltage. Also, consider the following. A current transformer S is shorted, 0 ohms, with 0 V and 1 A in S. The primary current then doubles. In S we now have 0 V, 0 ohms, and 2 A. In the first case we have 0 ohms and 0 volts, with 1 amp. In the second case we have 0 ohms and 0 volts with 2 amps. If transformers output voltage, with current determined by load resistance, we have a paradox here. If 0 volts is induced in S, and the terminating resistance is 0 ohms, what is the current? Is it 1 A or 2 A? It can be either. The voltage and resistance remained zero, yet the current changed. Why? Because the 0 ohm secondary resistance was reflected back to the primary. With the original primary current, and 0 ohms, in S, there is a primary (and secondary) voltage drop of 0 volts. Doubling the primary current results in a 0 volt drop as well. The S current is determined by P current, which is determined by the load resistance of S reflected back into P.
Along that line, consider the following. If the generator has 100 ohm resistance, and the secondary is loaded with 100 ohms, we get 60 V and 0.60 A in S. What determines the voltage and current in P and S? For those who think that flux produces S voltage (cause), and load determines S current (effect), how do you explain the fact that 60 V and 0.60 A is presented to the P? The generator outputs 120 V, so what determines the 60 V and 0.60 A in the P? The only answer is that the 100 ohm load is reflected back to the primary, 1:1, and it combines in series with the generator's 100 ohms, and the total resistance in the primary circuit is 200 ohms. A voltage divider determines the 60 V across P, and the 120 V and total resistance of 200 ohms determines 0.60 A (Ohm's law). Any other explanation makes no sense. For those who still insist that a transformer is an I/V converter, or that the induced quantity is voltage, with current being an effect of load resistance, please explain this 100 ohm / 100 ohm scenario.
To summarize, a transformer requires BOTH primary voltage and primary magnetizing current in order to produce flux in its core. This ac flux produces induction in the secondary. The induced quantities are BOTH voltage and current. When the secondary load impedance reflected back to the primary by the turns ratio squared is much smaller the the internal resistance of the source driving the primary, an ILLUSION of constant current (cause) is created, with voltage (effect) being determined by the load resistance. Such is the case with current transformers. In the opposite case, with reflected secondary load impedance being much greater than the source resistance, an ILLUSION of constant voltage (cause) and current (effect) being determined by load resistance takes place. This happens with voltage transformers. Since everybody is familiar with voltage xfmrs, and very few are familiar with current xfmrs, it is easy to understand why constant voltage or the "voltage is causal" myth is much more prevalent than the current counterpart. The primary current and voltage are determined by the reflected secondary impedance combined with all impedances in the primary circuit.
This is getting too long. I'll discuss displacement current tomorrow. I hope this helps. Best regards.
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