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23rd September 2003, 11:10 PM  #51  
diyAudio Moderator

Quote:
__________________
"Pity, wrath, heroism, filled them, but the power of putting two and two together was annihilated." EM Forster 

23rd September 2003, 11:15 PM  #52  
diyAudio Senior Member
Join Date: Aug 2002
Location: Belgium

Hi,
Quote:
It's exactly the same priciple of operation as is used for moving coil stepup xformers. Cheers,
__________________
Frank 

12th February 2004, 12:16 AM  #53  
diyAudio Member

What !!!
Quote:
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 ? 

12th February 2004, 07:57 AM  #54  
diyAudio Member
Join Date: Apr 2002
Location: *

Re: What !!!
Quote:
flavour of IV. If you compare just R IV, transformer after R, GBS IV 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 IV" and others prefer just R into a low noise hi gm tube. Make one and see what you like. Cheers, Terry 

12th February 2004, 08:01 AM  #55 
diyAudio Senior Member
Join Date: Aug 2002
Location: Belgium

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 midhighs with a 12dB slope....Yuk. But some people may like that anyway. Cheers,
__________________
Frank 
12th February 2004, 08:24 AM  #56  
diyAudio Member

Re: Re: What !!!
Quote:
Yes I have been looking around at the different options. rudolf's latest seems to be a good bet . What would you recommend ? I'm not familiar with all the EE acronyms GBS... ? Is this Grounded Base Something...? cheers mike 

12th February 2004, 06:22 PM  #57  
diyAudio Member
Join Date: Apr 2003
Location: NE Ohio

re : What!!!
Quote:
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 IV. 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 shortcircuited, so that zero volts exists across the S, and the P. A voltmeter 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 IV 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 IV 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 VV. 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 II conversion, but it does not include the P magnetizing current. This current is not subject to the law of balancing ampere turns. The II 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 VV conversion. The P current is determined by the S current, or II 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.
__________________
"We wish to reach the moon, not because it is easy, but because it is hard." John F. Kennedy, 19171963, US President 

12th February 2004, 11:34 PM  #58 
diyAudio Member

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 ? 
13th February 2004, 12:01 AM  #59 
diyAudio Member

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 
13th February 2004, 02:15 AM  #60 
diyAudio Member
Join Date: Apr 2003
Location: NE Ohio

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 shortcircuited (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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"We wish to reach the moon, not because it is easy, but because it is hard." John F. Kennedy, 19171963, US President 
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