Here's a description from _Electronic Designers' Handbook_ Landee, Davis and Albrecht 1957, p. 14-10:
If a voltage is applied to the primary winding of the transformer, primary current will flow. If the secondary is open, the current which flows in the primary is the exciting current i-sub-theta. The exciting current is composed of a current in phase with the voltage induced in the primary winding and a current which is lagging the induced voltage by 90 degrees. The first current is the core loss current i-sub-c, and the second current is the magnetizing current i-sub-m. The primary induced voltage is equal to the applied voltage less the vector (i-sub-theta)(R), where R is the primary winding resistance. It is the magnetizing current which causes magnetic flux to exist in the core, and it is the magnetic flux which causes a voltage to be induced in the primary winding. The magnitude of the primary induced voltage is usually very nearly equal to the applied voltage and can be determined from
e-sub-g = (R)(i-sub-theta) + e-sub-i
where e-sub-g = generator voltage (here called applied voltage) and
i-sub-theta = exciting current
e-sub-i = induced voltage and is proportional to d-theta/dt
The secondary induced voltage is of the same phase as the voltage induced in the primary and is related in amplitude to the primary induced voltage by the ratio of the turns in the secondary winding to the turns in the primary winding. The application of a load to the transformer secondary causes secondary current to flow and is reflected to the primary as a decrease in the primary input impedance and, hence, causes an increase in the primary current. In an ideal transformer, loading the secondary will cause an increase in the primary ampere-turns which is equal and opposite to the secondary ampere-turns caused by the loading. The net result is that there is no change in the core flux because of load currents in the windings.
Fairly dense, but the last sentence is what DF96 is trying to get folks to appreciate.
All good fortune,
Chris
If a voltage is applied to the primary winding of the transformer, primary current will flow. If the secondary is open, the current which flows in the primary is the exciting current i-sub-theta. The exciting current is composed of a current in phase with the voltage induced in the primary winding and a current which is lagging the induced voltage by 90 degrees. The first current is the core loss current i-sub-c, and the second current is the magnetizing current i-sub-m. The primary induced voltage is equal to the applied voltage less the vector (i-sub-theta)(R), where R is the primary winding resistance. It is the magnetizing current which causes magnetic flux to exist in the core, and it is the magnetic flux which causes a voltage to be induced in the primary winding. The magnitude of the primary induced voltage is usually very nearly equal to the applied voltage and can be determined from
e-sub-g = (R)(i-sub-theta) + e-sub-i
where e-sub-g = generator voltage (here called applied voltage) and
i-sub-theta = exciting current
e-sub-i = induced voltage and is proportional to d-theta/dt
The secondary induced voltage is of the same phase as the voltage induced in the primary and is related in amplitude to the primary induced voltage by the ratio of the turns in the secondary winding to the turns in the primary winding. The application of a load to the transformer secondary causes secondary current to flow and is reflected to the primary as a decrease in the primary input impedance and, hence, causes an increase in the primary current. In an ideal transformer, loading the secondary will cause an increase in the primary ampere-turns which is equal and opposite to the secondary ampere-turns caused by the loading. The net result is that there is no change in the core flux because of load currents in the windings.
Fairly dense, but the last sentence is what DF96 is trying to get folks to appreciate.
All good fortune,
Chris
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It is good that now DF96 has an interpreter.
Leaving aside that he is trying to convince us that power transformers and SE output transformers are the same thing, I cannot even remotely see the similitude, he does not mention any flux at all, but surely is my Tarzan-English.
I don't usually see popilin's posts, as he is in my ignore list due to previous experience with him.
I am not trying to convince anyone that power transformers and SE OPTs are the same thing - that would be daft. I did use a comparision of them to demonstrate that maximum primary current (AC+DC) is not limited by saturation. In a power transformer the maximum DC current is quite small; much smaller than the AC current during normal operation. In an SE OPT all that has happened is that the core has a gap, so the DC current can be greater. The AC current can still exceed the maximum allowed DC current, which is what some people seem unable to grasp.
As I said, if you want to talk about the details of audio transformers then first you need to understand the basics of all transformers. Otherwise you are just building on sand.
I am not trying to convince anyone that power transformers and SE OPTs are the same thing - that would be daft. I did use a comparision of them to demonstrate that maximum primary current (AC+DC) is not limited by saturation. In a power transformer the maximum DC current is quite small; much smaller than the AC current during normal operation. In an SE OPT all that has happened is that the core has a gap, so the DC current can be greater. The AC current can still exceed the maximum allowed DC current, which is what some people seem unable to grasp.
As I said, if you want to talk about the details of audio transformers then first you need to understand the basics of all transformers. Otherwise you are just building on sand.
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Yes you are but you are not aware that your analisys is poor. The real problem is that you have NO exprience at all with SE transformes.
This is clear also from small details.
For example, recommended max DC current in a datasheet IS an operative current at which the rated power will be achieved down to a certain FREQUENCY. It's not just a max DC current. The word "recommended" means something.
The worst thing however is that you ignore that a power transformer will work at FIXED frequency (50/60Hz) while a SE transformer is a WIDEBAND transformer. This has HUGE practical implications....
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Balance
All,
For a transformer with an open secondary the source sees the transformer as a primary inductance. With the secondary shorted the source sees the transformer’s leakage inductance. Connect an 8 ohm resistor to the secondary winding because of the turn ratio, the source sees a 5000 ohm connected load. From an AC point of view power going in the primary winding equals the power out the secondary winding, with AC power input there is no change in magnetic energy stored in the iron of the transformer in the process. Yes there is balance, like balance on a playground teeter totter.
https://www.youtube.com/watch?v=DQmuUNI7Zmk
8 Volt*amps input = 8 volt amps output.
Or a banana in, is a banana out.
DT
All,
For a transformer with an open secondary the source sees the transformer as a primary inductance. With the secondary shorted the source sees the transformer’s leakage inductance. Connect an 8 ohm resistor to the secondary winding because of the turn ratio, the source sees a 5000 ohm connected load. From an AC point of view power going in the primary winding equals the power out the secondary winding, with AC power input there is no change in magnetic energy stored in the iron of the transformer in the process. Yes there is balance, like balance on a playground teeter totter.
https://www.youtube.com/watch?v=DQmuUNI7Zmk
8 Volt*amps input = 8 volt amps output.
Or a banana in, is a banana out.
DT
It is sad that someone has an ignore list, even more on this forum, one of these days we have to make peace.
This is not a good idea for a demonstration, maximum DC bias current is given by manufacturer, maximum AC current is not; on the other hand, primary AC current must include the magnetizing current which is not just a small fraction of total primary AC current as you always argue, let me explain
Bac = μ Hac, Hac = f(Imagnetizing)
In an ungapped power transformer, μ is usually in the order of 10⁴, then, magnetizing current will be rather small, but the gap in a SE OPT usually reduces μ about two orders of magnitude, then magnetizing current will be greatly increased, and unfortunately magnetizing current is limited by saturation.
Agree.
Unfortunately, despite the much lower magnetizing current, saturation also happens in PP amplifiers within the audio range.
The only real advantage is that it usually happens at lower frequency respect to typical SE amps but in most cases still in the audio range. In lower power amplifiers it also happens and actually is just as common, this time because these are generally made with economy in mind and thus use relatively small transformers.
Here is an example:
http://www.tube-amps.net/images/Hashimoto_Specs/HW-100-5_1024.jpg
If you make a 100W amp with it you will get saturation just below 30Hz. And this is quite BIG!
The only real advantage is that it usually happens at lower frequency respect to typical SE amps but in most cases still in the audio range. In lower power amplifiers it also happens and actually is just as common, this time because these are generally made with economy in mind and thus use relatively small transformers.
Here is an example:
http://www.tube-amps.net/images/Hashimoto_Specs/HW-100-5_1024.jpg
If you make a 100W amp with it you will get saturation just below 30Hz. And this is quite BIG!
An interesting thing about saturation in output transformers is that it only occurs in the magnetising inductance, which is in shunt with the signal. This means it has its audible effect by nonlinearly loading the stage which is driving the transformer. (The OT tries to draw excessive current peaks when it is saturated.) In other words, if the driving stage had zero output impedance then the saturation would have no audible effect!
Valve circuits of course tend to have rather high output impedance (although some more than others).
Leakage inductances maintain a high reluctance because, by definition, their flux path includes a significant part that is not through the iron.
Valve circuits of course tend to have rather high output impedance (although some more than others).
Leakage inductances maintain a high reluctance because, by definition, their flux path includes a significant part that is not through the iron.
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Another difference is that power transformers work at fixed AC voltages, then magnetizing current is almost constant (It is maximum with no load); in a SE OPT, primary AC voltage can go from zero to Uac(max) and also does magnetizing current (I hate the term!) from zero to Im(max)
If you want a good transformer, you must winding it for yourself.
Now Hashimoto San is stingy, or he does not like good bass.
Now Hashimoto San is stingy, or he does not like good bass.
I have the smaller ones and they are excellent. In my case they do not saturate until 16 Hz. But that's because my amp in not very powerful.
They do not hate the bass but for a given core size they always try to find the best balance between power rating and losses. This one should in the 0.15-0.2 dB range (i.e. what people call transparent) and as you can see pretty well behaved up to high frequency roll-off. Not an easy job when the transformer as to be as universal as possible. If need more then you custom order.
It's ok. I had a thought about this and discussed with popilin . It will not cross the axis but what will happen if Bdc<Bac is that the tube will not be able to sustain the current required by the load at that frequency. Being low frequency the load won't be the nominal. In the example I made with LL1682/50mA used with 34 mA Bdc<Bac at 30Hz for 3.6W output (while nominally into a resistive load you would get 5.6W). So, despite the fact the transformer might seem fine, the tube will distort more in comparison to "another" transformer with same ratio but where Bdc>Bac. "Another" means you just need to re-gap the transformer for 34 mA dc so that Bdc = 0.8T and of course get more inductance. Probably it won't be clean 5.6W anyway but very close.
It's 45 DHT.
Dual Triode is nothing emotional. Me and Popilin are good friends.
Back to subject: it's the way it works!
It's not just math, I have done several SE transformers for myself. Never thought about the zero-crossing before because the Bdc>Bac condition firstly is required to make the tube work in optimal conditions as low as possible in frequency domain.What I told you are the conditions to optimize a transformer or, if you have a transformer, to find an optimal use for it. Bdc>Bac means that L*Idc>Vrms/4.44f (where 4.44 is true in sinusoidal regime). You have to satisfy this condition to get the most out of it at lower end otherwise the tube will cut-off soon.
If you re-gap the LL1692 to get 0.9T @34 mA instead of 50 mA it means that the gap will be smaller and the inductance will be in the region of 51-52H instead of 35. So at 30Hz the effective impedance will be higher and you will get some 4.2W instead of 3.5W before the cut-off. From this point plenty of distortion despite the transformer is fine. It's the tube!
Of course a better transformer can be made but at some point it will become unpractical. In other words you will never be able to achieve what can be done with a push-pull stage and its optimal (and practical) transformer.
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