Good to see you again! I hope you are staying well and carrying on with life. It sounds like you are and that is good! You have so much knowledge and experience, and it is very nice you are sharing it with us. Thank you! Take care of yourself!
Great video as always, David! As for @16:40 it doesn't matter how the coils are wound as long as they are wound on the same coil and share the same magnetic field your statement is correct. I will try to find and scan a nice excerpt from an old transformer book where it is very nicely explained with detailed formulas and share it.
@@Delatsch Great - thank you... you have made that clear. I wasn't sure as I thought, maybe, they had to be physically wound on top of one another. Always good to learn something.
Thanks for another great video David. I think it would be interesting if you got your hands on a OTL type amplifier and did the "El Paso TubeAmps once (or twice) over" on it showing how the design gets around impedance matching to the speakers without an output transformer. I am a fan of transcendent sound, but I do not have an amplifier I could share (yet).
Dave, The OTL design has banged around in my head for decades and it has been brought up a few times here on YT. I have to admit, it could be fun. I have a good many 6AS7 and even some other of those super low mu triodes with monster filament current requirements. I think it might end up as hot as the SV572-10 amplifier but with a lot lower plate voltage. After I finish my friends Theremin and the 1619 amplifier, I may crawl around in the attic and see if I can find enough tubes to make it happen. I have gotten into stereo amplifier building so I can, and will, actually use and listen to them and it isn't just another lab experiment. Building stereo amplifiers is a lot more work... Here is something I just found - www.bonavolta.ch/hobby/en/audio/6as7_2.htm I also am remembering what scares me about these amplifiers - direct coupling to the speakers - a plate to cathode short would put plate voltage across the speakers. Maybe not likely to happen but it would only take once. I will have to think about it and maybe it should be a one channel for experimenting - at least in the beginning. Thanks for your thoughts.
@@ElPasoTubeAmps Bruce (the designer of amplifier you linked to) has a book on Amazon with some OTL amp designs that have full speaker protection. Though I am thinking the one you linked to might as well. I think it could be an interesting video series.
I thought pri Z was 1/4 in Class B, but 1/2 in Class A, so the Z as seen by the valve or tube is dependent on it's operation. Another thing about rp or ra is that it forms a potential divider with the Z of the OPT, so a valve with a high rp needs an OPT with a high Z pri. Thanks for this video, it's always good to have another POV on the subject.
I will need to study rp and ra a bit further but it seems like this is taken into consideration in cascading voltages stages - I looked up several tubes, 4-400A, 833A and 4CX250B but I don't see any specs for A and then B ratings. I am sure that is possible but I need to learn more about that also. I might assume class-B would need a different plate load than class-A ?? I do know that in class-C RF amplifiers, the load the class-C tube presents to the modulation transformer is simply Z = E/I as the current drawn by the class-C amplifier is essential steady even under modulation. However, this same tube, at the same time, needs a RF load as is presented by the resonate plate tank circuit of about 1/2 that. Some formulas say the RF load is Z = E/(1.8*I) so there can be some nitpicking about that but for most practical purposes it is 1/2 the modulator load. Always something to learn - I suppose that is what keeps us going.
If the transformers are CT push pull for both classes of operation, the inductance and impedance will always be 4 to 1 which is due to the impedance being related to the turns ratio squared or N^2 where the turns ratio N is 2/0.5 =4. To make sense of this, realize that the primary total number of turns is twice that of either primary, so N = 2 and N^2 = 2^2 = 4. So that is why if Zp-p is 5k, Zp-CT is 1 25k. It's purely turns ratio related and nothing to do with the class of operation. Hope that makes sense
@@markg1051 This from the Valve Wizard website "While both valves are conducting the amplifier operates in Class A, and both valves 'see' a load of ½ the anode-anode impedance of the transformer (1/2 Za-a). However, as soon as one valve cuts off, that half of the transformer's primary is no longer part of the circuit. Because the impedance ratio is the square of the turns ratio, the load presented to the remaining 'on' valve must be only ¼ Za-a." Which if I understand correctly in Class A each valve sees 1/4 of the Z pri - 1/2 Z a-a, hence my confusion. So in essence your right. It is a confusing subject sometimes, apologies if I've added to the confusion.
@@diabolicalartificer Not a problem... Thank you for commenting. From reading what is quoted from the Valve Wizard site, the first sentence would lead one to believe the transformer impedance (Z) from CT to one plate is, 1/2, the Z of the transformer, plate-to-plate, which we know is not correct. Also the statement thrown in about Class-A is unnecessary, as the transformer Z is independent of any class of operation, and would add to the confusion but then the site starts to get it right at the end of the statement but now it has been said two different ways and everyone is likely to be confused... lol... 🙂 I did not understand this until a few years ago and to add to confusion let's say, for example, if measured with an very good inductance/impedance meter, a transformer that measures exactly, 20.0 H or 5000.0 ohms Z, plate-to-plate, the H or Z will not be exactly 5.0 H, or exactly 1250.0 Z ohms (but sort-of close...) from CT to one side of the transformer in a real-world measurement but in theory it is to be treated as as such... I think it is these "close" but not perfectly 1/4 values that can challenge an initial understanding and this is true with most real-world measurements in electronics of not being exact as the mathematical approach might lead us to believe. I hope I said it in a way that does not add to the confusion... Thanks again.
Thanks for the great videos. I'm about to build my first tube stereo amp. It will be a 6SL7 driver and 6V6 SE OP. I have picked some published schematics to work from. What I don't understand is the coupling of my other components which all output 2 Volts rms. I plan to use an ALPS style 100K pot for volume control at the amp input. How do I determine what component network I need between the 100K pot wiper and the grid of the 6SL7? THANKS CHEERS.
Two volts input to power amplifiers is a good level to start with. Your 100K pot is fine and nothing special has to be done. The 6SL7 is a high gain tube and the predecessor to the 12AX7 so you might actually have too much gain and want to have a 6SN7 handy to swap in and try. Your input signal to the amplifier goes to one end of the 100K pot, the other end goes to ground and the wiper goes to the grid of the input tube. Sometimes people use a "grid stopper" resistor between 1K and maybe 100k from the wiper to the grid as it can improve stability. There actually is an effect called the Miller Effect that occurs at some critical value of input R and internal capacitance that can cause parasitic oscillation at some input pot setting without the grid-stopper. Some people will add a capacitor (maybe 0.1 uF or so) in series with the wiper and the grid (and the grid-stopper resistor) but if you do this you must add a resistor to ground (470K or so...) from the grid side of the input capacitor to ground as you must provide some high resistance DC path to ground for the grid or a charge will build up on the grid at some random point and it will just stop working as a bias voltage will build on the grid inside the tube. It won't hurt anything it is just one of those funny things about making sure capacitors have a DC path around them. Hope this helps and makes sense. Here is a link to a good example of Williamson amplifier. www.vintage-radio.info/heathkit The amplifiers like the W-2 and W5 or any of the Williamson Type Amplifiers, are good examples.
@@ElPasoTubeAmps YES, that does help. I have built guitar amps but always from proven schematics. I understand the biasing but was unclear on grid circuitry and couldn't find any explanations on line. Many thanks.
I think that what matters is the reflected impedance under load, which ideally would be resistive but of course never is, speakers and transformers being what they are. Is that what Zobel networks are for? An attempt to cancel out some of the reactive components? Probably gm is more relevant when driving through a transformer. A thought provoking video. Greetings from across the pond.
I think it is important to know these characteristics of a tube, and how they relate to each other, before diving into it and after the tube is chosen, the more common parameters become important like plate dissipation, etc. The Zobel network is something that has caught my attention before. Here is the simplest explanation I can find that I think it is intended for, on the output of an audio transformer. "A Zobel can be used to restore the impedance of the driver to the nominal value in the high frequency range." The "driver", in this case, being the output transformer. www.omnicalculator.com/physics/cutoff-frequency#what-is-cutoff-frequency-cutoff-frequency-definition Being a series RC network it will have a cut-off (sometimes called a corner-frequency) (3 dB point) and a typical value is a 47 ohm resistor and 0.1 uF capacitor in series across the OPT. This is what is in the McIntosh MA230. Calculating this cutoff frequency, comes out to be 33.8 KHz .... I think you can see why I can be confused about it also. Why so high in frequency? These simple series RC circuits are common in Williamson design audio amplifiers right off the plate of the first stage of the 6SN7 or 12AU7 but the purpose they serve there, depending on their value, is to either suppress AM broadcast stations (design for about 1 MHz) from getting into the amplifier (that was a potential problem 50+ years ago), or around 75KHz to suppress a tendency for the amplifier to have parasitics at that frequency mostly caused by the OPT and global NFB issues. I have actually seen this issue on a spectrum analyzer in some, but not all, of the amplifiers I have built. Adding this circuit in the Williamson design will increase THD up at the 20 KHz range (slightly) but it does its job of suppressing this parasitic. Thanks for your thoughts and comments.
Yes, on some windings but not all, it's called bi-filar winding. This is usually only done on the secondary as a bi-filar primary has to have very good insulation, there is very high voltage between the two windings meaning there's a higher chance of breakdown between the two.
I see you got the answer from Diabolical Artificier. McIntosh does it this way and is one reason their amplifiers are so good. I think there are also trifilar windings with three wires wound at a time. McIntosh calles Unity wound transformers. th-cam.com/video/W58Rqx_4u2w/w-d-xo.html
David, very good. Two questions if I may. 1. EF86: rp=2M5 and gm=1.8mA/V, so mu=4500? Just checking my number crunching. 2. Why do the tube books and datasheets generally give mu for triodes but gm for pentodes? Happy birthday! EJP
Hi Esmond - thanks for remembering my birthday. I am not sure if I am older or younger than I think... hmmmmm.... Well, your calculations are what the formulas say so. I can't remember right off the top of my head - I should consult RDH or something reliable but I seem to remember they use gm in multigrid tubes as the mu can be very high (and variable as mu can also) and it seems like it has something to do with screen grid (amplification ?). I will look into it tonight. I posted a video on how I rescued my HV transformer by cleaning up an arc point and then "floating" it above ground so I could use it again.
Very interested in your knowledge and decades of tube experience. Great videos. However, I can't find your direct contact information. I would like to contact you directly for a possible consultation work, if you are interested.
Hi James, You can email me directly at my YT account: TheAudioShop@SBCGlobal.Net. Not sure what you are looking for but I will listen. I haven't done "work" since letting go of an after-retirement, hobby-job (don't ever say hobby-job to your tax accountant... :-) but, what do you have in mind?
No, 2 inductances in series are simply the sum, ie, L(total) = L1 + L2, just like with resistances. I think your confusing that with how Z ratio relates to turns ratio.
It is true that inductance in series is just like resistors if they are separate inductors but if coils are wound on the same core on top of each other, like in a common CT output transformer, the total inductance is the two values in series plus the mutual inductance they gain by being wound on top of each other on the same core and the total inductance is four times. Turns ratios and voltages ratios are the same and inductance and impedance ratios are the same.
Two separate inductors with two separate cores are L1 + L2. Wound on a single core, the inductance is proportional to the square of the amount of turns. So two inductors on the same core connected together square, they do not add. Even the turns ratio is turns squared over turns squared which is where you get Zo/Zi. I didn't know until I looked at some inductance equations and remembered.
No, it is related to N^2 where in this case N is 2 and N^2 is 4, so if half primary is half the total primary number of turns that gives an N of 2. So if you start with knowing half primary inductance, total primary will be 4x that and v.v.
@@markg1051 You suggest if I use my LCR meter and measure from the center-tap to each outer connection I measure 5H each but if I measure from outer connection to outer connection I will measure 20H?
@@MrTalaue That is correct assuming your transformer is 20H plate-to-plate. May not be 100% exactly but you will see it is essentially 1/4 the, inductance and impedance, from CT to either plate lead compared plate-to-plate.
@@ElPasoTubeAmps That is very interesting as I would expect half of the inductance when I measure a coil from one end to the middle, so it is up to the magnetic coupling which is different with the transformer.
Good to see you again!
I hope you are staying well and carrying on with life. It sounds like you are and that is good!
You have so much knowledge and experience, and it is very nice you are sharing it with us. Thank you!
Take care of yourself!
I glad you’re doing good
I start my new job today after 6months
thanks David this it the math answer I've been looking for 300 B amp i built is my best amp I've ever don for my Altec 604-8h
1/2 squired is 1/4 thus the 1.25 k to ct and thus 4 times one side of the ct to plate giving the 5k for the whole. Nice video. Good example shown . 😄
Thank you very much for this video, this information has been of great help for me!
Great video as always, David! As for @16:40 it doesn't matter how the coils are wound as long as they are wound on the same coil and share the same magnetic field your statement is correct. I will try to find and scan a nice excerpt from an old transformer book where it is very nicely explained with detailed formulas and share it.
*wound on the same core
@@Delatsch Great - thank you... you have made that clear. I wasn't sure as I thought, maybe, they had to be physically wound on top of one another. Always good to learn something.
I'm very glad you're doing well :)
Excellent explanation👌🏾
Thanks for another great video David. I think it would be interesting if you got your hands on a OTL type amplifier and did the "El Paso TubeAmps once (or twice) over" on it showing how the design gets around impedance matching to the speakers without an output transformer. I am a fan of transcendent sound, but I do not have an amplifier I could share (yet).
Dave,
The OTL design has banged around in my head for decades and it has been brought up a few times here on YT. I have to admit, it could be fun. I have a good many 6AS7 and even some other of those super low mu triodes with monster filament current requirements. I think it might end up as hot as the SV572-10 amplifier but with a lot lower plate voltage. After I finish my friends Theremin and the 1619 amplifier, I may crawl around in the attic and see if I can find enough tubes to make it happen. I have gotten into stereo amplifier building so I can, and will, actually use and listen to them and it isn't just another lab experiment. Building stereo amplifiers is a lot more work...
Here is something I just found - www.bonavolta.ch/hobby/en/audio/6as7_2.htm
I also am remembering what scares me about these amplifiers - direct coupling to the speakers - a plate to cathode short would put plate voltage across the speakers. Maybe not likely to happen but it would only take once.
I will have to think about it and maybe it should be a one channel for experimenting - at least in the beginning.
Thanks for your thoughts.
@@ElPasoTubeAmps Bruce (the designer of amplifier you linked to) has a book on Amazon with some OTL amp designs that have full speaker protection. Though I am thinking the one you linked to might as well. I think it could be an interesting video series.
@@Dave_____ Seconded. I'd love to see a vid on OTL design.
I thought pri Z was 1/4 in Class B, but 1/2 in Class A, so the Z as seen by the valve or tube is dependent on it's operation. Another thing about rp or ra is that it forms a potential divider with the Z of the OPT, so a valve with a high rp needs an OPT with a high Z pri. Thanks for this video, it's always good to have another POV on the subject.
I will need to study rp and ra a bit further but it seems like this is taken into consideration in cascading voltages stages -
I looked up several tubes, 4-400A, 833A and 4CX250B but I don't see any specs for A and then B ratings. I am sure that is possible but I need to learn more about that also.
I might assume class-B would need a different plate load than class-A ??
I do know that in class-C RF amplifiers, the load the class-C tube presents to the modulation transformer is simply Z = E/I as the current drawn by the class-C amplifier is essential steady even under modulation. However, this same tube, at the same time, needs a RF load as is presented by the resonate plate tank circuit of about 1/2 that. Some formulas say the RF load is Z = E/(1.8*I) so there can be some nitpicking about that but for most practical purposes it is 1/2 the modulator load.
Always something to learn - I suppose that is what keeps us going.
If the transformers are CT push pull for both classes of operation, the inductance and impedance will always be 4 to 1 which is due to the impedance being related to the turns ratio squared or N^2 where the turns ratio N is
2/0.5 =4.
To make sense of this, realize that the primary total number of turns is twice that of either primary, so N = 2 and N^2 = 2^2 = 4. So that is why if Zp-p is 5k,
Zp-CT is 1 25k. It's purely turns ratio related and nothing to do with the class of operation.
Hope that makes sense
@@markg1051 This from the Valve Wizard website "While both valves are conducting the amplifier operates in Class A, and both valves 'see' a load of ½ the anode-anode impedance of the transformer (1/2 Za-a). However, as soon as one valve cuts off, that half of the transformer's primary is no longer part of the circuit. Because the impedance ratio is the square of the turns ratio, the load presented to the remaining 'on' valve must be only ¼ Za-a." Which if I understand correctly in Class A each valve sees 1/4 of the Z pri - 1/2 Z a-a, hence my confusion. So in essence your right.
It is a confusing subject sometimes, apologies if I've added to the confusion.
@@diabolicalartificer Not a problem... Thank you for commenting. From reading what is quoted from the Valve Wizard site, the first sentence would lead one to believe the transformer impedance (Z) from CT to one plate is, 1/2, the Z of the transformer, plate-to-plate, which we know is not correct.
Also the statement thrown in about Class-A is unnecessary, as the transformer Z is independent of any class of operation, and would add to the confusion but then the site starts to get it right at the end of the statement but now it has been said two different ways and everyone is likely to be confused... lol... 🙂
I did not understand this until a few years ago and to add to confusion let's say, for example, if measured with an very good inductance/impedance meter, a transformer that measures exactly, 20.0 H or 5000.0 ohms Z, plate-to-plate, the H or Z will not be exactly 5.0 H, or exactly 1250.0 Z ohms (but sort-of close...) from CT to one side of the transformer in a real-world measurement but in theory it is to be treated as as such... I think it is these "close" but not perfectly 1/4 values that can challenge an initial understanding and this is true with most real-world measurements in electronics of not being exact as the mathematical approach might lead us to believe. I hope I said it in a way that does not add to the confusion... Thanks again.
@@ElPasoTubeAmps Thanks for clearing that up, much appreciated.
Thanks for the great videos. I'm about to build my first tube stereo amp. It will be a 6SL7 driver and 6V6 SE OP. I have picked some published schematics to work from. What I don't understand is the coupling of my other components which all output 2 Volts rms. I plan to use an ALPS style 100K pot for volume control at the amp input. How do I determine what component network I need between the 100K pot wiper and the grid of the 6SL7? THANKS CHEERS.
Two volts input to power amplifiers is a good level to start with. Your 100K pot is fine and nothing special has to be done. The 6SL7 is a high gain tube and the predecessor to the 12AX7 so you might actually have too much gain and want to have a 6SN7 handy to swap in and try. Your input signal to the amplifier goes to one end of the 100K pot, the other end goes to ground and the wiper goes to the grid of the input tube. Sometimes people use a "grid stopper" resistor between 1K and maybe 100k from the wiper to the grid as it can improve stability. There actually is an effect called the Miller Effect that occurs at some critical value of input R and internal capacitance that can cause parasitic oscillation at some input pot setting without the grid-stopper. Some people will add a capacitor (maybe 0.1 uF or so) in series with the wiper and the grid (and the grid-stopper resistor) but if you do this you must add a resistor to ground (470K or so...) from the grid side of the input capacitor to ground as you must provide some high resistance DC path to ground for the grid or a charge will build up on the grid at some random point and it will just stop working as a bias voltage will build on the grid inside the tube. It won't hurt anything it is just one of those funny things about making sure capacitors have a DC path around them. Hope this helps and makes sense.
Here is a link to a good example of Williamson amplifier.
www.vintage-radio.info/heathkit The amplifiers like the W-2 and W5 or any of the Williamson Type Amplifiers, are good examples.
@@ElPasoTubeAmps YES, that does help. I have built guitar amps but always from proven schematics. I understand the biasing but was unclear on grid circuitry and couldn't find any explanations on line. Many thanks.
I think that what matters is the reflected impedance under load, which ideally would be resistive but of course never is, speakers and transformers being what they are. Is that what Zobel networks are for? An attempt to cancel out some of the reactive components? Probably gm is more relevant when driving through a transformer. A thought provoking video. Greetings from across the pond.
I think it is important to know these characteristics of a tube, and how they relate to each other, before diving into it and after the tube is chosen, the more common parameters become important like plate dissipation, etc.
The Zobel network is something that has caught my attention before. Here is the simplest explanation I can find that I think it is intended for, on the output of an audio transformer.
"A Zobel can be used to restore the impedance of the driver to the nominal value in the high frequency range." The "driver", in this case, being the output transformer.
www.omnicalculator.com/physics/cutoff-frequency#what-is-cutoff-frequency-cutoff-frequency-definition
Being a series RC network it will have a cut-off (sometimes called a corner-frequency) (3 dB point) and a typical value is a 47 ohm resistor and 0.1 uF capacitor in series across the OPT. This is what is in the McIntosh MA230. Calculating this cutoff frequency, comes out to be 33.8 KHz .... I think you can see why I can be confused about it also. Why so high in frequency?
These simple series RC circuits are common in Williamson design audio amplifiers right off the plate of the first stage of the 6SN7 or 12AU7 but the purpose they serve there, depending on their value, is to either suppress AM broadcast stations (design for about 1 MHz) from getting into the amplifier (that was a potential problem 50+ years ago), or around 75KHz to suppress a tendency for the amplifier to have parasitics at that frequency mostly caused by the OPT and global NFB issues. I have actually seen this issue on a spectrum analyzer in some, but not all, of the amplifiers I have built. Adding this circuit in the Williamson design will increase THD up at the 20 KHz range (slightly) but it does its job of suppressing this parasitic. Thanks for your thoughts and comments.
Thanks for your share
Thanks 😁
well said very interesting
Are there any OPtransfomers that the two wires are wound together, at the same time? Then the wire length -resistance would be the same.
Yes, on some windings but not all, it's called bi-filar winding. This is usually only done on the secondary as a bi-filar primary has to have very good insulation, there is very high voltage between the two windings meaning there's a higher chance of breakdown between the two.
I see you got the answer from Diabolical Artificier. McIntosh does it this way and is one reason their amplifiers are so good. I think there are also trifilar windings with three wires wound at a time. McIntosh calles Unity wound transformers. th-cam.com/video/W58Rqx_4u2w/w-d-xo.html
@ around 19:30 is this typical insertion losses? Or am I way off base
David, very good. Two questions if I may.
1. EF86: rp=2M5 and gm=1.8mA/V, so mu=4500? Just checking my number crunching.
2. Why do the tube books and datasheets generally give mu for triodes but gm for pentodes?
Happy birthday!
EJP
Hi Esmond - thanks for remembering my birthday. I am not sure if I am older or younger than I think... hmmmmm....
Well, your calculations are what the formulas say so. I can't remember right off the top of my head - I should consult RDH or something reliable but I seem to remember they use gm in multigrid tubes as the mu can be very high (and variable as mu can also) and it seems like it has something to do with screen grid (amplification ?). I will look into it tonight. I posted a video on how I rescued my HV transformer by cleaning up an arc point and then "floating" it above ground so I could use it again.
Pentodes are more like current sources, so gm is more meaningful, while triodes are voltage sources, so mu is more useful.
More precisely mu is the voltage gain, gain can also mean current gain which would be transconductance or gm
Gm is not current gain. It's d(Ip)/d(Vg), a change in plate current, with respect to a change in grid VOLTAGE.
Very interested in your knowledge and decades of tube experience. Great videos.
However, I can't find your direct contact information. I would like to contact you directly for a possible consultation work, if you are interested.
Hi James,
You can email me directly at my YT account: TheAudioShop@SBCGlobal.Net. Not sure what you are looking for but I will listen. I haven't done "work" since letting go of an after-retirement, hobby-job (don't ever say hobby-job to your tax accountant... :-) but, what do you have in mind?
No, 2 inductances in series are simply the sum, ie, L(total) = L1 + L2, just like with resistances. I think your confusing that with how Z ratio relates to turns ratio.
It is true that inductance in series is just like resistors if they are separate inductors but if coils are wound on the same core on top of each other, like in a common CT output transformer, the total inductance is the two values in series plus the mutual inductance they gain by being wound on top of each other on the same core and the total inductance is four times. Turns ratios and voltages ratios are the same and inductance and impedance ratios are the same.
Two separate inductors with two separate cores are L1 + L2. Wound on a single core, the inductance is proportional to the square of the amount of turns. So two inductors on the same core connected together square, they do not add. Even the turns ratio is turns squared over turns squared which is where you get Zo/Zi. I didn't know until I looked at some inductance equations and remembered.
The inductances of 5H each combine into 10H, not into 20H!
No, it is related to N^2 where in this case N is 2 and N^2 is 4, so if half primary is half the total primary number of turns that gives an N of 2. So if you start with knowing half primary inductance, total primary will be 4x that and v.v.
@@markg1051 You suggest if I use my LCR meter and measure from the center-tap to each outer connection I measure 5H each but if I measure from outer connection to outer connection I will measure 20H?
@@MrTalaue That is correct assuming your transformer is 20H plate-to-plate. May not be 100% exactly but you will see it is essentially 1/4 the, inductance and impedance, from CT to either plate lead compared plate-to-plate.
@@ElPasoTubeAmps That is very interesting as I would expect half of the inductance when I measure a coil from one end to the middle, so it is up to the magnetic coupling which is different with the transformer.
@@MrTalaue that is correct with regards to your question on measurements with LCR meter..