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transformer test

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very foolish to say this.
Probably you have the winding schematic too, wire diameters etc etc.



Bla, bla, bla you change opinion according to how the wind is blowing! You really don't know what you have in hand, I am afraid.
If your transformer is in line with those in that link then your inductance is rather low for the size. Frequency response and square wave response at low level is the only thing I can compare. And that is.... Most of those transformers at least are small in size! Your transformer is heavier than mine on EI120 40x50 core that is already not small at all but reasonable for certain applications....
Compare your transformer to the Tango FC30 in the same conditions and you will see that is not better at low frequency. The FC30 has 20H minimum inductance and 25-26H already at a fraction of 1W. I say that you don't get here with you transformer!
As I told in the beginning there is nothing you can teach Japanese about transformers expecially considering your level of knowledge of the subject and inexperience with amplifiers....

There is no mistake. It's just that you do not consider I could use a different model.
I don't want to be specific as I don't want to tell about my winding technique just like the others do!
However if you had some real practical experience with this issue you would know that you don't need that complicated model for capacitance as it won't work so well in general. For the SE case the effective shunt capacitance will be about 1/2 of the total static capacitance with the secondary grounded and 1/4 for floating secondary. My practical model is basically made of two nearly identical (because the contribution of the single windings is not precisely the same) shunt capacitances equal to 1/2 of the total static capacitance C (that you can measure directly). So when you have the secondary grounded the effective capacitance is 1/2C and when floating they are "all is series" and you get 1/4 of total static capacitance! This will be quite general behaviour. For example you can check the Monolith Magnetics transformers where you get the shunt capacitance value in the specs with and without secondary grounded and you will see that the ratio between the two values is nearly equal to 2 in all cases with very good approx.
Single Ended Output Transformers - Summit product line

Anyway, the best way to know the precise value of shunt capacitance is measure the resonant frequency and the leakage inductance and derive it from the formula that defines the resonance. Until now no-one has bothered in developing an accurate model for capacitances in my knowledge for this tipe of transformers, firstly I think because different winding techniques results in different total capacitance overall and secondly because HiFi OPT's represent a small fraction of the entire transformer market....

There is nothing to explain but only some fundamental things you should understand. Your specs are wrong somewhere as you can't have 30W at 18Hz with 70-120 mA DC current and 30H at the same time with that core and those few turns. This is not alchemy....

You better stop with your nonsense!
You wrote " The transformer is a single ended double HiB c-core 3500 Ohm / 8 Ohm 60mA-120mA 30W at 18Hz. Copperloss 0,14dB. "

If you have 0.76T Bdc with 70 mA, it means that Bdc with 120 mA will be 1.3T and you cannot have 30W at 18Hz with 1860 turns and that core size (15.32 cm^2 and 25.9 cm path lenght) in BOTH CASES. With 120mA you cannot even have 13W @ 18Hz which require 0.93T.
This is not opinion!

I told you how to calculate it in the previous post and also the equations have been already written by Popillin. They are the same!!
It is you having problems with the subject.....30W at 18Hz means 1.42T Bac so your total B at 70 mA is already beyond saturation if Bdc is 0.76T.
If 30W at 18Hz is true then your Bdc can only be 0.58T and so your inductance cannot be 30H. They are all interdependent quantities you cannot pick them up arbitrarily.
From your frequency response measurement I say that you don't have 30H inductance and I could bet on this.

Yes I do and in fact I know your transformer better than you do!:hohoho:
 
it can be closed because nobody give the measurements i asked for and 45 is still talking rubbish.

esltransformer everytime you have to come out judging the others when you have no arguments. This is typical of people who don't know and it is certainly your case EVERY TIME. It's a fact.
If you just left aside certain arbitrary comparisons just because your only aim was to show off I wouldn't ever bother to respond. The problem is that you are not in the position to make comparisons. You already make mistakes at design level which firstly means you also are not in the position to judge someone else transformer.

At this point no measurements are actually needed if you want to know where you are. You can HONESTLY and PROPERLY measure your transformer as see if your specs are all true (they aren't!) and how they compare to those of the FC-30 that are of public domain!
If you want to see mines you have to wait. This thing has become quite boring now.
 

Quod erat demonstrandum

Sorry to say that, but so far you have not demonstrated anything, your transformer data are so confused, e.g. your core on post#3 was EI 150/70, then reduced to EI 120/50 on post#111, then reduced even more to EI 120/40 on post#114.

Unlike us, you have not posted neither the number of primary turns, so you can criticize our designs but yours are untouchable...:rolleyes:
 
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There is no mistake. It's just that you do not consider I could use a different model.
I don't want to be specific as I don't want to tell about my winding technique just like the others do!
However if you had some real practical experience with this issue you would know that you don't need that complicated model for capacitance as it won't work so well in general. For the SE case the effective shunt capacitance will be about 1/2 of the total static capacitance with the secondary grounded and 1/4 for floating secondary. My practical model is basically made of two nearly identical (because the contribution of the single windings is not precisely the same) shunt capacitances equal to 1/2 of the total static capacitance C (that you can measure directly). So when you have the secondary grounded the effective capacitance is 1/2C and when floating they are "all is series" and you get 1/4 of total static capacitance! This will be quite general behaviour. For example you can check the Monolith Magnetics transformers where you get the shunt capacitance value in the specs with and without secondary grounded and you will see that the ratio between the two values is nearly equal to 2 in all cases with very good approx.
Single Ended Output Transformers - Summit product line

Your model is totally wrong, sorry to say that your vision of real world is very naive, in real world transformers, capacitance is a nightmare to calculate, so a more complex model is needed.

For a transformer with grounded short circuited secondaries, interwinding capacitance is distributed as

Ce ≈ C ∑ (nı/N)² ı=1,...,m

Suppose now that secondaries are left floating, then

Ce ≈ C ∑ [(2nı-N)/2N]² ı=1,...,m

Now we must consider capacitance of each primary winding

Cwı = (4/3nı) [1 - (1/nı)] Clı ı=1,...,NP

Where Clı is the capacitance between two adjacent layers, then total winding capacitance

(Cw)⁻¹ = ∑ {(4/3nı) [1 - (1/nı)] Clı}⁻¹ ı=1,...,NP

If we take Cl as the average interlayer capacitance

(Cw)⁻¹ ≈ (Cl)⁻¹ ∑ {(4/3nı) [1 - (1/nı)]}⁻¹ ı=1,...,NP

Also

C = (ε A) / (4πd)

Cl = (εl A) / (4πdl)

For properly made transformers as I do, need at least 0.4mm to 0.56mm insulation between windings of NOMEX 410 (ε=1.6) and 0.05mm polyester with epoxy (ε=4.4) between layers, then say d=0.5mm and dl=0.01mm (0.05mm polyesrter and wire coat) we obtain

Cl ≈ 13.75 C

Given that high leakage inductance of transformers from your link, I strongly suspect that manufacturer lowers capacitance with a more simple winding pattern as 2P-S-4P-S-4P-S-2P, without any vertical sectioning, then

Ce ≈ 1.44 C (grounded secondaries)

Ce ≈ 0.36 C (floating secondaries)

Cw ≈ 0.98 C

Then, for grounded secondaries

CTg = Ce + Cw ≈ 2.42 C

For floating secondaries

CTf = Ce + Cw ≈ 1.34 C

Et voilà

CTg / CTf ≈ 1.8

This result match reasonably well with most of tranformers from your link: S-5 845 (1.78), S-7 211 (1.79), S-9 300B (1.72), S-15 GM70 (1.99), S-17 PSE 211 (2.06), S-19 PSE 300B (1.78), S-25 LP 845 (1.97), except for model S-833 (2.85), but this works with 2KV!

We need a different winding pattern, say 5P-S-10P-S-5P, more insulation, say 1mm Mylar between windings and 0.1mm Mylar between layers, also triple insulated magnet wire, so

Cl ≈ 5.5 C

Then

Ce ≈ 1.25 C (grounded secondaries)

Ce ≈ 0.25 C (floating secondaries)

Cw ≈ 0.31 C

Then, for grounded secondaries

CTg = Ce + Cw ≈ 1.56 C

For floating secondaries

CTf = Ce + Cw ≈ 0.56 C

Et voilà

CTg / CTf ≈ 2.78

Very close ah ? :cool:

BTW. The open poster OPT has far better specs and frequency plot than those from your link. :D

As you can see, this is not alchemy, it is necessary to pose equations and sweat!

Every winding pattern, every insulation pattern and every insulation material will change the results, things are not that easy as you think. ;)

P.S. You can measure Ciw with reasonable accuracy, with a capacimeter between P and S, so

C ≈ Ciw/m
 
It is not a mistake! The effective shunt capacitance for PP is 1/4 of the total static capacintance already with secondary grounded. So according to my model is like the capacitance in the SE with secondary floating are like if there were all in series as I use the same simple empirical model for both....you don't need complicated models. If you always leave a little bit of room for more insulation you can start with one and then increase it if you need.

Oh no!!! Again with that...how did you said?

is just hogwash...

I did explained you twice...:headbash:

If then you do the same with PP OT and balanced generator you won't get that increase! Because of the geometry the distribution is already as if they were all in series.

For any respectably constructed PP OPT, regardless of vertical sectioning, if short circuited secondaries are left floating, they still remain at 0 VAC, because each extreme of primaries beside secondaries, is at +U V, and the corresponding symmetric opposite extreme referred to +B (ground) at -U V, push-pull works thereby. AFAIK
So no redistributed voltages, so capacitance remains unchanged, that is!

As I showed before, your model is wrong, shunt capacitance doesn't depends on oil prices, between certain limits you can choose it by design, and that chosen capacitance doesn't change because at every time t, you have a simple electrostatic problem, and (+U)+(-U)=0, regardless of grounded or floating secondaries.

What university had the courage to grant you a PhD? :confused:

Maybe that's time to change your model, maybe your transformers would has a better frequency response...

The bandwidth (-3 dB) is 3Hz-67KHz...with no peaks and deeps.

67KHz is very poor, even for most Monolith Magnetics standards... ;)

Beware that the minimum insulation you need is not just that related to DC voltage.

Duh! :D
 
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