i am trying to wind a PP output trans for a tube power amp.
the the transformer needs to be around 1.7k p-p and around 15H to support the bass till ~25hz at 90ma bias.
i`ve done some simple calculations that were based for SE trans and have come up with this. please help me check my work.
i need a turns ratio of 14.5:1 for a 8 ohm tap.
To support the tube`s 90ma bias, i need to use ~0.25mm of copper enamel wire on the primary and ~ 0.75mm copper enamel wire on the secondary to support my tubes output of around 6w.
I intend to use Z11 cores but my transformer material supplier only as 66mm laminations and a standard sized bobbin a 35mm stack. it works out i can only get 4 watts single ended but since this is a push pull transformer and i only need a maxium of 8 watts. i was thinking i could get away with it. what are your opinions?
would H14 cores be better compared to limiting myself to 66mm Z11 laminates?
how many rounds do i require for my primary to achieve the 20H?
thank you for reading through!
the the transformer needs to be around 1.7k p-p and around 15H to support the bass till ~25hz at 90ma bias.
i`ve done some simple calculations that were based for SE trans and have come up with this. please help me check my work.
i need a turns ratio of 14.5:1 for a 8 ohm tap.
To support the tube`s 90ma bias, i need to use ~0.25mm of copper enamel wire on the primary and ~ 0.75mm copper enamel wire on the secondary to support my tubes output of around 6w.
I intend to use Z11 cores but my transformer material supplier only as 66mm laminations and a standard sized bobbin a 35mm stack. it works out i can only get 4 watts single ended but since this is a push pull transformer and i only need a maxium of 8 watts. i was thinking i could get away with it. what are your opinions?
would H14 cores be better compared to limiting myself to 66mm Z11 laminates?
how many rounds do i require for my primary to achieve the 20H?
thank you for reading through!
You need to decide the maximum flux density that you can tolerate in the core. For that, you need the B/H curves for the magentic material and you need to know the cross-sectional area of the core. The primary inductance will be dominated by the air gap. In theory, when you interleave E I laminations, you don't have an air gap. In practice, you do because they can never butt together perfectly.
Output transformer design has many variables (that's why it's hard). Try a search for output transformer design on this site. I seem to remember a post that had a great number of excellent references for transformer design.
Output transformer design has many variables (that's why it's hard). Try a search for output transformer design on this site. I seem to remember a post that had a great number of excellent references for transformer design.
i did do a search before posting and came up with a lot of single ended designs.
i`m more interested in push pull designs. anyway, where do i find the BH curves for the core material, all i seem to find is a bunch of numbers telling me the number of watts lost at 1.7T @50hz and the 800A/m is 1.8T. what does this mean?
i`m more interested in push pull designs. anyway, where do i find the BH curves for the core material, all i seem to find is a bunch of numbers telling me the number of watts lost at 1.7T @50hz and the 800A/m is 1.8T. what does this mean?
I'm not suggesting that you didn't do a search before posting, but I know that there was a thread/post on output transformer design, so you just need to find it.
Tesla (T) is the unit of flux density (the B bit of the B/H curves). H is the magnetising force, and this is usually measured in Amperes/turn. When you apply an alternating magnetic field to a magnetic material, you are alternately flipping magnetic dipoles within that material from NS magnetisation to SN. This takes energy, and the more times you do it, the more energy you require. The amount of energy used in a second is power, and this is measured in Watts, so core losses are given in Watts at a specified frequency, such as 50Hz. You can expect that core loss to rise with frequency because you are flipping the magnetic dipoles more times per second, but the relationship is not necessarily linear because as frequency rises the dipoles tend to become decoupled.
Tesla (T) is the unit of flux density (the B bit of the B/H curves). H is the magnetising force, and this is usually measured in Amperes/turn. When you apply an alternating magnetic field to a magnetic material, you are alternately flipping magnetic dipoles within that material from NS magnetisation to SN. This takes energy, and the more times you do it, the more energy you require. The amount of energy used in a second is power, and this is measured in Watts, so core losses are given in Watts at a specified frequency, such as 50Hz. You can expect that core loss to rise with frequency because you are flipping the magnetic dipoles more times per second, but the relationship is not necessarily linear because as frequency rises the dipoles tend to become decoupled.
Hi !
I've no data about Z11 cores, but assuming they accept to be run at 1 Tesla, here is a posible solution with the lamination size you have:
Primary: 1160 turns, 0.25mm wire, split in four sections.
Secondary: 5 sections each 79 turns, 0.315mm wire, tied in paralell (beware than they have EXACT the same turns number)
Doing so, the induction will be 1.1 tesla for 8W at 30Hz, and less than 1 at 6W.
The inductance will be 15 Hy if the permeability of your core is at least 2200, wich is quite common (M6X may reach 15000 in that configuration).
Start with a secondary section, then put 0.3mm spacer, then a primary section, and so on.
Each secondary section should fit in 1 layer, and primary in 3.
Such interleaving promises a leakage induction lower than 1mH.
The four primary sections must be tied in serie, and since the inner one has less resistance than the outer, it is good to use the inner one in serie with the outer for an anode, and the two remaining for the other to balance the resistance.
My 2 pence !
Yves.
I've no data about Z11 cores, but assuming they accept to be run at 1 Tesla, here is a posible solution with the lamination size you have:
Primary: 1160 turns, 0.25mm wire, split in four sections.
Secondary: 5 sections each 79 turns, 0.315mm wire, tied in paralell (beware than they have EXACT the same turns number)
Doing so, the induction will be 1.1 tesla for 8W at 30Hz, and less than 1 at 6W.
The inductance will be 15 Hy if the permeability of your core is at least 2200, wich is quite common (M6X may reach 15000 in that configuration).
Start with a secondary section, then put 0.3mm spacer, then a primary section, and so on.
Each secondary section should fit in 1 layer, and primary in 3.
Such interleaving promises a leakage induction lower than 1mH.
The four primary sections must be tied in serie, and since the inner one has less resistance than the outer, it is good to use the inner one in serie with the outer for an anode, and the two remaining for the other to balance the resistance.
My 2 pence !
Yves.
For 12 mil (.3mm) M-6, the initial permeability for DC is about 350, although for practical purposes in an output transformer with some DC inductance and an intrinsic gap, initial permeability is closer to 6600 at 100 guasses and .015 oerstads. The knee in the magnetization curve occurs at about the same place for all audio frequencies, i.e. 6000 guasses and .15 oerstads. DC permeability there is about 40,000 and decreasing to about 15,000 at 2khz and 3000 at 20khz. The figures for permeability for AC were extrapolated from magnetization curves for much thinner material and may not be accurate. You can operate in the region between 100 and 8000 guasses (.015 and .15 oerstads) without too much distortion.
At 30hz it is easily over 15000 at 1.1 tesla.
John
Uh ! LOOOOOOONG !
Because I'm not a physician nor a fluent mathematician, and since understanding magnetism laws is not trivial, and also because it seems that so many units and differents formulaes float around to finally describe the same things in different ways, I simplifed somewhat by using only that is revealant for designing an audio tranny, specially an output tranny.
Anyone intersted by a more complete approach could find a lot on the web, this site:
http://www.ee.surrey.ac.uk/Workshop/advice/coils/
learns me more than a lot !
All that follws is the result of compiling informations from various sources and personal experimentations. Usual disclaimers about possible errors, mistakes and typos apply !
FIRST
The iron core is needed here only to acheive a sufficient value of primary inductance to do not short out the signal at low end of the audio range.
It is said that its reactance, at the lowest frequency of interest, should be at least equal to the reflected impedance.
The tube's Rp had to be considered too, but this is another looooooooooong story.
The turns number and the knowledge of the core permeability (Mu) allow to compute that inductance as follow:
L = Mu x N x N x A / MPL / 10e8
N being turns, A the core section and MPL the Magnetic Path Lenght (i.e. related to the core size), both in centimeters.
Because permeability seldomly (if ever !) appears clearly in steel manufacturers data (yeah, I know: Mu = B/H but !), I simply wound some test units and measured the AC current (I) at mains (50Hz) frequency for various voltages (Vac), then it is easy to compute inductance and, in turn, obtain a curve of the effective Mu versus AC induction values (Bac):
the induction is:
Bac = Vac x 10e4 / 4.44 / N / F / A
the inductance is
L = Vac / I / 6.28 / 50
and thus Mu is
Mu = 10e8 x L x MPL / N / N / A
(An Excel Worksheet is of great help !)
For M6X lams, the Mu shows a peak (25000) around 0.8T and decreases quickly (10000 at 1.2T, 5000 at 1.6T).
Of course, the primary inductance varies in the same proportions.
Standards lams shows roughly one fifth theses values, the peak being less sharp.
So, for a given frequency, induction being dependent of the AC voltage, the inductance decreases when AC voltage accross the winding increases, this represents the saturation effect of the iron.
Induction increases also when frequency decreases, thus decreasing the inductance too.
The effects of iron saturation are in fact a bit more complex, but since it MUST be avoided, don't care, just staying below 1.4 Tesla is safe when using M6X (or below 1 Tesla using standard lams).
When some DC current (Idc) flows in one winding (if not cancelled by the same DC current in another IDENTICAL winding, as in a classic Push Pull design), the total induction rises to unnaceptable levels if the permeability is not artficially reduced by inserting a gap in the magnetic path, making the inductance to fall down too.
The Mu becomes:
GapedMu = Mu / (1 + (Mu x (Gap / MPL))) where Gap and MPL are in centimeters.
The DC induction is :
Bdc = GapedMu x N X Idc / MPL
It is remarkable that Bdc increases with turns number altought Bac decreases ! So, when designing a tranny for Single Ended usage, DO NOT simply add turns to increase inductance.
It is necessary to iterates with core size, gap and turns to obtain the desired inductance without exceeding the maximum permissible induction (same as above, but Bdc added to Bac one).
A tranny designed for PP usage exhibit usually more inductance that strictly needed and it is common to put a small magnetic gap to prevent bad effects caused by DC unbalance between tubes.
NEXT
At other end of audio range, (perhaps above 3 or 5 Khz) the iron core serves almost nothing.
First, note that, if iron losses are know to increase with frequency, this is only true at constant induction which fall down quickly with frequency.
Here there are two difficulties, the first is leakage inductance:
Since the magnetic field is no longer concentrated in the core, the distance between various windings or parts of windings induces imperfect coupling between them.
To cope with that, the windings are divided in multiple sections and interleaved so that the coupling factor is better.
Here again, I combined what can be found in various litterature and experimental/empirical approach.
Most cames from:
www.turneraudio.com.au/htmlwebpgs02/optdesigncalcs.htm
(found this link dead today, but may be temporary ?)
First, the thickness of the primary and secondary winding are computed using the wire diameter, the number of layers needed according to the bobin width and the total turns number.
Then when knowing Pthick (primary thickness) and SThick (secondary thickness), compute the total thickness including the thickness of all the insulations between winding section.
The number of insulations (NI, sometimes called interleaving factor) and theirs thickness are limited by the available space on the bobin, some iterations must be expected again !
Now, leakage inductance will be this:
LL = 0.417 x Pturns x Pturns x MLTurn x (( 2 x NI x TI) + TT) / 2 (NI x NI x BW) / 10000
Pturns is the primary turns number, MLTurn is the mean length of a turn, NI is the number of insulations, TI is the thickness of each insulation, BW is the width of the bobin, LL is in mH.
I can't demonstrate, I just can tell that this works !
Having interleaved the windings, some shunt capacitance (Csh) was added over the primary because the secondary is more than ever tied to ground.
Here again, a "cooking receipt" type of computation is used:
Csh = 10e6 x 11.5 x Dk x NI x BW x MLTurn / TI
where Dk is the dielectric constant of the insulation material (for paper use 4)
Now, the -3dB point can be computed as follow:
F3 = 1,000,000 / 6.28 / Rb / Csh
Rb is the tube's plate resistance in parallel with the reflected load.
The second difficulty is that Csh and LL form a resonnant circuit (sometimes called "pole") where phase will severely shift, producing ringing and other forms of instabilities, specially when feed back will be applied to the amp !
The resonant frequency is:
Fr = 5035 / Sqr (LL x C) where C is in µF.
The goal is to push FR as far as possible over of the audio frequency range.
To do that Csh must be kept low, i.e. using a material with lower dielectric constant or reducing the interleaving factor or increasing insulation thickness, but unfortunatly, this increases LL thus lowering Fr.
Iterates again.
))Thanks for thoses who read up to there, and now a good new: if, like me, you are too lazy to perform manually all the necessary iterations to reach a satisfactory design, have a look at:
http://geek.scorpiorising.ca/yves.html
And have fun !
Yves.
Yves,
I'd first like to thank you for taking the trouble to type such a long post - especially when it isn't even in your own language. I've read quickly through your post, and can follow it, but you left the best bit to the end! I will be downloading your program and doing some serious playing.
Thanks again.
I'd first like to thank you for taking the trouble to type such a long post - especially when it isn't even in your own language. I've read quickly through your post, and can follow it, but you left the best bit to the end! I will be downloading your program and doing some serious playing.
Thanks again.
Hey-Hey!!!,
Thanks also for the informative post, and magnetics program. There is another interesting means of confirming your design calculations.
Measure an existing OPT. Might as well pick a good one...
and see why the interleaving was done in the order it was, confirm measurements with expected results. Good Lab and skilled operator required.
I am going to do this with a Peerless S-265-Q.
I have its winding geometry, and core size. Thomas and Skinner's catalog has a remarkable amount of data on some of their offerings, so I can get some reasonable Iron data to stat playing.
From a bit of experimenting on a (by comparison ) simple center tapped inductor, I can tell you that capacitance is not to be taken lightly. Depending on application, either max coupling or minimum capacitance is required, and they are mutually exclusive...
Things like winding direction can reduce capacitive effects. If there is no AC voltage, the capacitance is a null issue. The voltage gradient on a winding across two layers can be quite substantial.
Consider this: wind in one direction, from one end of the bobbin to the other, lay down the film, and wind back to the starting point.
The AC voltage at one end is quite large compared to the other, and if you wind from one end to the other, lay down the film, stretch back to the starting end, and wind the second layer, the AC voltage is near constant across the entire winding.
The area under the curve, so to speak, is the same but depending on the service, one way will be more useful than the other, especially if two different windings are being wound. Something like McIntosh's Bifilar, Trifilar, and the Dynaco 441 and Acro 350 come to mind. What do you want ot do with the beastie when it is done?
Nothing magical, and not even too complicated, take one apart, and see how they are made, and see why they were made exactly that way. If it is good, copy it, improve it, learn from it. It is easy to duplicate any of the legendary ones, and at cost quite lower than acquiring original ones.
regards,
Douglas
Thanks also for the informative post, and magnetics program. There is another interesting means of confirming your design calculations.
Measure an existing OPT. Might as well pick a good one...
and see why the interleaving was done in the order it was, confirm measurements with expected results. Good Lab and skilled operator required.
I am going to do this with a Peerless S-265-Q.
I have its winding geometry, and core size. Thomas and Skinner's catalog has a remarkable amount of data on some of their offerings, so I can get some reasonable Iron data to stat playing.
From a bit of experimenting on a (by comparison ) simple center tapped inductor, I can tell you that capacitance is not to be taken lightly. Depending on application, either max coupling or minimum capacitance is required, and they are mutually exclusive...
Things like winding direction can reduce capacitive effects. If there is no AC voltage, the capacitance is a null issue. The voltage gradient on a winding across two layers can be quite substantial.
Consider this: wind in one direction, from one end of the bobbin to the other, lay down the film, and wind back to the starting point.
The AC voltage at one end is quite large compared to the other, and if you wind from one end to the other, lay down the film, stretch back to the starting end, and wind the second layer, the AC voltage is near constant across the entire winding.
The area under the curve, so to speak, is the same but depending on the service, one way will be more useful than the other, especially if two different windings are being wound. Something like McIntosh's Bifilar, Trifilar, and the Dynaco 441 and Acro 350 come to mind. What do you want ot do with the beastie when it is done?
Nothing magical, and not even too complicated, take one apart, and see how they are made, and see why they were made exactly that way. If it is good, copy it, improve it, learn from it. It is easy to duplicate any of the legendary ones, and at cost quite lower than acquiring original ones.
regards,
Douglas
Mechanical data are the same that the "EI66B" already in the database, but you may perhaps need to alter bobin size ?
For now, the program assume M6X iron grade, and can't be changed.
I've a more recent version where you may change the "MU vs B" table.
This is the allowed current density in the wires and used to compute wire diameter, in fact you may either alter this value, or directly change the proposed dia value for each winding by typing over.
When doing that, observe the copper losses !
Since an OPT seldomly runs continuously at full power, heathing is not the main limiting factor, but efficiency is ! Watts from tubes are somewhat precious !
Yves.
hacknet said:
Sorry, I don't know
Z11 looked good !
I just can suggest you wound a test coil, say 1000 turns and measure current at various voltage, the permeability values you'll be able to derive will tell you more than a short data sheet.
Usually, the thinner the lams, the better !
Regards, Yves.
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