PP OUTPUT TRANSFORMER CALCULATIONS.

FOR  PUSH PULL AMPLIFIERS WITHOUT NET DC FLOW IN ONE DIRECTION.
For calculations for Single Ended Output Transformers go to 'SE OPT calculations'
 
This page contains :-
Brief reference to Radiotron Designer's Handbook,  4th Ed, 1955
Fig 1.  OPT No1 from page on 'Output Transformer Theory'  
Design logic method, steps 1 to 45 for designing a push pull OPT using OPT No1 as the example.
No apologies for the complexities involved, go fishing if its all too hard!
Design method contains lots of calculations and list of Primary and Secondary interleaving configurations likely to be used with
tube audio PP OPT.
Fig 2.  OPT Secondary sub-sections for load matches with 2 and 3 secondary layers.
Fig 3.  OPT Secondary sub-sections for load matches with 4 secondary layers.
Fig 4.  OPT Secondary sub-sections for load matches with 5 secondary layers.
Fig 5.  OPT Secondary sub-sections for load matches with 6 secondary layers.
More checks of final design, calculation of leakage inductance, acB, dcB, inductance
and final winding height.
Fig 6.   OPT bobbin winding details for OPT No1.
Calculation of shunt capacitance.
Metric winding wire size chart for grade 2 polyester-imide wire.
                            --------------------------------------------------------------------------------------

Where does one go to start with output transformer design?  Well, there is a lot of good design advice in the Radiotron Designer's Handbook, 4th Edition, 1955. If you have no copy on hand, there is a CD you can acquire but its more awkward to take to bed to read. However, chapter 5 from page 199 to page 253 should be read repeatedly until what the authors are saying sinks in.  Its not easy to understand if you have no idea about basic tube operation and other basic electronic
behaviours, so the reading and learning experience will be very difficult for the novice.
Unfortunately, the associated reference material listed on pages 252 and 253 has mainly been lost or thrown out of many library archives to make way for  the huge mountain of more modern knowledge within which there is very little on output transformers because mainstream development for tube OPTs stopped in about 1959 when the onslaught of transistor amps was well underway.
 
Not many folks will have RDH4 in their library nor will there be a nearby technical library with easy browsing access so I shall try to unfold the design method I have evolved based on information in the RDH or other sources I have collected over the last 10 years. After having wound many very fine PP output transformers with bandwidth as wide as 10 Hz to as high as 270 Khz, I feel well qualified to speak from experience.
The list of logic steps involved in producing the best possible OPT is based on designing for low winding losses,
core saturation at full power at 14Hz, and adequate interleaving to extend the HF response up to at least 70kHz
by keeping both the leakage inductance and shunt capacitances to low quantities. The end result gives a well filled winding window with several impedance matches possible without having wasted sections of any secondary winding
so the winding losses and response is the same for any of the chosen load matches.

Let us examine the output transformer No1 which was given as an a recipe for a good design for a push pull amp of approximately 60 watts in my page on 'Output Transformer Theory'.
First I will display the schematic of OPT No1 :-
Fig1
Schematic, OPT No1.

In the older 2002 edition of my website I list steps 1 to 10 to design OPT No1. The method was rather vague
and in February 2005 I rationalised the design process to take all that RDH4 says about OPT design, plus create
a design flow that lists each step actually required to get to a working design that any transformer  winder can actually
use in his workshop without any further calculations to produce an OPT.

The main differences between the 2002 method and the 2006 method is that winding resistance losses are
factored in early after calculating the primary turns.  The primary winding fill factor of the actual core window size is
limited to to 0.28 x S x T which appears to work with most OPTs.
Until May 2006, I have not seen any computer program which achieves what I attempt to achieve by following logic steps and using one's brain, some intuitive imagination and a pocket calculator with easy equations.

The design logic flow could be used to construct a computer program where one would enter the design requirements such as power, secondary load, tube Ra, core dimensions,  then with a click on a "design" button, out would come a terrific design including all the exact wire sizes and a cross sectional drawing of the bobbin windup details that could be understood easily by anyone with some winding experience.
Alas, I am not a computer expert, but i can give you the flow of logic used to get a design finalised.

I invite anyone interested to prepare a PC program to encompass all the horrible fiddly details of fitting the wire
that is available from the suppliers into the cores available for superb OPTs.


OUTPUT TRANSFORMER No1 DESIGN EXAMPLE FOR 60 WATT PP OPT for 5,000 to 6 ohms.      

1. Choose tubes, operating conditions and primary load for the tubes applied across the
full primary, known as the anode to anode load, PRL, ohms.
**Note. Careful loadline analysis is required for accurate loading and
is not in this list of steps.

OPT1, One pair of 6550, UL AB1, Ea = 520V, Ia q = 60mA approx,
so allow for RL = 5ka-a...........................................................................5,000ohms

2.  Choose the secondary nominal speaker load value.
allow a default value of 5 ohms, SRL, ohms.

OPT1.....................................................................................................5ohms

3.  Choose the maximum power at clipping for the
PRL chosen above. Max PO for any class of operation, A1, AB1, AB2, watts

**Note. The anode to anode voltage swings can be read off the PP load line
analysis, and PO = ( Vrms anode to anode ) squared / RL

OPT1. PO = 60w.....................................................................................60w

4. Calculate the minimum required core centre leg cross sectional area, Afe,
for a nearly square core cross section.

Afe = 300 x sq.rtPO sq.mm

**Note.  This formula has been derived from a basic formula for
core size used for mains transformers, Afe = sq.root power input / 4.4
where the Afe is in sq inches. This ancient formula is based on B being about
1 Tesla at 50Hz but we would want B max < approx 0.5 Tesla for an OPT.
After considerable trials the above formula is a good guide for audio OPT.

OPT1, Afe = 300 x sq.rt 60 = 300 x 7.75 = 2,323 sq.mm........................2,323sq.mm

5.  Calculate the core tongue dimension, T.
For a square core section, tongue dimension = stack height, ie T = S.
T x S = Afe, sq.mm
Therefore theoretical T dimension = sq.rt AFe  =  th T, mm

eg, thT = sq.rt 2,323 = 48.2 mm............................................................48.2mm

Choose suitable standard T size from list of available wasteless E&I lamination core materials.

T sizes commonly available for OPTs :-
20mm, 25mm, 32mm, 38mm, 44mm, 51mm, 63.5mm

**Note.  Choosing a standard T size above thT gives lower copper winding losses, higher weight,
and choosing T below thT gives higher losses and lower weight. Afe will be the same for either 44mm
or 51mm chosen from above so the LF response won't change with tongue size. HF peformance
depends entirely upon the interleaving geometry and insulations.

OPT1 choose core T = 44mm..................................................................44mm

6.  Calculate theoretical stack height, thS using the chosen T size.
thS = Afe / T, then adjust to a larger height to suit nearest standard plastic bobbin size if available, mm.

OPT1, S = 2,323 / 44 = 52.7mm, but choose 51mm........................................................51mm

7.  Adjusted Afe = chosen T x chosen S, sq.mm

OPT1, Adjusted Afe = 44 x 51 = 2,244sq.mm.................................................................2,244sq.mm

**Note. Some constructors will be using non wasteless E&I lams,
or C cores which do not have the same relative dimensions as E&I Wasteless Pattern cores.
The actual sizes of the T, S, H, & L  of the core to be used must be confirmed.

8. Confirm the height of the winding window, H, mm.

OPT1, 44T wasteless material has H = 22...........................................................................22mm

9.  Confirm the length of the winding widow, L, mm.

OPT1, Wasteless material has  L = 66mm............................................................................66mm

10.  Calculate the theoretical primary winding turns, thNp

Np = sq.rt( PRL x PO) x 10,000 / Afe = thNp, no of turns.

**Note.  The formula here is derived from more complex and complete formula taking B and F
into account. If we assume magnetic field strength B = 1.6 Tesla, and F = 14 Hz, which is a suitably
low F for where saturation is commencing, and express V in terms of load and power,
we get the above short easy equation for primary turns required.
The full formula for calculating B is in step 40 below. The V factor can be expressed as
sq.root of ( Primary RL x power output ) as in the above simplified equation.

OPT1, RL = 5,000 ohms, PO = 60w, Afe = 2,244sq.mm from above,
 thNp = sq.rt ( 5,000 x 60 ) x 10,000 / 2,244 = 2,440 turns..............................2,440 turns

11.  Calculate theoretical Primary wire dia, thPdia.

**Note 4. The Primary wire used for the transformer will occupy a portion
of the window area = 0.28 x L x H. The constant of 0.28 works for 99% of OPT.
Each turn of wire will occupy an area = oa dia squared.  
Overlall or oa dia is the dia including enamel insulation.
Therefore theoretical over all dia of P wire including enamel insulation
= sq.rt ( 0.28 x L x H / Np ), mm.

OPT1, thoadia P wire = sq.rt ( 0.28 x 66 x 22 / 2,440 )
                         = sq.rt 0.167
                         = 0.4082 mm...............................................................................0.4082mm

12. Find nearest suitable oa wire size from the tables, oaPdia, mm
 
OPT1, try oa wire size = 0.414mm, ( for Cudia = 0.355 mm. )..............................0.414mm

13.  Establish the bobbin winding traverse width, Bww, in mm.
**Note 5. Bobbin traverse width is the distance between the cheek flanges and varies depending
on who made the bobbin, but each flange thickness = 2mm maximum is common but could be less.
Where bobbin flanges are not used, and insulation is simply extended to the full window length L,
the traverse width will be the same as in the case of of where bobbin does have flanges.
Ie, the winding will  traverse a distance = L - 4mm.

OPT1, Bww = 66 - 4 = 62 mm...........................................................................62mm

14. Calculate no of theoretical P turns per layer, thPtpl, turns.

thPtpl = 0.97 x Bww / oa dia from step 12.
**Note. The constant 0.97 factor allows for imperfect layer filling.
Leave out fractions of a turn.

OPT1, thPtpl = 0.97 x 62 / 0.414 = 145.26.........................................................145 turns

15.  Calculate theoretical  number of primary layers, thNpl,
then round down or up to convenient even number of layers.

Theoretical Npl = thNp / Ptpl, then round up/down.

OPT1, thNpl = 2,440 / 145 = 16.68 layers; round down to 16................................16 layers

**Note. Rounding down may reduce the Npl needed for Fs = 14 Hz.
But the actual turns used will still allow Fs = approximately 15 Hz, which is ok.
For those wanting to maintain Fs, or have Fs marginally lower than 14 Hz,
the Afe can be increased by increasing S from say 51 mm to 62 mm, and still use a standard
sized bobbin, and have Fs at 12Hz.
The calculated no of primary layers should be an even number to avoid a CT in the middle of a layer,
and because each 1/2 primary winding must have an equal number of turns and a symetrical geometric layout
either side of the CT.

16. Calculate actual Np.

Np = P layers x thPtpl

OPT1, Np = 16 x 145 = 2,320 turns...............................................................2,320 turns

17. Calculate average turn length, TL
 TL = ( 3.14 x H  ) + ( 2 x S ) + ( 2 x T ), mm.

OPT1, TL =  ( 3.14 x 22 ) + ( 2 x 51 ) + ( 2 x 44 ) = 259 mm............................259mm

18. Calculate primary winding resistance, Rwp.

Rwp = ( Np x TL ) / ( 44,000 x Pdia x Pdia ), ohms.
where 44,000 is a constant, and P dia is the copper dia from the wire tables.

OPT1, Rwp = 2,320 x 267 / ( 44,000 x 0.355 x 0.355 ) = 111.7 ohms................112ohms

19. Calculate primary winding loss %,

P loss % = 100 x Rwp / ( PRL + Rwp ), %.

OPT1 P loss = 100 x 112 / ( 5,000 + 112 ) = 2.19%..............................................2.19%

20.  Is the winding loss more than 2.5%? ..................................................yes or no.
If yes, the design calcs must be checked again, if no, proceed to 21.

OPT1, P winding loss is less than 2.5%.

**Note.   If the P winding losses are less than 2%, there is a possibility that the wire size could be reduced
to increase the turns per layer, and possibly reduce the number of P layers by say 2.
However I rarely find P winding losses will be less than 2% with the rated load, and one must allow for
where RL = 1/2 the design RL which will double P winding losses.
  
21.  Choose the interleaving pattern from the list below for the wattage of the transformer.

All OPT will have the secondary sections containing only one layer of wire.
While this may be subdivided into further secondary sub sections, there are no designs here which require
bifilar or trifilar winding or rectangular wire.

A section of a winding is defined as a layer or group of layers devoted solely to P or S.
The term "section" is not to be confused with "layer". For tube OPT, most P sections will have
more than one layer of wire.

In general, all OPT should comply with the following P&S layer number relationships :-

Where the first and last winding on is a primary section, then these sections should have near 1/2 the layers of the inner sections, hence if there are 3 outer p layers in a P section, the inner sections might be either 5, 6 or 7 p layers. When this
guide is adhered to there is the best HF response because the leakage inductance is fairly evenly and symetrically distributed. 

When starting and finishing with an S section all internal P sections should have the same number of p layers
but it is not always possible and having say 2 sections of 4 p layers and 2 sections of 5 p layers is OK.
The size of such "internal sections" should not vary more than 25%. Where such guidelines are adhered to along
with equal thickness insulations between P and S sections, there are minimal problems with resonances at HF.

For transformers to suit low drive devices such as mosfets or transistors, the same amount of interleaving is required for a
a given power level. The number of p layers will be reduced as Primary RL becomes low, and wire dia will increase.
An 8 ohm : 8 ohm OPT with very low dc voltage differences between P and S would have equal numbers of turns for P and S and perhaps be simply interleaved so each layer of thick wire is alternatively devoted to either P or S.
The bandwidth can then be very easily made to go up to 500kHz. As the primary or secondary load is reduced, the effect of shunt capacitance diminishes, so insulation thickness can be reduced. Transformers for electrostatic speakers which step up the
amplifier voltage between 50 and 100 times need to have good insulation for voltages involved and need to have lower capacitance. They resemble PP OPTs powered "backwards" and can be designed with the method here.

But for matching tubes to normal 3 to 9 ohm speaker loads, the interleaving list below with the number of primary layers per section possible will give at least 70 kHz of bandwidth, and where there is a highest number of interleavings the bandwidth
can be 300kHz. Using more interleaving than listed leads to less available room on the bobbin for wire due to too many layers of insulation, and poor HF due to high shunt capacitances, and higher winding losses.
For lower Primary RL and higher amplifier power the larger the OPT becomes and for a given number of interleavings the
HF response becomes less due to increasing leakage inductance so the larger the OPT becomes, the number of interleaved sections increases. So a small 15 watt OPT may only need
3S + 2P sections for 70kHz, but a 500 watt OPT may need 6S + 6P sections.

LIST OF PRIMARY AND SECONDARY WINDING  SEQUENCE ON THE BOBBIN :-

Up to 15W, 10 to 20 p layers.....S - 5p to 10p - S - 5p to 10p - S                                      3S + 2P sections
                                                  2p to 4p - S - 4p to 8p - S - 4p to 8p - S - 2p to 4p         3S + 4P
15W to 35W, 12 p layers .........2p - S - 4p - S - 4p - S - 2p                                              3S + 4P
                                                 S - 4p - S - 4p - S - 4p - S                                                4S + 3P
                       16 p layers..........3p - S - 5p - S - 5p - S - 3p                                              3S + 4P
                                                  S - 4p - S - 4p - S - 4p - S - 4p - S                                  5S + 4P
                                                  2p - S - 4p - S - 4p - S - 4p - S - 2p                                4S + 5P

                       18 p layers..........3p - S - 6p - S - 6p - S - 3p                                              3S + 4P
                                              
                        20 p layers.........3p - S - 7p - S - 7p - S - 3p                                               4S + 3P
                                                  S - 5p - S - 5p - S - 5p - S - 5p - S                                   5S + 4P
                                                  2p - S - 5p - S - 6p - S - 5p - S - 2p                                 4S + 5P

35W to 120W, 14 p layers...........S - 3p - S - 4p - S - 4p - S - 3p - S                                 5S + 4P
                                                    2p - S - 3p - S - 4p - S - 3p - S - 2p                                4S + 5P

                       16 p layers............S - 4p - S - 4p - S - 4p - S - 4p - S                                  5S + 4P
                                                    2p - S - 4p - S - 4p - S - 4p - S - 2p                                4S + 5P

                       18 p layers............S - 4p - S - 5p - S - 5p - S - 4p - S                                  5S + 4P
                                                    2p - S - 5p - S - 4p - S - 5p - S - 2p                                4S + 5P

                        20 p layers............S - 5p - S - 5p - S - 5p - S - 5p - S                                5S + 4P
                                                    2p - S - 5p - S - 6p - S - 5p - S - 2p                               4S + 5P
                       
                        22 p layers............S - 5p - S - 6p - S - 6p - S - 5p - S                                5S + 4P
                                                     2p - S - 6p - S - 6p - S - 6p - S - 2p                              4S + 5P

120W to 500W, 10 p layers.........2p - S - 2p - S - 2p - S - 2p - S - 2p                               4S + 5P
                                                    S - 2p - S - 3p - S - 3p - S - 2p - S                                 5S + 4P
                                                   1p - S - 2p - S - 2p - S - 2p - S - 2p - S - 1p                   5S + 6P
                                                    S - 2p - S - 2p - S - 2p - S - 2p - S - 2p - S                    6S + 5P

                         12 p layers...........2p - S - 3p - S - 2p - S - 3p - S - 2p                               4S + 5P
                                                     S - 3p - S - 3p - S - 3p - S - 3p - S                                5S + 4P
                                                    1p - S - 2p - S - 3p - S - 3p - S - 2p - S - 1p                  5S + 6P
                                                     S - 2p - S - 3p - S - 2p - S - 3p - S - 2p - S                   6S + 5P

                                                     
                                                     S - 2p - S - 2p - S - 4p - S - 2p - S - 2p - S                    6S + 5P

                          14 p layers..........2p - S - 3p - S - 4p - S - 3p - S - 2p                                4S + 5P
                                                     S - 3p - S - 4p - S - 4p - S - 3p - S                                  5S + 4P
                                                     1p - S - 3p - S - 3p - S - 3p - S - 3p - S - 1p                   5S + 6P
                                                      S - 2p - S - 3p - S - 4p - S - 3p - S - 2p - S                    6S + 5P

                         16 p layers............2p - S - 4p - S - 4p - S - 4p - S - 2p                                4S + 5P
                                                      S - 4p - S - 4p - S - 4p - S - 4p - S                                  5S + 4P
                                                      2p - S - 3p - S - 3p - S - 3p - S - 3p - S - 2p                   5S + 6P
                                                      S - 3p - S - 3p - S - 4p - S - 3p - S - 3p - S                    6S + 5P

                         18 p layers............2p - S - 5p - S - 4p - S - 5p - S - 2p                                4S + 5P
                                                      S - 5p - S - 4p - S - 4p - S - 5p - S                                  5S + 4P
                                                      2p - S - 4p - S - 3p - S - 3p - S - 4p - S - 2p                   5S + 6P
                                                      S - 3p - S - 4p - S - 4p - S - 4p - S - 3p - S                     6S + 5P

                         20 p layers............3p - S - 5p - S - 4p - S - 5p - S - 3p                                4S + 5P
                                                      S - 5p - S - 5p - S - 5p - S - 5p - S                                  5S + 4P
                                                      2p - S - 4p - S - 4p - S - 4p - S - 4p - S - 2p                   5S + 6P
                                                      S - 4p - S - 4p - S - 4p - S - 4p - S - 4p - S                     6S + 5P

                        22 p layers.............3p - S - 5p - S - 6p - S - 5p - S - 3p                                 4S + 5P
                                                      S - 5p - S - 6p - S - 6p - S - 5p - S                                  5S + 4P
                                                      2p - S - 5p - S - 4p - S - 4p - S - 5p - S - 2p                   5S + 6P
                                                      S - 4p - S - 6p - S - 4p - S - 6p - S - 4p - S                     6S + 5P

Record the choice of primary layers in step 15.
choose a suitable P& S interleaving pattern from the above list.

OPT1, 16 primary layers, 60 watts, choose ...............................................................4S + 5P.

22. Choose insulation, i, in mm used between primary layers, mm.
**Note.   Usually p to p insulation for all OPT needs to only be 0.05mm thick.
OPT1, i = 0.05mm....................................................................................................0.05mm

23. Choose insulation, I, in mm used between Primary and Secondary layers, mm.
**Note.   Usually, for where Ea is above 450V, and RL above 1k, the p to S
insulation is first reckoned = 0.6mm to keep shunt capacitance low with good enough
insulation.
OPT1,  I = 0.6 mm ...................................................................................................0.6mm

24.  Calculate portion of bobbin winding height comprising primary wire layers, p to p insulation,
and p to S insulation.
height of P+I+i = ( no of layers x oadia P wire ) + ( no x i ) + ( no x I ), mm.

OPT1, P = 16 x 0.414 = 6.624 mm,
 i ht =  11 x 0.05 = 0.55 mm,
 I ht =  8 x 0.6 = 4.8 mm,
Total ht of above = 11.974 mm............................................................................11.98mm

25. Calculate maximum total available height of all windings on bobbin.

Max wind ht = 0.8 x H, mm.
**Note.    The constant of 0.8 will suit most OPT.
OPT1, Wht = 0.8 x 22 = 17.6 mm...................................................................................17.6mm

26.  Calculate the max theoretical oa dia of the the secondary wire in secondary layers, thSoadia.
max thSoadia =  ( max wind ht - [ P+i+I ht ] ) / no of S layers, mm
**Note.  The available height for secondary layers =  maximum bobbin winding height - ( primary + all insulations ).

OPT1, We have chosen 4 layers of secondary wires. Height from step 24 = 11.98mm
thSoadia = ( 17.6 - 11.98 ) / 4 = 5.62 / 4 = 1.405 mm..............................................1.406mm

27.  Find nearest oa dia wire size less than thSoadia calculated in step 26.

OPT1, Try 1.351mm oa dia, which is 1.25 mm Cu dia ...............................................1.351mm

28.   Calculate the theoretical S turns per layer, to nearest turn, thStpl.
Get the Bobbin winding width from step 13, Bww.
Theoretical S turns per layer, thStpl = Bww / thSoadia from step 27

OPT1, thStpl = 62 / 1.351 = 45.89, choose 45...........................................................45 turns per layer

**Note.   The calculated turns per layer are for the thickest wire possible but could be more turns of a smaller dia
to obtain the wanted turn ratios to to give the wanted load matches.
No less than 45 turns per S layer can be used if the traverse width is kept full because the increase in wire size will give a total height of the winding which exceeds the allowable total winding height.

29.   Calculate the nearest full S turns needed for loads of 3.5 ohms, 5 ohms and 7 ohms.
Secondary turns = primary turns / square root of impedance ratio.

OPT3, 2,320 P turns. PRL = 3.1k ,
for 3.5 ohms want 61 turns,
for 5.0 ohms want 73 turns,
for 7.0 ohms want 87 turns.
 
30.  Choose a pattern of Secondary winding sections from Fig 2, 3, 4, 5 below to possibly give a suitable variety
of at least two secondary load matches of between 3 and 9 ohms to suit most modern speakers
while the Primary load is considered to be 5,000 ohms.

OPT1, From step 15 we caculated 16 primary layers.
In the P&S winding list in step 21 we selected the 5S + 5P winding pattern. 
See Fig3 below.

Reading the charts below could be confusing!!!
Each rectangle represents a given separate OPT . The figure of N and its multiples are shown to give the
relationship between numbers of turns in each winding shown as a thick line.
Consider example 2A in Fig2..
There are two layers, each divided into 2N and N turns, which means there could be 50 and 25 turns respectively.
Where it says "3 @ 2N" means there are 3 parallel windings of 2N turns each
In the case of 2A, it means there are in fact 2 windings of 2N each and the third is made up of N+N in series.
Ns, or the secondary turn number for the transformer = 2N turns, since paralleling any number of same turn windings
does not alter Ns .
"2 @ 3N consists of two parallel windings each consisting of 2N + N turns in series.
In the case of all transformers the first line of impedance loads are listed for a given number
of N as  "Z = 1.0   1.7   3.0   5.0" , and the figures are starting reference impedances for each transformer.
The next line below for an increased number of N give the relative values of Z for that number of N
and the vertical columns of Z values give the relative impedance relationships for the various numbers of N
in windings.
So reading 2A, if we have 1.0 ohms as the match possible for 2N turns, then for 3N turns
the match is for 2.3 ohms, and for 6N turns the match is to 9 ohms.
There could also be a match where 2N = 3.0 ohms, and reading down the figures 3N gives 6.8 ohms, 6N gives 27 ohms.

In 2B, there is 4 @ 1N, Z = 1.0, etc, but the figure of N for 2A and 2B have no relationship.
I hope I have made it easy for everyone to get easy valuable winding information.

Fig 2.

OPT sub sections 2 & 3 sec layers.

Fig 3.
OPT secondary sub sections, 4 S layers.

Fig 4.
OPT secondary sub sections, 5 S layers.

Fig 5.
OPT secondary sub-section patterns.

Here we have 17 possible arrangements for secondary windings for many different OPTs.
Each rectangle represents the secondaries in a given OPT. There is no need to include the primary layers because
the number of primary layers could vary hugely without any change to the secondary layout ans sub section divisions.
The secondaries are always going to have turns appropriate to relatively low loads in the majority of OPT.

For the OPT No1 example we have chosen to use 4S + 5P so we can trial the secondary subsections
from any of the 3 examples of 4A, 4B, or 4C shown in the above Fig 3. We have already calculated in step 28
that we could possibly have 45 turns per layer.
Compare all available arrangements of secondary sub-sections and decide which arrangement offers the most useful
range of load matches.

OPT1, Examine 4A from Fig3.
There is a total of 6 windings with the numerical relationships of 3 windings of N turns, and 4 windings of 3N turns.
So we can have the top S layer divided into 3 windings of 15 turns each and 3 other layers of 45 turns each.

This allows the connection of windings to be as follows :-
4 parallel windings of 45 turns as calculated in step 29.........................................1.88 ohms
3 parallel windings of 60 turns consisting of 45 + 15 turns each............................3.34 ohms
2 parallel windings of 90 turns each consisting of  2 x 45 turns in series.................7.52 ohms
4 series windings of 45 turns each = 180turns .....................................................30.1 ohms

The above selection has two load matches between 3 ohms and 9 ohms.

**Note.   The calculated number of turns per layer of 45 just happens to be exactly divisible by 3.
If the layer was not exactly divisible, say it was 41, 44, 46, or 47 turns, we would be forced to adjust the turns
to the next highest number of turns divisible by 3, and revise our load match calculations.

Examine 4B.
There is a total of 8 windings, 4 @ 4N, 4 @ 1N.
So we could have 4 @ 36 turns and 4 @ 9 turns.
This allows the connection of windings to be as follows :-
5 parallel windings of 36 turns each.....................................................................1.20 ohms
4 parallel windings of 45 turns as calculated in step 29.........................................1.88 ohms
2 parallel windings of 90 turns each consisting of  2 x 45 turns in series.................7.52 ohms
4 series windings of 45 turns each = 180turns .....................................................30.1 ohms

Only one load load match that is suitable.

Examine 4C.
There is a total of 8 windings, 4 @ 2N, 4 @ 1N.
So we could have 4 @ 45 turns and 4 @ 15 turns.
This allows the connection of windings to be as follows :-
6 parallel windings of 30 turns each.....................................................................0.84 ohms
4 parallel windings of 45 turns as calculated in step 29.........................................1.88 ohms
3 parallel windings of 60 turns consisting of 45 + 15 turns each............................3.34 ohms
2 parallel windings of 90 turns each consisting of  2 x 45 turns in series.................7.52 ohms
4 series windings of 45 turns each = 180turns .....................................................30.1 ohms

Choose between available options.
4A and 4C give 2 load matches between 3 and 9 ohms and thus either could be used.
4B only gives one load match within 3 and 9 ohms and cannot be chosen.

**Note.   4C has a match to 0.84 ohms not available in 4A, but this is unlikely to ever be used.
4A would be easier to wind, another reason to select 4A.

**Note.    When 4A is selected, the number of secondary turns per layer could be increased to possibly keep within the
limits for wanted low winding losses.
48 turns give 2.1 ohms, 3.8 ohms and 8.6 ohms.   OK, two matches between 3 and 9 ohms.
51 turns give 2.4 ohms, 4.3 ohms and 9.64 ohms, NOT OK , only one match between 3 and 9 ohms.
57 turns give 3.0 ohms, 5.4 ohms and 12.0 ohms, OK, two matches between 3 and 9 ohms.
60 turns give 3.4 ohms, 5.9 ohms and 13.4 ohms, OK, two matches between 3 and 9 ohms.

Choose final option to suit nominal 4 ohms and 8 ohm speakers which can often have average Z =3 or 6 ohms.

**Note.  All turns per layer which are more the calculated minimum tpl
will have higher winding losses.

The 4A secondary pattern with 57 turns per layer seems suitable, but the 1.0mm wire will be a tight squeeze to
get 57 turns between the cheeks.

31.  Confirm the turn selection step 30 with regard to actual available wire size.
Theoretical  oa wire dia = bobbin winding width / S turns per layer.
Select wire from tables, Cu wire dia, mm

OPT1, oa wire dia = 62mm / 57 = 1.088 mm, so from wire tables select 1.00mm wire
because oa dia = 1.093 approx.
57 turns occupies 62.3 mm but we would fit the turns if the entry/exit of wire ends is staggered slightly
on each side of the bobbin.

Cu dia ...........................................................................................................................1.0mm

32.  Calculate chosen secondary option winding resistance.
List the following :-
Primary turns chosen for Ns, no.
S wire Cu dia for chosen tpl, from wire tables, mm.
Turn length is from step 17, mm.
No of parallel sections from option chosen to make up Ns turns, no.
ZR for chosen S turns, no.
S winding resistance
 = Ns x TL / ( No of parallel S sections x 44,000 x wire dia x wire dia ), ohms.

OPT1, Consider 57 turns per layer, arranged for 3 parallel windings of ( 57 + 19 ) turns each.
Ns .......................................................................................................................76
Eg, wire Cu dia mm...............................................................................................1.0
TL from step 17, mm.............................................................................................267
No parallel S sections, no.......................................................................................3
ZR = 2,320 / 76 squared = 932:1...........................................................................932


Swr = 76 x 259 / ( 3 x 44,000 x 1.0 x 1.0 ) = 0.149 ohms......................................0.149 ohms.

33. Calculate S loss % for chosen option for chosen Ns turns

Nominate the SRL, ohms.
SRL = PRL / ZR from step 32.

OPT1, SRL = 5,000 / 932 = 5.4 ohms..................................................................5.4 ohms

S loss % = 100 x Swr / ( SRL + Swr ), %. 

OPT1 S loss  = 100 x 0.149 / ( 0.149 + 5.4 ) = 2.69%..............................................2.69%

**Note.  If the option where 45 turns per secondary layer was chosen the wire size is 1.25mm dia
and the secondary winding losses would be lower at 2.4%. 

34.   Calculate total winding losses, chosen option,

Total winding loss % = P loss + S loss.....................................................ttl loss, %

P loss from step19, %
S loss from step 33, %

OPT1, P loss from step 19, %........................................................................2.19%
S loss from step 33, % ..................................................................................2.69%
Total loss = 2.19 + 2.69 = 4.96%...................................................................4.88%

35. Are total winding losses acceptable?............................................................yes/no

OPT1, yes,.......................................................................................................yes.

36. If yes to step 35, proceed to check if the winding height really is practical.

37.   Check the final winding height and bobbin thickness to make sure
the completed wound bobbin will fit into the core window.

OPT1,
Primary layers, 16 x 0.414mm = 6.624mm.
Insulation, p to p layers, 4 x 3 x 0.05mm = 0.6mm.
Secondary layers, 4 x 1.09mm = 4.36 mm.
Insulation P to S layers, 8 x 0.6mm = 4.8mm.
Insulation over top of last on primary, 0.6mm.
Bobbin bass thickness, 2mm.

Total winding height including bobbin bass = 18.96.

Remaining clearance = window height of 22mm - 18.96mm = 3.04mm = OK


38.  Calculate leakage inductance, where is is considered to be
an equivalent quantity of inductance in series with the primary load
looking into one end of the primary, with the other end grounded.

LL  =    0.417 x Np squared x TL x { ( 2 x n x c ) + a }               
                 1,000,000,000 x n squared x b

Where LL = leakage inductance, in Henrys,
0.417 is a constant for all equations to work,
Np = primary turns,
TL = average turn length around bobbin,
2 is a constant, since there is an area at each end of a layer where leakage occurs,
n = number of dielectrics, ie, the junctions between layers of P and S windings,
c = the dielectric gap, ie, the distance between the copper wire surfaces in P and S windings,
a = height of the finished winding in the bobbin,
b = the traverse width of the winding across the bobbin.

Distances are all in mm!

  OPT1,

  LL  =        0.417 x 2,320 x 2,320 x 267 { ( 2 x 8 x 0.7 ) + 17.7 }  
                            1,000,000,000 x 8 x 8 x 62  

           =  0.00435 Henry = 4.35 mH.

39.   Is the leakage inductance low enough?
Calculate reactance of LL at 100 kHz.

ZLL at 100kHz = L in henrys x 2 x pye x F
                          = L x 6.28 x 100,000Hz, ohms

OPT1,
ZLL at 100kHz = 0.00435 x 6.28 x 100,000 = 2,731 ohms .................2,731 ohms

Is ZLL less than PRL at 100 kHz?.............................................................yes/no

Eg, we have RL = 5,000 ohms,  ZLL = 2,731 ohms at 100 kHz...................yes.

40. If answer to step 38 is yes, leakage inductance is low enough.

41. Check that calculation of primary turns gives less than 1.6 Tesla magnetic field strength, B,
at full power and at 14 Hz.

 B   =    22.6 x V x 10,000 
               S x T x Np x F 

where B is in Tesla,
22.6 and 10,000 are constants for all transformer equations,
V = Vrms signal voltage across the primary, or sq.rt ( PO x PRL )
S = core stack height,
T = core tongue width,
Np = primary turns,
F = frequency at which B is to be measured.

All dimensions in mm!!

Eg, for 60 watts into 5k, Va-a = 547vrms.

B at  14Hz  =  22.6 x 547 x 10,000      =  1.57 Tesla = OK
                       55 x 44 x 2,320 x 14

**Note.  The above wind up is for plain UL. If CFB windings are used, there will need to be some increase in
some of the primary to primary insulations because the CFB windings will be at 0V potential.
so instead of at least 2 x 0.05mm p-p insulation layers there may be extra 2 x 0.6mm.

**Note. There is some room for bulge in the windings as they are wound on.
Wire will not lay tightly as layers are put on and will tend to spring up across the rectangular core.
The bulge will distribute itself from start to finish and wires will not be naturally tight in the vertical direction
hence the importance of impregnation with varnish to make windings adhere to each other.

42. Have all parameters been satisfied?...............................................yes/no

43. If yes to step 42, Design is OK and sourcing materials can be undertaken.

44.  Are the wire sizes available? .......................................................yes/no

45.  If  no to step 42, find out what sizes are available, and design to suit
these sizes without compromises!!!

46.  Draw up the bobbin winding details for the proposed OPT No 1 ready for
the guy who is going to wind the OPT.

Fig 6.
bobbin winding detail opt no1

47. Shunt capacitance of an OPT.

There are several areas in an audio transformer where capacitance exists, and with an OPT we are primarily interested
in the total measured capacitances when we measure the capacitance at the anode terminals of the OPT.
There is capacitance between primary wires in the form of the "self capacitance" of the primary layers of wire
and between  layers of wire adjacent to secondary sections which have much lower signal voltages
and are effectively at 0V potential. 

To calculate the primary shunt capacitance in an audio transformer such as OPT No1, refer to the above bobbin winding
layout. Neglect the self capacitance of the primary windings; it will be such a small amount compared to the
main shunt C between adjacent P to S interfaces.

The distance between the copper surfaces of primary and secondary layers including the insulation thickness
of 0.6mm and the wire enamel of about 0.05mm = approx 0.7mm.
Then you must allow for the curved surface of the wire turns so total distance = approx 0.75mm.

Capacitance between two metal plates =   ( A x K ) / ( 113.1 x d )   
where
Capacitance is in pF,
A is the area in square millimetres of the plates assumed to be of equal size,
K is the dielectric constant of the material between the plates, air being = 1.0,
113.1 is a constant for all equations to work,
d is the distance in millimetres between the plates and is the same for the area of the plates.

For example, if the turn length around the wound bobbin nearest 'anode 1' = 250mm, and winding traverse width
= 62mm, then area = 250 x 62 = 15,500 sq.mm.
Let us say the K for the polyester = 2.
( The C can be measured if unknown using metal plates of known area, and using a sample of polyester
clamped tight between the plates for the whole area. Once the C measurement has been recorded,
the plates are set up with a very small width strips of polyester leaving the plates the same distance apart
but with mostly air between the plates, and the C measured again.
K = C measured with full amount of polyester / C measured with just air. )

The d we calculated above = 0.75mm.

C in pF = 15,500 x 2 / ( 113.1 x 0.75 ) = 365.5pF.

The amount of capacitance in each P to S interface varies with turn length so that nearest 'anode 2'
the C would be less.

But for a simple estimate of the C at each P to S interface we could say C simply = 360pF.

The first P-S interface down from 'anode 1', or the top of the wind-up at above the GH-IJ-KL
is at a position of 6.5 layers / 8 layers along the P winding from the CT where the signal voltage is zero.
This positioning results in the capacitance being subject to the impedance ratio at this position.

Therefore the C is transformed to 360pF x (6.5 / 8) squared, = 360 x 0.66 = 237pF.

The next area of 360pF down from anode 1 appears below the sec layer GH-IJ-KL and the impedance ratio
is (5.5 / 8) squared = 0.47 so the C due to this interface at anode 1 = 360pF x 0.47 = 170pF.
Next down the Z ratio is above the EF sec and = (2.5 / 8) squared = 0.098, so C = 35pF at anode 1,
then below EF the Z ratio reduces the 360pF by (1.5 / 8) squared = 13pF.
The total C appearing at the anode 1 connection is the sum of all these transformed capacitances =

237pF + 170pF + 35pF + 13pF =  455pF, which is approximately what we would measure with a capacitance
meter connected to anode 1 with CT grounded and one end of all the secondary layers all grounded.

If the OPT No1 bobbin was used for an SE amplifier design the capacitance would be calculated similarly
but have a total of many more calculated values of effective capacitances seen at the ONE end of the P winding.
Cathode Feedback use further complicates the capacitance calculation but the effect of the capacitance
on amplifier bandwidth is effectively reduced by the NFB because the NFB reduces the Ra of the
tubes.

If the above transformer No1 is used with a pair of KT88 in beam tetrode mode then if
the Ra is simply about 18,000 ohms at each anode and without a load the gain of the tube will reduce
-3dB from approximately being equal to µ of the tube at say 500Hz at where the
capacitive reactance = Ra. Since C = approx 445pF, then this -3dB pole
is found easily at a frequency = 159,000 / ( Ra x C in uF) = 159,000 / ( 18,000 x 0.000445 ) = 19.8kHz.
In practice this would be about correct, and one way to measure the C shunt with a tube
is to use a high Ra tube unloaded such as a tetrode or pentode, and work from the observed -3dB point.
The leakage inductance will have little effect on the unloaded response.

The capacitance and leakage inductance will react together to form a tuned circuit and
low pass filter with an ultimate slope of more than 6dB/octave.
So rapid phase shift increase occurs as F becomes high so it is important to minimise C and LL
to force the frequency of resonance to be as far as possible above the audio band and where
the phase shift with loop NFB does not cause oscillations.
Trying to establish an equivalent model of the complex LCR offered by such a simple OPT as No1
is beyond my abilities and there is little point to achieve such modelling. It is simply easier to
establish low values of C and LL by empirical methods and then critically damp the HF gain
of the amp to achieve low overshoot on square waves with a 0.22 uF across the output without any R load,
while maintaining a maximal HF pole with a solely R load.

--------------------------------------------------------------------------------------------------------------------------                                                     

METRIC WINDING WIRE SIZE CHART
The metric winding wire sizes were kindly given to me by a local Sydney wire and transformer parts supplier.
The original chart contained the same copper sizes as shown for grade 1 with less enamel thickness
and grade 3 with more enamel thickness. I only use grade 2 which is the only grade shown in the chart below.
Grade 2 is the only grade stocked by my supplier because it is the industry norm for 99% of high temperature rated winding wire for electric motors and stressful industrial applications.
The range of sizes shown are not all obtainable off the shelf, and to get some sizes a wait for an order is involved,
so I sometimes have to design around the wire size available, which adds to the challenge.
Anyone not used to measuring in millimetres better start getting used to metric because here the
diameter measurement matters more than the wire guage, and there are is AWG, SWG, BS, all very confusing,
and I don't have conversion charts so if you work in guages and inches and feet, provide your own solutions.
Before winding anything, make sure you have an accurate micrometer to confirm that the size is correct.

Metric winding wire sizes.

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