SE OUTPUT TRANSFORMER
For parallel KT120, KT88, KT90, 6550, EL34, 6CA7, 6LGC, 807, 5881 6CM5
or a single 13E1.

CONTENT :-

Fig 1. OPT5 for SE 1.8k : 5 ohms with 44T x 50S core and details of bobbin layers,
insulation, wire etc.

Fig 2. OPT6 for SE 1.8k : 2.2, 3.5 6.2, 13.9 ohms with 51T x 51S core and details
of bobbin layers, insulation, wire etc.

Design flow includes checks on dc current ability, leakage inductance,
primary inductance
calculations, check of ac Bmax, dc B, calculations of the
µe and air gap.

Readers can refer to my 3 pages on SE OPT CALCULATIONS to be able to
better follow
the steps as listed below because of the explanation notes.
The design examples should provide a template for anyone with high school
math ability,

some basic intelligence, a pocket calculator and with keen commitment to
calculate their way to better music.

Metric winding wire size chart for grade 2 polyester-imide wire.

OPT5

1.Tubes can be 1 x 13E1,  3 or 4 x  6550, KT88, 6550, KT90, KT120, 4 x EL34,
KT66, 5881, 6L6GC, 3 x 300B, etc.
Ea = 375V, total Ia = 190mA. Primary RL = 1.8k

2. Secondary RL = 5 ohms, with options in website text for other load matches.

3. PO = 25 watts

4. Afe = 450 x sq.rtPO in sq.mm = 450 x sq.rt 25 = 2250 sq.mm

5. Core tongue dimension, theoretical T = sq.root 2250 = 47.4mm
Choose core T = 44mm.

6. Calculate theoretical stack height, thS using the chosen T size.
S = 2,250 / 44 = 51.1mm,
choose 51mm to suit bobbin for 44 x 51mm, or T =1.75" and S = T = 2"

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

8. Confirm the height of the winding window, H = 22 mm.

9. Confirm the length of the winding widow, L, = 66 mm.

10.  Primary winding turns, thNp,

Np = sq.rt( PRL x PO) x 20,000 / Afe, turns.

= sq.root ( 1,800 x 25 ) x 20,000 / 2,244 =  1,890

11. Theoretical over all dia of P wire including enamel insulation
= sq.rt ( 0.28 x L x H / Np )

= sq.root ( 0.28 x 66 x 22 / 1,890 ) = 0.464 mm

12. oa wire size from the tables, try oa wire size = 0.462 mm, for Cudia = 0.400 mm.

13.  Bobbin winding traverse width = 66 - 4 = 62 mm

14. No of theoretical P turns per layer, thPtpl, = 0.97 x 62 / 0.462 = 130 turns

15. Number of primary layers = 1,890 / 130  = 14.5, round up to15 layers

16. Actual Np = P layers x Ptpl, turns = 15 x 130 = 1,950 turns.

17. TL = ( 3.14 x H  ) + ( 2 x S ) + ( 2 x T ) = 259mm

18. Rwp = ( Np x TL ) / ( 44,000 x Pdia x Pdia )
= 1,950 x 259 / ( 44,000 x 0.4 x 0.4 ) = 72 ohms

19. P loss % = 100 x Rwp / ( PRL + Rwp ) P loss
= 100 x 72 / ( 1,800 + 72 ) = 3.9%

20. P winding loss is less than 4%, OK

21.  Choose the interleaving pattern from the list below for the wattage of the
transformer.

Choose 4S x 5P but with following interleaving pattern :-
2p - S - 4p - S - 3p - S - 4p - S - 2p                          

Draw the bobbin wind up on paper or in MS Paint similar to what I have done.
Insulation details can be added when known and confirmed for dimensions during
final checks on winding height.
The subdivision of secondary layers, the thick lines, can be included when the
full details of the secondaries are known after step 30.
I have also included the strapping details for CFB windings.
It is always assumed in all the OPTs on this website that the winding lathe need
only run in one turning direction.
The start and finish points of each winding are important, and each consecutive
primary layer is wound on
so that its finish point is the same side as the start point for the next primary layer.
Where one winds across right to left then returns with a layer on top left to right
then voltages automatically add for the two layers.
All secondary layers start on the left side and proceed to the right side of the bobbin.
If there OPT is used with local CFB, the cathode winding consists in this case of
2 of 15 primary layers which are then seriesed to produce the correct phase and
voltage for connection between cathode and 0V to give 13.3% of local CFB.
The screen taps available for UL are not critical for small multigrid tubes,
and in the case below are at 10-11 = 33%, 8-9 = 40%, 6-7 = 60%.
60% is the right value for 13E1, and perhaps 6550/KT88/KT90.
The insulation between possible CFB windings and anode windings is the same
as between anode windings and earthy secondary windings.

I hope this keeps things simple for everyone, only about 50 things to worry about!

Fig1, possible design for OPT6
  OPT bobbin
          details, SEUL 13E1, etc.

22. Choose thin insulation i = 0.05mm

23. Choose thick insulation I = 0.5 mm
 
24. Height of P+I+i = ( no of layers x oadia P wire ) + ( no x i ) + ( no x I )
= ( 15 x 0.462mm ) + ( 10 x 0.5mm ) + ( 8 x 0.05mm ) 
= 6.93mm + 5.0 + 0.4mm = 12.33mm.

25. Maximum total available height of all windings on bobbin
= 0.8 x 22 = 17.6mm.


26. Max thSoadia =  ( max wind ht - [ P+i+I ht ] ) / no of S layers
= ( 17.6 - 12.33 ) / 4 = 5.27 / 4 = 1.317mm

27. Wire tables give 1.279mm oa dia for 1.18mm Cu dia wire

28. Theoretical S turns per layer, thStpl = Bww / thSoadia from
step 27 = 62 / 1.279 = 48.47 turns, so minimum turns per layer = 48 turns.

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.

OPT5, 1,950 P turns. PRL = 1.8k ,
for 3.5 ohms want 86 turns,
for 5.0 ohms want 103 turns,
for 7.0 ohms want 121 turns.

30.  From step 15 we calculated 15 primary layers.
In the P&S winding list in step 21 we selected the 4S + 5P winding pattern.
Inspect the range of secondary arrangements where there are 4 secondary
layers shown on website pattern list in pages on SE OPT Calculations-2,
see selection Figs 6 to 10 for secondary sub-section patterns.

31. Confirm the turn selections in step 30.

See SE OPT Calculations-2 and examine sub-section pattern 4A
from Fig 8 choice of secondary winding patterns. This appears in step 32
at SE OPT calcs-2 page.
There is a total of 6 windings with the numerical relationships of
3 windings of N turns, and 4 windings of 3N turns.
For this to be possible, the number of turns per full secondary layer
must be evenly divisible by 3.

Using 48 tpl we could have :-
4 @ 48  = 1.091 ohms,
3 @ 64  = 1.93 ohms,
2 @ 96  = 4.36 ohms.

These ratios are too low, since two load matches between
3 and 9 ohms are wanted.

Try 51 tpl, 1.12mm copper dia,

4 @ 51t in parallel, = 1.23 ohms,
3 @ 68t in parallel, = 2.18 ohms,
2 @ 102t in parallel = 4.92 ohms.

This gives one usable 5 ohm load match with low winding losses which
is a compromise between having a match to both 4 and 8 ohms, but we
do not have to subdivide the secondary layers.

The Fig1 above is for OPT5 to simply suit 1.8k to 5 ohms only.

To get two load matches between 3 and 9 ohms to suite 4 and 8 ohm speakers,
we could increase the S turns per layer.
To match to 7 ohms, then Ns = 121.6 turns, so the the
Stpl = 60, so wire size would be 0.9mm Cu dia instead of the 1.18mm Cu size
we began with to give 48 tpl.

Choose option of 60 turns per layer.
4 @ 60t in parallel, = 1.7 ohms,
3 @ 80t in parallel, = 3.0 ohms,
2 @ 120t in parallel = 6.8 ohms.

For 6.8 ohms, want 2 x 60 turns x 0.9mm Cu dia wire.

32. Calculate chosen secondary option winding resistance, 6.8 ohm option,
Ns ..............................................................................................................120
Eg, wire Cu dia mm................................................................................0.9mm
TL from step 17, mm.............................................................................259mm
No parallel S sections, no..............................................................................2
SRL.....................................................................................................6.8ohms
Swr = 120 x 259 / ( 2 x 44,000 x 0.9 x 0.9 ) = 0.436 ohms.

33. S loss for 6.8 ohm,
S loss % = 100 x Swr / ( SRL + Swr ) = 100 x 0.436 / 7.24 = 5.8%

Calculate 32 and 33 again but for 4.9ohm option,
S loss % = 100 x Swr / ( SRL + Swr ) =
100 x 0.239 / ( 0.239 + 4.92 ) = 4.63%

34.  Total winding losses, chosen options :-

OPT5 P loss from step 19, %.................................................................3.9%
S loss from step 33, 6.8 ohm / 0.9 / 60 tpl option...................................5.8%
Total loss................................................................................................9.7%

OPT5 P loss from step 19, % ................................................................3.9%
S loss from step 33, 4.9 ohm / 1.12 / 51 tpl option.................................4.7%
Total  loss ............................................................................................. 8.6%.


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

At this point we have to make a decision whether its worth allowing up to
2.4 watts of the tube power to be wasted and settle for 22.6 watts at the
secondary output, and proceed to build the OPT.

Using 60 tpl for the secondary of OPT5 instead of 51 tpl will cost us
slightly more in wasted power and use slightly more labour, but
perhaps its worth it!! leakage inductance and other checks will be
about the same for all of the transformers we are exploring.

The high secondary winding losses are simply a function of the absence
of sufficient copper wire dia to fill the available winding space to get low
winding losses. Eg, there are 4 layers of 0.9 mm wire instead of 4 layers
of 1.4mm Cu dia wire.

LET US TRY using a core with a larger window, call it OPT No 6.
SE OPT6.
Repeat steps from step 5 above......

5.  Core tongue dimension, theoretical T = sq.root 2250 = 47.4mm
Choose core T =51mm

6.  Calculate theoretical stack height, thS using the chosen T size.
S = 2,250 / 51 = 44.1mm.
44.1 bobbins are unavailable, but bobbins for 51mm x 51mm
( 2" square ) are available, so choose 51mm stack.

7.  Adjusted Afe =  51 x 51 = 2,601 sq.mm

8.  Confirm the height of the winding window, H = 25 mm.

9.  Confirm the length of the winding widow, L, = 76 mm.

10.  Primary winding turns, thNp,

Np = sq.rt( PRL x PO) x 20,000 / Afe, turns.

= sq.root ( 1,800 x 25 ) x 20,000 / 2,601 = 1,631 turns.

11. Theoretical over all dia of P wire including enamel insulation
= sq.rt ( 0.28 x L x H / Np )

= sq.root ( 0.28 x 76 x 25 / 1,631 ) = 0.571mm

12.  oa wire size from the tables, try oa wire size = 0.569 mm,
for Cudia = 0.500 mm.


13. Bobbin winding traverse width = 76 - 4 = 72 mm

14. No of theoretical P turns per layer, thPtpl, = 0.97 x 72 / 0.569
= 122 turns


15.  Number of primary layers = 1,631 / 122  = 13.36, so round up to14 layers

16.  Actual Np = P layers x Ptpl, turns = 14 x 122 = 1,708 turns.

17.  TL = ( 3.14 x H  ) + ( 2 x S ) + ( 2 x T ) = 282mm

18. Rwp = ( Np x TL ) / ( 44,000 x Pdia x Pdia )
= 1,708 x 282 / ( 44,000 x 0.5 x 0.5 ) = 44 ohms

19. P loss % = 100 x Rwp / ( PRL + Rwp ) P loss
= 100 x 44 / ( 1,800 + 44 ) = 2.4%

20.  P winding loss is less than 4%, OK

21. Choose the interleaving pattern from the list below for the
wattage of the transformer.

Choose 5S x 4P but with following interleaving pattern :-
S - 3p - S - 4p - S - 4p - S - 3p - S                             

Draw the bobbin wind up on paper or in MS paint similar to what I have done.
Insulation details can be added when known and confirmed for dimensions
during final checks on winding height. The subdivision of secondary layers,
the thick lines, can be included when the full details of the secondaries are
known after step 30.
Fig2.
  13E1 SEUL
          OPT bobbin layers.

22. Choose i = 0.05mm

23. I = 0.5 mm
 
24. Height of P+I+i = ( no of layers x oadia P wire ) + ( no x i ) + ( no x I )
= ( 14 x 0.569mm ) + ( 10 x 0.5mm ) + ( 8 x 0.05mm )
= 7.96mm + 5.0 + 0.4mm = 13.36mm.

25. Maximum total available height of all windings on bobbin
= 0.8 x 25 = 20.0mm.


26.  max thSoadia =  ( max wind ht - [ P+i+I ht ] ) / no of S layers
= ( 20.0 -13.36 ) / 5 = 6.64 / 5 = 1.328 mm

27.  From the wire tables try 1.18mm Cu dia wire which is 1.279mm oa dia.

28. Theoretical S turns per layer, thStpl = Bww / thSoadia from step 27
= 72 / 1.279 = 56.29 turns so try 56 turns per layer.

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.

OPT6, 1,708 P turns. PRL = 1.8k ,
for 3.5 ohms want 75 turns,
for 5.0 ohms want 90 turns,
for 7.0 ohms want 106 turns.

30. From step 15 we calculated 14 primary layers.
In the P&S winding list in step 21 we selected the 5S + 4P winding pattern.
Inspect the range of secondary arrangements where there are 4 secondary
layers shown on website pattern list in page on SE OPT Calculation-2,
see step 32, Fig 9 Tube OPT secondary subsections.

31. Examine 5A from Fig 9, SE OPT Calculation-2
There is a total of 5 layers with 8 windings with the numerical relationships of
4 windings of N turns, and 4 windings of 4N turns. For this to be possible,
the number of turns per full secondary layer must be evenly divisible by 4.
56 turns is divisible by 4.
we could have :-
5 @ 56 = 1.95 ohms,
4 @ 70 = 3.05 ohms,
2 @ 140 = 12.2 ohms.
 
Unfortunately 3.0 and 12.2 are too low and a little too high,
since 3.5 and 7 would be the ideal preference.
For 3.5 ohms we need 75 turns, and for 7  there are 106 turns needed.

Examine 5B from Fig4,'SE OPT Calculation'
Each layer must have its total turns divisible by 6,

Tpl could be 54 turns 1.18mm Cu dia wire for a full layer,
6 @ 45 = 1.25 ohms,
5 @ 56 = 1.93 ohms,
3 @ 90 = 5.0 ohms.
This is the ideal solution if we wanted a match to only 5 ohms.

Tpl = 66 turns of 1.0mm Cu dia,
6 @ 55 = 1.86 ohms,
5 @ 66 = 2.68 ohms,
3 @ 110 = 7.46 ohms.
This arrangement does give us two load matches between 3 and 9 ohms.

Try 5C. Tpl must be divisible by 12, so try 60 tpl, 1.0mm Cu dia.
The wire would have to be slightly spread out across the bobbin because
the oa dia = 1.093, and 60 turns tightly packed is 66mm, and the bww = 72 mm.
But 1.06 mm Cu dia wire is a size on my charts, and should be used;
its oa dia = 1.155mm so 60 turns = 69.3mm, which is OK.
5 @ 60 = 2.22 ohms,
4 @ 75 = 3.47 ohms,
3 @ 100 = 6.17 ohms,
2 @ 150 = 13.9 ohms.
This gives the best range of loads. 

32.  Calculate chosen secondary option winding resistance,
tpl = 60 turns / layer, 1.00mm Cu dia wire.
Ns ..................................................................................................75
Eg, wire Cu dia mm..................................................................1.0mm
TL from step 17, mm...............................................................282mm
No parallel S sections, no................................................................4
SRL.....................................................................................3.47ohms
Swr = 75 x 282 / ( 4 x 44,000 x 1.0 x 1.0 ) = 0.12 ohms.

33. S loss for 72 turns, 3.21 ohm,
S loss % = 100 x Swr / ( SRL + Swr ) = 100 x 0.12 / 3.59 = 3.30%

34.  Total winding losses, 60 tpl option.

OPT6, P loss from step 19........................................................2.4%
S loss from step 33, 3.21 ohm / 1.0 / 60 tpl option....................3.3%
Total P + S loss %.....................................................................5.7%

35. Are total winding losses acceptable?..............................yes/no
Check the dc current rating for primary.
Ia = 190mA, 0.5mm Cu dia wire is rated for 588mA at 3A/sq.mm, so
actual DC is 1/3 of rated amount, OK.

36. If OPT6 is so far ok , check the final winding height and bobbin
thickness to make sure the completed wound bobbin will fit into the core
window without difficulty.

37.  OPT6,
Primary layers, 14 x 0.569mm = 7.97mm.
Insulation, p to p layers, 8 x 0.05mm = 0.4mm.
Secondary layers, 5 x 1.155 = 5.78 mm.
Insulation P to S layers, 10 x 0.5mm = 5.0mm.
Insulation over top of last on secondary, 0.2mm.
Bobbin bass thickness, 2mm.

Total winding height including bobbin base and tape over completed
wind up = 21.35mm

Remaining clearance = window height of 25.3mm - 21.35 = 3.95mm = OK

At this point we could say the losses lower than 6% would be acceptable,
so we can consider the above OPT6 is a possible design.

38.  OPT7,Calculate leakage inductance,  

LL  =   0.417 x 1,708 x 1,708 x 282 { ( 2 x 8 x 0.6 ) + 20 }  
                            1,000,000,000 x 8 x 8 x 72 

           =  0.0022 Henry = 2.2 mH.

39.  Is the leakage inductance low enough?
OPT7, ZLL = 0.0022 x 6.28 x 100,000 = 1,256 ohms at 100 kHz
Is ZLL less than PRL at 100 kHz?.....yes, PRL = 1,800 ohms.

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

41. Check that calculation of primary turns gives less than
0.8 Tesla magnetic field strength, B, 14Hz.

B at 14Hz  =  22.6 x 212 x 10,000     =  0.77 Tesla  = OK
                       51 x 51 x 1,708 x 14

42. Calculate the wanted minimum primary inductance, Lp, Henrys.

Wanted Lp will have reactance in ohms = nominal Primary RL at no
higher than 20 Hz.
Minimum Lp =       PRL             
                             6.28 x F 

OPT3, Min Lp = 1,800 / ( 6.28 x 20 ) = 14.33H

42.  Calculate the effective permeability, µe.

µe =  1,000,000,000 x Lp x mL
        1.26 x Np squared x S x T
Where  µe is effective permeability of core with an air gap, and is just
a number, no units. 1,000,000,000 and 1.26 are constants for all
equations to work,
Lp = primary inductance,
mL = magnetic path length of the iron,
S = stack height,
T = tongue width.
All dimensions in mm !!!

OPT7,  µe =    1,000,000,000 x 14.33 x 240    =  360.
                      1.26 x 1,708 x 1,708 x 51 x 51

43.  Calculate the air gap required.

Air gap = mL x ( µ - µe )
                   µ x µe u
Where gap is the air gap distance placed into the iron magnetic circuit,
mL = the iron magnetic path length,
µ = iron permeabilty maximum,
µe = effective permeability with air gap.
all dimensions in mm !!!

OPT7, Air gap =  280 x ( 17,000 - 360 ) / ( 17,000 x 360 ) = 0.76mm.

NOTE.  The air gap is the total gap. In an E&I or C-core which is air gapped,
there  are TWO gaps inserted in the magnetic path around each rectangle
of two which form the core.

Therefore the dimension of the plastic material or paper used to make the
gap will be HALF that calculated above.

44.  Calculate the dc field strength created in the core by the dc flow,
in Tesla.

Bdc = 12.6 x µe x Np x I
              mL x 10,000
Where Bdc is in Tesla, 12.6 and 10,000 are constants for all equations
to work, µe = effective permeability,
Np = the primary turns,
I = dc current in AMPS,
mL is the iron magnetic path length.

OPT7,  Bdc = 12.6 x 360 x 1,708 x 0.19 / ( 280 x 10,000 )  = 0.525 Tesla.

45. Calculate maximum total ac + dc magnetic field strength at 14Hz.

Maximum B = acB + dcB.

OPT6, 
From step 40, ac B max = 0.77Tesla.
From step 44, dc B =  0.525 Tesla.

Total max B = 1.3Tesla.

46.  Is the total B max below the core material maximum at saturation?

OPT No3 core material is GOSS which will saturate at approx 1.6 Tesla
Total B max = 1.3 Tesla at 14Hz, therefore design is OK, and in fact the
core will not saturate until acB max reaches 1.0T which is at 11 Hz.


48. Have all parameters been satisfied?......................................yes,

49. If yes to step 48, Design is OK and sourcing materials can be undertaken.

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

51.  If  no to step 50, find out what sizes are available, and design to suit
these sizes without compromises, which usually means the OPT will be
larger and heavier!!!
The slightest change to wire sizes shown without making calculations
carefully will usually lead to a completely useless OPT.


Conclusions.
OPT5 with 51S x 44T core, minimum losses possible = 8.5%
OPT6 with 51S x 51T core, minimum losses possible are 5.7%,
and worth the price of the extra iron and wire.

By simply increasing the stack height of OPT6, the allowable applied
anode signal load voltage can be increased proportionately for the same
LF cut off point providing the Idc is constant.
However where Ia was raised by 25%, the Bdc would rise and tend to reduce
the cut off frequency marginally. But listeners would not use any more power
than with the 25 watt setting so the increased power ceiling would still be
effective from above 16Hz.

OPT7 allows for 190mA and Ea = 375V, and RL = 1.8k.

13E1 are uncomfortable with Ea above 375V where Eg2 is the same voltage
as Ea.

But Ea could be raised to 500V for 3 x KT120 or 4 x KT90 operation with
Ia = 250mA, for Pda = 31 watts each and at 35% efficiency in CFB or SEUL
the Po = 43 watts and anode load would be about the same 1.8k.

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 gauge,
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.

table-wire-sizes.GIF

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