OPT5
1.

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

2.
SRL = 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 the
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%, 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.

Fig1, possible design for OPT6
 

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 )
= ( 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 caculated 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 page on 'SE OPT Calculation' / Tube OPT secondary subsections, Fig2,3,4&5.

31.  Confirm the turn selections in step 30.

 Examine 4A from Fig3/'SE OPT calculations'.
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.

The 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 don't 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 absense 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.

But let's try using a core with a larger window, and we'll 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.
 

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.

30. From step 15 we caculated 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' / Tube OPT secondary subsections, Fig2,3&4.

31.Examine 5A from Fig4,'SE OPT Calculation'
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.

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 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, 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.

OPT3, 
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.

Buy 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 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 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.

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