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

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.

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.

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.

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