SE OPT CALCULATIONS PAGE 2.

Continued from SE OPT Calc Page 1,
SE OPT calc page 2 Contents :-

31.  Calculate sec winding losses for turns in Step 28,
       Calculate total P and S winding losses. 
32.  Choose Secondary Winding Sub Section pattern.
       Fig 6, 7, 8, 9, 10 Secondary sub-section patterns.
       Fig 11. Explanation for chosen 4A sub-section pattern.
33.  Calculate Sec turns to suit a chosen sub-section pattern.
       Table for total turns vs configurations possible.
       Avoiding "illegal" configurations.
       Summary & Conclusions, 2 questions, A and B.

Re-calculate from step 24 to get better outcome :-

24A.  Calculate insulation.
25A.  Calculate height of Primary layers plus all insulation.
26A.  Find nearest oa dia wire size from wire size tables.
27A.  Calculate theoretical oa dia Secondary wire.
28A.  Calculate Th sec oad Tpl.
29A.  Calculate load matches with Sec turns from Step 28A.
         Tables of matches including legal and illegal configurations.
30A.  Calculate Sec turns for matches to 3, 6 and 12 ohms.
31A.  Calculate sec winding losses for turns in Step 28A,
         Calculate total P and S winding losses.
32A.  Choose the same sub-section pattern as for Step 32.
         Fig 12. showing chosen 4A interleaving pattern.
         Table of resulting matches.
        
Conclusion that design is OK.
33A.  Check total height of all bobbin contents.

Proceed from Step 33....

34.  Bobbin winding details.
       Fig 13. OPT4 bobbin winding details.
35.  Tapped Secondary windings.
       Fig 14. OPT4TS bobbin winding details.
       Notes and conclusions.
36.  Calculate leakage inductance.
37.  Is the leakage inductance low enough?
38.  Shunt capacitance and Primary Inductance of SE OPT.
       Fig 15. Test schematic for Csh and Lp. 
       How to measure shunt capacitance and primary inductance, + notes.
       Fig 16. F response from test measurements.
       More notes.
       Table of Csh at points along primary winding. 
39.  Calculate wanted µe.
40.  Calculate Lp.
41.  Calculate XLp = RLa at F cut-off, Fco.
42.  Calculate Fsat.
43.  Calculate the air gap, Ag.
       Fig 17. Graph of Air gap sizes for µe Vs grade of iron.
44.  Notes on testing OPT4.
       Fig 18. Photo of sample OPT and the home made pot.
       Notes about potting OPTs.

Continued at SE OPT calc Page 3.
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Beginning of page 2 content :-

31. Calculate sec winding losses for turns in Step 28,
Calculate total P and S winding losses.


Calculate secondary winding resistance =
Rws = Ns x TL / ( [ No of parallel windings of Ns ] x 44,000 x Cu dia squared. )

OPT4. Rws = 37 x 297 / ( 4 x 44,000 x 1.8 x 1.8 ) = 0.0192 ohms.
Load match for Ns = 37 turns = 0.66 ohms.

Calculate Rws % losses = 100 x Rws / ( Rws + sec RL )
= 100 x 0.0192 / ( 0.0192 + 0.66 ) = 2.83%

From Step 18, primary winding losses = 3.42%

Total P and S winding losses = 3.8% + 2.83% = 6.63%

Is this MORE than 10% above 7.0%? If YES, consider revising design
with higher core stack, lower Ns.
If NO, then Sec losses could be higher if Sec turns per layer were
increased.

OPT4, Total P & S losses for the design so far = 6.63%.
Therefore Sec losses may be slightly increased.

NOTE.   Where the Sec loss % is less than 3.5%, the secondary turns per layer
could be increased, and another tedious formula could be developed, but it is
better to choose a range of higher sec turns per layer up to 1.4 times those so
far calculated.

OPT4. Sec turns per layer from Step 27 = 37 turns.
Maximum possible Sec tpl = 1.4 x 37 = 52 turns.

NOTE. The original formula for calculating the core Afe in sq.mm is
Afe = 450 x sq.rt Output Power max.
450 is a constant which gives a good range of acceptabe sizes for Afe.
This formula suits most SE OPT to give low winding losses. There is probably
a better formula which takes into account the load value and the tendency of
the formula to be inaccurate where RLa is very high, say 20k0 for an SE amp
with one 211, or the OPT is very small, say for an SE amp with 2A3, or a single
EL84 in pentode mode.
I have not done 30 possible designs just to establish tables to adjust the constant
of 450 to better suit the max audio power or the RLa value.
The above formula does give everyone a place to begin from, and practice will
soon teach anyone that the calculations so far indicate a core could be larger
or smaller.
Total winding losses for SE OPT tend to be higher than for the same sized core
size/weight used in PP designs. I do not think total losses should exceed 7%.
7% may seem high but many SE OPTs I have examined have total losses
exceeding 20%, which is very poor.

32. Choose Secondary Winding Sub Section pattern.

Confirm choice of interleaving pattern from Step 21,
Interleaving pattern = 5P + 4S.


Inspect Figs 6, 7, 8, 9, 10 patterns of Secondary Winding Sub Sections and
choose from one
which lists secondary sub-sections for the chosen number of
secondary sections.

OPT4.  Fig 8 below has sub-sections for 4 sections of secondary wire.

Choose a pattern from those shown, 4A, 4B or 4C from Fig 8.

Fig 6.
opt-sec-2ABC-sub-sectionsX.GIF

Fig 7.
opt-sec-3ABC-sub-sectionsX.GIF

Fig 8.
opt-sec-4ABC-sub-sectionsX.GIF

Fig 9.
opt-sec-5ABCDE-sub-sectionsX.GIF

Fig 10.
opt-sec-6ABCD-sub-sectionsX.GIF

Step 32 continued...Confirm choice of 4A from Fig 8.

NOTE. The chosen winding pattern requires some further explanations.
The division of the single layers of secondary windings into exact fractions of
a layer allows the windings to be arranged to suit a number of secondary
load values yet maintain the same OPT primary load value connected to
anodes :-

Fig 11.
opt-sec-4A-sub-section-connectionsX.GIF

NOTE. Fig 11 above shows how winding pattern 4A may be connected to give 3
different load matches, so the primary load remains constant for 3 different
secondary loads.
The RELATIVE NUMBERS of S turns varies between 2N, 3N and 6N.
This gives RELATIVE Sec LOADS of 1ohm, 1.78 ohms, 4 ohms,
or as they are listed in the table within Fig 11.

NOTE. For the arrangement in 4A to work properly, it is necessary that the
TOTAL number of all S turns in all layers is exactly divisible by all selected
numbers of paralleled windings.

For example, in 4A, the total turns = 4 x 3N = 12N. Each choice of parallel
windings has a different fraction of the total number of all secondary
winding turns in all secondary layers.
Their relative turn number are described as 3N, 4N or 6N turns.
The total number of all secondary turns must be divisible by 3, 4 or 6.
Fortunately, it is easy to check if any number of total S turns are divisible
by 3, 4, or 6, and if not, it is easy to increase the total turns to fit the
"divisibility" rule.

NOTE. If the calculation of turns per layer in step 28 gives a total number
of S turns which cannot be exactly divided by the wanted numbers of paralleled
windings, it must be increased to comply with this rule.
For this design example, if in step 28, 47 turns were calculated, and pattern
4A is chosen where the secondary = 4 layers of wire, then total S turns
= 4 x 47 = 188t.
188 divided by 3 = 62.66, and divided by 6 = 31.33, both unacceptable because
the division produces a fraction of a turn.
But if the Tpl is increased to 48t, total S turns = 192, and is exactly divisible by
3, 4, and 6.
In fact Tpl from step 28 = 37tpl. Total turns are 148t, and for pattern 4A the
totals could be 156t, 168t, 180t, 192t, 204t, 216t, 228t, 240t and each are
able to be divided by 3, 4 and 6 to give
Tpl of 39t, 42t, 45t, 48t, 51t,
54t, 57t, 60t.

The increase in secondary winding resistance may be marginal and acceptable
if total of P and S winding resistance losses are less than 7%. The most efficient
OPT have nearly equal winding loss % in both P and S windings, and if the P
losses are say 2.2%, then S losses may be allowed to be 4.8% for total of 7%.

NOTE. Consider that you may have selected a 5S + 4P interleaving pattern              
from Table 4, and then chosen sub section arrangement 5A from Fig 9.
There are 4N turns for each of 5 layers giving a total of 20N turns. The windings
are grouped either as 4 paralleled windings with 5N turns or 5 paralleled windings
of 4N turns. The total turns must be divisible by 20. Ie, by 4 and 5.
The available winding height for secondary windings may be 5.4mm, so each
of the 5 layers has theoretically 1.08mm oa dia wire, and we may select
0.95mm Cu dia with oa dia = 1.04mm, giving 60t per layer.
5 Sec layers = 300t, and divisible by 20, ( and are we not lucky? ).
The possible total S turns = 300, 320, 340, 360, giving Tpl of 60, 64, 68, 72.
The Tpl will be then exactly divisible by 4 to give the 5A pattern.
Where Tpl = 60t, there are 4 layers at 60t each and one layer of 4 x 15t.
Thus Ns may be 5 // 60t or 4 // 75t.
Load matches with RLa = 1,213 ohms and 1,586 P turns are :-
60t = 1.74 ohms, 75t = 2.7 ohms.
With Tpl = 72t, the two S loads will be:-
72t = 2.5 ohms, 96t = 3.9 ohms.
This 5A pattern does not include a wide range, so the choice of some other
pattern for 5 Sec layers must be tried.

NOTE. Consider selection of 5B, with windings of 5N, 2.5N and N turns,
with turns in each layer = 6N.
If one layer has 60t, total turns for 5 layers = 300t and it is divisible by 30, 
ie, by 5, 7.5 and 10. The Sec may be :-
6 // 50t = 1.2 ohms, 4 // 75t = 2.7 ohms, 3 // 100t = 4.8 ohms, 2 // 150t = 10.8 ohms.
There is only one load match between 3 ohms and 9 ohms, and more trials
of other patterns should be done, with careful attention to dimensions and
and winding losses.

When considering designs with 5 Sec layers there is more difficulty
finding a useful range of load matches fairly well spaced apart.
Ideal load match spacing would be 3 ohms, 6 ohms, 12 ohms, but for this you
need sec turns with relationship of 1N, 1.414N and 2N turns.
Nice convenient fractions of layers are 1/2, 1/3, 1/4 and 1/5, and combinations
of fractions of layers will never give the 0.414 fraction, but give 0.50N, 0.333N,
0.25N, or 0.20N. So a popular range of relative N might be 1N, 1.5N, 2N, which
would match 3 ohms, 6.75 and 12 ohms which is ideal for nominal speakers
of 4, 8 and 16 ohms.
The aim is to find :-
A. Easy fractions allowing a good spread of loads,
B. E
qual current density in each and every secondary winding wire,
C. Low enough secondary winding resistance.
D. Not too many secondary connections,
E. No winding taps unless the tapped secondary option is initially wanted.

However, Some slight imperfect arrangement with non equal current density
in wires will be tolerated without causing significant extra winding losses or
changed frequency behaviour. For example, in Fig 9, 5E, each layer of 5 is
divided to give N, 2N and 3N sub-sections.
The total sum of N = 5 x ( N + 2N + 3N ) = 30N, and this cannot be exactly
divided by 4. But from the winding arrangement there may be 7 windings of 4N
made up as follows :-
5 x ( N + 3N ), 2 x ( 2N + 2N ), and this leaves a spare 2N section which is
then connected in parallel to one of the other 2N windings. The arrangement
will work OK because there are 4N turns with less Iac than the other 30N turns
and fraction of total N with changed current density = 4N / 30N, or less than
15%. Such calculations of imperfect but tolerable solutions to the Puzzle Of
Turns are probably beyond the capability of a PC which is a rather dumb
brute, and it could never manage a robot to tie my shoelaces without
mincing my foot!

33. Choose sec turns to suit chosen sub-section pattern.

Confirm choice of pattern = 4A.

Confirm range of Sec turns per layer possible between Tpl in Step 28
up to 1.414 x Tpl in Step 28.

OPT4.  Range of tpl = 37 turns to 52 turns.

Calculate range of Total sec turns for all Sec layers.

OPT4.   No of layers = 4, Range of total turns = 148 turns to 208 turns.

Calculate all numbers of turns within the range which may be divisible by the
number given for the chosen pattern.

OPT4.  The pattern 4A shows total turns to be divisible by 12.

Numbers of turns within range divisible by 12 are 156, 168, 180, 192, 204.

List the possible winding arrangements with sec loads with the total turns.

OPT4, Primary RLa = 1,213 ohms, Np = 1,586 turns. 

Total turns :-
156
168
180
192
204
Configuration,

ohms

ohms

ohms

ohms

ohms
4 parallel
39
0.73
42
0.85
45
0.98
48
1.11
51
1.25
3 parallel
52
1.30
56
1.51
60
1.74
64
1.97
68
2.22
2 parallel 
78
2.93
84
3.40
90
3.90
96
4.44
102
5.01
Illegal 
117
6.59
126
7.65
135
8.82
144
9.99
153
11.25
4 series 156 11.72 168 13.60 180 15.60 192 17.76 204 20.04

NOTE. The "Illegal" configuration is where 3 complete secondary layers are
connected in series with one remaining layer strapped across one of the the
other 3 layers. Therefore the current flows in 2 series layers is twice the current
flow in each of the 2 parallel layers. Such a configuration may create unwanted
effects at HF, and slightly increased winding losses.

Are there 3 wanted load matches close to what is listed in Step 30
for 3, 6 and 12 ohms without inclusion of illegal configurations?

OPT4.  There are only 2 matches close to 3 and 12 ohms with 39 turns per layer. 
There is a useful but illegal match close to 6 ohms.

33.  Summary & Conclusion.

How are illegal configurations avoided?

A. Try doubling the number of Sec turns per layer and using two layers
of wire of 1/2 the diameter for each Sec section for sub-section pattern
already chosen.
This involves minor re-design calculations from Step 24, and they
will be shown below as 24A, 25A etc, to a conclusion.

B. Can an interleaving pattern be chosen with more secondary sections
but with higher sec turns per layer? This question is answered after the
re-design process from Step 24A has reached a conclusion below.
The choice of different interleaving pattern involves complete re-design
after Step 20.

Go back to re-calculate from step 24..........

24A. Calculate insulation.........

OPT4.  Confirm choice of Sec sub-section pattern, 4A.

OPT4.  Interleaving Pattern = 2p - 2S - 3p - 2S - 3p - 2S - 3p - 2S - 2p.

This is the same 5P + 4S pattern as chosen before but with 2 layers
of Sec wire
per Sec section.

Insulation :-
p-p 8 x 0.05mm insulation
= 0.40mm.
S-S 4 x 0.05mm insulation = 0.20mm.
8 x 0.50mm insulation = 4.00mm.
Total thickness of all insulation layers = 4.60mm.

25A.  Calculate height of Primary layers plus all insulation.

OPT4.  13 layers of P wire at oa dia of 0.569mm = 7.4mm,
Total height of all insulation = 4.6mm,

Total height of P wire + all insulation = 12.0mm.

26A.  Calculate maximum theoretical oa dia of the
secondary wire.

Calculate available height for layers of secondary wire :-
Available Sec height = ( Available height in bobbin ) - ( Height P + all Insulation ).
Available height in Bobbin = 0.8 x H window dimension.

OPT4. Avail Sec height = ( 0.8 x 25mm ) - 12.0mm = 8.0mm.

Th Sec oa dia = ( avail Sec height ) / no of sec layers of wire,
OPT-1A, Th oa dia sec = 8.0 / 8 = 1.00mm
.

27A.  Find nearest oa dia wire size from wire size tables.
OPT4   Inspect wire size table in step 12.
Try 0.990mm oa dia, which is 0.90 mm Cu dia = 0.99mm oa dia.

28A.   Calculate theoretical S turns per layer, thStpl.

Theoretical S turns per layer, thStpl = Bww / thSoadia from step 27

ThStpl = 72 / 0.99 = 72.7, round down to 72 turns per layer.

NOTE.  These calculated turns per layer are for the thickest wire possible, and
fewer turns per layer are forbidden because the increased wire size to fill a layer
would make the winding height unable to fit onto the bobbin. Wires should never
be wound on and spread apart so that the Tpl is reduced while keeping wire size
the same, lest secondary winding resistance losses be increased too much.

29A. Calculate load matches with Sec turns from Step 28A.

These load matches will be theoretical and secondary turns
and winding pattern will probably require adjustments.

Nominate Ns, number of secondary turns consisting of paralleled or seriesed
sec layers calculated in Step 28A.

Primary RLa 1,213 ohms
Primary turns, Np 1,586t
Ns TR

ZR
Secondary
RL ohms
4 parallel layers 72t 22.02
485 2.50 ohms
2 parallel x ( 2 series layers )
144t
11.01
121
10.00 ohms
4 series layers 288t
5.50
30.25
40.00 ohms

Are there 2 secondary load matches between 3 ohms and 9 ohms?

OPT4. NO, there are NO useful matches between 3 and 9 ohms.

NOTE. All configurations tabled above are "legal" and should give low
total winding loss % and the current density of all wires is equal in
each configuration. Analysis of currents and voltages will proove this.

If NO, secondary turns per layer will need to be revised, and sec layers
will have to divided into sub-windings to allow a greater range of load
matches which are useful.

Are there any **illegal** ways of gaining other load matches between
3 and 9 ohms?

OPT4. NO. but 3 layers may be connected in series with one layer
connected in parallel to any one of the 3 layers, thus giving an
"illegal" match :-

3 series layers, with 2 parallel 216t
TR 7.34
ZR 53.9
22.5 ohms
 
But the "illegal configuration" of S windings is not useful, as it matches
to a load which is far above any load commonly expected.

NOTE. The illegal connection causes twice the current flow in two layers in
series as flows in each winding in parallel. The illegal connection may be
ignored while investigation of other possibilities continues.........
 
NOTE.  So far, the turns available give 3 LEGAL choices for Ns which do
not include illegal configurations. The doubling of sec turns in each Ns choice
gives an awkward and useless fourfold difference between each choice.


The IDEAL load matches would be for 3 ohms, 6 ohms and 12ohms which
would have the amplifier cope well with real world impedances of 4, 8 and
16 ohm speakers.

30A. Calculate Sec turns for matches to 3, 6 and 12 ohms.


Calculate turns for any chosen secondary load.
Ns = Primary turns x square root of [chosen secondary load / RLa].

OPT4,  Np = 1,586 turns, RLa = 1,213 ohms.
For 3 ohms, Ns = 1,586 x sq.rt [3 / 1,213] = 78.8 turns.
For 6 ohms, Ns = 111.5 turns,
For 12 ohms, Ns = 157.6 turns.


31A. Calculate sec winding losses for turns in Step 28A,
Calculate total P and S winding losses.
 

Calculate secondary winding resistance =
Rws = Ns x TL / ( [ No of parallel windings of Ns ] x 44,000 x Cu dia squared. )

OPT4. Rws = 72 x 297 / ( 8 x 44,000 x 0.9 x 0.9 ) = 0.075 ohms.
Load match for Ns = 72 turns = 2.5 ohms.

Calculate Rws % losses = 100 x Rws / ( Rws + sec RL )
= 100 x 0.075 / ( 0.075 + 2.5 ) = 2.91%

From Step 18, primary winding losses = 3.42%

Total P and S winding losses = 3.8% + 2.91% = 6.71%

Is this MORE than 10% above 7.0%? If YES, consider revising design
with higher core stack, lower Ns.
If NO, then Sec losses could be higher if Sec turns per layer were
increased.


OPT4, Total P & S losses for the design so far = 6.25%.
Maximum losses permitted = 7%.
Therefore losses may be increased by the difference =
Maximum permissible loss - total loss calculated so far = 7.0 - 6.33 = 0.67%.

Therefore sec winding resistance losses may be increased from from 2.91%
to ( 2.91 + 0.67 ) % = 3.58%.

NOTE.   Where the Sec loss % is less than 3.5%, the secondary turns per layer
could be increased, and another tedious formula could be developed, but it is
better to choose a range of higher sec turns per layer up to 1.4 times those so
far calculated.

OPT4. Sec turns per layer from Step 28A = 72 turns.

Maximum possible Sec tpl = 1.4 x 72 = 101 turns.


32A. Choose the same sub-section pattern as for Step 32.
Fig 12.
bobbin-details-basic2-SE-OPT4-cfb.GIF

OPT4.  Sub section pattern is 4A, shown above in Fig 12.

Prepare a table of possible winding winding arrangements with all complete Sec
sections and and subdivided sections having identical double layer windings.
Total turns will be number of Sec sections x double layer windings.

OPT4, turns per double layer = 72t, Total turns = 288t. Total turns are
divisible by 12 exactly.

Primary RLa 1,213 ohms
Primary turns, Np 1,586t
Ns TR

ZR
Secondary
RL ohms
4 parallel sections, 72t 22.02
485 2.50 ohms
3 parallel sections, ( 72t + 24t series )each
96t
16.52
272
4.44 ohms
2 parallel sections, ( 72t + 72t series )each 144
11.01
121.2
10.00 ohms

Are there 2 secondary load matches between 3 ohms and 9 ohms?

OPT4. NO. But there is ONE match between 3 and 9 ohms.

Is it possible to vary the turns per layer to obtain load matches
closer to the ideal stated in Step 30A?

OPT4.  Np = 1,586 turns, RLa = 1,213 ohms. Ideal matches are

For 3 ohms, Ns = 1,586 x sq.rt [3 / 1,213] = 78.8 turns.
For 6 ohms, Ns = 111.5 turns,
For 12 ohms, Ns = 157.6 turns.

Consider using 84turns per layer. Total turns = 336t and divisible by 12.

With Sub-section Pattern 4A, matches for the turns will be 

Primary RLa 1,213 ohms
Primary turns, Np 1,586t
Ns TR

ZR
Secondary
RL ohms
4 parallel sections, 84t 18.88 356 3.40 ohms
3 parallel sections, ( 84t + 28t series )each
112t
14.16
200
6.05 ohms
2 parallel sections, ( 84t + 84t series )each 168t
9.44
89.1
13.60 ohms

Are there 2 secondary load matches between
3 ohms and 9 ohms?

INDEED THERE ARE.


Calculate wire size for double layers of 84t per Sec section.
OPT4.  Theoretical wire oa dia = Bww / tpl = 72 / 84 = 0.857mm.

From wire table, choose nearest wire size less than th oa dia,

Try 0.832 oa dia wire for 0.71mm Cu dia.

Calculate secondary winding resistance,

OPT4. Rws with Cu dia = 0.71, Sec RL = 3.4 ohms,

Rws =
84 x 297 / ( 8 x 44,000 x 0.71x 0.71) = 0.14 ohms.
Load match for Ns = 84 turns = 3.4 ohms.

Calculate Rws % losses = 100 x Rws / ( Rws + sec RL )
= 100 x 0.14 / ( 0.14 + 3.4 ) = 3.95%

From Step 18, primary winding losses = 3.8%

Total P and S winding losses = 3.8% + 3.95% = 7.75%


Is this MORE than 10% above 7.0%? If YES, consider revising design
with higher core stack, lower Ns.

OPT4, Total P & S losses for the design so far = 7.75%.
Maximum losses permitted = 7.7%.
Therefore losses may not be increased.

If slightly more than 7% winding losses are acceptable, this design
is complete and a diagram drawn for the winding tradesman or
tradeswoman to follow in their workshop.

NOTE.
Winding losses can be confusing. In all my examples the OPT is
assumed to have zero resistance, and the output power at the anode load
of RLa = OPT ZR x secondary load. One could include winding resistance
in all calculations but it would not make much difference to the final design.
But it would complicate design formulas so much that everyone using my
logic flow would be driven insane within a week.

So far, OPT4 is to have RLa = 1,213 ohms. The winding loss total = 7.8%
of 1,213 ohms = 94 ohms. So the REAL load will be 1,307 ohms. But if the
RLa was reduced to 1,119 ohms, by means of reducing the speaker load by 7.8%,
then the anode load would remain at 1,213 ohms. 44 Watts may be generated
at the anodes, which is 231Vrms and 190mArms, and thus 213Vrms is effectively
working across the 1,119 ohms, and output power at the secondary must be
40.4 Watts. The initial design called for 35 Watts and this aim has been
realized.

32A. Summary & Conclusion.
The design of the OPT4 is now acceptable.

Answer to question in B above for more possibilities for OPT4........
Is it possible to choose a different interleaving pattern with a higher number of
secondary sections with more turns per layer than calculated in Step 28?
Yes, and it would be possible to use a
6S + 5P interleaving pattern from
Fig 6 for OPT from 100W to 250W.
For OPT4, the primary winding turns and wire size MUST stay the same for
the core size, maximum audio power, and RLa for the tubes chosen.

The Sec turns per single layer of wire would have to be the same as calculated
in this Step 32A above as 84 turns and when all 6 single layer sections are
paralleled the secondary winding loss increases from 3.95% to 5.26%, and
Total winding losses = 8.69%, rather too high. If 6 sections of secondary are
chosen instead of 4 sections, the number of insulation layers between P and S
sections increases from 8 to 10 and shunt capacitances begins to be a problem.


Therefore I see little merit in trying to use more interleaving than I have
but anyone is welcome to explore further by recalculating the whole
design with a core size of say 75mm high which would reduce Np to
1,226 turns allowing only 10 layers of primary winding wire of the
same size as now calculated.


33A. Check total height of all bobbin contents.

Primary wire 13 layers x 0.569mm = 7.397mm.
Secondary wire 8 layers 0.832 = ......6.656mm.
Insulation, 8 x 0.5mm = ......................4.0mm.
Insulation, 12 x 0.05mm = ..................0.6mm.
Insulation wound over final layer = ......0.5mm.
Total of Bobbin contents = ............19.153mm.

Bobbin base thickness, .......................1.5mm.

Total content in winding window = 20.653mm.

Winding window Height = 25mm, so all windings will fit in window OK.

34. Bobbin winding details.

Fig 13.
bobbin-details-SE-1k2-44W-OPT4.GIF
--------------------------------------------------------------------------------------------------------------------------------------
35. Tapped Secondary windings.
Fig 14.
bobbin-details-Tapped-Sec-SE-1k2-44W-OPT4.GIF

Fig 14 shows OPT4TS with Tapped Secondaries, for those who like to have
4 output terminals labelled Com, 4, 8 16, to suit 3 ranges of speaker impedances
without soldering any connections anywhere, and to suit real speaker load ranges
of 2.5 -- 5 ohms, 5 -- 10 ohms and over 10 ohms.

There are some technical characteristics to be considered.

Winding losses will vary for each of the 3 outlets. The only load which has
identical winding losses to the Wasteless Secondary method described above
is the 13.6 ohm outlet where ALL Sec turns are used in all layers to make
4 parallel windings of 168 turns.

Calculate 13.6 ohm Sec winding resistance, 4//168 turns.
Rws = 168 x 297 / ( 4 x 44,000 x 0.71x 0.71) = 0.56 ohms.
13.6 ohm Sec losses = 100 x 0.56 / ( 13.6 + 0.56 ) = 3.95%.
Total P & S losses = 3.8% + 3.95% = 7.75%.

Calculate 6.8 ohm Sec winding resistance, 4//( 84t+35t ).
Rws = 119 x 297 / ( 4 x 44,000 x 0.71x 0.71) = 0.40 ohms.
6.8 ohm Sec losses = 100 x 0.40 / ( 6.8 + 0.40 ) = 5.56%.
Total P & S losses = 3.8% + 5.56% = 9.36%.

Calculate 3.4 ohm Sec winding resistance, 4//84t.
Rws = 84 x 297 / ( 4 x 44,000 x 0.71x 0.71) = 0.28 ohms.
6.8 ohm Sec losses = 100 x 0.28 / ( 3.4 + 0.28 ) = 7.6%.
Total P & S losses = 3.8% + 7.6% = 11.4%.

With Secondary Taps there is a load match for 1.16 ohms between the 13.6
ohm and 6.8 ohm outlets for 4//49t.

Calculate 1.16 ohm Sec winding resistance, 4//49t.
Rws = 49 x 297 / ( 4 x 44,000 x 0.71x 0.71) = 0.16 ohms.
1.16 ohm Sec losses = 100 x 0.16 / ( 1.16 + 0.16 ) = 12.1%.
Total P & S losses = 3.8% + 12.1% = 15.8%.


Tapped Secondary Conclusions.
A.   Tapped Secondaries waste more output power than Wasteless Secondaries
because not all secondary wire is used.
B.   In OPT4 is is co-incidental that the same secondary winding wire and layers
for Wasteless Secondaries may be used for Tapped Secondaries.
C.   OPT4 will probably provide the same sound quality with either Tapped Secs
or Wasteless Secs, and the problems of amplifier stabilization at HF will be similar.
D.   Despite having the luxury of a nicely labelled outlet which says "4 ohm",
Some Hi-Fi enthusiasts will probably make a silly mistake and plug a 4 ohm
speaker into an amp outlet labelled "16 ohms" where there is a choice between
4, 8 or 16 ohms with Tapped Secondaries. Don't blame me for the smoke!
C.   With Wasteless Secs, a similar mistake could be made, and the amp owner
may be reluctance to change the impedance setting to suit a pair of speakers.
Laziness can also cause smoke.
D.   If an owner was likely to never want to change the outlet settings then the
Secondary need only be set up for ONE load match for between 5 and 6 ohms.
This will allow use of any speaker of nominal Z of 4 ohms or more, but the
maximum power available is reduced. If only low average power of 1 Watt
is needed then the single load match is OK for most listeners with any kind
of speakers above 3 ohms.

36.  Calculate leakage inductance.
Leakage inductance is considered to be an equivalent quantity of inductance in
series with the primary load RLa looking into the anode end of the primary,
with the other end grounded via large value capacitance to 0V.

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,
1,000,000,000 is a constant for all equations,
a = height of the finished winding in the bobbin from bottom of first
winding to top of last winding,
b = the traverse width of the winding across the bobbin.

Distances are all in mm!

OPT4, 
LL  =  0.417 x 1,586 x 1,586 x 297 [ ( 2 x 8 x 0.6 ) + 18.7 ]  
                           1,000,000,000 x 8 x 8 x 72  

      =  0.00191Henry = 1.91mH.

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

ZLL at 100kHz
= LL in Henrys x 2 x pye x F = LL x 6.28 x 100,000Hz,

OPT4.  ZLL = 0.00191 x 6.28 x 100,000 = 1,201 ohms at 100 kHz

Is ZLL less than PRL at 100 kHz?

OPT4, We have RLa = 1,213 ohms, ZLL at 100 kHz is less, OK.

38. Shunt capacitance and Primary Inductance of SE OPT.

There are several areas in an audio transformer where capacitance exists, and
with any OPT the measured capacitance between anode terminals of the OPT
and 0V are of main interest. The measurement of the capacitance may be done
with measurement of Primary inductance.

To measure the shunt C between anode and 0V and away from the amplifier,
and measure primary inductance set up the OPT as follows :-

Fig 15.
schem-opt-test-Csh-Lp.GIF

Fig 15 shows a signal source with a known low source resistance of less than
600 ohms and capable of a known flat F response between 10Hz and at least 50kHz
and capable of up to 25Vrms output. Such a source could be an audio amp fed
by a signal generator with well calibrated frequency adjustment and flat response
ranges fron say 2Hz to 250kHz.

There must be a series resistance between the signal source and OPT which
between 5 and 10 times the Primary RLa value for the OPT. Above I show
5k0, which includes the amplifier source resistance. The highest possible voltage
from the source should be initially be used as long as the signal THD < 1%.

Signal voltage measurements may be most reliably done by setting the oscilloscope
trace to occupy exactly 1/2 the screen height as shown and at 1kHz to get what
is known as the 0dB reference Vo level. Once this is done, frequencies on the sig
gene should be varied each side of 1kHz from about 300Hz and 3kHz to ensure the
1kHz level is indeed in the middle of the expected bandwidth to be observed.

In the case of OPT4, I have assumed the primary inductance = 9Henrys, and
shunt capacitance = 1,840pF.

Fig 16.
graph-OPT4-Rsource5k.GIF

Fig 16 shows a carefully drawn graph of the response you should be able to plot
on a graph sheet. There are TWO POLES of interest, LF and HF.

To plot the LF pole, the frequency is lowered until the oscilloscope level falls by a
factor of 0.707. At this level, the signal response = -3dB. It is wise to measure the
Vac with a reliable meter with wide bandwidth. Digital meters are accurate below
1kHz, but are highly inaccurate above 1kHz so a well constructed voltmeter is
essential for HF measurements.
However, the oscilloscope will be found to have a flat response between 2Hz
and say 20MHz, so once you have set the LF level on the screen then you will
easily measure the relative levels at HF.
The oscilloscope screen may not have a graticule line at the 0.707 point below
half way up the screen. If not, place masking tape on the side of the
screen and plot points where the reference 0dB signal appears, and -3dB.
If the 0dB level = 25Vrms, then the -3dB level = 17.7Vrms, and it may be
set using a voltmeter. -6dB = 12.5Vrms, and -12dB = 6.25Vrms
and -18dB = 3.12Vrms.
As the frequency is reduced to -3db, -6dB, -12dB and -18dB levels, the
frequencies are recorded or plotted directly onto a graph sheet copy as dots.
The dots are joined by drawing a curved line through all dots, and you should
see a response similar to what I have drawn. On the graph, I have also plotted
the boundary lines for the curve. There is a thinner and sloped straight line drawn
from the LF pole F on the 0dB line and sloping down at 6dB/octave. The actual
response curve should begin to follow this straight line if the inductance is linear
with applied voltage. But iron cored inductance windings usually have less
with lesser applied voltage and more inductance with lower frequency, so do not
expect the idealistic curve I have shown below the -6dB graph point.

The HF response is similarly plotted at points between 0dB and -18dB and the
slope of the curve may not follow the straight line boundary of -6dB per octave
exactly, because the leakage inductance may have some effects at HF.

Where LL is found to be high, then the R-source resistance may be increased
so that the series LL reactance becomes a negligible quantity compared to the
source resistance, and the shunt C should then enable a more reliable HF
curve to be drawn to plot the important -3dB frequency.

From the above graph we can say the LF pole is at 88Hz.
To calculate Lp,
LP = R / (F x 6.28) = 5,000 ( 88 x 6.28 ) = 9.047H.

The inductance measurement with SE OPTs may be unreliable because
the presence of DC flow and partial magnetization of the iron core
will reduce the effective iron permeability to an unknowable extent.
Therefore the most accurate Lp measurement is gained with the OPT
within an amplifier, and this is described here later.   

From the above graph we can say the HF pole is at 17.5kHz.
To calculate Cshunt,

Csh = 159,000 / ( F x R ) = 159,000 / ( 17,500 x 5,000 ) = 0.001817uF.
This is a reliable measurement and unaffected by DC flow in the windings.

The shunt capacitance of the oscilloscope and probe cable probe must be
subtracted from the Cshunt measurement. If the probe and cable has C =
20pF, it may be ignored.

To calculate the primary shunt capacitance.
In the OPT4 example, refer to the bobbin winding layout. Self capacitance
of the primary winding may be neglected as it is a negligible compared to
the main C between adjacent P to S winding layer surfaces.

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.


The P and S wire layers act as metal plate capacitors.

In OPT4, the average distance between the copper surfaces of primary and
secondary layers including the enamel insulation thickness of 0.1mm and the
P-S insulation of 0.5mm is approximately 0.65mm.including for the curved
surface of the wire.

The average turn length around the wound bobbin = 297mm, and traverse
width = 72mm, so the Area of capacitance = 297 x 72 square mm = 21,384
sq.mm. The K for polyester insulation is approx 2.5.  d = 0.65mm.

C in pF = 21,384 x 2.5 / ( 113.1 x 0.65 ) = 727pF.

The amount of capacitance in each P to S interface is proportional to turn
length so the P-S C nearest the core is slightly less than at the top of the
bobbin. For OPT4, minimum C = 616pF and maximum C = 763pF.  

To get lowest Csh, the anode should be connected to the innermost
primary end.

But for a simple calculation of the Total effective Csh at the anode,
consider every P to S interface has the average C value = 727pF.
There are 8 P-S interfaces so summed 8 x C = 5.8nF. ( 0.0058uF ).

The actual measured total Csh between anode and 0V will be less.

The 8 capacitances act as though there was a simple winding with 8 tap
points and from each one there is 727pF to 0V. One end of the winding
is at 0V ac. The other end at anode swings the anode Vac, and away
from the anode at all the tap points the Vac becomes lower so current
flow in each C becomes lower and the C loading is thus reduced by
the turn ratio difference. The OPT4 Csh is maximum when the OPT
is set up for pentode, UL, triode and without CFB. The 8 capacitances
effective value at the cathode may be tabled with the positions of the
C in terms of the winding layers :-

P-S Csh at
each interface, pF
winding
position
Fraction
Fraction
Squared
Factor
Effective C
at anode, pF
727
11.5 / 13
0.88
0.78
567
727
10.5 / 13
0.81
0.65
472
727 8.5 / 13
0.65
0.42
305
727 7.5 / 13
0.57
0.33
240
727 5.5 / 13
0.42
0.18
130
727 4.5 / 13
0.35
0.12
87
727 2.5 / 13
0.19
0.04
29
727 1.5 / 13
0.11
0.01
9
Totals sum 5,816



1,839

The Capacitance table for OPT4 shows the positions of each C along the
primary winding and is expressed in layers-along / total P layers, and the
fractions are listed. Each fraction is squared because the impedance ratio
is the relative turn ratio squared, thus each C has a Fraction Squared
Factor number which is multiplied by the actual C to give the amount
of C to which it is transformed at the anode connection.
The Total C at the anode is simply the sum of all 8 transformed
values of each 727pF.

So Total C at anode = 1,839 pF. In practice it may be about 1,750pF
due to shorter turn length near the anode connection.

But let us allow for the worst amount of Csh, and let C = 1,839pF,
( 0.00184uF. )

Cathode Feedback windings
CFB windings complicates the capacitance calculation.
The effect of the capacitance on the output stage bandwidth is effectively
reduced by the CFB because the CFB effectively reduces the Ra of the
tubes, and there are only 6 x C in the anode circuit and 2 x C are in the
low impedance cathode circuit. The CFB acts like all NFB and at some
HF probably above 80kHz there can be oscillations due to phase shifts.
Such oscillations may be damped with a Zobel R&C network between
anode and cathode, or across the anode portion of primary, or wherever
it is found to be most effective without adversely loading the output stage
at HF. 

If the OPT4 example is used with 4 x EL34 in pure pentode mode then
the Ra may be about 3,000 ohms total. Without any resistance load
connected, the gain of the tubes will be close to µ at about 1kHz, ie,
about 125x. The gain will reduce -3dB where the capacitive reactance
= Ra. With Ra = 3,000 ohms and C = 1,839pF, The frequency for -3dB
cut off = 159,000 / ( 3,000ohms x 0.001839uF ) = 28.82kHz.
In practice this would be only approximately correct because there is
leakage inductance involved which forms an undamped resonant and
usually causes a peak in the response which may be calculated,
Fo = 5,035 / square root of ( L x C ), L in mH, C in uF.
In this case,
Fo = 5,035 / sq.rt( 1.91 x 0.00184 ) = 84.9kHz.

Since this F is so much higher than the HF cut off calculated at 28.82kHz,
then the effect of LL with pentode mode is not much as 28.82kHz, and
the high Ra damps the peak at 84.9kHz where XC = XLL = 1,017 ohms.
If the EL34 are connected for 61% SEUL, then Ra is about 800 ohms and the
HF cut off will theoretically be at 108kHz, but then the effect of LL and Csh
resonance will be seen as a peak at around 85kHz with rapid phase shift above
110kHz.
Usually the resonance between leakage inductance and capacitances will not
cause any problems with HF stability once the correct critical damping networks
are in place within the completed amplifier.

Trying to establish an exact equivalent LCR model of the OPT designed here
is beyond my abilities and there is little point to achieve such modelling, and
even if all the equivalent C and L amounts were shown for each layer of wire
then nobody would understand the complexity. 
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 resistive R load.

If the rules for P to S insulation thicknesses have been adhered to, ie, P-S
insulation is more than 0.5mm and the insulation material dielectric constant
is less than 2.5, the shunt capacitance should not cause undue HF attenuation
for an ideal number of interleavings.

Although it may be possible to reduce LL to half the value calculated in Step
35, the number of required interleavings becomes excessive because there
would be more P-S interfaces and C would increase too much, quite negating
the effect of reducing the LL.

39. Calculate wanted effective permeability, µe.

NOTE.  The core of the SE OPT will have an air gap which will need to greatly
reduce the maximum permeability, µ, when the laminations are fully interleaved.
This prevents the core from becoming magnetically saturated by the dc current.
The primary inductance is proportional to the effective permeability, µe, the
permeability with a gapped core. The permeability varies slightly with frequency
and Vac and DC magnetization, but the DC magnetization may be considered to
be half the max B allowed for the core material.

The Bac max and Bdc max may each be equal and each be less than half the 
allowed total B max for the core material.
For GOSS, total Bmax = 1.5Tesla, for NOSS it may be 1.3Tesla, and for
some other lower grade iron it may be 1.1Tesla.

If the SE OPT is wisely designed the iron distortions will be much less than
produced by the vacuum tubes where total B < 1Tesla, and Lp reactance
> 2 x RLa. The distortion becomes worst at very low F and when
operating near the saturation point of the core which is at just less than total
of max Bac + Bdc. 

OPT4. The core material will be GOSS with allowed Bmax = 1.5Tesla.
The Bdc max = 0.75T and Bac = 0.75T.

SE OPT Primary inductance depends largely on core µe with an air gap.

µe = Bdc x mL x 10,000
        12.6 x  x Np x Idc
            

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

OPT4, Np = 1,586t, Idc = 0.272Adc, Iron mL = 280, Bdc = 0.75T. 

µe  =   0.75 x 280 x 10,000
          12.6 x 1,586 x 0.272
 = 386.

Is the µe < 750? If so, proceed, but if µe > 750, it may not be possible to adjust
the air gap to a small enough value to increase the µe to the wanted amount.
To reduce µe to a commonly seen value between 300 and 500, more primary
turns might be used.

40. Calculate primary inductance.

Lp  =  1.26 x Np squared x S x T x µe
              1,000,000,000 x mL
Where  µe is effective permeability of core with an air gap, and is just a number,
no units, ie, the number of times the the Lp is increased by the presence of iron
over having no iron present.
1,000,000,000 and 1.26 are constants for all equations to work,
Lp = primary inductance,
mL = magnetic path length of the iron only, and for wasteless pattern E&I lams =
2 x ( L + H ) + ( 3.14 x H ) where L & H are window dimensions, 3.14 = pye, 22/7.
S = stack height,
T = tongue width.
All dimensions in mm !!!

OPT4,  Np = 1,586t, S = 58mm, T = 51mm, µe = 386, mL = 280mm.

OPT4, Lp =  1.26 x 1,586 x 1,586 x 51 x 58 x 386
                              1,000,000,000 x 280      
Lp = 12.92H

41. Calculate Frequency where XLp = RLa, or Fco.

NOTE. Fco is considered as cut off frequency below which it is impossible
to maintain the Vo at 0.0dB level without distortion rapidly increasing
due to Lp inductance shunting RLa thus reducing total RLa to 0.707 x RLa.

Fco = RLa / ( 6.28 x Lp )
Where Lp = Henrys, 6.28 = a constant of 2pye for all equations, F = frequency.

OPT4, Fco = 1,213 / ( 6.28 x 12.9 ) Hz = 14.97Hz.

Is Fco < 20Hz? if yes, then proceed, if no, perhaps the stack height might
be increased.

42. Calculate Fsat.

Fsat will occur when Bac is at maximum allowed.
For SE OPT, Maximum Bac = Bdc = 1/2 total allowed B for material

Fsat  =   22.6 x Va x 10,000  
           T x S x Np x Bac max

Where B is in Tesla for ac signals,
22.6 and 10,000 are constants for all transformer equations,
V = Vrms signal voltage across the primary,
S = core stack height,
T = core tongue width,
Np = primary turns,
F = frequency at which B is to be measured.

All dimensions in mm!!

OPT4. Bac = 1.5T / 2 = 0.75T.
Fsat  =   22.6 x 233 x 10,000  
            51 x 58 x 1,586 x 0.75


Fsat = 14.96Hz.

43. Calculate the air gap to suit calculated µe.

Air gap = mL x ( µ - µe )
                   µ x µe
Where gap is the air gap distance placed into the iron magnetic circuit,
mL = the iron magnetic path length,
µ = iron permeability maximum with laminations maximally interleaved,
or in the case of C-cores, with polished cuts tight together.
µe = effective permeability with air gap.
all dimensions in mm !!!

OPT4, mL = 280mm, assume µ = 5,000 with no air gap for the sample
of GOSS E&I lams, µe is the wanted µe with air gap = 386.

Air gap =  280 x ( 5,000 - 386 ) / ( 5,000 x 386 ) = 0.67mm.

NOTE.  The air gap is the total gap. In an E&I core or pair of double
C-cores there are TWO gaps inserted in the magnetic path around each
rectangular window of two which form the core. Therefore the thickness
of the plastic gapping material to make the gap will be HALF the above
calculated gap size.

NOTE.  See below for practical method for measuring Lp to confirm
that primary inductance inductance is sufficient
, and checking that
the air gap is correct and how to adjust the gap size.

Fig 17.
air-gapped-core-permearbility.GIF

Fig 17 above shows the variation of air gap size to give the same µe for
values of maximum possible µ between 1,000 and 40,000. For material
with mL = 280mm, a variation of µ between 2,500 to 10,000 will result
in an air gap difference between 0.56mm and 0.65mm. This range of µ
covers the range of most E&I laminations between low grade NOSS
and high grade GOSS.


44. Notes about testing OPT4 for 4 x EL34.

From Step 31A, P:S TR = 1,586t : 96t

ZR = 1,213 ohms : 6.04 ohms.

The 4 x EL34 should produce 233Vrms between anode and cathode, and
thus give over 44 Watts into RLa = 1,213 ohms, with Sec load = 6 ohms,
Vo sec = 14.1Vrms at 1kHz.
The 0.0dB Vo reference level may be taken as 14.0Vrms when loaded
by 6 ohms or without any load. The air gap should be set at the calculated
value from Step 42 above.
5 Watt x 10 ohm current sensing resistances are used to confirm AC and
DC currents between anodes and OPT anode winding and between bottom
of cathode windings and 0V, if there is a CFB winding.

Using the same input level signal needed to generate the 0dB Vo level
the amp should be tested at all frequencies between 2Hz and 1MHz, and
the response plotted
with and without any RL, and both with and without
the loop of GNFB connected. The LF response need only be plotted down
to where THD begins to exceed 5% due to core saturation or loading effects
caused by low Lp reactance. Graphs should be drawn for the sine wave
response for Vo levels of -6dB and -12dB.
Any peaks in the sine wave response or instability tendencies should be
countered by shelving networks to reduce gain and phase shift at the
extreme ends of the bandwidth between DC and
20Hz and over 15kHz. The onset of core saturation is most easily seen
on the oscilloscope when no load is connected. Lp should be measured
at Fsat for 0.0dB, but the measurements should be done at the -6dB level
to prevent distortion causing Vrms measurement errors.
Amp tests at very low F where Fsat is occurring or where low Lp reactance
exists should be kept brief lest tubes overheat. 

The XLp reactance is calculated as Va-k / Ia, ohms.

Lp = XLp / ( 6.28 x F ) Henrys.


OPT4. Summarizing the calculations above,
Step 40, Lp calculated at 12.9Hz.
Step 41, Fco where XLp = RLa-a = 14.97Hz,
Step 42, Fsat =14.96 Hz at Vo = 0.0dB level = 233Vrms, with no RL connected.

Before measuring Lp the amp must be tested at 0dB level without any RL
to measure Fsat. The actual Fsat may be above of below what was calculated
in Step 42. Let us assume for example OPT4 that Fsat = 14.96Hz, or say 15Hz
+/- 1Hz. Then it should be found that with level set at -6dB there will be very
low Vo and Ia distortion and thus measurements of Lp may be undertaken.
Let us assume Lp = 12.9Henrys.
Then when measuring Vac the Va = 116Vrms, and the voltage across the 10 ohms
in anode circuit should be 0.957Vrms. The signal Ia = 0.0957Amps rms.
ZLp = 116V / 0.0957A = 1,112 ohms.
Lp = 1,112 / ( 6.28 x 15Hz ) = 12.87H.

If the Fco and Fsat are both measured to be close to 14Hz then the air gap may be
assumed to be optimized.

If the air gap is too small, Lp will be higher and Fco will lower than 14Hz.
Fsat at 0dB will be above 14Hz. 

If the air gap is too large, Lp will be lower and Fco will be higher than 14Hz
Fsat at 0dB will be below 14Hz, although when measuring the lower Fsat
the lower Lp reactance will be a low load on the tube thus causing higher THD
due to loading before the Fsat appears.

CONCLUSION. If the Fco and Fsat are both measured to be nearly equal
and
both are below 20Hz, the design should be fairly well optimized for
hi-fi
applications.

The air gap can be set at first by using the calculated gap using sheets of paper
placed on both sides of the OPT core so that if the air gap was calculated at
0.7mm, there will be 0.35mm on each side of the core. The actual gap need
not actually be of air, but consist of sheets of paper or other non-magnetic
and non-metal material.

One sheet of notebook paper may be about 0.05 to 0.07mm thick.
To determine the paper thickness, measure the thickness of 100 sheets of
a notebook not including its cardboard covers and divide the total thickness
by the number of sheets.

It is most important to know what size of gap you are trying set.

The rectangular yokes to hold E and I together should be in place with retaining
bolts but not fully tightened. Yokes for E&I laminations on SE OPT should be
made from aluminium so the magnetic flux acting across the gap does not suffer
interference.

When the air gap is optimized for low Fco and low Fsat, the total thickness of
temporary gap material is measured and replaced with what can be the same
thickness but permanent. Before tightening the bolts and yokes, it is good practice
to apply slow setting epoxy resin such as Araldite to all surfaces of materials
within the gap. When yoke bolts for E&I laminations are tightened the piles of
Es and Is should be clamped together with an external G-clamp and timber
blocks to ensure the gap size does not increase during this mechanical assembly
work.
After the epoxy has cured for a couple of days, the G-clamps are removed
any gaps between bobbin and core material should be shimmed tight with plastic
sheet off cuts so that the wound core cannot move on the core material.
The transformer terminal board must be prepared to allow the fully assembled
OPT to be placed into its prepared mild steel sheet metal container, or "pot".

There should be at least 5mm clearance between the pot surfaces to core material
or to any windings. When all is ready, the OPT may be surrounded with a potting
compound. The best potting compounds are very expensive but are very effective
to stop noise and keep moisture out while seeping into every void before curing to
their rubbery nature. DIYers may find a 50-50 volume mix of fine dry clean sand
and fibre glass resin mix will make a good concrete mix for potting. 

In the distant past before modern chemistry brought the best potting mixes to the
world, transformers were potted in molten tar which may have a high melting
point between 35C and 70C. For most people it is very difficult, messy, and
smoky and toxic to use because it tends to harden as soon as it cools against
core material, thus not being able to penetrate small gaps. Road tar may become
non solid at only 35C, and may run out of a pot if ever the OPT becomes
overheated. Cheap candle wax has melting point too low and shrinks as it cools.
Beeswax is better with MP of 62C, and probably micro-crystalline petroleum
wax is better, if ever one might buy a small quantity. Indramic 82-T is a brand
name which may be used, MP = 76C minimum. Canauba wax might also be
considered which has a MP = 82C, but I found it shrinks when it cools and thus
the transformer will become loose within its pot after such wax has solidified.
A 5050 mix of canauba + bees wax plus dry clean sand may work well to make
a wax concrete which will not shrink much. The pot and transformer may be
preheated using an old electric frypan to slow the solidification of wax concrete
and allow settlement of the sand so it occupies as much volume as possible while
surrounded by wax. 
Pot shapes maybe tailored to the shape of the transformer such as shown here :-

Fig 18.
OPT-sample-potted.jpg

Fig 18 shows a small sample OPT with C-cores and with flying leads.
Clamps using  3mm thick Al plates and 4mm threaded rod are used.
The pot has been home made using 0.6mm steel color bond sheet
found in a dumper bin at a building site. Pop rivets hold the "lid" on,
and some silicone is wiped across gaps inside the pot so it seals the pot.
With the pot upside down, the transformer is placed inside with a small
wood block under the Al plates to get some clearance. The volume of
epoxy mixed plus sand is calculated roughly, and less than required
is used initially, and poured into the pot and allowed to cure. A
amount of potting mix and sand may then be mixed to top up with
less danger of wasting valuable potting mix.
After several attempts and mistakes with OPTs and potting,
one eventually learns to do it right.
The pot made here is roughly equal in size to a cube with 110mm sides.
I used scrap timber to make a 300mm length of wood 110mm x 110mm
and then planed it down to give the octagonal shape to hug the sides of
the OPT. I cut a length of 110mm wide sheet metal and bent it neatly
around the octagonal wooden "mandrel" held in a vice.
The metal bent well at 45 degrees, and was clamped as I proceeded.
With finished sides, the mandrel section was scribed onto the sheet metal,
and thickness of bent sides allowed for so that the lid could be formed
with its turned down lip of 15mm all around. The 8 straight lips of the
lid were bent with wide nose pliers and all adjusted and done so the
appearance is as neat as seen here.
When potting mix has all be poured in and cured for a day, the gaps
around the lid are filled with epoxy panel beater putty, and when all is
hard the whole lot carefully sanded down in a belt sander. Matt paint
may be rolled on with a very small roller and later given a coat of semi
gloss polyurethane and although curing is slow, the finish will end up
hard enough for home use amps, and look well.

The potted transformer is held to the chassis by the 4 rods through the
chassis. Leads should be prepared and strapped into their wanted
position before potting so leads come to wanted positions under the
chassis.

The method of confirming the gap size is one requiring natural skills in
perceiving what is really happening in a given amplifier circuit expressed
in disciplined engineering terms. Many DIY enthusiasts make terrible
mistakes with SE OPTs. Bass performance is poor, and silly reputations
are given to SE tube amps without justification such as "300B have no
top end or bass; just beautiful midrange." This is a complete myth
perpetuated by people who don't understand what they are doing
and don't have any ideas about how to make OPTs or amplifiers.
There is no reason why a 300B cannot give excellent performance
between 20Hz and 20kHz, but remember, you only get 8 Watts
with one 300B.


Many amplifier builders who may not have much experience at designing
or building tube amplifiers may find the next section of interest.
This next section deals with examination of an amp with a single EL34
which should be built before trying to copy or design and build something such
as the SE35 amp shown at http://www.turneraudio.com.au/se35cfb-monobloc.htm

Back to SE OPT calc Page 1.

Forward to SE OPT calc Page 3.