SE OPT CALCULATIONS PAGE 1.

FOR  SINGLE ENDED AMPLIFIERS WITH
DC FLOW IN OPT
IN ONE DIRECTION ONLY.

For calculations for Push Pull Output Transformers go to
'PP OPT calculations'


My 2006 SE OPT calculation pages have been revised during 2011.
There are 3 separate pages to enable easier browsing.
I have used SeaMonkey, WYSIWYG for editing website pages.


You are at Page 1,

Content List of the 3 pages......

SE OPT Calcs, Page 1.

RDH4, SE amp history, trends and preferences.
Table 1.  SE Pentodes, Beam Tetrodes, CFB, UL,
For operating conditions of common tubes.
Table 2.  SE Triodes.
For operating conditions of common tubes.
Interpreting Table 1 and Table 2.

SINGLE ENDED OPT4 DESIGN EXAMPLE.
1.  DEFINE THE AIM OF THIS PROJECT.
     Fig 1. Graph for EL34 pentode PO vs RLa.
     HOW TO DRAW THE PO vs RLa GRAPH FOR ANY TUBE.
     Calculate Center Value RLa and PO for all RLa.
     A for pentodes and beam tetrodes only,
     B for triodes only. 
2.  LOADLINE ANALYSIS FOR ONE SE PENTODE.
     Fig 2. Graph Loadlines for EL34 pentode. How to draw loadlines,
     Steps A, B, C, D.
     LOADLINE ANALYSIS CONCLUSION.
3.  Calculate Center Value RLa for OPT4, the design project, and confirm
     maximum PO and Va max.
4.  Calculate theoretical core cross sectional area, Afe,
     for a nearly square core centre leg cross section.
5.  Calculate the core tongue dimension, T.
6.  Calculate theoretical stack height for chosen T
     from step 5.
7.  Calculate usable Afe = chosen T x chosen S.
    
Table of possible T and S for output powers. 
8.  Confirm L and H of the winding window.
9.  Calculate theoretical primary winding turns, thNp.
10.  Calculate theoretical Primary wire dia, thPdia.
11.  Find nearest suitable theoretical oa wire size for Primary.
       Wire size table for Grade 2 winding wire, metric sizes.
12.  Calculate the bobbin winding traverse width, Bww.
13.  Calculate no of theoretical P turns per layer.
14.  Calculate theoretical  number of primary layers.
16.  Calculate average turn length.
15.  Calculate actual Np. Np = P layers x Ptpl, turns
17.  Calculate Primary winding resistance.
18.  Calculate primary winding loss %.
19.  Is the primary winding close to 3.5%?
20.  Is the Primary wire able to carry the
       intended Idc current?
21.  Choose the winding interleaving pattern.
       Tables 2, 3, 4, 5, Pri and Sec interleaving patterns.
       Fig 3. Section view through sample wound bobbin.
       Fig 4. Bobbin winding detail for Fig 3 bobbin.
22. 
Calculate exact CFB winding percentage.
23.  Choose insulation thickness used between primary
       layers with the same Vdc potential.
24.  Choose insulation thickness used between Primary and
       Secondary layers, or between primary Anode layers
       and Cathode layers with full Vdc potential difference.
       Table 6. Insulation thickness Vs Voltage.
       Fig 5. OPT4 basic bobbin winding details.
25.  Calculate height of Primary layers plus all insulation.
26.  Calculate oa dia Secondary wire size.
27.  Select select Secondary wire size from wire size table.
28.  Calculate theoretical S turns per layer, thStpl.
29.  Calculate load matches with Sec turns from Step 28.
       Table for turns, TR, ZR, Sec RL.
       Ask for 2 matches between 3 and 9 ohms.
30.  Calculate Sec turns Ns for matches to 3, 6 and 12 ohms.

Notes about wire size table.
    Table for Metric Wire Sizes, Grade 2, magnetic winding wire.

Blank graph sheets :-
  
Fig 25. Graph for Cu wire dia required for working idle DC current.
   Fig 26. Log scale F response sheet 15Hz to 25kHz.
   Fig 26. Log scale F response sheet 10Hz to 200kHz.
   Fig 27. Log scale F response sheet 4.7Hz to 320kHz.


SE OPT Calcs, Page 2.
Content :-
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.

SE OPT Calcs, Page 3. 
Content :-
Practical Testing of 8 Watt SEUL OPT for 1 x EL34 for old AM radio.
       Fig 19. Schematic for rather good 8 Watt SE amp.
       Notes about schematic etc.
       Fig 20. Photo of old radio after restoration.
       Notes on testing amp and OPT.

Oscilloscope pictures of waveforms produced with 8 Watt SEUL OPT.
CRO 1. Healthy wave at clipping power at 1kHz.
CRO 2. Measuring Lp.
      How to adjust the air gap. 
      Fig 21. Wasteless pattern lamination dimensions.
ALTERNATIVE METHOD to calculate max PO of any pentode
CRO 3, 4, 5, Waves during core saturation phenomena.
      Fig 22. Graph of Air gap Vs Fsat and Lp.    
CRO 6, 7, 8, Waves at -6dB level for 47Hz, 20Hz, 16Hz.
     Conclusions about SEUL for 1 x EL34,
     Notes and calculation checks for all.

SE OPT Easy Method for calculating any SE OPT.
For Basic Parameters only.
1E.  Calculate RLa and PO.
2E.  Calculate Afe.
3E.  Calculate required Lp.
4E.  Calculate Np.
5E.  Calculate minimum primary wire size.
6E.  Calculate minimum core window size L x H.
7E.  Calculate Core T + S.
8E.  Calculate µe.
9E.  Check Lp, Fsat.
10E. Calculate Air gap.

High Voltage testing of transformers.
     Fig 23. Schematic for testing insulation of transformer.
    
OPT3 bobbin details for 25W OPT from web pages created May 2006.

OPT3 from 2006.
     Fig 24. Details of OPT3.


END OF LIST OF CONTENTS FOR 3 PAGES
.
---------------------------------------------------------------------------------------------------------

Beginning of Page 1 content :-


RDH4, SE amp history, trends and preferences.
There is much good advice in the Radiotron Designer's Handbook,
4th Edition, 1955, and most effort seemed to focus on PP amplifiers. Hardly
anyone in 1955 thought that real hi-fi could be had from a single ended design.

But millions of Single Ended amplifiers with SE OPT were installed into radios
and radio-grams for the 40 years prior to 1960. Triodes were used for all audio
power amps up to about 1933 when power beam tetrodes and pentodes were
invented and which gave nearly twice the power from a given amount of idle power
dissipation. In 1960, many people listened to radio or TV broadcasts where one
6V6, EL84, 6M5, 6BM8 gave up to about 3 Watts of power from a class A
single ended audio amp. The THD & IMD was tolerable at low levels with
sensitive speakers using very low power.
Very occasionally a manufacturer may have used an 807 which had been bought
in bulk from military disposal stores after WW2 when millions of 807 were
over produced, delighting many ham radio operator and audio DIYers until
about 1965 when old WW2 stocks finally ran out. The brighter audio DIYers
found the 4 Watts available from an SE 807 in triode mode sounded far better
than 4 Watts from a 6V6 beam tetrode. I agree with them, and when repairing
or restoring old radios I often use an EL34 in triode mode instead of a 6V6 in
beam tetrode without NFB.

In 1955, whenever more than 4 watts was required, a PP circuit was employed
for 99% of applications and 10 watts AB1 became possible from a pair of 6BM8
or 6GW8 in fairly linear UL configuration, and it seemed cheaper to make than
a single 807, 6L6, or EL34.
PP began to be used in deluxe TV sets and some hi-fi amps; SE was considered
old fashioned.

Despite commercially profitable trends and fickle public tastes, a few hi-fi
fanatics were not concerned about the costs, and found their ears told them them
an 807, 6L6, EL34 strapped as a triode could make a far better sounding 4 watts
than the first 4 watts from any pair of little tetrode or pentode tubes in PP.
The best 4W of SE power can be had from a lone 2A3, and the 300B can
make a magnificent 9 Watts.

Now it is 2011, and I have had an AM tube radio in my kitchen since I designed
and built it in 1999. It has never missed a beat, although I now have an old Marantz
FM tuner sitting on top which gives me mono FM, quite good enough for the kitchen.
The AM tube part has wide AF bandwidth and has a single EL34 in triode for the
output tube, and speaker is a nice two way thing with dome tweeter. So the AM
often sounds better than FM. If one does all the things which commercial
manufacturers leave out, it is remarkable just how good AM radio can sound.

In Japan after WW2, there were quite a number of audio enthusiasts who liked SET
amplifiers. A Mr Kondo San of Audio Note in Japan brought renewed interest in
SE triode amps in the 1990s. He died in Jan 2006 after 30 years of building amplifiers,
and he left a legacy worth preserving. Since I last edited this website in 2006, I have
built several high power SE amps, one with a pair of parallel 845 for 55 watts, another
with 4 x 6CA7 in parallel for 35 Watts, and yet another with 6 x 6550 in parallel for
60 Watts.  

My method for SE OPT design is based on the theory in the RDH4 or
other sources
I have collected over the last 15 years. After having wound many
very fine SE output transformers with full power bandwidth from 20Hz to 70kHz,
I feel well qualified to speak from experience. The list of logic steps involved in
producing the best possible OPT is based on designing for low winding losses, core
saturation at full power at 20Hz or lower, and adequate interleaving to extend the
HF response up to at least 70kHz by keeping both the leakage inductance and shunt
capacitances to low quantities. The end result gives a well filled winding window
with several impedance matches possible without having wasted sections of any
secondary winding so the winding losses and response is the same for any of the
chosen load matches.

The design for SE OPT has some common features to Push Pull OPTs,
but I have tried to give enough info in this page without need to change to other
pages for reference.
SE OPT design for purely class A1 operation does not have to cope with
complexities of PP class AB operation.

I have made Tables 1a and 1b for expected ranges of performance
from some common power tubes :-

Table 1a.   SE Pentode, Beam Tetrode, CFB, UL.

Tube type Single Ended
Mode
Max
Pda
at idle,
Watts
Ea
+Vdc
a to k
Ia
mA
dc
Max
Audio
Power
Watts
Effici-
ency
%
Max
Eg2
g2
to k
Ig2
mA
dc
RL
for
max
PO
EL84 and 6V6 P, BT, CFB,
Eg2 < Ea.
12
12
12
300
275
250
40
43
48
5.0
4.8
4.5
42
40
38
270
235
200
4
5
6
6k7
5k7
4k7
EL84 and 6V6 UL,
Eg2 = Ea
11
11
11
300
275
250
37
40
44
4.4
4.2
4.0
40
38
36
300
275
250
4
5
6
7k3
6k2
5k1
EL86 P,CFB,
Eg2 < Ea.
12
12
12
250
225
200
48
53
60
5.0
4.8
4.5
42
40
38
200
200
200
4
5
6
4k7
3k9
3k0
EL86 UL,
Eg2 = Ea
12
12
12
250
225
200
48
53
60
4.8
4.5
4.3
40
38
36
250
225
200
5
6
7
4k7
3k9
3k0
807, 6L6,
KT66, 5881
BT, CFB,
Eg2 < Ea.
21
21
21
400
350
300
53
60
70
9.2
8.8
8.4
43
41
39
300
270
250
4
5
6
6k8
5k3
3k8
807, 6L6,
KT66, 5881
UL,
Eg2 = Ea
20
20
20
400
350
300
53
60
70
8.6
8.4
8.0
43
41
39
400
350
300
4
5
6
6k8
5k3
3k9
EL34, 6CA7 P, BT,
CFB,
Eg2 < Ea
24
24
24
420
350
270
57
68
88
10.3
9.8
9.3
43
41
39
270
270
270
7
8
9
6k6
4k7
2k8
EL34, 6CA7 UL,
Eg2 = Ea
23
23
23
420
350
270
54
65
85
9.2
8.7
8.2
40
38
36
420
350
270
8
9
10
7k0
4k9
2k9
6550, KT88 BT, UL,
CFB,
Eg2 < Ea
30
30
30
30
450
390
330
270
66
76
90
110
12.9
12.3
11.7
10.5
43
41
39
35
300
270
270
270
4
5
6
7
6k2
4k7
3k3
2k2
6550, KT88 UL,
Eg2 = Ea
30
30
30
30
450
390
330
270
66
76
90
110
12.3
11.7
11.1
10.5
41
39
37
35
450
390
330
270
4
5
6
7
6k2
4k7
3k3
2k2
KT90, KT120 BT, UL,
CFB,
Eg2 > Ea
35
35
35
35
450
390
330
270
77
89
106
129
15.5
14.3
13.6
12.2
43
41
39
35
300
270
270
270
5
6
7
8
5k3
3k9
2k8
1k9
KT90, KT120 UL,
Eg2 = Ea
35
35
35
35
450
390
330
270
77
89
106
129
14.8
13.6
13.0
12.2
41
39
37
35
450
390
330
270
5
6
7
8
5k3
3k9
2k8
1k9
13E1 BT, CFB,
Eg2 < Ea
72
72
72
72
550
460
380
300
130
156
189
240
31.0
30.0
28.1
25.2
43
41
39
35
220
200
180
160
6
5
5
4
3k2
2k7
1k9
1k2
13E1 UL,
Eg2 = Ea
70
70
70
375
350
325
186
200
215
27.3
25.9
22.1
39
37
35
375
350
325
13
12
10
1k8
1k6
1k4

Table 2.   SE TRIODE.
Tube type
Single Ended
Triode Mode
Max
Pda +
Pg2
at idle,
Watts
Ea
+Vdc
a to k
Ia +
Ig2
mA
dc
Max
Audio
Power
Watts
Eff-
icient
%
Max
Ra at
Eg1=
0Vdc
RL
for
max
PO
EL84,
Triode
12
12
12
330
300
280
36
40
43
3.1
2.5
2.0
26
21
17
2k2
2k2
2k2
4k8
3k1
2k2
EL86
Triode
12
12
12
250
225
200
48
53
60
4.3
4.1
3.6
36
34
30
0k7
0k7
0k7
3k8
2k9
2k0
6V6
Triode
12
12
12
370
340
317
32
35
38
3.1
2.5
2.0
6
21
17
k8
2k8
2k8
k0
4k2
2k8
807, 6L6,
KT66, 5881
Triode 22
22
22
22
500
450
400
350
44
49
55
63
8.6
8.1
7.4
6.3
39
37
33
28
2k4
2k4
2k4
2k4
9k0
6k8
4k9
3k2
6CM5, EL36 Triode 18
18
18
375
350
325
48
51
55
7.8
7.5
7.4
43
41
40
0k5
0k5
0k5
6k8
5k8
4k9
EL34, 6CA7 Triode 24
24
24
24
450
420
390
360
53
57
61
66
9.1
8.7
8.1
7.6
37
36
33
31
1k0
1k0
1k0
1k0
6k5
5k4
4k4
3k5
6550, KT88 Triode 33
33
33
33
500
450
400
350
66
73
82
94
13.0
12.0
11.0
9.6
39
36
33
29
0k8
0k8
0k8
0k8
6k0
4k6
3k3
2k2
KT90 Triode 37
37
37
37
500
450
400
350
74
82
92
105
14.7
13.7
12.6
10.8
39
37
34
29
0k7
0k7
0k7
0k7
5k4
4k1
3k0
2k0
KT120 Triode 40
40
40
40
500
450
400
350
80
88
100
114
15.6
14.7
13.5
11.0
39
36
33
27
0k7
0k7
0k7
0k7
4k9
3k8
2k7
1k7
45
Triode 8
8
8
270
240
210
29
33
38
2.4
2.0
1.4
30
25
17
1k8
1k8
1k8
5k7
3k7
1k9
2A3 Triode 12
12
12
300
275
250
40
43
48
4.4
4.2
3.9
36
35
32
0k9
0k9
0k9
5k6
4k6
3k4
300B Triode 28
28
28
420
380
340
65
73
82
10.6
10.0
9.3
37
35
33
0k7
0k7
0k7
5k1
3k8
2k8
845 Triode 75
75
75
1,100
950
800
68
78
93
27.1
23.7
18.2
36
31
24
2k2
2k2
2k2
11k8
7k8
4k2
GM70
Triode
75
75
75
1,100
950
800
68
78
93
29.1
26.1
22.0
38
34
29
1k8
1k8
1k8
12k6
8k6
5k1
13E1
Triode 70
70
70
375
350
325
186
200
215
24.0
23.0
22.0
34
32
31
0k3
0k3
0k3
1k4
1k2
1k0

NOTE.   Interpreting the tables above.
Tube Type
. Includes those most commonly used plus a few which are
uncommon, like 13E1 which is no longer made.

Single Ended Mode.
"P" and "BT" = Pentode and Beam Tetrode, with fixed Eg2 between screen
g2 and cathode k.
CFB = Cathode feedback windings are used on OPT. CFB windings have
portion of total primary turns up to 50% of total primary turns. Screen g2
is taken to fixed Vdc supply often lower than anode B+ supply. Such a
connection is known as Acoustical, as used in early Quad-II amps.
Sometimes g2 is taken to taps on anode winding for some screen feedback
like the UL connection.
UL = Ultralinear, where g2 is taken to taps on anode winding where the
portion of B+ to screen winding is not more than 66% of total anode turns.
Triode = Triode strapped multigrids where g2 and perhaps g3 is tied to anode
or where the tube is a real triode like 300B.

Max Pda + Pg2 at idle, Watts = Total of product of ( Ea-k x Iadc ) +
( Eg2-k x Ig2dc ) at idle condition. The Pda is the maximum recommended
and may be lower to allow longer tube life and reliability. The higher Pda,
the higher the operating temperature and shorter the tube life.
If Pda is
from what is listed, all information in following table columns must be
re-calculated.

Ea +Vdc a to k and Ia + Ig2 mAdc = Range of maximum and minimum
Ea and Ia values which could be used. A graph may be made on an Ia / Ea
graph to determine other values between those shown.

RL is calculated for all tubes except triode as RL = 0.9 x (Ea / Ia)

RL for triode is calculated as RL = (Ea / Ia) - (2 x Ra) where Ra is the straight
line calculation for the first part of the Ra curve for where Eg1 bias = 0Vdc.

Max Audio Power Watts at Anode For pure class A1 at clipping and where
THD < 5% for the load value capable of producing the maximum possible PO
for the Ea and Ia conditions at idle.

Beam Tetrodes and Pentodes,
POa = [ 0.5 x ( Iadc at idle - 5% )squared ) x anode RLa ] Watts.
With applied NFB the THD may permit more PO with less THD and higher
maximum anode efficiency.

Efficiency % Max = Anode efficiency,
calculated as = ( 100 x maximum possible anode audio power / Pda at idle ) %

The tables are a GUIDE for OPT design. The Ea and Ia determine the load
value which gives the highest power at clipping. Raising Ea and lowering Ia
for the sam
e Pda will allow a higher RL, and lowering Ea and raising Ia will
allow a lower RL. Thus the Ea and Ia may be adjusted to suit a given RL
or to suit an OPT which is available.


NOTE. For Parafeed connection where a high value choke inductance is used to
bring the Idc to the tubes, and the OPT primary is coupled to the anode with a
capacitance, the OPT may be designed using the method for PP OPT calcs.

SINGLE ENDED OPT4 EXAMPLE.

1. DEFINE THE AIM OF THIS PROJECT.

OPT4 is to be designed for use to allow up to 35 Watts of pure class A1 audio
power over a frequency range between 20Hz and 65kHz using a selection of
tubes from SE tube tables above.


The range of anode load values to be considered are nominated as :-
Minimum RLa for less than maximum class A1 power, but useful,
Middle RLa for maximum class A1 power,
Maximum RLa for less than maximum class A1 power but useful.


OPT4 is to have secondary winding layers sub-divided into enough
separate windings to allow a number of winding arrangements made up with
seriesed and or paralleled windings useful range of speaker loads to suit
optimal tube operation, usually about 3 matches between 1.5 ohms and 16 ohms.


THE CALCULATIONS FOR OPT4 WILL ALLOW SE TRIODE 
OPERATION.
BUT FOR ONLY TRIODE OPERATION A SEPARATE DESIGN
PATH MUST BE FOLLOWED.


Choose a suitable choice of tube from above tables and which may be paralleled
to give wanted total power output, PO.

Total PO = PO for one tube x number of paralleled tubes - will depend on mode. 

OPT4, for 35W, List some choices.
7 x EL84/6BQ5 in pentode mode, 10 for triode operation.
3 x 6550/KT88 in beam tetrode, 5 for triode operation.
4 x 6L6GC, 5881, KT66, 807, EL34, 6CA7 in beam tetrode, pentode,
5 for triode. 

Note. An example of a real amp already exists this website at SE35 monobloc.  

Let OPT4 be the design example for 4 x EL34/6CA7 in pentode with 
CFB, ( SE Acoustical ).
Nominate audio power required from 1 tube = Max PO / No of tubes = 35W / 4
= 8.75 Watts. 

Choose from above tables and centre Ea values listed.

Tube
SE Pentode
Pda Ea Ia PO
Eff
Eg2 Ig2 RLa
EL34
CFB, Eg2 < Ea  24W  +350V  68mA 9.8W 41%  +270V  8mA
4k7

 The choice will produce 9.8W from one tube so therefore 39.2W from 4 tubes and allowing
for winding losses there will probably be 35 Watts at the secondary speaker terminal.

Graph 1.
graph-EL34-se-pentode-po-vs-rl-ea-350V.GIF

Graph 1 shows the maximum power levels for various RLa loads available from
ONE SE EL34 or 6CA7 in Pentode mode, UL or CFB mode. Everyone should
consider what ONE tube will do before considering what 4 tubes in parallel may do.
Fig 1 tells us we have an OPT with ZR = 1,285:1, and with 3.5 ohms at the
secondary the RLa will be 1,285 x 3.5 ohms = 4,495 ohms. The arrangement
would suit a speaker with "nominal" Z = 4 ohms.
Maximum PO is only possible at one value of RL for a given set of idle conditions
for Ea, Ia and Eg2 and the graph curve is only valid for one idle condition.
Most SE amps are designed to give the maximum SE power to either 4, 8 or
perhaps 16 ohms. But if we consider the OPT is set up for max PO at 3.5 ohms
as in Fig1, in fact the tube will drive a range of loads below and above 3.5 ohms
and produce useful power. From the graph we see that 6 watts or more is available
at all RLa loads between 2k5 and 8k6 which corresponds to speaker loads between
about 2 ohms and 6.7 ohms. All the power is pure class A so there is no THD
generated as a result of devices switching on or off as with PP amps. So as long
as there is sufficient NFB the result is listenable. If a horn loaded speaker
Z = 8 ohms with 100dB/W/M sensitivity and never needed more
than 1 watt, then the amp will still give 5W into 8 ohms.

If the ZR of the OPT was 4,500 : 5.6ohms, ie, 803:1, then the range of anode
RLa values would also be 2k5 and 8k6 and the speaker load could be between
3.1 ohms and 10.7 ohms for 6 watts of output. Most SE amps will withstand
the load variations with ease because the power is still all class A. If there are
4 parallel tubes each making 6 useful Watts then you have 24 Watts total
which is enough for most people. 

So when considering SE PO, the Center Value RLa for Maximum PO is
worked out, then the OPT ratio calculated for this and to give a range of
loads from below and above which are suitable.

Fig 1 below will introduce everyone to the very basic pentode and triode
Ra curves for EL34. Only the Ra curves for Eg = 0V are shown because
these are all that is needed for OPT design.

Fig 1.
EL34-6CA7-Ra-curve-Eg1=0V+280ohms.GIF

Fig 1 shows the EL34 Pentode Ra curves for Eg1=0V, Eg2 = 350V,
and
for Eg1=0V, Eg2 = 250V.

Pda is shown at 25Watts for EL34 or 6CA7. The Pg2 is included.
EL34 data indicates max Pda = 28W, but I do not recommend that
any EL34 should ever have total Pda plus Pg2 idle power > 25W.

Ra curve for Eg1 = 0V for triode connected EL34 or 6CA7 is also shown.
 
There is also a straight line with slope of 280 ohms shown which is the Ea
swing "limiting line" value which may be used for all pentodes and beam power
tetrodes where the Ultralinear screen taps up to 60% taps or CFB windings are
used.

The slope of the 280 ohm line is "less steep" than the Ra curves for below
Ea = 50V which have been copied from old tube data sheets.
Ra for below Ea = 50V may be 140 ohms for EL34 in pentode mode, but
when UL connection or CFB is used there is some reduction of possible Ea
swing and the line for 280 ohms is a conservative value which may be used
for all power pentodes and beam tetrodes.

The Ra curves for pentodes and tetrodes seem strange because between the
Ea swing between 0V to 50V, the Ra appears to be very low, only 140 ohms,
yet Ra increases to many thousands of ohms between 50V and say 500V.
Pentodes and beam tetrodes produce all their power in the region where
Ra is many thousands of ohms, but their power is slightly limited by the
low Ra region where Ea < 50V.

Now if one assumes the pentode Ra Limiting Line = 280 ohms, and that the
knee of Ra curves will always be well above 2 x Ia at idle for an appropriate
Eg2 value shown in the tables above, then one may calculate RLa after
choosing Ea and Ia values without using any load line analysis providing Ea is
within the ranges shown in Table 1 above.

The Ra curve for EL34 Triode strapped for Eg1 = 0V is also shown.
For calculations of triode PO, the Ra may be read off the data as follows :-
Step 1. Read the Ea and Ia where the chosen Pda curve intersects the Ra triode
curve for Eg1 = 0V. This is POINT H, where Ea = 180V and Ia = 140mA.

Step 2. Calculate Ra = Ea / Ia at point H = 180V / 0.14A = 1,285 ohms.

The Ra curve is the limiting line for Ea minimum for any value of RL for
SE Class A1 triode operation. EL34 do not respond well when being driven
into class A2.

In this case I show a Pda curve at 25W, but if you have Pda = 20W, then the
position of Pda curve will be lower on the graph. Pda should not be lower than
20W, lest the amp become based on an a non-optimal operating point.


But 807, 6L6, KT66, 5881 6550, KT88 may all be driven into class A2 if desired,
but increase in amp parts is not worth the possible 20% increase of triode power.

Calculating Center Value RLa and the maximum
output power, and calculating power for RLa lower
and higher than Center Value :-


A. Pentodes and Beam Tetrodes only :-

Calculate Center Value RLa for maximum possible PO,
Formula 1:- RLa =
0.9 x Ea / Ia  ohms. 

Formula 2 :- RLa = Ea / Ia - ( 2 x 280 )  ohms.
where 280 ohms is the limiting low Ra for where Eg = 0V,
and Ea < 50V, which prevents Ea swinging to a lower voltage because of
grid current.

NOTE. The curve data for most power pentodes and tetrodes may show the Ra
limiting line for Ea < 50V having a slope of say 150 ohms, but when such tubes
are used with UL taps or with CFB windings the Ra line may be assumed to be
280 ohms.

PO at clipping = 0.5 x RLa x Ia squared, where Ia = Iadc at idle.

Example 1:- EL34 with Ea = 350V, Ia = 68mA.

Formula 1, RLa = 0.9 x 350/0.068 = 4,632 ohms.

Formula 2, RLa = ( 350/0.068 ) - 560 = 4,587 ohms.

Both formulas give near the same answer.

PO at clipping = 0.5 x 4,600 x 0.068 x 0.068 = 10.6 Watts.

For any selected RLa Lower than Center Value for max possible PO,

PO =
RLa x ( Ia squared / 2 ), RLa Lower than Center Value.

Example 2 :- RLa = 3,000 ohms, Ea 350V, Ia 68mA,
PO = 0.5 x 3,000 x 0.068 x 0.068 = 6.94 Watts. 

For any selected RLa Higher than Center Value for max possible PO,

PO = 0.5 x ( Ea - [ 280 x Ia ] ) squared    x RLa
Higher than Center Value.
                     ( 280 + RL ) squared

Example 3 :-
RLa = 9,000 ohms,
Ea 350V, Ia 68mA,
PO = 0.5 x ( 350 - [ 280 x 0.068 ] ) squared x 9,000
                      ( 280 + 9,000 ) squared
= 0.5 x ( 350 - 19 ) x ( 350 - 19 ) x 9,000  
              ( 9,280 x 9,280 )
= 5.72 Watts.

B. Triodes only :-

Calculate Center Value RLa for maximum possible PO,
RLa = ( Ea / Ia ) - ( 2 x Ra ) ohms.

Where Ra is the limiting Ra value where Eg = 0V which prevents Ea
reaching any lower voltage because of grid current.
If the above equation gives RLa less than twice Ra, the Ea should be raised
and Ia reduced so Pda stays constant or slightly less so RLa will then be a
preferred minimum of 3 x Ra, if not more.

Example 4 :- Triode connected EL34, Ra = 1,285ohms, Ea = 350V, Ia = 68mA.

RLa = ( 350 / 0.068 ) - ( 2 x 1,285 ) = 5,147 - 2,570 = 2,577 ohms.

PO max = 0.5 x 2,577 x 0.068 x 0.068 = 5.9 Watts.

For any selected RLa Lower than Center Value for max possible PO,

PO =
RLa x ( Ia squared / 2 ), RLa Lower than Centre Value.

Example 5 :- RLa = 1,500 ohms, Ea 350V, Ia 68mA,
PO = 0.5 x 1,500 x 0.068 x 0.068 = 3.5 Watts. 

For any selected RLa Higher than Center Value for max possible PO,

PO = 0.5 x ( Ea - [ Ra x Ia ] ) squared    x RLa Higher than Center Value.
                     ( Ra + RLa ) squared

Example 6 :-
RLa = 4,600 ohms,
Ea 350V, Ia 68mA,
PO = 0.5 x ( 350 - [ 1,285 x 0.068 ] ) squared x 4,600
                      ( 1,285 + 4,600 ) squared
= 0.5 x ( 350 - 87 ) x ( 350 - 87 ) x 4,600  
              ( 5,885 x 5,885 )
= 4.6 Watts.

NOTE. These sample calculations for EL34 pentode and triode are for the same
Ea and Ia conditions, and Example 1 for pentode with Centre Value RLa gives
maximum PO = 10.6W and the same load for triode gives 4.6W. It will
be found that the SEUL connected EL34 with 40% to 50% screen tap or with
15% CFB will give higher power and better sound than triode for the Ea and Ia
conditions. But when Ea Is raised and Ia reduced, the triode can then perform
much better.

Example 7 :- Ea = 400V, Ia+Ig2 = 57mA, Pda = 23W, Ra = 1,300 ohms.
Center Value RLa = ( 400 / 0.057 ) - ( 2 x 1,300 ) = 4,417 ohms.

PO = 0.5 x 4,417 x 0.057 x 0.057 = 7.17 Watts.

Example 8 :-  RLa = 2,557 ohms,
PO = 0.5 x 0.057 0.057 x 2,557 = 4.15 Watts.

Example 9 :-  RLa = 9,000 ohms,
PO = 0.5 x ( 400 - [ 1,300 x 0.057 ] ) squared x 9,000
                      ( 1,285 + 4,600 ) squared
= 0.5 x ( 400 - 74 ) x ( 400 - 74 ) x 9,000  
= 4.78 Watts.
              ( 10,300 x 10,300 )

NOTE. The above calculations usually give slightly less PO than indicated
with load line analysis. In practice, slightly more power is available providing
there is enough loop NFB to keep THD < 2% at onset of clipping.

 
2.  LOADLINE ANALYSIS FOR ONE SE EL34 PENTODE.

The most accurate way to see what is calculated is to work from
graphical load line analysis.


Fig 2.
graph-EL34-6CA7-3-loadlines-Eg2-250-350V.GIF

Fig 2 shows the Ra curve for EL34/6CA7 at Eg1 = 0V copied from tube
data sheets in old data books. There are THREE Ra curves given for Eg1
= 0V because there are THREE values of Eg2 also shown.
Fig 2 shows that for SE operation the Eg2 does not have to be as high as Ea.
The ideal Eg2 for many SE pentodes and beam tetrodes is well below Ea,
but not often below 200V, which might be the minimum Ea anyone would
consider.
If Eg2 = 250V for EL34, the Ig2 is low, and the Eg1 needed for biasing is also
low so little heat is wasted in a cathode resistor in cathode biasing. This means
that where Ea = +350V, and Ek = +17V, and Vdc drop across OPT Rw = +15V,
then total B+ supply rail to OPT would be +382Vdc, so 450Vdc rated electrolytics
may be used without fear of them being subject to too high a voltage when
no tubes were present in the amp. This assumes that Si diodes and not tube
diodes have been selected for the rectifier.
In Fig 2, the curves for Ra for many values of Eg1 have not been shown
because they are not needed, and establishing the cathode bias must be done by
experiment when the amp is first tested, and with shunt regulated screen supply
established.

HOW TO DRAW THE LOADLINES.
The Fig 2 loadines are shown here to save everyone the trouble of going to
another page on load line analysis, and this graph is relevant to the OPT4 design.
I have chosen idle conditions with Ea = +350Vdc, Ia dc = 68mA, Pda = 24W.

A.   Plot point Q, at Ea = 350V and Ia = 68mA.

NOTE.  There are "knees" in the 3 curves for Ra at Eg1 = 0V for each value of
Eg2. All knees are well above 2 x Ia which is the maximum load current which
can ever exist in pure class A amplifiers.

NOTE.  For UL use of pentodes or beam tetrodes, Eg2 = Ea. The knee of the
Ra curve is high **if** the UL tapping point is less than 60% of the total anode
turns in most pentodes or tetrodes. UL operation with UL tap up to 60% allows
max class A1 PO up to the same as for pure pentode or beam tetrode, but
distortion will become similar to triode, mainly 2H. The best UL tap position
for SEUL operation is that which is the closest to the anode terminal but which
still allows PO to be close to full pentode or tetrode power.

B. Draw the load line which will give maximum audio power. This load line
is for the CENTER VALUE RLa = 4,853 ohms, shown as line AQB.

Calculate 2 x Ia = 2 x 68mA = 136mA.
Plot point A on the Ra curve at Ia = 136mA.

Draw a vertical line thru A and thru Ea axis line, and record Ea.
The vertical line is at Ea = 20V, and is the Ea minimum peak voltage during
each wave cycle for 4,853 ohms load value.

NOTE.   The value of Ra between 0V and 50V is very low and in this example
may be calculated as Ra = Ea minimum / ( 2 x Ia ) =  20V / 0.136A = 147 ohms.
( This will become handy to know in Step D below. )

Calculate peak Ea load voltage swing = Ea - Ea minimum = 350V - 20V = 330V.
This is the negative going peak load voltage swing, VRLa pk. 
Calculate RLa = VRLa pk / Ia = 330V / 68mA = 4,853 ohms.

Calculate Ea maximum peak voltage when Ia reaches zero mA, or tube cut off.
Ea max = Ea + VRLa pk = 350V + 330V = 680V.

Plot Point B on the Ea axis for Ea maximum pk swing.
B is at Ea = 680V.

Draw a straight line from A to B and it should pass thru Q exactly.
The straight line AQB is the load line for 4,853 ohms.

Calculate the theoretical maximum possible PO from the tube with this load.
Peak Ia swing = Ia dc at idle.
PO = 0.5 x Ia peak swing in load squared x RLa.
= 0.5 x 0.068A x 0.068A x 4,853ohms = 11.22 Watts.

C.  Draw the load line for RLa =
1/2 Center Value.
Calculate 1/2 CV RLa = 4,853 / 2 = 2,426 ohms.

For all RLa LESS than Center Value, the Ia pk swing is always +/- idle Ia
above and and below the idle Ia, ie, between 0.0mA and 2 x Iadc at idle.
 
Calculate Ea pk load swing = Ia x RLa =
68mA x 2,426 ohms = 165Vpk.
Ea min = Ea - Ea pk swing = 350V - 165V = 185V.
Plot Point C at Ea = 185V and Ia at 2 x Ia = 136mA.

Ea max = Ea + Ea pk swing = 350V + 165V = 515V.
Plot point D at Ea = 515V on Ea axis, Ia = 0.0mA.
Draw straight line from C to D and it should pass through Q exactly.

Calculate the theoretical maximum possible PO for this load.
Peak Ia swing = Ia dc at idle.
PO =
0.5 x Ia peak swing in load squared x RLa.
= 0.5 x 0.068A x 0.068A x 2,426ohms = 5.61 Watts.

D.
Draw the load line for RLa = 2 x Center Value.
Calculate 2 x CV RLa = 2 x 4,853 = 9,706 ohms.

For all RLa MORE than the Center Value, the max Ia pk swing is always
less than Ia.
Ia max = Ia pk load swing + Ia, Ia min = Ia - Ia pk load swing.

Calculate Ia maximum.
 
Ia max = Iadc at idle + ( Ia pk load swing for higher RLa )
= Ia +[    Ea - ( Ra x Ia )  ]
            higher RLa + Ra

Ra was calculated in Step B above as 147 ohms.

So Ia max for RLa = 9,706 ohms
= 0.068 + [ 350 - ( 147 x 0.068 ) ]
                      9,706 + 147
= 0.068 + ( 340 / 9,853 ) = 0.068 + 0.0344 = 0.1024A = 102.4mA.

Ea minimum for higher RLa = Ia max x Ra = 0.1024 x 147 = 15.05V.

Ea pk load swing = Ea - Ea min = 350V - 15V = 335V.

Plot Point E for Imax for the higher RLa on the Ra curve at Ea minimum.
Point E will be at Ia = 102mA and Ea = 15V.

Ea maximum for higher RLa = Ea + Ea pk load swing
= 350V + 335V = 685V.
Ia minimum for higher RLa = Ia - ( Ia swing x higher RLa )
Ia swing was calculated above as Ia pk load swing
 
Ea - ( Ra x Ia )  
    higher RLa + Ra
= 0.0344A = 34.4mA.

Ia minimum for RLa 9,706 ohms = 68mA - 34.4mA = 33.6mA.

Plot Point F for Ea max and Ia min at 685V and 33.6mA.

Draw a straight line between point E and F and it should pass thru Q exactly.

Calculate the theoretical maximum possible PO for this load.
Peak Ia swing is NOT Ia dc at idle.
PO =
0.5 x Ia peak swing in load squared x RLa.
= 0.5 x 0.0344A x 0.0344A x 9,706ohms = 5.74 Watts.

LOADLINE ANALYSIS CONCLUSION.

The loadline analysis shows Center Value RLa for one EL34 = 4,853 ohms.
Maximum output power is up to 11.2Watts.

NOTE. The load range between 0.5 and 2.0 times the Center Value RLa
give a useful amount of audio power above approximately 5.6 watts.
While this does not seem like much power, with 4 parallel output tubes
PO ranges between 22.4Watts and 44.9Watts and for a wide load range.


NOTE. Ra curve data shown in old data books may be optimistic.
High amounts
of THD in pentodes and tetrodes usually reduces the available
max PO unless
enough NFB is used to reduce it to less than 2% at 1dB
below clipping. All the
above PO loadline analysis and calculations are
theoretical only and for a guide
to OPT design. The OPT should be
designed to allow for the theoretical results
from loadline analysis.
Output power is considered at the anode without any
allowances for
OPT primary and secondary winding resistance losses.


3. Calculate Centre Value RLa for OPT4 design project,
and confirm maximum PO, Va max, Ea and Iadc.

Calculate RLa = RLa for ONE tube / No of parallel tubes.
 
OPT4. CV RLa = 4,853 ohms / 4 parallel tubes = 1,213 ohms.

Calculate maximum PO = PO max for one tube x No parallel tubes
= 11.22 x 4 = 44.9 Watts.

Calculate maximum Va at max PO = 0.707 x peak Ea swing
= 0.707 x 330V = 233 Vrms.


Ea = +350V
from load line analysis.

Calculate Ia for 4 parallel tubes = 4 x 68mA = 272mA.

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

OPT4, nominal PO max = 44W with NFB.

Afe = 450 x sq.rtPO in sq.mm, where 450 is a constant, PO in Audio Watts.


NOTE.  This formula has been derived from the basic formula for core size
used for mains transformers, Afe = sq.root power input / 4.4 where the Afe
is in sq inches. This ancient formula is based on signal Bac max being about
1 Tesla at 50Hz, but wanted Bac max = approx 0.33 Tesla for an SE OPT at
50Hz, or 0.8T at 20Hz. After considerable trials the above formula is a good
guide for SE audio OPT.

OPT4 theoretical Afe = 450 x sq.rt 44 = 450 x 6.63 = 2,984sq.mm

5.  Calculate the core tongue dimension, T.

For a square core centre leg section,
tongue width = stack height, ie T = S, and Afe = T x S, = T squared.

Therefore theoretical T dimension = square root AFe, mm. 

OPT4, thT = sq.rt 2,984 = 54.6mm.

Choose nearest T size less than thT from list of available wasteless GOSS
E&I laminations or C-cores.


NOTE. T sizes commonly available for wasteless pattern E&I laminations :-
20mm, 25mm, 32mm, 38mm, 44mm, 51mm, 63.5mm.

NOTE.  C-cores are normally used to make a double C core, and a wide variety
of strip build up heights and strip widths are available. Usually the C-core
ratio of window area to a square section AFe is larger than for E&I laminations.
The T dimension for double C-cores is twice the strip build up of one C-core
and the S dimension, or stack, is the strip width of the C-core. 

NOTE. Choosing a standard T size above thT gives a larger window area and
lower copper winding losses and possibly a stack height less than the T.
Choosing T below thT gives a smaller window size and higher winding losses
but lower weight and smaller overall size.
If T is chosen below thT, Afe may need to be increased so less turns may be used
to keep winding losses low and to maintain the wanted bass response

NOTE. In general, S should not exceed 2 x T.

NOTE. HF performance depends entirely upon the interleaving geometry and
insulation.

For OPT4, choose core T = 51mm

6.  Calculate theoretical stack height S for chosen T from step 5.

thS = Afe / T, then adjust to a larger height to suit nearest standard plastic bobbin
size if available.
NOTE, standard bobbin sizes are usually available for all square core leg shapes
for T = S. There are other S heights usually always greater than the T size, at
convenient larger sizes. For T = 51mm, S might be 51mm, 62.5mm, 76mm, and
102mm. Sometimes it is necessary to make the whole bobbin from fibreglass
sheeting cut neatly and glued together.

OPT4.  thS = 2,984 / 51 = 58.5mm, say 58mm, not a standard S size.

Bobbin would be handmade at S = 58mm, but a premade bobbin with S = 62.5mm
could be used.

S = 58mm will be used for OPT4.

7. Calculate usable Afe = chosen T x chosen S.

OPT4.  Confirm Afe = 51 x 58 = 2,958sq.mm.

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

For wasteless pattern E&I, where the centre leg tongue dimension = T,
the window Length L = 1.5T and window Height H = 0.5T.
Overall dimensions of
plan area of assembled E&I lamination is 2.5T x 3T.
Plan area of assembled laminations = 6Tsquared, not including both windows.
From this the volume of the iron used in the wasteless core
= 6 x Tsquared x Stack height.
Core weight in Kg = Volume x 0.0084 where volume is in cubic centimetres.  

Some other E&I lamination patterns and most C-cores have a much larger
window area for their effective T dimension so that larger wire sizes for less
copper loss may be employed or to give more room for more turns and
insulation layers. Regardless of the core type and pattern, the ratio of Afe
size relative to Bac max must be maintained. Some C-cored OPT look very
 impressive and physically large but Afe might be much too small for the
number of turns used to keep Fsat low enough.
If the Afe is reduced by 50% then Np must be increased by 50%. Then
winding resistance rises unless thicker wire is used. If one thing is changed,
many other things will have to change to comply with all design rules. I prefer
wasteless E&I patterns, and most designs have an Afe aspect ratio where
S = 1.5T approximately. Suitable C-cores are rarely ever in stock with any
transformer parts suppliers and are more expensive than GOSS E&I lams.
The design process may be written out step by step using one choice of
core and it is never too much trouble to repeat the design process 4 or five
times to choose alternative core sizes to compare results which will offer the
best result.

Table of possible T and S for output powers. 

OPT
Power
Watts
AFe
sq.mm
Theoretical
T, mm.
Choose
T x S,
mm.
 
Choose
T x S,
mm.
Wasteless
E&I Core
Weight
Kilograms
100
4,500
67.0
62.5 T
75 S
50 T 100 S 14.8
70
3,764
61.4
50
80
44
90
10.1
50
3,181 56.4
50
65
44
80
8.2
35
2,662
51.6
50
52
44
62
6.6
28
2,381
48.8
50
47
44
55
5.9
20
2,012
44.9
44
46
38
55
4.5
14
1,683
41.0
38
45
32
57
3.3
10
1,423
37.7
38
38
32
48
2.8
7
1,190
34.4
32
38
25
60
2.0
5
1,006
31.7
32
32
25
50
1.7

8.  Confirm L and H of the winding window.

OPT4,
L = 76mm for 51mm wasteless E&I lams.

H = 25mm for 51mm wasteless E&I lams.

9.  Calculate theoretical primary winding turns, thNp.

Th Np = Va max x 20,000 / Afe
or = square root of ( anode RL x PO) x ( 20,000 / Afe ).

NOTE.  The formula here is derived from more complex and complete formula
taking Bac max and F into account. For GOSS cores, maximum ac magnetic field
strength, Bac cannot be allowed above 0.8 Tesla at 14Hz. This is based on the
maximum dc magnetic field being 0.8 Tesla, so the combined maximum
Bac and Bdc = 1.6T. For lesser grades of steel it is safer to work with Bac = Bdc
= 0.6T with totla of both = 1.2T.

OPT4.   RL = 1,213 ohms, PO = 44 Watts, Va = 233Vrms, Afe = 2,958sq.mm.

Th Np = 231 x 20,000 / 2,958 = 1,562 turns.

10.  Calculate theoretical Primary wire dia, thPdia.

NOTE.   The Primary wire used for the transformer will occupy a portion
of the window area = 0.28 x L x H. The constant of 0.28 works for most  OPT.
Each turn of wire will occupy an area = oa dia squared. Overall or oa dia is the
copper dia PLUS enamel insulation. Therefore theoretical over all dia of P,
thoaPdia, of wire including enamel insulation
= square root ( 0.28 x L x H / Th Np ), mm.

OPT4.  Th oa dia P wire = sq.rt ( 0.28 x 76 x 25 / 1,562 ) = sq.rt 0.341

= 0.584mm.

11. Find nearest suitable theoretical oa wire size for Primary.

Wire Size Table of available wire sizes.
table-wire-sizes.GIF

OPT4, Try oa wire size = 0.569mm, ( for Cudia = 0.475 mm. ) 

12.  Calculate the bobbin winding traverse width, Bww.

NOTE. Bobbin traverse width is the distance between the cheek flanges and
varies depending on who made the bobbin, but 2mm for each flange thickness
is common but could be more or less, and this affects the number of turns
per layer. Where bobbin flanges are not used, and inter-layer insulation is
simply extended to the full window length L, the traverse width will be the
same as in the case of of where bobbin does have flanges.
So the winding will traverse a distance = L - 4mm.

OPT4. Bww = 76 - 4 = 72 mm.

13. Calculate no of theoretical P turns per layer.

Th Ptpl = 0.97 x Bww / oa dia from Step 11.
 
NOTE.   The constant 0.97 factor allows for imperfect layer filling with
such fine wire. Leave out fractions of a turn.

OPT4. Th P tpl = 0.97 x 72 / 0.569 = 122 turns

14.  Calculate theoretical  number of primary layers.

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

OPT4. ThNpl = 1,562 / 122 = 12.8 layers; round up to 13 layers

NOTE. Rounding down to 12L may reduce the Np needed for Fsat = 20 Hz.

NOTE. Fsat for SE amps at full PO may be allowed to be up to 20Hz and not
14Hz for PP OPT because it will be found the design size, weight and cost will
rapidly increase if Fsat is much below 20Hz.  

Rounding up No of layers increases Np which gives Fsat marginally lower than
20 Hz, which is OK.
Fsat may be lowered if Afe is increased by increasing S from say 51 mm to
62mm and be able to use a standard sized bobbin, and the change lowers Fsat
from 20Hz to 16Hz and only slightly extends low bass response. 

15.  Calculate actual Np. Np = P layers x Ptpl, turns

OPT4. Np = 13 x 122 = 1,586turns.

16.  Calculate average turn length.

TL = ( 3.142 x H  ) + ( 2 x S ) + ( 2 x T ), mm.
NOTE. 3.14 is pye, or 22/7.
OPT4. TL =  ( 3.142 x 25 ) + ( 2 x 51 ) + ( 2 x 58 ) = 297mm.

17.   Calculate Primary winding resistance.

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

OPT4, Rwp = 1,586 x 297 / ( 44,000 x 0.475 x 0.475 ) = 48 ohms.

18.  Calculate primary winding loss %.
P loss % = 100 x Rwp / ( PRL + Rwp ), %.

OPT4 P loss = 100 x 48 / ( 1,213 + 48 ) = 3.80%.

19.  Is the primary winding close to 3.5%?
If 15% more than 3.5%, the design calculations must be checked and perhaps
a larger core stack or window size chosen. If no, proceed to Step 21.

P winding loss exceeds 3.5% by less than 15%, so proceed...

NOTE. If the P winding losses are less than 2.5%, there is a possibility that
the P wire size could be reduced to increase the turns per layer, and possibly
reduce the number of P layers from say 14 to 13. However I rarely find SE
OPT P winding losses will be less than 2.5% with the rated load, and one
must allow for where the speaker load used maybe is half the RL used for
the design. Where the load used = 1/2 the design RL,  P winding losses
will double. It is better to have winding losses lower than required so that
the windings are unlikely to overheat if a tube malfunctions and draws
excessive Idc during a "bias failure event".

20.  Is the Primary wire able to carry the intended Idc current?

Calculate copper wire section area and allowed Idc based on maximum current
density of 2A per sq.mm. Allowed Idc should be well above intended Idc. 

OPT4. Primary wire section area = ( 22/7 ) x ( Cu dia squared / 4 )
= 3.142 x ( 0.475 x 0.475 / 4 ) = 0.177 sq.mm.
Allowed Idc = Cu Area x 2A / sq.mm current density = 0.177 x 2A = 0.354 amps.
Idle DC current from Step 4 = 272mA.
Allowed Idc is well above idle Idc.

NOTE.  The primary wire heat dissipation = Idc squared x RwP
= 0.272 x 0272 x 48 = 3.55Watts.
This will cause a small temp rise but if the Idc was 1 Amp, the Pd = 48Watts,
and wire will become too hot. The wire may even fuse open if current density
reached 30A per square mm, ie, 5.3 Amps. Therefore a fuse of 0.25Amps
should be used between each EL34 cathode and OPT connection. Each
EL34 should have its own RC biasing network and series fuse. Active
protection to guard against bias failure in one or more OP tubes should be used.
If a tube is allowed to seriously over heat the internal grid wires may melt
and a short circuit may develop between anode and cathode or between anode
and grid which can cause expensive damage.  

21.  Choose the winding interleaving pattern.

Search the Tables 2, 3, 4, 5, below for the output power rating
of the transformer.

MANY NOTES :-
Only circular section copper magnet winding wire
, Grade 2, with high
temperature polyester-imide enamel coating
is recommended at this
website.
Transformers are wound onto moulded plastic or fabricated bobbins
with polyester sheet or other suitable inert insulation materials used between
all layers of neatly wound wire.
Layers of electrical paper insulation are not
used. There is no need for bifilar or trifilar winding.


To obtain good high frequency response in all audio frequency transformers, 
the primary and secondary windings need to be each subdivided and the parts
are interleaved with each other. As the height of the winding layers are built
up on the bobbin individual layers or groups of layers are parts of either the
Primary or Secondary windings. Each alternating part of either primary winding
or secondary winding consisting of one or more layers of wire is called either
a 'primary section' or 'secondary section' which may be connected in series
with or in parallel with other sections.

For nearly all tube amp OPTs to couple tubes to speakers, there will be usually
be more numerous layers of small diameter wire for the high voltage & low
currents in tubes than a lesser number of layers of large diameter wire for the
lower voltage but higher currents in the loudspeaker.

There is often only one layer of wire in any secondary section of most tube OPTs.
But when the ZR is relatively low, below say 1k5 : 8 ohms, there may be more
than one layer per secondary section. For tapped secondaries where there may
be 4 or 5 identical double sections each with enough turns to suit speakers from
4 to 16 ohms.

Any primary section may usually have more than one layer of wire.
So an OPT is most often built up with groups of layers of primary wire interleaved
with single layers of Secondary wire. Each layer of secondary wire may be
subdivided into further secondary sub-sections to allow a number of primary
to secondary turn ratios to give a number of "impedance matches." 

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

Where the first and last layer of wire wound onto the bobbin is in a primary
section, then these sections should have nearly 1/2 the layers of the inner
primary sections. So if there are 3 outer primary layers in a Primary section,
the inner P sections might be either 5p, 6p or 7p layers. When this guide ensures
the best HF response because the leakage inductance is evenly and symmetrically
distributed.

This rule applies so that the winding section pattern is a mirror image below and
above center of the bobbin, and should always be the case for PP OPTs. This
means that if the first on wire layer is a Primary, the last on layer will also be a
Primary.
However, for SE OPTs, it is possible to start with Primary layer and end with a
Secondary layer. The anode is usually connected to the innermost end of the
Primary where the average turn length is shortest hence giving lowest shunt
capacitance. This is aided by starting with a Primary layer so that the highest
anode signal voltage is not adjacent to earthy secondary layers.  

When starting and finishing with a secondary layer section, all internal primary
sections should have the same number of layers but it is not always possible
because the total number of primary layers may not be exactly divided by the
number of wanted primary sections. So where there were a total of 18 Primary
layers in 4 sections there may be 2 sections of 4 layers and 2 sections of 5 layers.
The size of such inner sections located between secondary sections should
not vary more than 25%.
Equal thickness of insulation should used between
all primary and secondary sections.
This gives minimal problems with
resonances at HF.
Alternatively, one might use an interleaving pattern with
4P and 4S sections, with the sections being 3p-S-5p-S-5p-S-5p-S, starting in
that order from the bobbin bottom. But where possible I try to start and finish
with layers of either Pri or Sec.

Capacitance. The insulation thickness between primary and secondary is
needed to prevent arcing between high primary voltages and earthy speaker
secondaries. As the tube amp primary load is reduced, the effect of shunt
capacitance diminishes, so insulation thickness can be reduced, but kept to
a minimum of about 0.4mm to prevent any arcing and help keep wire layers
neat and flat as the bobbin is wound.

Cathode FB windings will mostly comprise between 10% and 25% of all
primary turns. CFB windings should have only full layers of primary wire
and be distributed in the height of the bobbin winding and located between an
adjacent anode layer with the same thick insulation as used for between all
other primary layers and the speaker secondary layers.
For example, if there were 18p layers in a 4P + 5S pattern, and 3 P layers
were devoted to the CFB winding then CFB % = 100% x 3/18 = 16.7%.
the ideal pattern of layers will be :-
S-4p-c-S-4p-S-c-4p-S-c-4p-S.
Such a pattern ensures the CFB winding is well coupled to all other windings.
However, I have seen a number of OPT with all layers used for a CFB winding
put into one central section, located with Secondary layers on each side.
An example for a pattern of 6S x 5P might be:-
S-4p-S-4p-S-3C-S-3p-S-4p-S. Because the pattern has such a high number
of interleavings the coupling between CFB to S to anode P is good.
but in all cases of CFB windings it is often necessary for Zobel networks
to be placed to load the OPT at very HF at the best location for damping
transients.

Transformers for Electrostatic Speakers, ESL :-

These step up the amplifier voltage between 50 and 300 times.
Their primary winding resembles a tube amp secondary winding meant for
amp output voltages up up to about 30Vrms. Their secondary winding may
need to have greater insulation thickness for high voltages up to 6,000Vrms,
and to achieve lowest possible shunt capacitance. They resemble PP tube
OPTs powered "backwards" and can be designed with the method here.
The design of ESL step-up transformers requires serious additional
understanding of LCR modeling not within this website.

Solid State OPTs and speaker load matching transformers :-

For transformers to couple mosfets or transistors to loudspeakers, or to
couple a given amplifier output load outlet to suit a higher or lower load, the
same amount of interleaving is required for a given power level. When the
primary load becomes similar to the secondary load the number of primary
layers will become similar to number of secondary layers and wire size for
either will become similar. An 8 ohm : 8 ohm OPT with very low dc voltage
differences between primary and secondary would have equal numbers of turns
for primary and secondary and perhaps be simply interleaved so each layer of
thick wire is alternatively devoted to either primary or secondary.
The bandwidth can then be up to 250kHz.
Speaker matching transformers are useful where the amplifier has an outlet
labelled "8 ohms" but the loudspeaker is 4 ohms, and we wish to allow the
amp to experience 8 ohms load or higher. Externally located speaker matching
transformers are very useful for use in OTL tube amplifiers which may have
been designed with many parallel tubes to avoid an OPT, but which would
very much benefit the use of a load of say 32 ohms instead of 8 ohms.
Speaker matching transformers may be designed to be auto-transformers
with ONE combined primary and secondary winding but they must still be
interleaved to give wide bandwidth.

In General :-

But for matching tubes to normal speaker loads between 3 and 9 ohms, the
interleaving list below will give at least 50 kHz of bandwidth, and where there
is a highest number of interleavings the bandwidth can be 300kHz. Using more
interleaving than listed may lead to less available room on the bobbin for wire
due to too many layers of insulation giving poorer high frequency response due
to high shunt capacitance, even though the leakage inductance has been made
very low. The designs here give both low leakage inductance and low shunt
capacitance which are both required for optimal high frequency response and
amplifier stability and freedom from overloading across the widest possible
bandwidth.

The higher the amplifier power and the lower the primary RL becomes, and the
larger the OPT becomes, the number of interleaved sections increases.
So a small 15 watt OPT may only need 3S + 2P sections for 70kHz, but a
500 watt OPT may need 6S + 6P sections.

Tables 2, 3, 4, 5, show interleaving pattern possibilities for SE OPTs.

TABLE 2.
Total P
layers
Primary and Secondary layer distribution.  
P&S
section
pattern
0 to 7W
10p to 24p
S - 10p~24p - S
2S + 1P
7W to 15W 10p to 20p S - 5p~10p - S - 5p~10p - S 3S + 2P
7W to 15W
10 to 20 2p~4p - S - 4p~8p - S - 4p~8p - S - 2p~4p 3S + 4P

TABLE 3.
Total P
layers
Primary and Secondary layer distribution. P&S
section
pattern
15W to 30W 12p 2p - S - 4p - S - 4p - S - 2p 3S + 4P

12p S - 4p - S - 4p - S - 4p - S 4S + 3P

13p
2p - S - 3p - S - 3p - S - 3p - S - 2p 4S + 5P

14p 2p - S - 3p - S - 4p - S - 3p - S - 2p 3S + 4P

14p S - 5p - S - 4p - S - 5p - S 4S + 3P

15p
S - 5p - S - 5p - S - 5p - S 4S + 3P

16p 3p - S - 5p - S - 5p - S - 3p 3S + 4P

16p S - 4p - S - 4p - S - 4p - S - 4p - S 5S + 4P

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

18p 3p - S - 6p - S - 6p - S - 3p 3S + 4P

18p S - 4p - S - 5p - S - 5p - S - 4p - S 5S + 4P

20p 3p - S - 7p - S - 7p - S - 3p 4S + 3P

20p S - 5p - S - 5p - S - 5p - S - 5p - S 5S + 4P

20p 2p - S - 5p - S - 6p - S - 5p - S - 2p 4S + 5P

TABLE 4.
Total P
layers
Primary and Secondary layer distribution. P&S
section
pattern
30W to 100W 13p
2p - S - 3p - S - 3p - S - 3p - S - 2p 4S + 5P

14 p S - 3p - S - 4p - S - 4p - S - 3p - S 5S + 4P

14p
2p - S - 3p - S - 4p - S - 3p - S - 2p 4S + 5P

15p
S - 5p - S - 5p - S - 5p - S 4S + 3P

15p
2p - S - 4p - S - 3p - S - 4p - S - 2p 4S + 5P

16p
S - 4p - S - 4p - S - 4p - S - 4p - S 5S + 4P

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

18p S - 4p - S - 5p - S - 5p - S - 4p - S
5S + 4P

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

19p
2p - S - 5p - S - 5p - S - 5p - S - 2p 4S + 5P

20 p S - 5p - S - 5p - S - 5p - S - 5p - S 5S + 4P

20p
2p - S - 5p - S - 6p - S - 5p - S - 2p 
4S + 5P

21p
3p - S - 5p - S - 5p - S - 5p - S - 3p 4S + 5P

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

22p
2p - S - 6p - S - 6p - S - 6p - S - 2p 4S + 5P

TABLE 5.
Total P
layers
Primary and Secondary layer distribution. P&S
section
pattern
100W to 250W 10p 2p - S - 2p - S - 2p - S - 2p - S - 2p 4S + 5P

10p S - 2p - S - 3p - S - 3p - S - 2p - S 5S + 4P

10p 1p - S - 2p - S - 2p - S - 2p - S - 2p - S - 1p 5S + 6P

10p S - 2p - S - 2p - S - 2p - S - 2p - S - 2p - S 6S + 5P

12p 2p - S - 3p - S - 2p - S - 3p - S - 2p 4S + 5P

12p S - 3p - S - 3p - S - 3p - S - 3p - S 5S + 4P

12p 1p - S - 2p - S - 3p - S - 3p - S - 2p - S - 1p 5S + 6P

12p S - 2p - S - 3p - S - 2p - S - 3p - S - 2p - S 6S + 5P

12p
S - 2p - S - 2p - S - 4p - S - 2p - S - 2p - S 6S + 5P

13p
2p - S - 3p - S - 3p - S - 3p - S - 2p 4S + 5P

13p
S - 3p - S - 4p - S - 3p - S - 3p - S 5S + 4P

13p
S - 2p - S - 3p - S - 3p - S - 3p - S - 2p - S 6S + 5P

14p
2p - S - 3p - S - 4p - S - 3p - S - 2p 5S + 5P

14p S - 3p - S - 4p - S - 4p - S - 3p - S 5S + 4P

14p 1p - S - 3p - S - 3p - S - 3p - S - 3p - S - 1p 5S + 6P

14p S - 2p - S - 3p - S - 4p - S - 3p - S - 2p - S 6S + 5P

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

16p S - 4p - S - 4p - S - 4p - S - 4p - S 5S + 4P

16p 2p - S - 3p - S - 3p - S - 3p - S - 3p - S - 2p 5S + 6P

16p S - 3p - S - 3p - S - 4p - S - 3p - S - 3p - S
6S + 5P

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

18p S - 5p - S - 4p - S - 4p - S - 5p - S
5S + 4P

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

18p S - 3p - S - 4p - S - 4p - S - 4p - S - 3p - S  6S + 5P

19p
S - 3p - S - 4p - S - 5p - S - 4p - S - 3p - S  6S + 5P

20p 3p - S - 5p - S - 4p - S - 5p - S - 3p 4S + 5P

20p S - 5p - S - 5p - S - 5p - S - 5p - S 5S + 4P

20p 2p - S - 4p - S - 4p - S - 4p - S - 4p - S - 2p 5S + 6P

20p S - 4p - S - 4p - S - 4p - S - 4p - S - 4p - S
6S + 5P

21p
S - 4p - S - 4p - S - 5p - S - 4p - S - 4p - S 6S + 5P

21p
3p - S - 5p - S - 4p - S - 5p - S - 3p 4S + 5P

22p 3p - S - 5p - S - 6p - S - 5p - S - 3p 4S + 5P

22p S - 5p - S - 6p - S - 6p - S - 5p - S 5S + 4P

22p 2p - S - 5p - S - 4p - S - 4p - S - 5p - S - 2p
5S + 6P

22p S - 4p - S - 6p - S - 4p - S - 6p - S - 4p - S 
6S + 5P

Fig 3.
Bobbin-section-thru-windings.gif

Fig 3
shows the layers of wire in small transformer bobbin.
The winding pattern is 2S + 1P, with layers
S - 4p - S.

Fig 4.
Bobbin-section-thru-windings-schematic.GIF

Fig 4 shows the bobbin winding diagram as it is drawn up and detailed so that
a tradesman or trades-woman can wind it for the engineer who has designed it.
For those unfamiliar with how transformers are constructed, I suggest the
reader try to dismantle an old transformer, or perhaps actually wind a transformer.
After a few days practice with a winding lathe the items wound will be much
better than a first attempt.


21 Cont'd....  For OPT4. Choose from Table 4, 30W-100W. 

30W to 100W 13p
2p - S - 3p - S - 3p - S - 3p - S - 2p 4S + 5P

Confirm choice, interleaving pattern = 4S + 5P,
Layers are 2p - S - 3p - S - 3p - S - 3p - S - 2p.

NOTE. Above pattern tables include for odd numbers of p layers because
unlike a PP OPT, the SE OPT does not require a CT with the same number
of P layers each side of the CT. The SE transformer may have slight variations
of number of layers in the inner sections.
It is also possible to have an equal number of P and S sections but I favour
the arrangements shown.

Will a CFB winding be used? If so, CFB winding will comprise of one or more
whole layers of primary. 

If yes, proceed to Step 22, if no, proceed to Step 23.

22. Calculate exact CFB winding percentage.

OPT4. CFB winding percentage wanted is always between 10% and 25%
List possibilities :-  2 layers = 15.38%, 3 layers = 23.0%.

OPT4.   Choose 3 layers of primary, 23%, for the CFB windings.

NOTE. Ideally, CFB layers of the primary winding will be tightly magnetically
to all other windings, ie, have low leakage inductance to all other windings due
to the high number of interleaved sections in patterns at this website.


Where several CFB layers are to be distributed up the height of a wound bobbin
they should be placed between a secondary and anode primary layer. The
thickest insulation used between anode primaries carrying B+ and earthy
secondaries must also be used between anode layers and CFB layers.

It is possible to use one whole primary section of several layers as the CFB
winding and locate it between secondaries on each side.

Whatever way the CFB windings are arranged, it is a form of applied series
voltage feedback and there will always be some low amounts of leakage
inductance and shunt capacitance within the FB loop which may cause
oscillations above 50kHz.
In all cases, the instability effects may be damped by well chosen Zobel networks.
These must be designed to give resistive loading at HF, and cannot be easily
calculated and must be created and trialed to minimize square wave ringing with
any value of pure resistance or pure capacitance loading across the secondary.
The most commonly used C loading while stabilizing the HF is 0.22uF.

23. Choose insulation thickness used between primary
layers with the same Vdc potential.

NOTE.  Usually, p to p insulation for all OPT needs to only be 0.05mm thick,
where Vdc between layers is the same, to avoid the risk of shorted turns.
OPT4.  i = 0.05mm.

24. Choose insulation thickness used between Primary
and Secondary layers, or between primary Anode layers
and Cathode layers with full Vdc potential difference.

NOTE. For any interleaved audio coupling transformer, there will be a sum of
Vdc plus peak Vac between adjacent P and S windings and insulation must have
sufficient thickness and dielectric strength to prevent arcing between windings.
The insulation thickness selected will always be far more than required to prevent
arcing because low capacitance is so important. Insulation thicknesses should
be selected from the insulation thickness table 6 :-

Table 6.

Total Vdc + Vac pk,
working maximums 
Minimum thickness,
Polyester sheet.
0Vdc to 100Vac pk
0.1mm
0Vdc to 400Vac pk
0.2mm
300Vdc to 600Vac pk
0.4mm
450Vdc to 900Vac pk
0.45mm
600Vdc to 1,200Vac pk
0.5mm
1,200Vdc to 2,400Vac pk
0.7mm
2,400Vdc to 4,800Vac pk
1.4mm

OPT4.  Calculate probable peak Vac + Vdc between P&S windings allowing
for unknown future OPT use.
500Vdc + 500Vpk ac = maximum working condition = 1,000V.
P-S Insulation Minimum Thickness, I = 0.50mm

Confirm the interleaving pattern and list the layers of insulation which
will be used as calculated so far.

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

Show CFB windings if used :-
Fig 5.
bobbin-details-basic1-SE-OPT4-cfb.GIF

Fig 5 bobbin diagram shows the OPT4 bobbin wire layers with all insulation
nominated.

OPT4 listed Insulation :-
Between primary to primary layers, 8 x 0.05mm insulation
= 0.40mm.
Between primary and secondary layers, 8 x 0.5mm = 4.0mm.
Total thickness of all 20 insulation layers = 4.40mm.

NOTE. The Fig 5 bobbin diagram has 4 secondary sections each with 2
layers of wire with double layer secondaries. 

25.  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 = 4.4mm,

Total height of P wire + all insulation = 11.80mm.

26.  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 ) - 11.80mm = 8.2mm.

th Sec oa dia = ( avail Sec height ) / no of sec layers of wire,
OPT-1A, Th oa dia sec = 8.2 / 4 = 2.05mm
.

27.  Find nearest oa dia wire size from wire size tables.
OPT4   Inspect wire size table in step 12.
Try 1.916mm oa dia, which is 1.80 mm Cu dia = 1.916mm oa dia.

28.  Calculate theoretical S turns per layer, thStpl.

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

OPT4. ThStpl = 72 / 1.916 = 37.58, round down to 37 turns = 37tpl.

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.

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

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 27.
OPT4.  Ns may be 4 x parallel windings of 37t = 37t.
2 x parallel windings of ( 37t + 37t in series ) = 78t.
4 x 37t all in series = 156t. 

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

ZR
Secondary
RL ohms
4 parallel layers 37t 42.86
1,837 0.66 ohms
2 parallel x ( 2 series layers )
78t
21.43
459
2.64 ohms
4 series layers 156t
10.71
115
10.56 ohms

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

NOTE. All configurations tabled above are "legal" and acceptable because
the 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.

Usually the answer to the above question is NO.
OPT design outcomes are rarely determined by Lady Luck.

Are there any **illegal** ways of gaining other load matches between minimum
and maximum in table above?

OPT4.  YES 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 111t
TR 14.29
ZR  204
5.94 ohms
 
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.

30. 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.

Forward to SE OPT calc Page 2.

Forward to SE OPT calc Page 3.