PP OPT calcs Page 3.

Design of OPT-1A continued....

30.  Choose the interleaving pattern.

         Fig 10. Cross section through a hypothetical transformer with
         concentric layered windings. Basics explained.
         Tables 2, 3, 4, 5, showing interleaving pattern possibilities for PP OPTs.
31.  Choose insulation thicknesses.
         Table 6.
Insulation Thickness Vs Voltage.
32.  List layers of insulation which are to be used.
         Fig 11.
OPT-1A bobbin winding diagram for CFB use.
33.  Calculate height of Primary layers and all insulation.
34.  Calculate the max theoretical oa dia of secondary wire.
35.  Find nearest Sec oa dia wire size.
36.  Calculate the theoretical Sec turns per layer.

         Table for ZR, TR, conclusions.
37.  Choose Secondary sub-section pattern.
         Fig 12, Sec = 2L,
         Fig 13, Sec = 3L,
         Fig 14, Sec = 4L,
         Fig 15, Sec = 5L,
         Fig 16, Sec = 6L.
         Fig 17,
Sec sub-section pattern
4A explained.
         Table for TR, ZR and loads. Conclusions.
         Fig 18, Sec sub-section pattern 4C explained.
Table for TR, ZR and loads. Conclusions.
         Fig 19, Link pattern for 4C sub-sections.
         Fig 20, Link pattern for 4A sub-sections.
         Alternative Single Simple termination.
38.  Calculate secondary winding loss %.
39.  Calculate total winding losses.
40.  Calculate total height of bobbin contents.
41.  Draw sketches of bobbin details.
         Fig 21,
OPT-1A, Ultralinear config.
         Fig 22, OPT-1A, CFB config.
         Fig 23, OPT-1A,  Schematic of windings.
42.  Calculate Fsat with Middle RLa-a.
43.  Calculate minimum required Lp.
         Fig 24,
E&I core Lp + µ vs Bac + Vac.
         Many notes and calcs.
44.  Partial air gap for PP OPTs.
         Fig 25,
Gapping effects.
45.  Calculate leakage inductance.
Is LL low enough? 2 methods.
46.  Shunt capacitance of an OPT.
         12 Steps
to determine C, many notes, calcs.
30.  Choose the interleaving pattern.
Choosing an interlaving pattern may entirely bamboozle many readers
or designers who have not much experience with winding audio frequency
transformers for wide bandwidth between about 14Hz and at least 70kHz.

Fig 10.

Fig 10 shows a cross section through a hypothetical transformer with concentric
layered windings neatly wound with an interleaving pattern of 2P + S,
ie, with two primary sections and one secondary section located between
each primary section.

The Fig 10 is drawn to show the contents of a transformer with two windings,
a primary and secondary, arranged in 5 layers.
If we say the smaller wire size wire is all for one primary winding then there are
4 layers each with 12 turns, so Np = 48 turns.
The Secondary winding has 6 turns in one layer. There are two layers of P wire
wound on before the Sec, and two wound on after the Sec, and each pair of
P layers forms ONE section of P winding, so you have 2 P sections with one P
section each side of ONE Sec layer which is ONE section. The simple winding
pattern can be called a 2P + 1S pattern, or P-S-P pattern.
There is only one section of secondary, but there could have been more located
above and below the primary sections shown, if the bobbin space permitted.

This hypothetical transformer in Fig 10 has Turn Ratio = 48t : 6t, , ie, 8 : 1,
and Impedance Ratio = 64 : 1.

With OPT-1A, the number of turns and layers and sections is much greater than
shown in Fig 10.

The number of LAYERS must NOT be confused with the number of SECTIONS.

A winding "section" is one or more layers of wires devoted solely to either
anode primary current, cathode primary current or secondary speaker load
current. There is no direct connection between the sections designated for
anode, cathode or speaker, and the deignated sections are connected only
by way of magnetic coupling only.

I try to design all tube amp OPT with the Secondary arranged so there is never
more than one layer of circular section wire in one secondary "section."

The Primary may usually have more than one layer of wire in each Primary
So an OPT is built up with groups of layers of Primary wire interleaved
with single layer Sections of Secondary wire. Each layer of secondary wire may be
subdivided into "secondary sub sections" to allow varied series and parallel
connections to give variable load matches to suit a wide range of speaker ohm
load values, while keeping the anode load fairly constant and optimal.

There are no designs here which require rectangular section wire or bifilar or
trifilar winding winding which is most difficult for low batch number productions.

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

Where the first and last winding wound onto the bobbin is in a Primary section,
then these sections should have approximately 1/2 the number of layers of the
inner P sections.

If there are 3 outer primary layers in a P "outer" section, the inner P sections
may have 5, 6 or 7 p layers.
When this guide is adhered to there is the best HF response because the leakage
inductance is fairly evenly and symetrically distributed. 

When starting and finishing with an S section all internal P sections should have
the same number of p layers but it is not always possible and having say 2
sections of 4 p layers and 2 sections of 5 p layers is OK. The size of such
"internal sections" should not vary more than 25%. Where such guidelines
are adhered to along with equal thickness insulations between P and S sections,
there are minimal problems with resonances at HF.

The Push Pull OPT designed here using my methods should display adequate
magnetic coupling between each half of the primary winding so that distortion
resulting from current cut off in tubes in class AB is minimised, and need not
be worried about. In the 1930s, the absense of much interleaving in OPT
could create serious problems in class B amplifiers but between then and the
1940s it was realised that adequate interleaving is essential to avoid unwanted
magnetic phenomena.

For transformers to suit high current + low voltage drive devices such as mosfets
or transistors, the same amount of interleaving is required for a given power level
and desired bandwidth. When coupling mosfets or bjts The number of p layers
will be reduced as Primary RL becomes low, and wire dia will increase.

An 8 ohm : 8 ohm OPT with very low dc voltage differences between P and S
would have equal numbers of turns for P and S and perhaps be
simply interleaved so each layer of thick wire is alternatively devoted to either P
or S. The bandwidth can then be very easily made to exceed 250kHz.

As the tube amp primary load is reduced, the effect of shunt capacitance diminishes,
so insulation thickness can be reduced, but kept to a mimimum of about 0.4mm
so that the insulation prevents any arcing and helps to keep layers neat and flat
as the bobbin is wound. Transformers for electrostatic speakers which step up
the amplifier voltage between 50 and 300 times need to have greater insulation
thickness for higher voltages involved and to achieve lower capacitances between
adjacent windings and any other windings. ESL step up transformers resemble
PP OPTs powered "backwards" and can be designed with the method here.
The QUAD ESL57 has a large step up transformer with a primary very much
like the secondary of an OPT. The ESL secondary has 11,000 turns of very
fine wire all neatly layer wound and the total capacitance seen by the amplifier
must be less than the capacitance of the treble speaker panels. For much more
information about ESL step up transformers one should read the theoretical
modelling work by Peter Baxandal written in the 1950s.

But for matching tubes to normal 3 to 9 ohm speaker loads, the interleaving list
below with the number of primary layers per section possible will give at least
70 kHz of bandwidth, and where there is a highest number of interleavings the
bandwidth can be 300kHz.

Using more interleaving than listed leads to less available room on the bobbin
for wire due to too many layers of insulation, and poor HF due to high shunt
capacitances, and higher winding losses. The designs here give good balance
between total shunt capacitance and leakage inductance, so that neither is too high,
and that the resonant frequency generated between them is at a frequency
exceeding 70kHz, and thus able to be damped by R&C Zobel networks without
affecting the amplifier performance below 20 kHz.

For lower Primary RL and higher amplifier power the larger the OPT becomes
and for a given number of interleavings the HF response becomes less due to
increasing leakage inductance. So 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.

Inspect tables 2, 3, 4, 5 below for the power from the transformer.

Tables 2, 3, 4, 5, show interleaving pattern possibilities for PP OPTs :-

Total P layers
Primary and Secondary layer distribution.  
P&S section
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 10p to 20p 2p~4p - S - 4p~8p - S - 4p~8p - S - 2p~4p 3S + 4P

Total P layers Primary and Secondary layer distribution. P&S section
15W to 35W 12p 2p - S - 4p - S - 4p - S - 2p 3S + 4P

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

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

14p S - 5p - S - 4p - 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

Total P
Primary and Secondary layer distribution. P&S
35W to 120W 14 p S - 3p - S - 4p - S - 4p - S - 3p - S 5S + 4P

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

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

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

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

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

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

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

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

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

Total P
Primary and Secondary layer distribution. P&S
120W to 500W 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

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

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

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

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

This example, choose P& S interleaving pattern from 35W-120W table.
Confirm choice of primary layers in step 15, No P Layers = 16.

choose pattern :-   2p - S - 4p - S - 4p - S - 4p - S - 2p = 4S + 5P.

31. Choose insulation thicknesses.
Insulation is used between primary layers
to lessen possibility
of shorted turns where layers have the same Vdc potential, and
to facilitate winding with small diameter.

Usually p to p insulation for all OPT needs to only be 0.05mm thick
where layers have same Vdc. pri-pri insulation is nominated as "i". 
OPT-1A, i = 0.05mm.

For between Primary and Secondary layers, or between Primary anode
layers and cathode layers with full B+ Vdc potential difference between
adjacent windings, insulation will be thicker than pri-pri insulation.

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
0Vdc to 400Vac pk
300Vdc to 600ac Vpk
450Vdc to 900Vac pk
600Vdc to 1,200Vac pk
1,200Vdc to 2,400Vac pk
2,400Vdc to 4,800Vac pk

OPT-1A, Calculate probable peak Vac + Vdc between P&S windings :-
500Vdc plus 500peak Vac swing = 1,000V.

Insulation I minimum thickness = 0.5 mm.

NOTE. In fact Vac swing with no secondary load present may exceed +/- 1,500V
peak at each anode if clamping diodes are not used.  However arcing is unlikely
unless the excessive voltages are maintained for some time, or there is moisture
or pollution present or if poor insulation material is used.

32. List layers of insulation which are to be used.

Confirm interleaving pattern.
OPT-1A.    Interleaving pattern = 2p - S - 4p - S - 4p - S - 4p - S - 2p
Fig 11.

Fig 11 has been inserted here to allow easier determination of the winding insulation
layers required.
OPT-1A :-
0.05mm insulation pri-pri layers, i, height =  9 x 0.05 = 0.45mm.
0.5mm insulation between anode primary and cathode primary = 2 x 0.5 = 1.0mm.
0.5mm insulation between anode and cathode primaries and secondary = 8 x 0.5 = 4.0mm.

Total thickness of all insulation = 5.45mm.

33.  Calculate height of Primary layers and all insulation.

OPT-1A,    Height of 16 Layers of Primary wire of 0.414mm oa dia = 6.62mm.
Height of all insulation layers = 5.45mm.
Total height of Primary + all Insulation = 6.62mm + 5.45mm = 12.04mm.

34.  Calculate the max theoretical oa dia of secondary wire.

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

OPT-1A, Available bobbin height = 0.8 x 22mm = 17.6mm.
Height of all secondary wire layers =
= Available bobbin height - ( height of all insulation + primary wire layers )
= 17.6mm - 12.04mm (from Step 33) = 5.56mm.

Th Sec oa dia = ( Avail Sec height ) / no of sec sections with one layer each,
OPT-1A, Theoretical oa dia sec = 5.56 / 4 = 1.39mm

35.  Find nearest Sec oa dia wire size.
Actual chosen overall wire dia from Wire Size Table must be less than
calculated in Step 34.

Inspect wire size table in step 20.

OPT-1A. Try 1.35mm o/a dia wire, Copper dia = 1.25mm.

36.  Calculate the theoretical Sec turns per layer.

Theoretical S turns per layer, thStpl = Bww / thSoadia (from Step 35.)

OPT-1A. ThStpl = 62mm / 1.35 = 44 turns per layer, (omit fractions of a turn.)

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.

Calculate load matches available where Ns = Minimum turns per layer above,

OPT-1A, Simplest range of Load Matches possible are with all 4 sec
layers in parallel to give Ns = 44 turns,
Np = 2,320t, Sec 44t, TR = 52.7:1, ZR = 2,780:1.

Or with two pairs of paralleled windings in series to give
Ns = 88 turns,  TR = 26.36:1, ZR = 695.0:1 :-

Primary RLa-a, k-ohms
Primary turns, Np 2,320t
Secondary RL ohms
Secondary turns, Ns 44t 
1.61 ohms
9k0 3.23 ohms
6.47 ohms
Primary RLa-a, k-ohms
Primary turns, Np 2,320t
Secondary RL ohms,
Secondary turns 88t
6.47 ohms
12.95 ohms
25.90 ohms

What conclusions can be made?

How may useful load matches are there to the Middle RLa-a value
between 2.5 and 10 ohms ?

OPT-1A,  Middle RLa-a = 9k0, with Ns = 44t, load value = 3.23 ohms.

This will suit many "4ohm" speakers.

There are no other useful matches to the Middle RLa-a.

There is really only one useful load match.

NOTE. This is a typical conclusion which might be made where the initial
theoretical number of secondary layers are either all in parallel to give
Ns = 1 x Tpl calculated, or to give Ns = 2 x Tpl calculated.

  The the secondary turns per layer calculated above may be increased
up to 1.4 times using smaller dia wire without increasing winding resistance too
much, and to possibly allow a greater number of useful load matches to the
Middle RLa-a.

OPT-1A.  Tpl calculated = 44t, may be up to 44 x 1.4 = 61tpl.

NOTE.   In most OPTs, the turns in one or more layers of secondary wire
will have to be divided up to allow additional parallel or series combinations
of secondary turns to give at least 2 or 3 useful secondary load matches
which allow matching of Middle RLa-a to more values of speakers
between 2.5 and 10 ohms.

37. Choose Secondary sub-section pattern.

Inspect Fig 12, 13, 14, 15, 16 patterns of Secondary Winding Sub Sections,
and Choose the Fig which lists the secondary sub-sections for the
number of secondary layers chosen so far.

OPT-1A has 4 layers of secondary winding layers.
Therefore choose Fig 14 where secondary sub-sections are shown for 4
layers of secondary.

Chose a pattern from those shown, 4A or 4B or 4C from Fig 14.
Fig 12.

Fig 13.

Fig 14.

Fig 15.

Fig 16.

Step 37, Continued... Choose 4A from Fig 14.

NOTE.   The chosen sub-section winding pattern requires some futher explanations.

In Fig 17 below, The letter N is given to represent exactly 1/3 of the turns in
a layer of wire. Therefore 1 complete layer of wire has 3N turns.   

Fig 17.

In Fig 17, the pattern 4A has a window below the pattern where it says
"Total turns divisible by 12" and this means that total turns in all 4 layers
of secondary wire MUST be exactly divisible by 12, or else the pattern
will just not work properly.

OPT-1A.   In above steps, Minimum Tpl calculated = 44 turns.
4 layers of 44Tpl = 176 turns Total.
Now 176t is NOT exactly divisible by 12, but from NOTE above the tpl
could be increased by up to 61tpl if desired.

List the numbers above 176 which may be exactly divided by 12, and the
Tpl for each layer and impedance matches are :-
180t, get 45tpl,
192t, get 48tpl,
204t, get 51tpl,
216t, get 54tpl,
228t, get 57tpl,
240t, get 60tpl.

OPT-1A. Choose total secondary turns = 180 turns = 45turns per layer.

List impedance matches available to Middle RLa-a = 9k0.
Primary Np = 2,320 turns :-

Np = 2,320t,
RLa-a = 9k0
Ns = 4 // 3N
= 45t,
ZR = 2,658
Ns = 3 // 4N
= 60t,
ZR = 1,495
Ns = 2 // 6N
= 90t,
ZR =  665
Sec load, ohms :-

Conclusions. This shows that there are TWO secondary load matches between
2.5 and 10 ohms for Middle RLa-a = 9k0 and that they ideally suit speakers
which have a nominal value of 4 ohms and 8 ohms. The third match is good
for 16 ohm speakers.

For higher AB power a 3.38 load could be used with Ns = 60t and RLa-a
will then become 5k0.
A 6 ohm load could be used with Ns = 90t, giving RLa-a = 4k0. 
There is no way high AB power could be generated into a 16 ohm speaker.

For much more pure class A, the 6 ohm load can be used with Ns = 45t,
giving RLa-a = 16k0,
A 13.5 ohm load used with Ns = 60t gives RLa-a = 20k2, all pure class A.

But there is no way to all pure class A power for load of 3.38 ohms.

Let us consider using pattern 4C from Fig 14,

Fig 18.

In Fig 18, there are 4 impedance matches and a table may be drawn up :-

Np = 2,320t,
RLa-a = 9k0
Ns = 6 // 2N
= 30t
ZR = 5,980
Ns = 4 // 3N
= 45t,
ZR = 2,658
Ns = 3 // 4N
= 60t,
ZR = 1,495
Ns = 2 // 6N
= 90t,
ZR =  665
Sec load, ohms :-

This arrangement of windings allows all the connections that are available
with pattern 4A, plus an extra one which gives pure class A with 3.38 ohms,
using Ns = 30t.

Conclusion. Pattern 4C offers the following full range of loads to be used :-

Np = 2,320t
Ns = 30t, ohms
Ns = 45t, ohms
Ns = 60t, ohms
Ns = 90t, ohms
RLa-a = 4k5,
72 Watts max
RLa-a = 6k4
57 Watts max
RLa-a = 9k0
45 Watts max
1.50 ***
3.38 *** 
6.02 ***
13.52 ***
RLa-a = 12k7
33 Watts max
RLa-a = 18k0
24 Watts max
All ClassA1

I have marked the best matches available for most people with *** where
the Middle RLa-a = 9k0, and loads can be 1.5, 3.4, 6.0 or 13.5, depending
which winding arrangement is chosen.

OPT-1A. The pattern 4C from Fig 14 will require 16 terminals for the
secondary windings to be set out on a terminal board to allow soldered
wire links of the terminals in 4 different patterns to achieve the desired
load matching. To many audiophiles, such complexity is likely to lead to
a mistake being made when a new pair of speakers is purchased with
different impedance.

Fig 19.

Fig 19 shows how the sec windings should be labelled using alphabet capital
letters. This means one may have up to 26 single letter labels for any OPT.
It is unlikely that more than 26 terminals would be required.

NOTE. The finished OPT should have a removable cover on the rear panels
of OPT potted enclosures and at least both the strapping patterns should
be clearly shown on the covers to allow anyone to easily make a visual
check to see if the links are correctly wired.
The layout of terminals and links allows the same two terminals to be used
for the 0V grounding point and live active output point from the secondary

NOTE. Primary terminals should be on another board with restricted
access because there are high voltage connections.
Inevitably, someone will try to change load match links without remembering
to turn the amplifier off.
Primary Terminals should be numbered and not
lettered, to avoid confusion when winding or when servicing the amp.

Fig 20.
Fig 20 shows the terminal board 4A link patterns for 4 and 8 ohms.

NOTE. The Fig 20 arrangement does not show the links which would allow
"16 ohms" to be used with Ns = 90t. If this was never required, then only
8 terminals are required because windings AB, CD, and EF may be taken
to just two terminals. Therefore an octal socket could be used to terminate
winding wires and octal plugs may used with their pins strapped in different
ways to give the 4 ohm or 8 ohm matches. Two plugs are kept with wiring
for either 4 or 8 and some means of securing the unused plug to prevent
its loss must be devised. Octal plugs are fairly reliable for the job.

Alternative Single Simple termination.

If the amplifier is likely to be used with a wide variety of load values between
say 3 ohms and 9 ohms, then the secondary may be configured to give a
load match between Middle RLa-a to 5.0 ohms.

OPT-1A. Middle RLa-a = 9k0, and ZR required for Sec = 5.0 ohms
= 9,000 / 5.0 = 1,800 : 1.

Required Turn Ratio = square root of ZR = sq.rt.1,800 = 42.4:1.
Therefore Ns required = Np / TR = 2,320 / 42.4 = 54 turns.

Therefore a suitable arrangement of secondaries should chosen to
give paralleled windings so that Ns = 54 turns.

Above, we have worked out that the tpl may be between 45turns and 60 turns,
and so using 4 layers each with 54 turns will give a suitable load match for
5.0 ohms.

The following load matches are available using Np = 2,320 turns,
secondary of 4 parallel windings of 54 turns each, with ZR = 1,845:1:-

Np = 2,320 turns, k-ohms
Ns = 4 // 54turns, ohms
RLa-a = 4k5  2.44 ohms, 72Watts
3.45 ohms, 57 Watts
4.87 ohms, 45 Watts
6.89 ohms, 33 Watts
9.75 ohms, 24Watts

Conclusion. This table shows that useful load matches exist for all
loads between 2.5 ohms and 10 ohms, and that speakers with nominal
Z between 4 and 8 ohms will give RLa-a between 7k4 and 14k8
and the sound should be fine if average levels are about 1 watt
for speakers rated for 88dB SPL per watt.

Conclusion. The use of variable load match linked windings give the
same low winding losses for each pattern chosen for linked windings
where the RLa-a remains constant.

Conclusion. Where there is a single fixed secondary winding, the winding
losses vary proportionally to the load used at the secondary, so that if
total winding losses were 6% with sec load of 2.44 ohms, then at 4.87 ohms
the losses are 3%, and at 9.75 ohms they are 1.5%.

38.  Calculate secondary winding loss %.

Secondary winding loss % =  100 x Rws / ( SecRL + Rws ) %,
where the Sec RL may be any chosen value likely to be used
or nominated.

The number of turns per layer is more than the minimum Tpl calculated
in Step 36.

Therefore the wire size must be re-chosen to allow the increased number Tpl.

OPT-1A, Theoretical sec oa dia wire = Bww / ( revised Tpl )
= 62mm / 45 = 1.378mm.
Choose from wire tables, try wire oa dia = 1.351mm, = Cu dia = 1.25mm

NOTE. This is the size previously chosen, but for where the Tpl increases
further the wire dia will be smaller.

OPT-1A, Np = 2,320 turns, Ns = 45 turns, RLa-a = 9k0,
RL sec = 3.4 ohms.

Secondary Winding = 4parallel windings each with 45 turns of
1.25mm Cu dia wire.

Confirm Average Turn Length = 275mm from Step 26.

Sec winding resistance
= 2.26
( Ns x TL ) / [ 100,000 x No parallel secs x Sec Cu dia squared ] ohms.
where 2.26 and 100,000 are constants and Cu dia is the wire's copper dia from
the wire tables. ( The resistance of 100,000 mm of 1.0mm dia copper wire
is 2.26 ohms )

OPT-1A, Rws = 2.26 x 45 x 275 / [ 100,000 x 4 x 1.25 x 1.25 ] = 0.045 ohms.

Secondary winding loss % =  100 x Rws / ( SecRL + Rws ) %.
= 100% x 0.045 / ( 0.045 + 3.4 ) = 1.30%

NOTE. Winding loss % is only relevant to the load connected to the sec winding.
For example, if the sec load = 6.8 ohms, then winding losses are lower at 0.66%,
and if Sec load = 1.7 ohms the losses are  2.6%.

39.   Calculate total winding losses.
For the chosen winding pattern with stated primary and secondary loads.
Check that winding losses will be less than 7%.

OPT-1A.  OPT TR = 2,320t : 45t,
Load ratio RLa-a to sec load = 9k0a-a : 3.4 ohms.

From Step 28,
Rwp = 2.26 x ( Np x TL ) / ( 100,000 x Pdia x Pdia ), ohms.

OPT-1A, PRwp = 2.26 x 2,320 x 275 / ( 100,000 x 0.355 x 0.355 ) = 114 ohms.

Primary losses with RLa-a = 9k0 = 100% x 114 / 9,114 = 1.25%.

Rws from Step 38 = 0.045 ohms
Secondary losses with Sec RL
= 100% x 0.045 / ( 0.045 + 3.4 ) = 1.30%

Total winding loss % = P loss % + S loss %

= 1.25% + 1.30% = 2.55%

Total losses are less than 3% with RLa-a 9k0 : 3.4 ohms.

Total Losses vary inversely with the secondary load, ie, if the Sec load
is halved, the total winding losses are doubled.

If Sec RL = 1.7 ohms and RLa-a is 4k5, then with the same
OPT turn ratio, losses will be ( 3.4 / 1.7 ) x 2.55% = 5.1%.

NOTE. If total winding losses exceed 7% with minimum RLa-a and sec load,
a larger Core T size may be needed to allow thicker wire sizes, or higher
stack with fewer turns of thicker wire. The whole design would have to
be re-calculated.

40.   Calculate total height of bobbin contents.

This includes P and S windings, insulation and bobbin base thickness and total
height should be no more tha 90% of the winding window H dimension.

Primary wire layers, 16 x 0.414..........................6.624mm.
0.05mm insulation, p to p layers, 9 x 0.05mm......0.450mm.
0.5mm Insulation, p to p layers, 2 x 0.5mm.........1.000mm.
0.5mm Insulation p to S layers, 8 x 0.5mm .........4.000mm.
Secondary wire layers, 4 x 1.351mm...................5.404mm.
0.2mm Insulation wound over last on P layer.......0.200mm.
Bobbin base thickness........................................2.000mm.

Total winding height including bobbin base.........19.678mm.

Calculate 90% of window H dimension = 0.9 x H.

OPT-1A. 0.9 x 22mm = 19.8mm.
Total height of all bobbin content plus bobbin base is less than 19.8mm, OK.

NOTE.  If bobbin content + bobbin base height is more than 90% of H, it
may be difficult to insert E shaped laminations into wound bobbin.
If this is the case, it may be
necessary to revise all calculations after selecting a larger Tongue size or
larger Stack height with less turns for the core.

NOTE. Wire will not lay perfectly flat as layers are put on and will tend to
spring up across the rectangular core. The windings will develop a bulge
and apparent winding height will become higher than calculations predict.
Layers of wire and insulations will not be perfectly tight. I suggest it is
important to apply slow setting epoxy varnish by brush to all windings
and all surfaces of insulations as the bobbin is wound and then gently
cramp the wound bobbin with a g-clamp and carefully cut blocks of
plywood between bobbin cheeks to flatten the wind up. This will increase
the tightness of all bobbin contents and cause varnish to further penetrate
any voids.
The well varnished bobbin will need to have all layers of wire and
insulation glued well together to minimize audio frequency "howl noise"
from the OPT.
Before removing the bobbin from the winding lathe, the completed bobbin
is left cramped up for 2 days to give a long enough setting time for the
epoxy varnish
This prevents the wound bobbin shape deforming if it is not held
between plates on the lathe. Care must be taken to prevent adherence
of excessive varnish to parts of the lathe.
The practice of hand winding OPT can be extremely messy and the smelly
fumes given off by the epoxy varnish can be quite toxic. The winding work
should be done in a well ventilated workshop. I have used Wattyl 7008
two pack epoxy polyurethane floor varnish which has mixed pot life
long enough to allow a full day for winding an OPT like OPT-1A.
I found that to clean my hands I needed to use methylated spirits
and a clean cloth constantly throughout a winding process.

To gain winding skills, there is no substitute for practice, and I suggest
beginners learn by winding a perfectly layered and insulated choke
before they even think about an OPT.

41. Draw sketches of bobbin details.
To ensure whoever winds the transformer will be able to adhere to
the diagram without any confusion, guesswork, or needing further

Fig 21 and Fig 22 below have the same details for winding and insulation.

Fig 21 shows the connection of the windings if simple Ultralinear taps for
screen connection are used. 37.5% screen taps are points 5&16 and 50% at
points 4&17. 25% taps are available at 4&13.

Fig 22 shows the way cathode feedback windings are formed using two primary
layers, 6-7, 12-13. This CFB winding arrangement means that 12.5% of the Vac
between anode and cathode appears at the cathode of an output tube.
For 71 Watts into 4k5, Va-a = 566Vrms, Va-k for each 6550 = 283Vrms, with
+247.6 Vrms at anode and -35.4 Vrms at cathode. If the Vgk = -20Vrms, then the
grid input voltage to each 6550 = -55.4Vrms. 25% CFB could be used if all 4
central primary layers 6 to 13 were used for 25% CFB. But then grid input
signal will rise to 90.8Vrms, and be more difficult to produce without added THD.

Fig 21.

Fig 22.


Fig 23.

NOTE. If the design method here is followed so far, there should be no need to
make any further calculations of Leakage Inductance, Saturation Frequency at
Bac max, Shunt Capacitance, or Maximum safe Idc for the primary wire.
But it is prudent to check all thse things.

NOTE.  The design calls for partial air gapping to ensure the OPT does not
become saturated too easily at F below 20Hz. The Partial Air-gapping technique
is a mostly forgotten and ignored practice in 2011 because to get this done
correctly the permeability of the core material must be carefully measured and
a gap established experimentally, ie, by trial and error. 

42. Calculate Fsat with Middle RLa-a.

Fsat  = 22.6 x V x 10,000
            S x T x Np x B 

where Bac is in Tesla, with 1 Tesla = 10,000 gauss,
22.6 and 10,000 are constants for all transformer equations,
V = Vrms signal voltage across the primary, or sq.rt ( PO x PRL )
S = core stack height,
T = core tongue width,
Np = primary turns,
F = frequency at which B is to be measured.

All dimensions in mm!!

OPT-1A, For 45 watts into 9k0, Va-a = 636Vrms, Bac max = 1.6Tesla,
S = 59mm, T = 44mm, Np = 2,320 turns,

Fsat at B = 1.6Tesla  =  22.6 x 636 x 10,000     =  14.92Hz.
                                   59 x 44 x 2,320 x 1.6

NOTE. This check on Fsat is necessary in case a gross error may have been
made. The initial aim of the design was to achieve Fsat at 14Hz, and if Fsat
is not more than 25% of the design aim then the design will work flawlessly.

NOTE. The actual maximum Va-a at bass frequencies occurs when the
RLa-a is very high, often because at bass frequencies there are peaks in the
bass speaker impedance of perhaps 10 times the nominal speaker load Z.
However, most music has very little content below 32 Hz and bass signals
cannot be at the maximum possible Va-a level lest there be no headroom
for all other musical frequencies. Therefore core saturation is not a problem
in most hi-fi amps even where the Fsat occurs at max Va-a at say 32Hz.
I prefer all my OPTs to saturate at 14Hz if possible because there is less
bass distortion and the bass is subjectively superior. I make no apologies
for the size or weight of my designs.

43.  Calculate minimum required Lp.

For any PP OPT, the primary inductance Lp should have
XLp > 2 x RLa-a at Fsat at Va-a > 1/2 the maximum possible sine wave

Inductance in iron cored components is not a perfectly linear property
and varies with applied voltage, frequency, and the permeability, µ, of the
core material. The core material is usually grain oriented silicon steel, GOSS,
and its maximum µ may be at aproximately 30Hz. Most manufacturers of
GOSS cores do not ever state the µ of their GOSS, because most makers
cater for the manufacture of power transformers operating at 50 or 60Hz,
and the only relevant core information is the "loss per Kg" in Watts/Kg
for a given Bmax, usually at about 1.5Tesla.

But for a PP OPT, the µ needs to be known, and one can determine the
µ by using a test coil, Vac source and a sample of the core material.

I have drawn a graph of the results of such a test :-

Fig 24.

Fig 24
shows the graph gained after recording the Vac across a test coil with
430 turns around a GOSS core with a 10mm Stack x 28mm Tongue E&I lams.

The graph is on LOG SCALES in both axis.

If the maximum possible µ of laminations or C-cores selected is unknown,
it may be measured using the method to gain the graph in Fig 23 above.

I used a 10mm high sample Stack of 28mm Tongue size E&I lams which were
assembled into a temporary coil of 430 turns of 0.55mm wire on a convenient
bobbin into which the sample core fitted. The small stack of laminations was
maximally interleaved.

I used a sine wave sig gene set at 47Hz to feed a tube power amp capable of
20Vrms of output. A series resistor of 100 ohms was used to sense current.
An oscilloscope used to monitor the voltage across the 100 ohms and to view
the increase in the current wave distortions as the voltage to coil was increased
from 0.125Vrms up to 19Vrms when THD > 10%. I have recorded the measured
Vac and Iac on a table in Fig 23, and then calculated XL = V / I and calculated
the inductance L, and the Bac for various applied Vac.
Notice the graph curve for inductance & µ is NOT linear. If the inductance
remained constant for all applied Vac levels the graph would be a straight line.

Notice the L curve shows maximum L of about 2.1H at 12Vac at 0.48T.

µ =    Lp x 1,000,000,000 x ML  
         1.26 x Np squared x T x S

In this example,
µ = 2.14 x 1,000,000,000 x 156 
= 5,118.
      1.26 x 430 x 430 x 28 x 10

At only 0.125Vac the L has become 0.4H, and µ has become 956.

If we assume the material for OPT-1A is similar to the above sample for a
test, it is safe to say the maximum Lp will be when Bac = 0.5 Tesla at 50Hz.
In practice, µ may be slightly higher below 50Hz, and above 50Hz the µ
reduces and may be 1/10 of its maximum at 1 kHz, and negligible at 10kHz.
In fact, at above 10kHz the OPT does not need an iron core and it works
perfectly well without any iron core as an air cored OPT.

Most core material measurements are done at 50Hz. 

What we can say about the test results is that µ will reduce from its
maximum of approximately 5,000 at 0.5T down to approximately 1,000
at 0.005T. The Vac may be reduced by a factor of 0.01 to give a µ
reduction of about 0.2.

The maximum Va-a for 45Watts into 9k0 = 636Vrms. At 50Hz, the Bac
= 0.48 Tesla, so µ should be at near its maximum of 5,000 and so to
LP will be at its maximum = 359Henrys. Therefore if the Vac is reduced
from 636Vrms down to 6.36Vrms, the µ will be approximately 1,000,
so Lp will be 71.8 Henrys.

At Fsat = 14.9Hz, the 636Vrms produces Bac = 1.6Tesla with core
saturation. Not all E&I lams will reach up to 1.6T without saturation and
some cores saturate well below 1.6T. But now we are interested in inductance,
and at 0.5T at Fsat, Va-a is about 200Vrms and there is no saturation and µ should
be at its maximum of 5,000 giving maximum Lp of at least 359H.
At 0.005T, Va-a = 2.0Vrms, and Lp should be 71.8H.

My recomendation for Lp at high level operation is :-
Lp = 2 x RLa-a / ( 6.28 x Fsat ) Henrys.
where RLa-a = Middle RLa-a, 6.28 is a constant of 2 x pye,
ie, 2 x 3.1428, Fsat is at full clipping Va-a for max PO at 1kHz. 

For OPT-1A,  Wanted Lp at high level,
Lp = 2 x 9,000 / ( 6.28 x 14.9 ) = 204 Henrys.

Calculate core µ required to achieve the wanted Lp :-

µ =    Lp x 1,000,000,000 x ML  
         1.26 x Np squared x T x S
where µ = permeability, Lp is inductance in Henrys, 1,000,000,000 and 1.26
are constants, ML is the iron magnetic path length, Np is primary turns,
T = core tongue size, S = core stack. All dimensions in mm.

OPT-1A,  ML for wasteless core T = 44mm is 275mm.
µ =     204 x 1,000,000,000 x 275    3,186.
      1.26 x 2,320 x 2,320 x 44 x 59

Conclusion. If the core has max µ above 3,186, the Lp will be sufficient to
prevent overloading by inductive shunting of tube signal current above Fsat.

The reduction of µ above Fsat from a max = 3,186 but before saturation
commences may be approximately 30%, so at full PO load voltage the µ may
be 2,230. The inductance would then be down 30% and = 143Henrys.

XLp at full PO and at 14.9Hz = 13,362 ohms.
RLa-a = 9,000 ohms.
Combined impedance of RLa-a and Lp in parallel
           RLa-a x XLp                             
     sq.rt. ( RLa-a squared + XLp squared )
= 9k x 13.36k / sq.rt ( 81k + 178k ) = 7,470 ohms

Conclusion. The RLa-a at 14.9Hz at the full PO level will be slightly
less than the 9,000 ohms so there will be some tube overloading distortion
in addition to saturation distortion if the full PO level is maintained.
But at 0.7 x max PO, or 445Va-a, ie, 1/2 full PO, the should be only a
very slight reduction of the open loop response level, even with pure beam
tetrode operation when Ra-a is very high at about 60k ohms.
Therefore, the core µ may be allowed to be as low as 2,200 at high Bac
levels while having a maximum at 3,186 at some lower level

The use of any GOSS with µ above 3,186 is permissable. Even NON grain
oriented steel cores may be used, and although the LF distortion may be
much higher than GOSS, it is still in audible because it remains lower than
tube distortion above 20Hz, but only IF the design theory I have at this
website is followed, where my designs use more iron and turns of wire
than most other manufacturers.

The reduction of Lp at low levels of Va-a can lead to LF instablity, especially
where the tubes operate in pure beam tetrode and if they have no load
connected. This problem is entirely solved if a LF shelving network is placed
between the input and driver tube stages of the amplifier. See my many
power amp schematics for this application.

Some GOSS have a much higher maximum µ well above 5,000, and I have
used some E&I lams which have measured a max µ = 17,000. Some C-cores
reach 12,000, and toroidal cores using GOSS may have µ = 40,000.

I don't like toroidal OPTs which I have seen used in some very expensive
PP amps and which I qualify as "Glorified Garbage" Toroidal OPTs made by
Plitron would be the exception, and they are supposed to be excellent, and
are so highly priced that none of the low end makers such as made at
http://www.garbageaudio.com can afford to install them, even though they
want $5,000 for a typical generic 5050 amp with 2 x KT88 per channel.
The horrid samples of toroidal OPTs I have seen do not have enough
load matching ability, and the load matches they do have are plain wrong,
along with their Ea, and many other details. The OPTs usually have
been wound with very thin insulation used between anode primary windings
and the earthy speaker secondary and shunt C can be 10 times what it
should be.
The other big problem with toroidal cores using a strip of GOSS wound
in a tight spiral is the very high µ which can be 40,000. The slightest
difference in Idc in each 1/2 primary winding can magnetize the core so
much there is little magnetic headroom left for signal magnetization.
In any core, the sum of Bdc and Bac cannot exceed the maximum B for
the core. So excellent Idc balancing is needed for any PP amp with a
toroidal cored OPT. The other problem with high core µ in OPTs is that
the core saturation with Vac applied is extremely sudden, and stray Idc
non balance sways and very low F cause saturation all too easily.

Suppose OPT-1A had Np = 2,320 turns, and µe = 40,000,
and net Idc difference across the primary = 5mA, and ML = 245mm.
Then :-
The dc field strength, Bdc = 12.6 x µe x N  x Idc
                                                 ML x 10,000
where Bdc is in Teslas, ue is effective permeability,
N is the turns,
Idc in Amps dc,
ML is the magnetic path length of the iron in mm,
and 12.6 and 10,000 are constants for all equations to work.

Bdc for toroidal OPT1 =
12.6 x 40,000 x 2,320 x 0.005 / ( 245 x 10,000 ) = 2.38 Tesla.
This is a "silly formula result" because the GOSS cannot be magnetized
more than 1.6 Tesla. But we can conclude that very little Idc non-balance
is needed to magnetize the core fully, ruin the music, and cause the tubes to
over heat badly.

If the core material has µe = 3,186 which was calculated as the advisable µ
above, then Bdc with 5mA of non-balance = 0.189 Tesla, and despite some
reduction in distortion bass signals the music would be little affected.
In many amplifiers I have brought to me for servicing, Idc non balance is
often more than 20mAdc if the tubes are aged, or the fixed bias has been
very poorly adjusted.
To avoid this problem, I like to use self regulating cathode biasing with some
adjustablity of Idc balance with a pot, or have fixed bias with a balance adjust
pot and a pair of LEDS to indicate the balance status visually, so that when
the balance pot adjusts for LEDS to glow equally bright, the Idc is balanced
within about +/- 3mA dc, and nobody has to do any more than use a finger
nail in the pot shaft to make the adjustment. Hardly any commercial maker
includes such features as I have explained at my 5050 amp
and others
at my website.

The use of GOSS E&I cores or C-cores may result with µ being too high,
say over 4,000. To avoid the problem of the high Bdc and unwanted
saturation effects with very low F, the core may be air gapped.....

44.   Partial air gap for PP OPTs.

For GOSS E&I laminations or C-cores with excessively high permeability.

There is rarely any need for maximum µ to be higher than 3,000.

With most GOSS core E&I material, if the Es are ALL facing one way with a pile
of Is against them, it is a form of "normal air gapping" even though the Is are
hard against the Es. In such a case the µ may only reach a maximum of 800
and it is too low to obtain optimal primary inductance. The use of partial air
gapping and partial lamination interleaving gives µ max which is intermediate
between full interleavig and normal traditional air gapping.

To achieve a good reduction of maximum µ for high µ material, the Es and Is
may be arranged in sub-stacks, with each sub-stack of Es and Is butted together
but each set of butted Es and Is facing in opposite directions to give both partial
airgapping and a much reduced number of reverse direction interlavings.
I re-arranged the laminations used to draw Fig 24 so that the 20 Es and Is
were divided into 3 sub-stacks of 6+7+7 lams thick, and the top stack and
bottom stack faced east, while the middle stack faced west.
The same coil was used and test results shown on Fig 25 :-

Fig 25.

Fig 25 has Curve A being the same curve gained in Fig 24 for maximum E&I
interleaving plus the Curve B which was gained with the partial air gapping.
The reduction of maximum µ by partial air gapping is from 5,100 to 2,100.

The curves shown here are a guide to how the ardent audio enthusiast and
amplifier designer may find out the nature of the materials he chooses to use
and how to optimise his design for wide bandwidth, low THD/IMD and
freedom from adverse saturation effects at low frequencies.

Before anyone finalises their OPT construction the samples of the core should
be tested. This may be done using the actual wound bobbin with output tubes
as the Va-a voltage generator, and at least with tubes operating in Triode to
take advantage of the low Ra-a anode resistance to minimise distortion.

Tests of core material may also be done using a test coil fed with signals from
the output of a normal audio amp with low Rout. The Test coil may be the
secondary winding of the bobbin to be used, or a conveniently wound coil
with perhaps 500 turns of 0.5mm wire to allow a stack of say 20mm high
of proposed E&I to be tested.

With C-cores, the partial air gapping is really a normal proper air gap,
but it would be much smaller than that used for a similar power rated Single
Ended OPT with a high Idc flow. In a 50Watt mosfet powered amp I built
before 2000, I used C-cores for a PP OPT and the µ max was about 5,000
max with cut surfaces tightly together, but with a single layer of plastic about
0.02mm thick, the µ max was reduced by 1/2, and the saturation behaviour
became much less likely to cause excessive currents in the mosfets. Bass
performance is excellent, see the amp details at 50 WATT PP CLASS A AMP.    

45.  Calculate leakage inductance.

The leakage inductance in written specifications is sometimes described
as being "referred to the primary."
This means it is considered to be an
inductance in series with the primary load looking into one end of the
primary with the other end grounded. Typical good values of this inductance
would be 10 milliHenrys for where RLa-a = 10ka-a, ie, 1mH per 1k0 of RLa-a.

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
a = height of the finished winding in the bobbin,
b = the traverse width of the winding across the bobbin.

Distances are all in mm!

For OPT-1A, LL  = 
0.417 x 2,320 x 2,320 x 275 x [ ( 2 x 8 x 0.6 ) + 19.0 ]  
                            1,000,000,000 x 8 x 8 x 62  

           =  0.00444 Henry = 4.44 mH.

Is the leakage inductance low enough?

Method 1:-
The simplest way to answer is to see if there is at least
1mH of LL for each 1k0 of load RLa-a.

The leakage inductance causes the lowest HF pole
where RLa-a is the lowest value.

OPT-1A. Lowest RLa-a expected = 4k5.

There are 4.5 1k0 units of RLa-a

No of mH per 1k0 = 4.4mH / 4.5 = 0.977 mH per 1k0 .

This is slightly less than 1.0 and appears OK.

Method 2 :-
Calculate reactance, XLL, of LL at 100 kHz.

XLL at 100kHz = LL in Henrys x 2 x pye x F
, ohms.

OPT-1A, XLL = 0.00444 x 6.28 x 100,000Hz = 2,788 ohms.

Is XLL less than minimum RLa-a at 100 kHz?

OPT-1A, RLa-a min = 4,500 ohms,  ZLL = 2,788 ohms at 100 kHz.

XLL is lower than RLa-a at 100kHz. OK

46.  Shunt capacitance of an OPT.

Capacitance in tube  amp OPTs affects the HF response and C exists between
primary turns and between primary layers which all sums to an amount of
"self capacitance" of a primary winding. With a much interleaved primary
winding the self C of each primary section are effectively all in series and
the amount of total self C for the whole primary is a negligible amount with
very little effect on the response below 100kHz. But there is considerable C
between adjacent primary and secondary sections where there is high Vac
in the primary sections and negligible Vac in the earthy secondary sections.
The total measurable shunt capacitance across the whole primary is of great
importance. This is called the "lumped shunt C", or "total primary shunt C".

The easiest way to measure the shunt
C between anode and 0V is to set up the OPT as follows :-

1. Connect the secondary windings as they will be used in the amplifier and
connect one end of the two secondary terminals to 0V.
2. Do not have any resistance load connected across any windings.
3. Connect the primary winding CT to 0V.
4. Connect one end of the primary winding to a signal generator with
Rout = 600 ohms.
5. Connect a series resistance of 5 x RLa-a between sig gene output
and one primary end.
6. An oscilloscope with a high impedance input and low capacitance probe is
connected between the primary end and 0V.
7. 5Vrms at 1kHz is applied from the sig gene so you can see a good wave form.
8. While keeping the gene level at a constant 5Vrms, you should be able to plot
a graph of the frequency response at the OPT primary between 2Hz and 200kHz.
9. For the OPT-1A example, the series R between sig gene and OPT would be 22k0.
10. You should find that the response will show a flat central portion each side of
2kHz with the HF pole at -3dB at say 7.6kHz.
11. C may be calculated,

C = 159,000 / ( R x F ), where C is in uF, R in ohms and F in Hz.

For the example, C = 159,000 / ( 22,000ohms x 7,600Hz ) = 0.00095 uF
= 950pF.

12. Let us suppose you subtract the probe capacitance of say 40pF from the
calculated C value to give Shunt C = 910pF. Now the applied voltage to
one end of the OPT primary will appear at the other end of the primary, but of
opposite phase.
Regardless of phase, you have effectively applied signal to BOTH sides of the
OPT primary, and in fact the capacitance "looking into" one end of the OPT
primary is in fact the effects of TWO lots of capacitance, ie, 455pF which
exists between each end of the OPT and 0V.
Many people will try to measure capacitance with a DMM, or in ways other
than I have specified here and they will make very big mistakes and obtain
completely WRONG measurements.

To calculate the primary shunt capacitance in an audio transformer
such as OPT-1A, refer to the above bobbin winding layout.

The average distance between the copper surfaces of primary and secondary
layers including the insulation thickness of 0.5mm and the wire enamel of
about 0.05mm is approx 0.7mm. Then you must allow for the curved
surface of the wire turns so total distance = approx 0.75mm. In other words,
you may consider layers of wire as pieces of equivalent adajacent sheets of

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.

For OPT-1A, there is some variation of turn length but calculations will
be accurate enough if the TL is the average for all turns, ie, 275mm.
Traverse width across the bobbin = 62mm, so area of each interface
= 275 x 62 = 17,050sq.mm. Polyester insulation has K = 2.5 approx but
some of the gap between turns is air, so allow K = 2.5.

The d we calculated above = 0.75mm.

C in pF = 17,050 x 2 / ( 113.1 x 0.75 ) = 402pF.

The first P-S interface down from 'anode 1', or the top of the wind-up at
above the GH-IJ-KL is at a position of 6.5 layers / 8 layers along the P
winding from the CT where the signal voltage is zero. This positioning
results in the capacitance being subject to the impedance ratio at this position.
Therefore the C is transformed to [ 402pF x (6.5 / 8) squared ], = 402 x 0.66
= 265pF.

The next area of 360pF down from anode 1 appears below the sec layer
GH-IJ-KL and the impedance ratio is (5.5 / 8) squared = 0.47 so the C
due to this interface at anode 1 = 402pF x 0.47 = 189pF.
Next down the Z ratio is above the EF sec and = (2.5 / 8) squared = 0.098,
so C = 39pF at anode 1.
Below EF the Z ratio reduces the 402pF by (1.5 / 8) squared = 14pF.

The total C appearing at the anode 1 connection is the sum of all these
capacitances = 265pF + 189pF + 39pF + 14pF = 507pF.

507pF is the shunt C from one anode to 0V where the CT is connected to 0V.
The actual total C looking into ONE primary end is twice 507pF = 1,014pF.
We can consider that each tube powering an OPT in class A sees 507pF
between anode and 0V. In class AB, each tube sees 507pF in class A but
sees 1,014pF after the other tube cuts off.

Where the number of interfaces is 4 or more, then total C equals aproximately
1/3 of the tptal of all C simply added. In this example, total C = 4 x 402 pF / 3
= 1,608pF/3 = 536pF, fairly close to the 507 calculated before.

Cathode Feedback windings complicates the capacitance calculation.
The effect of the capacitance on amplifier bandwidth is effectively reduced
by the CFB because the CFB effectively reduces the Ra of the tubes.

The C at the anode is reduced, while C at the cathode has little effect
because the CFB winding is a much lower impedance winding than the
anode winding.

If OPT-1A is used with a pair of 6550 in pure class A beam tetrode mode
the Ra of each tube is 32,000 ohms.
Without any resistance load connected, the gain of the tubes will be close to
µ at about 1kHz but reduce by -3dB where the capacitive reactance = Ra.
If C = 507pF, the -3dB pole should occur at the calculated frequency
= 159,000 / ( 32,000 x 0.000507 ) = 9,800kHz.

In practice this would be only approximately correct, and the better way
to test for C is to use a known low Z gene and a real series resistance as
explained above. Usually the leakage inductance, LL, will have no effects on
any methods described so far because the LL should be such a small value.

The capacitance and leakage inductance will react together to form a tuned
circuit and low pass filter with an ultimate slope of more than 6dB/octave.
While measuring the shunt C of the OPT the series R used reduces the Q
of the tuned circuit to much less than 1.0. But when the OPT primary is
tested with a signal source of very low resistance, say 600 ohms, and
without any resistance loads, the plot of high frequency response may
often show a high peak above 20kHz before rolling off at 12dB/octance.
So rapid phase shift increase occurs as F becomes high so it is important
to minimise C and LL to force the frequency of resonance to be as far as
possible above the audio band and where the phase shift with loop NFB
does not cause oscillations.

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.
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.

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