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 PSP 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 OPT1A, 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
section.
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
:
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  10p to 20p  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 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 
TABLE 4. 
Total P layers 
Primary and Secondary layer distribution.  P&S section pattern 
35W to 120W  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  
16p 
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  
18p 
2p  S  5p  S  4p  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  
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 
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  
12p 
S  2p  S  2p  S  4p  S  2p  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  
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
35W120W
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. pripri insulation is nominated as
"i".
OPT1A, 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 pripri 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 
0.1mm 
0Vdc to 400Vac pk 
0.2mm 
300Vdc to
600ac Vpk 
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 
OPT1A,
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.
OPT1A.
Interleaving pattern = 2p

S
 4p  S  4p  S  4p  S 
2p.
Fig 11.
Total
thickness
of
all
insulation
=
5.45mm.
33. Calculate height of Primary layers and all
insulation.
OPT1A, 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.
OPT1A, 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,
OPT1A, 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.
36.
Calculate
the
theoretical
Sec
turns
per
layer.
Theoretical S
turns per layer, thStpl = Bww /
thSoadia (from Step 35.)
OPT1A. 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.
Or with two pairs of paralleled
windings in series to give
Ns = 88 turns, TR =
26.36:1, ZR = 695.0:1 :
Primary
RLaa,
kohms Primary turns, Np 2,320t 
Secondary
RL
ohms Secondary turns, Ns 44t 
4k5 
1.61
ohms 
9k0  3.23
ohms 
18k0 
6.47
ohms 
Primary
RLaa,
kohms Primary turns, Np 2,320t 
Secondary
RL
ohms, Secondary turns 88t 
4k50 
6.47
ohms 
9k0 
12.95
ohms 
18k0 
25.90
ohms 
What conclusions can be made?
How
may
useful
load
matches
are
there
to
the
Middle
RLaa
value
between 2.5 and 10 ohms ?
OPT1A, Middle RLaa =
9k0, with Ns = 44t, load value =
3.23 ohms.
This will suit many "4ohm"
speakers.
There are no other useful matches
to the Middle RLaa.
Step
37,
Continued...
Choose
4A
from
Fig
14.
NOTE. The chosen subsection 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.
Np
=
2,320t,
RLaa = 9k0 
Ns
=
4
//
3N
= 45t, ZR = 2,658 
Ns
=
3
//
4N
= 60t, ZR = 1,495 
Ns
=
2
//
6N
= 90t, ZR = 665 
Sec
load,
ohms
: 
3.38

6.02 
13.52 
Np
=
2,320t,
RLaa = 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
: 
1.50 
3.38 
6.02 
13.52 
Np
=
2,320t 
Ns
=
30t,
ohms 
Ns
=
45t,
ohms 
Ns
=
60t,
ohms 
Ns
=
90t,
ohms 
RLaa
=
4k5,
72 Watts max 
0.75

1.69 
3.01 
6.77 
RLaa
=
6k4 57 Watts max 
1.06 
2.38 
4.26 
9.52 
RLaa
=
9k0 45 Watts max 
1.50
*** 
3.38
***

6.02
*** 
13.52
*** 
RLaa
=
12k7 33 Watts max 
2.12 
4.78 
8.51 
19.12 
RLaa
=
18k0 24 Watts max All ClassA1 
3.01 
6.75 
12.08 
27.04 
Np
=
2,320
turns,
kohms 
Ns
=
4
//
54turns,
ohms 
RLaa = 4k5  2.44
ohms,
72Watts 
6k4

3.45
ohms,
57
Watts 
9k0

4.87
ohms,
45
Watts 
12k7 
6.89
ohms,
33
Watts 
18k0 
9.75
ohms,
24Watts 
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 )
OPT1A, 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%.
OPT1A. OPT TR = 2,320t : 45t,
Load ratio RLaa to sec load = 9k0aa : 3.4 ohms.
From Step 28,
Rwp = 2.26 x ( Np x TL ) / ( 100,000 x
Pdia x
Pdia ), ohms.
OPT1A, PRwp = 2.26 x 2,320 x 275
/ ( 100,000 x 0.355 x 0.355 ) = 114
ohms.
Primary
losses
with
RLaa
=
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 RLaa 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 RLaa 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 RLaa 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 recalculated.
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.
OPT1A,
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.
OPT1A. 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.
42.
Calculate
Fsat
with
Middle
RLaa.
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!!
OPT1A, For 45 watts into 9k0,
Vaa = 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 Vaa at bass frequencies occurs when the
RLaa 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 Vaa level lest there be no headroom
for all other musical frequencies. Therefore core saturation is not a
problem
in most hifi amps even where the Fsat occurs at max Vaa 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.
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 RLaa 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 OPT1A 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 OPT1A, refer to the above bobbin winding layout.
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 OPT1A, 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 PS interface down
from 'anode 1', or the top of the
windup at
above the GHIJKL 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
GHIJKL 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
transformed 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.
If OPT1A 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.
Forward to PP OPT Calcs Page 4.
Back to PP OPT Calc Main Page.