OPT4. Fig 8 below has
sub-sections for 4 sections of
Choose a pattern from those
shown, 4A, 4B or 4C from Fig 8.
continued...Confirm choice of 4A from Fig 8.
winding pattern requires some further
The division of the single layers of secondary windings into exact fractions of
a layer allows the windings to be arranged to suit a number of secondary
load values yet maintain the same OPT primary load value connected to
arrangement in 4A to work properly, it is necessary
TOTAL number of all S turns in all layers is exactly divisible by all selected
numbers of paralleled windings.
For example, in 4A, the total
turns = 4 x 3N = 12N. Each choice
windings has a different fraction of the total number of all secondary
winding turns in all secondary layers.
Their relative turn number are described as 3N, 4N or 6N turns.
The total number of all secondary turns must be divisible by 3, 4 or 6.
Fortunately, it is easy to check if any number of total S turns are divisible
by 3, 4, or 6, and if not, it is easy to increase the total turns to fit the
calculation of turns per layer in step 28 gives a
of S turns which cannot be exactly divided by the wanted numbers of paralleled
windings, it must be increased to comply with this rule.
For this design example, if in step 28, 47 turns were calculated, and pattern
4A is chosen where the secondary = 4 layers of wire, then total S turns
= 4 x 47 = 188t.
188 divided by 3 = 62.66, and divided by 6 = 31.33, both unacceptable because
the division produces a fraction of a turn.
But if the Tpl is increased to 48t, total S turns = 192, and is exactly divisible by
3, 4, and 6.
In fact Tpl from step 28 = 37tpl. Total turns are 148t, and for pattern 4A the
totals could be 156t, 168t, 180t, 192t, 204t, 216t, 228t, 240t and each are
able to be divided by 3, 4 and 6 to give Tpl of 39t, 42t, 45t, 48t, 51t,
54t, 57t, 60t.
The increase in secondary
winding resistance may be marginal
if total of P and S winding resistance losses are less than 7%. The most efficient
OPT have nearly equal winding loss % in both P and S windings, and if the P
losses are say 2.2%, then S losses may be allowed to be 4.8% for total of 7%.
Consider that you
may have selected a 5S + 4P
from Table 4, and then chosen sub section arrangement 5A from Fig 9.
There are 4N turns for each of 5 layers giving a total of 20N turns. The windings
are grouped either as 4 paralleled windings with 5N turns or 5 paralleled windings
of 4N turns. The total turns must be divisible by 20. Ie, by 4 and 5.
The available winding height for secondary windings may be 5.4mm, so each
of the 5 layers has theoretically 1.08mm oa dia wire, and we may select
0.95mm Cu dia with oa dia = 1.04mm, giving 60t per layer.
5 Sec layers = 300t, and divisible by 20, ( and are we not lucky? ).
The possible total S turns = 300, 320, 340, 360, giving Tpl of 60, 64, 68, 72.
The Tpl will be then exactly divisible by 4 to give the 5A pattern.
Where Tpl = 60t, there are 4 layers at 60t each and one layer of 4 x 15t.
Thus Ns may be 5 // 60t or 4 // 75t.
Load matches with RLa = 1,213 ohms and 1,586 P turns are :-
60t = 1.74 ohms, 75t = 2.7 ohms.
With Tpl = 72t, the two S loads will be:-
72t = 2.5 ohms, 96t = 3.9 ohms.
This 5A pattern does not include a wide range, so the choice of some other
pattern for 5 Sec layers must be tried.
selection of 5B, with windings of 5N, 2.5N and N
with turns in each layer = 6N.
If one layer has 60t, total turns for 5 layers = 300t and it is divisible by 30,
ie, by 5, 7.5 and 10. The Sec may be :-
6 // 50t = 1.2 ohms, 4 // 75t = 2.7 ohms, 3 // 100t = 4.8 ohms, 2 // 150t = 10.8 ohms.
There is only one load match between 3 ohms and 9 ohms, and more trials
of other patterns should be done, with careful attention to dimensions and
and winding losses.
When considering designs with 5
Sec layers there is more
finding a useful range of load matches fairly well spaced apart.
Ideal load match spacing would be 3 ohms, 6 ohms, 12 ohms, but for this you
need sec turns with relationship of 1N, 1.414N and 2N turns.
Nice convenient fractions of layers are 1/2, 1/3, 1/4 and 1/5, and combinations
of fractions of layers will never give the 0.414 fraction, but give 0.50N, 0.333N,
0.25N, or 0.20N. So a popular range of relative N might be 1N, 1.5N, 2N, which
would match 3 ohms, 6.75 and 12 ohms which is ideal for nominal speakers
of 4, 8 and 16 ohms.
The aim is to find :-
A. Easy fractions allowing a good spread of loads,
B. Equal current density in each and every secondary winding wire,
C. Low enough secondary winding resistance.
D. Not too many secondary connections,
E. No winding taps unless the tapped secondary option is initially wanted.
However, Some slight imperfect
arrangement with non equal
in wires will be tolerated without causing significant extra winding losses or
changed frequency behaviour. For example, in Fig 9, 5E, each layer of 5 is
divided to give N, 2N and 3N sub-sections.
The total sum of N = 5 x ( N + 2N + 3N ) = 30N, and this cannot be exactly
divided by 4. But from the winding arrangement there may be 7 windings of 4N
made up as follows :-
5 x ( N + 3N ), 2 x ( 2N + 2N ), and this leaves a spare 2N section which is
then connected in parallel to one of the other 2N windings. The arrangement
will work OK because there are 4N turns with less Iac than the other 30N turns
and fraction of total N with changed current density = 4N / 30N, or less than
15%. Such calculations of imperfect but tolerable solutions to the Puzzle Of
Turns are probably beyond the capability of a PC which is a rather dumb
brute, and it could never manage a robot to tie my shoelaces without
mincing my foot!
Choose sec turns to suit
chosen sub-section pattern.
pattern = 4A.
Confirm range of Sec
turns per layer possible
between Tpl in Step 28
up to 1.414 x Tpl in Step 28.
OPT4. Range of
tpl = 37 turns to 52 turns.
Calculate range of
Total sec turns for all Sec
OPT4. No of
layers = 4, Range of total turns
= 148 turns to 208 turns.
numbers of turns within the range
which may be divisible by the
number given for the chosen pattern.
OPT4. The pattern 4A shows total turns to be divisible by 12.Numbers of turns within range divisible by 12 are 156, 168, 180, 192, 204.
- 3p - 2S - 3p - 2S - 3p - 2S -
+ 4S pattern as chosen
before but with 2 layers
of Sec wire per Sec section.
p-p 8 x 0.05mm insulation = 0.40mm.
S-S 4 x 0.05mm insulation = 0.20mm.
8 x 0.50mm insulation = 4.00mm.
Total thickness of all insulation layers = 4.60mm.
13 layers of P wire at
oa dia of 0.569mm = 7.4mm,
Total height of all insulation = 4.6mm,
height of P wire + all
insulation = 12.0mm.
dia of the
Available Sec height = ( Available height in bobbin ) - ( Height P + all Insulation ).
Available height in Bobbin = 0.8 x H window dimension.
OPT4. Avail Sec height = ( 0.8 x 25mm ) - 12.0mm = 8.0mm.
Th Sec oa
dia = ( avail Sec
height ) / no of sec layers of wire,
OPT-1A, Th oa dia sec = 8.0 / 8 = 1.00mm.
Theoretical S turns per layer, thStpl = Bww / thSoadia from step 27
ThStpl = 72 / 0.99 = 72.7, round down to 72 turns per layer.
turns per layer are for the
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.
and winding pattern will probably require adjustments.
Ns, number of secondary
turns consisting of paralleled or seriesed
sec layers calculated in Step 28A.
Primary turns, Np 1,586t
|4 parallel layers||72t||22.02
|4 series layers||288t
|3 series layers, with 2 parallel||216t
Primary turns, Np 1,586t
|4 parallel sections,||72t||22.02
|2 parallel sections, ( 72t + 72t series )each||144
Primary turns, Np 1,586t
|4 parallel sections,||84t||18.88||356||3.40
|2 parallel sections, ( 84t + 84t series )each||168t
capacitance and Primary Inductance of
several areas in an
audio transformer where
with any OPT the measured capacitance between anode terminals of the OPT
and 0V are of main interest. The measurement of the capacitance may be done
with measurement of Primary inductance.
the shunt C
between anode and 0V and away from the
and measure primary inductance set up the OPT as follows :-
15 shows a signal source with
a known low source resistance of less than
600 ohms and capable of a known flat F response between 10Hz and at least 50kHz
and capable of up to 25Vrms output. Such a source could be an audio amp fed
by a signal generator with well calibrated frequency adjustment and flat response
ranges fron say 2Hz to 250kHz.
be a series resistance
between the signal source and OPT which
between 5 and 10 times the Primary RLa value for the OPT. Above I show
5k0, which includes the amplifier source resistance. The highest possible voltage
from the source should be initially be used as long as the signal THD < 1%.
voltage measurements may be
most reliably done by setting the oscilloscope
trace to occupy exactly 1/2 the screen height as shown and at 1kHz to get what
is known as the 0dB reference Vo level. Once this is done, frequencies on the sig
gene should be varied each side of 1kHz from about 300Hz and 3kHz to ensure the
1kHz level is indeed in the middle of the expected bandwidth to be observed.
16 shows a carefully drawn
graph of the response you should be able to plot
on a graph sheet. There are TWO POLES of interest, LF and HF.
To plot the
LF pole, the frequency
is lowered until the oscilloscope level falls by a
factor of 0.707. At this level, the signal response = -3dB. It is wise to measure the
Vac with a reliable meter with wide bandwidth. Digital meters are accurate below
1kHz, but are highly inaccurate above 1kHz so a well constructed voltmeter is
essential for HF measurements.
However, the oscilloscope will be found to have a flat response between 2Hz
and say 20MHz, so once you have set the LF level on the screen then you will
easily measure the relative levels at HF.
The oscilloscope screen may not have a graticule line at the 0.707 point below
half way up the screen. If not, place masking tape on the side of the
screen and plot points where the reference 0dB signal appears, and -3dB.
If the 0dB level = 25Vrms, then the -3dB level = 17.7Vrms, and it may be
set using a voltmeter. -6dB = 12.5Vrms, and -12dB = 6.25Vrms
and -18dB = 3.12Vrms.
As the frequency is reduced to -3db, -6dB, -12dB and -18dB levels, the
frequencies are recorded or plotted directly onto a graph sheet copy as dots.
The dots are joined by drawing a curved line through all dots, and you should
see a response similar to what I have drawn. On the graph, I have also plotted
the boundary lines for the curve. There is a thinner and sloped straight line drawn
from the LF pole F on the 0dB line and sloping down at 6dB/octave. The actual
response curve should begin to follow this straight line if the inductance is linear
with applied voltage. But iron cored inductance windings usually have less
with lesser applied voltage and more inductance with lower frequency, so do not
expect the idealistic curve I have shown below the -6dB graph point.
response is similarly
plotted at points between 0dB and -18dB and the
slope of the curve may not follow the straight line boundary of -6dB per octave
exactly, because the leakage inductance may have some effects at HF.
Where LL is
found to be high, then
the R-source resistance may be increased
so that the series LL reactance becomes a negligible quantity compared to the
source resistance, and the shunt C should then enable a more reliable HF
curve to be drawn to plot the important -3dB frequency.
above graph we can say
the LF pole is at 88Hz.
To calculate Lp,
LP = R / (F x 6.28) = 5,000 ( 88 x 6.28 ) = 9.047H.
the presence of DC flow and partial magnetization of the iron core
will reduce the effective iron permeability to an unknowable extent.
Therefore the most accurate Lp measurement is gained with the OPT
within an amplifier, and this is described here later.
above graph we can say
the HF pole is at 17.5kHz.
To calculate Cshunt,
Csh = 159,000 / ( F x R ) = 159,000 / ( 17,500 x 5,000 ) = 0.001817uF.
This is a reliable measurement and unaffected by DC flow in the windings.
capacitance of the
oscilloscope and probe cable probe must be
subtracted from the Cshunt measurement. If the probe and cable has C =
20pF, it may be ignored.
calculate the primary shunt
In the OPT4 example, refer to the bobbin winding layout. Self capacitance
of the primary winding may be neglected as it is a negligible compared to
the main C between adjacent P to S winding layer surfaces.
around the wound bobbin = 297mm, and traverse
width = 72mm, so the Area of capacitance = 297 x 72 square mm = 21,384
sq.mm. The K for polyester insulation is approx 2.5. d = 0.65mm.
C in pF =
21,384 x 2.5 / ( 113.1 x
0.65 ) = 727pF.
of capacitance in
each P to S interface is proportional to turn
length so the P-S C nearest the core is slightly less than at the top of the
bobbin. For OPT4, minimum C = 616pF and maximum C = 763pF.
lowest Csh, the anode
should be connected to the innermost
But for a
simple calculation of
Total effective Csh at the anode,
consider every P to S interface has the average C value = 727pF.
There are 8 P-S interfaces so summed 8 x C = 5.8nF. ( 0.0058uF ).
measured total Csh
between anode and 0V will be less.
capacitances act as though
there was a simple winding with 8 tap
points and from each one there is 727pF to 0V. One end of the winding
is at 0V ac. The other end at anode swings the anode Vac, and away
from the anode at all the tap points the Vac becomes lower so current
flow in each C becomes lower and the C loading is thus reduced by
the turn ratio difference. The OPT4 Csh is maximum when the OPT
is set up for pentode, UL, triode and without CFB. The 8 capacitances
effective value at the cathode may be tabled with the positions of the
C in terms of the winding layers :-
each interface, pF
at anode, pF
Capacitance table for OPT4
shows the positions of each C along the
primary winding and is expressed in layers-along / total P layers, and the
fractions are listed. Each fraction is squared because the impedance ratio
is the relative turn ratio squared, thus each C has a Fraction Squared
Factor number which is multiplied by the actual C to give the amount
of C to which it is transformed at the anode connection.
The Total C at the anode is simply the sum of all 8 transformed
values of each 727pF.
So Total C
at anode = 1,839 pF. In
practice it may be about 1,750pF
due to shorter turn length near the anode connection.
But let us
allow for the worst
amount of Csh, and let C = 1,839pF,
( 0.00184uF. )
If the OPT4
example is used
with 4 x EL34 in pure pentode mode then
the Ra may be about 3,000 ohms total. Without any resistance load
connected, the gain of the tubes will be close to µ at about 1kHz, ie,
about 125x. The gain will reduce -3dB where the capacitive reactance
= Ra. With Ra = 3,000 ohms and C = 1,839pF, The frequency for -3dB
cut off = 159,000 / ( 3,000ohms x 0.001839uF ) = 28.82kHz.
In practice this would be only approximately correct because there is
leakage inductance involved which forms an undamped resonant and
usually causes a peak in the response which may be calculated,
Fo = 5,035 / square root of ( L x C ), L in mH, C in uF.
In this case,
Fo = 5,035 / sq.rt( 1.91 x 0.00184 ) = 84.9kHz.
F is so much higher
than the HF cut off calculated at 28.82kHz,
then the effect of LL with pentode mode is not much as 28.82kHz, and
the high Ra damps the peak at 84.9kHz where XC = XLL = 1,017 ohms.
If the EL34 are connected for 61% SEUL, then Ra is about 800 ohms and the
HF cut off will theoretically be at 108kHz, but then the effect of LL and Csh
resonance will be seen as a peak at around 85kHz with rapid phase shift above
Usually the resonance between leakage inductance and capacitances will not
cause any problems with HF stability once the correct critical damping networks
are in place within the completed amplifier.
Although it may be possible to
reduce LL to half the value
calculated in Step
35, the number of required interleavings becomes excessive because there
would be more P-S interfaces and C would increase too much, quite negating
the effect of reducing the LL.
The core of
SE OPT will have an air gap which will need to greatly
reduce the maximum permeability, µ, when the laminations are fully interleaved.
This prevents the core from becoming magnetically saturated by the dc current.
The primary inductance is proportional to the effective permeability, µe, the
permeability with a gapped core. The permeability varies slightly with frequency
and Vac and DC magnetization, but the DC magnetization may be considered to
be half the max B allowed for the core material.
The Bac max and Bdc max may each be equal and each be less than half the
allowed total B max for the core material.
For GOSS, total Bmax = 1.5Tesla, for NOSS it may be 1.3Tesla, and for
some other lower grade iron it may be 1.1Tesla.
OPT4. The core material will be
GOSS with allowed Bmax =
The Bdc max = 0.75T and Bac = 0.75T.
SE OPT Primary inductance
depends largely on core µe with
an air gap.
= Bdc x mL
12.6 x x Np x Idc
Where Bdc is in Tesla, 12.6 and 10,000 are constants for all equations to work,
µe = effective permeability,
Np = the primary turns,
Idc = dc current in AMPS,
mL is the iron magnetic path length.
OPT4, Np = 1,586t, Idc =
Iron mL =
280, Bdc = 0.75T.
µe = 0.75 x 280 x 10,000
12.6 x 1,586 x 0.272
the µe < 750?
If so, proceed, but if µe > 750, it may not be possible
the air gap to a small enough value to increase the µe to the wanted amount.
To reduce µe to a commonly seen value between 300 and 500, more primary
turns might be used.
1,000,000,000 x mL
Where µe is effective permeability of core with an air gap, and is just a number,
no units, ie, the number of times the the Lp is increased by the presence of iron
over having no iron present.
1,000,000,000 and 1.26 are constants for all equations to work,
Lp = primary inductance,
mL = magnetic path length of the iron only, and for wasteless pattern E&I lams =
2 x ( L + H ) + ( 3.14 x H ) where L & H are window dimensions, 3.14 = pye, 22/7.
S = stack height,
T = tongue width.
All dimensions in mm !!!
OPT4, Np =
1,586t, S = 58mm, T
mL = 280mm.
OPT4, Lp = 1.26
1,000,000,000 x 280
Lp = 12.92H
41. Calculate Frequency
where XLp = RLa, or Fco.
Fco is considered as
cut off frequency below which it is
to maintain the Vo at 0.0dB level without distortion rapidly increasing
due to Lp inductance shunting RLa thus reducing total RLa to 0.707 x RLa.
All dimensions in mm!!
gap = mL x (
- µe )
µ x µe
Where gap is the air gap distance placed into the iron magnetic circuit,
mL = the iron magnetic path length,
µ = iron permeability maximum with laminations maximally interleaved,
or in the case of C-cores, with polished cuts tight together.
µe = effective permeability with air gap.
all dimensions in mm !!!
OPT4, mL = 280mm, assume µ
= 5,000 with no air gap for
of GOSS E&I lams, µe is the wanted µe with air gap = 386.
Air gap = 280 x ( 5,000 -
386 ) / ( 5,000 x 386 )
The air gap
the total gap. In an E&I core or pair of double
C-cores there are TWO gaps inserted in the magnetic path around each
rectangular window of two which form the core. Therefore the thickness
of the plastic gapping material to make the gap will be HALF the above
calculated gap size.
method for measuring Lp to confirm
that primary inductance inductance is sufficient, and checking that
the air gap is correct and how to adjust the gap size.
ZR = 1,213 ohms : 6.04 ohms.
The 4 x EL34 should produce
233Vrms between anode and cathode,
thus give over 44 Watts into RLa = 1,213 ohms, with Sec load = 6 ohms,
Vo sec = 14.1Vrms at 1kHz.
The 0.0dB Vo reference level may be taken as 14.0Vrms when loaded
by 6 ohms or without any load. The air gap should be set at the calculated
value from Step 42 above.
5 Watt x 10 ohm current sensing resistances are used to confirm AC and
DC currents between anodes and OPT anode winding and between bottom
of cathode windings and 0V, if there is a CFB winding.
One sheet of notebook paper may
be about 0.05 to 0.07mm thick.
To determine the paper thickness, measure the thickness of 100 sheets of
a notebook not including its cardboard covers and divide the total thickness
by the number of sheets.
It is most important to know
what size of gap you are trying
The rectangular yokes to hold E
and I together should be in
bolts but not fully tightened. Yokes for E&I laminations on SE OPT should be
made from aluminium so the magnetic flux acting across the gap does not suffer
When the air gap is optimized
for low Fco and low Fsat, the
total thickness of
temporary gap material is measured and replaced with what can be the same
thickness but permanent. Before tightening the bolts and yokes, it is good practice
to apply slow setting epoxy resin such as Araldite to all surfaces of materials
within the gap. When yoke bolts for E&I laminations are tightened the piles of
Es and Is should be clamped together with an external G-clamp and timber
blocks to ensure the gap size does not increase during this mechanical assembly
After the epoxy has cured for a couple of days, the G-clamps are removed
any gaps between bobbin and core material should be shimmed tight with plastic
sheet off cuts so that the wound core cannot move on the core material.
The transformer terminal board must be prepared to allow the fully assembled
OPT to be placed into its prepared mild steel sheet metal container, or "pot".
There should be at least 5mm
clearance between the pot surfaces
to core material
or to any windings. When all is ready, the OPT may be surrounded with a potting
compound. The best potting compounds are very expensive but are very effective
to stop noise and keep moisture out while seeping into every void before curing to
their rubbery nature. DIYers may find a 50-50 volume mix of fine dry clean sand
and fibre glass resin mix will make a good concrete mix for potting.
In the distant past before
modern chemistry brought the best
potting mixes to the
world, transformers were potted in molten tar which may have a high melting
point between 35C and 70C. For most people it is very difficult, messy, and
smoky and toxic to use because it tends to harden as soon as it cools against
core material, thus not being able to penetrate small gaps. Road tar may become
non solid at only 35C, and may run out of a pot if ever the OPT becomes
overheated. Cheap candle wax has melting point too low and shrinks as it cools.
Beeswax is better with MP of 62C, and probably micro-crystalline petroleum
wax is better, if ever one might buy a small quantity. Indramic 82-T is a brand
name which may be used, MP = 76C minimum. Canauba wax might also be
considered which has a MP = 82C, but I found it shrinks when it cools and thus
the transformer will become loose within its pot after such wax has solidified.
A 5050 mix of canauba + bees wax plus dry clean sand may work well to make
a wax concrete which will not shrink much. The pot and transformer may be
preheated using an old electric frypan to slow the solidification of wax concrete
and allow settlement of the sand so it occupies as much volume as possible while
surrounded by wax.
Pot shapes maybe tailored to the shape of the transformer such as shown here :-
18 shows a small
sample OPT with C-cores and with flying
Clamps using 3mm thick Al plates and 4mm threaded rod are used.
The pot has been home made using 0.6mm steel color bond sheet
found in a dumper bin at a building site. Pop rivets hold the "lid" on,
and some silicone is wiped across gaps inside the pot so it seals the pot.
With the pot upside down, the transformer is placed inside with a small
wood block under the Al plates to get some clearance. The volume of
epoxy mixed plus sand is calculated roughly, and less than required
is used initially, and poured into the pot and allowed to cure. A
amount of potting mix and sand may then be mixed to top up with
less danger of wasting valuable potting mix.
After several attempts and mistakes with OPTs and potting,
one eventually learns to do it right.
The pot made here is roughly equal in size to a cube with 110mm sides.
I used scrap timber to make a 300mm length of wood 110mm x 110mm
and then planed it down to give the octagonal shape to hug the sides of
the OPT. I cut a length of 110mm wide sheet metal and bent it neatly
around the octagonal wooden "mandrel" held in a vice.
The metal bent well at 45 degrees, and was clamped as I proceeded.
With finished sides, the mandrel section was scribed onto the sheet metal,
and thickness of bent sides allowed for so that the lid could be formed
with its turned down lip of 15mm all around. The 8 straight lips of the
lid were bent with wide nose pliers and all adjusted and done so the
appearance is as neat as seen here.
When potting mix has all be poured in and cured for a day, the gaps
around the lid are filled with epoxy panel beater putty, and when all is
hard the whole lot carefully sanded down in a belt sander. Matt paint
may be rolled on with a very small roller and later given a coat of semi
gloss polyurethane and although curing is slow, the finish will end up
hard enough for home use amps, and look well.
The potted transformer is held
to the chassis by the 4 rods
chassis. Leads should be prepared and strapped into their wanted
position before potting so leads come to wanted positions under the