OPT4. Fig 8 below has
subsections for 4 sections of
secondary wire.
Choose a pattern from those
shown, 4A, 4B or 4C from Fig 8.
Step
32
continued...Confirm choice of 4A from Fig 8.
NOTE.
The chosen
winding pattern requires some further
explanations.
The
division of the single layers of secondary windings into exact
fractions of
a layer
allows the
windings to be arranged to suit a number of secondary
load values yet maintain the
same OPT primary load value connected to
anodes :
Fig
11.
NOTE.
For the
arrangement in 4A to work properly, it is necessary
that the
TOTAL number of all S turns in all layers is exactly divisible
by all selected
numbers of paralleled windings.
For example, in 4A, the total
turns = 4 x 3N = 12N. Each choice
of
parallel
windings has a different fraction of the total number of all
secondary
winding
turns in
all secondary layers.
Their relative turn number are described as 3N, 4N or
6N turns.
The
total
number of all secondary turns must be divisible by 3, 4 or
6.
Fortunately, it is easy to check if any number of total S turns
are
divisible
by 3, 4, or 6, and if not, it is
easy to increase the
total turns to fit the
"divisibility" rule.
NOTE.
If the
calculation of turns per layer in step 28 gives a
total
number
of S turns which cannot be exactly divided by the wanted numbers
of
paralleled
windings,
it must be increased to comply with this rule.
For this design example, if in step 28, 47 turns were
calculated, and
pattern
4A is
chosen where the secondary = 4 layers of wire, then total S
turns
= 4 x
47 =
188t.
188 divided by 3 = 62.66, and divided by 6 = 31.33, both
unacceptable
because
the division produces a fraction of a turn.
But if the Tpl is
increased to
48t, total S turns = 192, and is exactly divisible by
3, 4, and 6.
In fact Tpl from step 28 = 37tpl. Total turns are 148t, and for
pattern
4A the
totals could be 156t, 168t, 180t, 192t,
204t, 216t, 228t, 240t and each are
able to be divided by 3, 4 and 6 to give Tpl of 39t,
42t,
45t, 48t, 51t,
54t, 57t, 60t.
The increase in secondary
winding resistance may be marginal
and
acceptable
if total of P and S winding resistance losses are less than 7%.
The
most
efficient
OPT have nearly equal winding loss % in both P and S windings,
and if
the P
losses are say
2.2%, then S losses may be allowed to be 4.8% for total of 7%.
NOTE.
Consider that you
may have selected a 5S + 4P
interleaving
pattern
from Table 4, and then
chosen
sub section arrangement 5A from Fig 9.
There
are 4N turns for
each of
5 layers giving a total of 20N turns. The windings
are grouped either
as 4 paralleled windings
with 5N turns or 5 paralleled windings
of 4N turns. The total turns
must be divisible by 20. Ie, by 4 and
5.
The available winding height for secondary windings may be
5.4mm, so
each
of the 5 layers has theoretically 1.08mm oa dia wire, and we may
select
0.95mm Cu dia with oa dia = 1.04mm, giving 60t per layer.
5 Sec layers = 300t, and divisible by 20, ( and
are we not lucky? ).
The possible total S turns = 300, 320, 340, 360,
giving Tpl of 60, 64, 68, 72.
The Tpl will be then exactly divisible by 4
to give the 5A
pattern.
Where Tpl = 60t, there are 4 layers at 60t each and one layer of
4 x
15t.
Thus Ns may be 5 // 60t or 4 // 75t.
Load matches with RLa = 1,213 ohms and 1,586 P turns are :
60t = 1.74 ohms, 75t = 2.7 ohms.
With Tpl = 72t, the two S loads will be:
72t = 2.5 ohms, 96t = 3.9 ohms.
This 5A pattern does not include a wide range, so the choice of
some
other
pattern for 5 Sec layers must be tried.
NOTE.
Consider
selection of 5B, with windings of 5N, 2.5N and N
turns,
with turns in each layer = 6N.
If one layer has 60t, total turns for 5
layers = 300t
and it is divisible by 30,
ie, by 5, 7.5 and 10. The Sec may
be :
6 // 50t = 1.2 ohms, 4 // 75t = 2.7 ohms, 3 // 100t = 4.8 ohms,
2 //
150t = 10.8 ohms.
There is only one load match between 3 ohms and 9 ohms, and more
trials
of other patterns should be done, with careful attention to
dimensions
and
and winding losses.
When considering designs with 5
Sec layers there is more
difficulty
finding a useful range of load matches fairly well spaced apart.
Ideal
load match
spacing would be 3 ohms, 6 ohms, 12 ohms, but for this you
need sec turns with relationship of 1N,
1.414N and 2N turns.
Nice convenient fractions of layers are 1/2, 1/3, 1/4 and 1/5,
and
combinations
of fractions of layers will never give the 0.414 fraction, but
give
0.50N, 0.333N,
0.25N, or 0.20N. So a popular range of relative N might be 1N,
1.5N,
2N, which
would match 3 ohms, 6.75 and
12 ohms which is ideal for nominal speakers
of 4, 8 and 16 ohms.
The aim is to find :
A. Easy fractions allowing a good spread of loads,
B. 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
current
density
in wires will be tolerated without causing significant extra
winding
losses or
changed
frequency behaviour. For example, in Fig 9, 5E, each layer of 5
is
divided to give N, 2N and
3N subsections.
The total sum of N = 5 x ( N + 2N + 3N ) = 30N, and this cannot
be exactly
divided by 4. But from the winding arrangement there may be 7
windings
of 4N
made up as follows :
5 x ( N + 3N ), 2 x ( 2N + 2N ), and this leaves a spare 2N
section
which is
then connected in parallel to one of the other 2N windings. The
arrangement
will work
OK because there are 4N turns with less Iac than the other 30N
turns
and fraction of total N
with changed current density = 4N / 30N, or less than
15%. Such calculations of
imperfect but tolerable solutions to the Puzzle Of
Turns are probably beyond the capability of a PC
which is a rather dumb
brute, and it could never manage a robot to tie my shoelaces
without
mincing my foot!
33.
Choose sec turns to suit
chosen subsection pattern.
Confirm
choice of
pattern = 4A.
Confirm range of Sec
turns per layer possible
between Tpl in Step 28
up to 1.414 x Tpl in Step 28.
OPT4. Range of
tpl = 37 turns to 52 turns.
Calculate range of
Total sec turns for all Sec
layers.
OPT4. No of
layers = 4, Range of total turns
= 148 turns to 208 turns.
Calculate all
numbers of turns within the range
which may be divisible by the
number given for the chosen pattern.
OPT4. The pattern 4A shows total turns to be divisible by 12.
Numbers of turns within range divisible by 12 are 156, 168, 180, 192, 204.Total
turns
: 
156  168  180  192  204  
Configuration, 
ohms 
ohms 
ohms 
ohms 
ohms 

4
parallel 
39 
0.73 
42 
0.85 
45 
0.98 
48 
1.11 
51 
1.25 
3
parallel 
52 
1.30 
56 
1.51 
60 
1.74 
64 
1.97 
68 
2.22 
2
parallel 
78 
2.93 
84 
3.40 
90 
3.90 
96 
4.44 
102 
5.01 
Illegal 
117 
6.59 
126 
7.65 
135 
8.82 
144 
9.99 
153 
11.25 
4 series  156  11.72  168  13.60  180  15.60  192  17.76  204  20.04 
OPT4.
Interleaving
Pattern =
2p

2S
 3p  2S  3p  2S  3p  2S 
2p.
This
is
the
same
5P
+ 4S pattern as chosen
before but with 2 layers
of Sec wire per Sec
section.
Insulation
:
pp 8 x 0.05mm insulation = 0.40mm.
SS 4 x 0.05mm insulation = 0.20mm.
8 x 0.50mm insulation = 4.00mm.
Total thickness of all insulation
layers = 4.60mm.
25A.
Calculate
height
of
Primary
layers
plus
all
insulation.
OPT4.
13 layers of P wire at
oa dia of 0.569mm = 7.4mm,
Total height of all insulation
= 4.6mm,
Total
height of P wire + all
insulation = 12.0mm.
26A.
Calculate
maximum
theoretical
oa
dia of the
secondary wire.
Calculate
available
height
for
layers
of
secondary
wire
:
Available Sec height = ( Available height in bobbin )  (
Height P + all Insulation ).
Available height in Bobbin = 0.8 x H window dimension.
OPT4. Avail Sec height = ( 0.8 x 25mm )  12.0mm = 8.0mm.
Th Sec oa
dia = ( avail Sec
height ) / no of sec layers of wire,
OPT1A, Th oa dia sec = 8.0 / 8 = 1.00mm.
28A.
Calculate theoretical
S
turns
per
layer,
thStpl.
Theoretical S
turns per layer, thStpl = Bww /
thSoadia from step 27
ThStpl = 72 / 0.99 = 72.7, round down to 72
turns per layer.
NOTE.
These calculated
turns per layer are for the
thickest
wire
possible, and
fewer turns per layer are forbidden because the increased wire
size to
fill a
layer
would make the winding height unable to fit onto the bobbin.
Wires
should never
be
wound on and spread apart so that the Tpl is reduced while
keeping wire
size
the
same, lest secondary winding resistance losses be increased too
much.
29A.
Calculate
load
matches
with
Sec
turns
from
Step
28A.
These
load
matches
will
be
theoretical
and
secondary
turns
and winding pattern will probably require adjustments.
Nominate
Ns, number of secondary
turns consisting of paralleled or seriesed
sec layers calculated in Step 28A.
Primary
RLa
1,213
ohms Primary turns, Np 1,586t 
Ns  TR 
ZR 
Secondary RL ohms 
4 parallel layers  72t  22.02 
485  2.50
ohms 
2
parallel
x
(
2
series
layers
) 
144t 
11.01 
121 
10.00
ohms 
4 series layers  288t 
5.50 
30.25 
40.00
ohms 
3 series layers, with 2 parallel  216t 
TR
7.34 
ZR
53.9 
22.5
ohms 
Primary
RLa
1,213
ohms Primary turns, Np 1,586t 
Ns  TR 
ZR 
Secondary RL ohms 
4 parallel sections,  72t  22.02 
485  2.50
ohms 
3
parallel
sections,
(
72t
+
24t
series
)each 
96t 
16.52 
272 
4.44
ohms 
2 parallel sections, ( 72t + 72t series )each  144 
11.01 
121.2 
10.00
ohms 
Primary
RLa
1,213
ohms Primary turns, Np 1,586t 
Ns  TR 
ZR 
Secondary RL ohms 
4 parallel sections,  84t  18.88  356  3.40
ohms 
3
parallel
sections,
(
84t
+
28t
series
)each 
112t 
14.16 
200 
6.05
ohms 
2 parallel sections, ( 84t + 84t series )each  168t 
9.44 
89.1 
13.60
ohms 
38. Shunt
capacitance and Primary Inductance of
SE OPT.
There are
several areas in an
audio transformer where
capacitance
exists, and
with any OPT the measured capacitance between anode terminals of
the OPT
and 0V are of main interest. The measurement of the capacitance
may be
done
with measurement of Primary inductance.
To measure
the shunt C
between anode and 0V and away from the
amplifier,
and measure primary inductance set
up the OPT as follows :
Fig
15.
Fig
15 shows a signal source with
a known low source resistance of less than
600 ohms and capable of a known flat F response between 10Hz and
at
least 50kHz
and capable of up to 25Vrms output. Such a source could be an
audio amp
fed
by a signal generator with well calibrated frequency adjustment
and
flat response
ranges fron say 2Hz to 250kHz.
There must
be a series resistance
between the signal source and OPT which
between 5 and 10 times the Primary RLa value for the OPT. Above
I show
5k0, which includes the amplifier source resistance. The highest
possible voltage
from the source should be initially be used as long as the
signal THD
< 1%.
Signal
voltage measurements may be
most reliably done by setting the oscilloscope
trace to occupy exactly 1/2 the screen height as shown and at
1kHz to
get what
is known as the 0dB reference Vo level. Once this is done,
frequencies
on the sig
gene should be varied each side of 1kHz from about 300Hz and
3kHz to
ensure the
1kHz level is indeed in the middle of the expected bandwidth to
be
observed.
Fig
16 shows a carefully drawn
graph of the response you should be able to plot
on a graph sheet. There are TWO POLES of interest, LF and HF.
To plot the
LF pole, the frequency
is lowered until the oscilloscope level falls by a
factor of 0.707. At this level, the signal response = 3dB. It
is wise
to measure the
Vac with a reliable meter with wide bandwidth. Digital meters
are
accurate below
1kHz, but are highly inaccurate above 1kHz so a well constructed
voltmeter is
essential for HF measurements.
However, the oscilloscope will be found to have a flat response
between
2Hz
and say 20MHz, so once you have set the LF level on the screen
then you
will
easily measure the relative levels at HF.
The oscilloscope screen may not have a graticule line at the
0.707
point below
half way up the screen. If not, place masking tape on the side
of the
screen and plot points where the reference 0dB signal appears,
and
3dB.
If the 0dB level = 25Vrms, then the 3dB level = 17.7Vrms, and
it may
be
set using a voltmeter. 6dB = 12.5Vrms, and 12dB = 6.25Vrms
and 18dB = 3.12Vrms.
As the frequency is reduced to 3db, 6dB, 12dB and 18dB
levels, the
frequencies are recorded or plotted directly onto a graph sheet
copy as
dots.
The dots are joined by drawing a curved line through all dots,
and you
should
see a response similar to what I have drawn. On the graph, I
have also
plotted
the boundary lines for the curve. There is a thinner and sloped
straight line drawn
from the LF pole F on the 0dB line and sloping down at
6dB/octave. The
actual
response curve should begin to follow this straight line if the
inductance is linear
with applied voltage. But iron cored inductance windings usually
have
less
with lesser applied voltage and more inductance with lower
frequency,
so do not
expect the idealistic curve I have shown below the 6dB graph
point.
The HF
response is similarly
plotted at points between 0dB and 18dB and the
slope of the curve may not follow the straight line boundary of
6dB
per octave
exactly, because the leakage inductance may have some effects at
HF.
Where LL is
found to be high, then
the Rsource resistance may be increased
so that the series LL reactance becomes a negligible quantity
compared
to the
source resistance, and the shunt C should then enable a more
reliable
HF
curve to be drawn to plot the important 3dB frequency.
From the
above graph we can say
the LF pole is at 88Hz.
To calculate Lp,
LP = R / (F x 6.28) = 5,000 (
88 x
6.28 ) = 9.047H.
The
inductance
measurement
with
SE
OPTs
may
be
unreliable
because
the presence of DC flow and partial magnetization of the iron
core
will reduce the effective iron permeability to an unknowable
extent.
Therefore the most accurate Lp measurement is gained with the
OPT
within an amplifier, and this is described here later.
From the
above graph we can say
the HF pole is at 17.5kHz.
To calculate Cshunt,
Csh = 159,000 / ( F x R )
=
159,000 / ( 17,500 x 5,000 ) = 0.001817uF.
This is a reliable measurement and unaffected by DC flow in the
windings.
The shunt
capacitance of the
oscilloscope and probe cable probe must be
subtracted from the Cshunt measurement. If the probe and cable
has C =
20pF, it may be ignored.
To
calculate the primary shunt
capacitance.
In the OPT4 example, refer to the bobbin winding layout. Self
capacitance
of the primary winding may be neglected as it is a negligible
compared
to
the main C between
adjacent P
to S winding layer surfaces.
The average
turn length
around the wound bobbin = 297mm, and traverse
width = 72mm, so the Area of capacitance = 297 x 72 square mm =
21,384
sq.mm. The K for polyester insulation is approx 2.5. d =
0.65mm.
C in pF =
21,384 x 2.5 / ( 113.1 x
0.65 ) = 727pF.
The amount
of capacitance in
each P to S interface is proportional to turn
length so the PS C nearest the core is slightly less than at
the top
of the
bobbin. For OPT4, minimum C = 616pF and maximum C = 763pF.
To get
lowest Csh, the anode
should be connected to the innermost
primary end.
But for a
simple calculation of
the
Total effective Csh at the anode,
consider every P to S interface has the average C value = 727pF.
There are 8 PS
interfaces so summed 8 x C = 5.8nF. ( 0.0058uF ).
The actual
measured total Csh
between anode and 0V will be less.
The 8
capacitances act as though
there was a simple winding with 8 tap
points and from each one there is 727pF to 0V. One end of the
winding
is at 0V
ac. The other end at anode swings the anode Vac, and away
from the anode at
all the tap points the Vac becomes lower so current
flow in each C becomes
lower and the C loading is thus reduced by
the turn ratio difference. The OPT4 Csh is maximum when the OPT
is set up for pentode, UL, triode and without CFB. The 8
capacitances
effective value at the cathode may
be tabled with the positions of the
C in terms of the winding layers :
PS
Csh
at each interface, pF 
winding position 
Fraction 
Fraction Squared Factor 
Effective
C at anode, pF 
727 
11.5
/
13 
0.88 
0.78 
567 
727 
10.5
/
13 
0.81 
0.65 
472 
727  8.5
/
13 
0.65 
0.42 
305 
727  7.5
/
13 
0.57 
0.33 
240 
727  5.5
/
13 
0.42 
0.18 
130 
727  4.5
/
13 
0.35 
0.12 
87 
727  2.5
/
13 
0.19 
0.04 
29 
727  1.5
/
13 
0.11 
0.01 
9 
Totals
sum
5,816 
1,839 
The
Capacitance table for OPT4
shows the positions of each C along the
primary winding and is expressed in layersalong / 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.
Since this
F is so much higher
than the HF cut off calculated at 28.82kHz,
then the effect of LL with pentode mode is not much as 28.82kHz,
and
the
high Ra damps the peak at 84.9kHz where XC = XLL = 1,017 ohms.
If the EL34 are connected for 61% SEUL, then Ra is about 800
ohms and
the
HF cut off will theoretically be at 108kHz, but then the effect
of LL
and Csh
resonance will be seen as a peak at around 85kHz with rapid
phase shift
above
110kHz.
Usually the resonance between leakage inductance and
capacitances will
not
cause any problems with HF stability once the correct critical
damping
networks
are in place within the completed amplifier.
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 PS interfaces and C would increase too much, quite
negating
the
effect of reducing the LL.
NOTE.
The core of
the
SE OPT will have an air gap which will need to greatly
reduce the
maximum permeability, µ, when the laminations are fully
interleaved.
This prevents the core from
becoming magnetically saturated by the dc current.
The primary inductance is proportional to the effective
permeability,
µe, the
permeability with a gapped core. The permeability varies
slightly with
frequency
and Vac and DC magnetization, but the DC magnetization may be
considered to
be half the max B allowed for the core material.
The Bac max and Bdc max may each be equal and each be less than
half
the
allowed total B max for the core material.
For GOSS, total Bmax = 1.5Tesla, for NOSS it may be 1.3Tesla,
and for
some other lower grade iron it may be 1.1Tesla.
OPT4. The core material will be
GOSS with allowed Bmax =
1.5Tesla.
The Bdc max = 0.75T and Bac = 0.75T.
SE OPT Primary inductance
depends largely on core µe with
an air gap.
µe
= Bdc x mL
x 10,000
12.6
x
x
Np
x
Idc
Where Bdc is in Tesla, 12.6 and 10,000 are constants for
all
equations to work,
µe = effective permeability,
Np = the primary turns,
Idc = dc current in AMPS,
mL is the iron magnetic path length.
OPT4, Np = 1,586t, Idc =
0.272Adc,
Iron mL =
280, Bdc = 0.75T.
µe = 0.75 x 280 x 10,000
12.6
x 1,586 x 0.272
= 386.
Is
the µe < 750?
If so, proceed, but if µe > 750, it may not be possible
to
adjust
the air gap to a small enough value to increase the µe to
the
wanted amount.
To reduce µe to a commonly seen value between 300 and 500,
more
primary
turns might be used.
Lp
= 1.26
x
Np
squared
x
S
x
T
x µe
1,000,000,000
x
mL
Where µe
is effective permeability of core with an
air gap, and is just a number,
no units, ie, the number of times the the Lp is increased by the
presence of iron
over having no iron present.
1,000,000,000 and 1.26 are constants for all equations to work,
Lp = primary inductance,
mL = magnetic path length of the iron only, and for wasteless
pattern
E&I lams =
2 x ( L + H ) + ( 3.14 x H ) where L & H are window
dimensions,
3.14 = pye, 22/7.
S = stack height,
T = tongue width.
All dimensions in mm !!!
OPT4, Np =
1,586t, S = 58mm, T
=
51mm,
µe
=
386,
mL = 280mm.
OPT4, Lp = 1.26
x
1,586
x
1,586
x
51
x
58
x
386
1,000,000,000 x
280
Lp = 12.92H
41. Calculate Frequency
where XLp = RLa, or Fco.
NOTE.
Fco is considered as
cut off frequency below which it is
impossible
to maintain the Vo at 0.0dB level without distortion rapidly
increasing
due to Lp inductance shunting RLa thus reducing total RLa to
0.707 x
RLa.
All dimensions in mm!!
Air
gap = mL x (
µ
 µe )
µ
x
µe
Where gap is the air gap distance placed into the iron magnetic
circuit,
mL = the iron magnetic path length,
µ = iron permeability maximum with laminations maximally
interleaved,
or in the case of Ccores, with polished cuts tight together.
µe = effective permeability with air gap.
all dimensions in mm !!!
OPT4, mL = 280mm, assume µ
= 5,000 with no air gap for
the
sample
of GOSS E&I lams, µe is the wanted µe with air
gap =
386.
Air gap = 280 x ( 5,000 
386 ) / ( 5,000 x 386 )
=
0.67mm.
NOTE.
The air gap
is
the total gap. In an E&I core or pair of double
Ccores there are
TWO gaps inserted in the magnetic path around each
rectangular window of two
which
form the core. Therefore the
thickness
of the plastic gapping material to make the gap will be
HALF the above
calculated gap size.
NOTE.
See below
for practical
method for measuring Lp to confirm
that primary inductance inductance is sufficient, and
checking that
the air gap is correct and how to adjust the gap size.
ZR = 1,213 ohms : 6.04 ohms.
The 4 x EL34 should produce
233Vrms between anode and cathode,
and
thus give over 44 Watts into RLa = 1,213 ohms, with Sec load = 6
ohms,
Vo sec = 14.1Vrms at 1kHz.
The 0.0dB Vo reference level may be taken as 14.0Vrms when
loaded
by 6 ohms or without any load. The air gap should be set at the
calculated
value from Step 42 above.
5 Watt x 10 ohm current sensing resistances are used to
confirm AC and
DC currents between anodes and OPT anode winding and between
bottom
of cathode windings and 0V, if there is a CFB winding.
One sheet of notebook paper may
be about 0.05 to 0.07mm thick.
To determine the paper
thickness, measure the thickness of 100 sheets of
a notebook not including its cardboard
covers and divide the total
thickness
by the number of sheets.
It is most important to know
what size of gap you are trying
set.
The rectangular yokes to hold E
and I together should be in
place with
retaining
bolts but not
fully tightened. Yokes for E&I laminations on SE OPT should
be
made from aluminium so the magnetic flux acting across the
gap does not suffer
interference.
When the air gap is optimized
for low Fco and low Fsat, the
total thickness of
temporary gap material is measured and replaced with what can be
the
same
thickness but permanent. Before tightening the bolts and yokes,
it is
good practice
to apply slow setting epoxy resin such as Araldite to all
surfaces of
materials
within the gap. When yoke bolts for E&I laminations are
tightened
the piles of
Es and Is should be clamped together with an external Gclamp
and
timber
blocks to ensure the gap size does not increase during this
mechanical assembly
work.
After the epoxy has cured for a couple of days, the Gclamps 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 5050 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 microcrystalline
petroleum
wax is better, if ever one might buy a small quantity. Indramic
82T is
a brand
name which may be used, MP = 76C minimum. Canauba wax might also
be
considered which has a MP = 82C, but I found it shrinks when it
cools
and thus
the transformer will become loose within its pot after such wax
has
solidified.
A 5050 mix of canauba + bees wax plus dry clean sand may work
well to
make
a wax concrete which will not shrink much. The pot and
transformer may
be
preheated using an old electric frypan to slow the
solidification of
wax concrete
and allow settlement of the sand so it occupies as much volume
as
possible while
surrounded by wax.
Pot shapes maybe tailored to the shape of the transformer such
as shown
here :
Fig
18.
Fig
18 shows a small
sample OPT with Ccores and with flying
leads.
Clamps using 3mm thick Al plates and 4mm threaded rod are
used.
The pot has been home made using 0.6mm steel color bond sheet
found in a dumper bin at a building site. Pop rivets hold the
"lid" on,
and some silicone is wiped across gaps inside the pot so it
seals the
pot.
With the pot upside down, the transformer is placed inside with
a small
wood block under the Al plates to get some clearance. The volume
of
epoxy mixed plus sand is calculated roughly, and less than
required
is used initially, and poured into the pot and allowed to cure.
A
amount of potting mix and sand may then be mixed to top up with
less danger of wasting valuable potting mix.
After several attempts and mistakes with OPTs and potting,
one eventually learns to do it right.
The pot made here is roughly equal in size to a cube with 110mm
sides.
I used scrap timber to make a 300mm length of wood 110mm x 110mm
and then planed it down to give the octagonal shape to hug the
sides of
the OPT. I cut a length of 110mm wide sheet metal and bent it
neatly
around the octagonal wooden "mandrel" held in a vice.
The metal bent well at 45 degrees, and was clamped as I
proceeded.
With finished sides, the mandrel section was scribed onto the
sheet
metal,
and thickness of bent sides allowed for so that the lid could be
formed
with its turned down lip of 15mm all around. The 8 straight lips
of the
lid were bent with wide nose pliers and all adjusted and done so
the
appearance is as neat as seen here.
When potting mix has all be poured in and cured for a day, the
gaps
around the lid are filled with epoxy panel beater putty, and
when all
is
hard the whole lot carefully sanded down in a belt sander. Matt
paint
may be rolled on with a very small roller and later given a coat
of
semi
gloss polyurethane and although curing is slow, the finish will
end up
hard enough for home use amps, and look well.
The potted transformer is held
to the chassis by the 4 rods
through the
chassis. Leads should be prepared and strapped into their wanted
position before potting so leads come to wanted positions under
the
chassis.