CALCULATIONS PAGE 3.
Continued from SE
OPT Calc Page 2,
SE OPT calc page 3 Contents :-
Practical Testing of 8 Watt SEUL OPT for 1 x EL34 for old
rather good 8 Watt SE amp.
Notes about schematic etc.
Fig 20. Photo of old radio after restoration.
Notes on testing amp and OPT.
Oscilloscope pictures of
produced with 8 Watt SEUL OPT.
CRO 1. Healthy wave at clipping power at 1kHz.
CRO 2. Measuring Lp.
How to adjust the air gap.
Fig 21. Wasteless pattern
ALTERNATIVE METHOD to calculate max
PO of any
CRO 3, 4, 5, Waves during core saturation phenomena.
Fig 22. Graph of
Air gap Vs Fsat and Lp.
CRO 6, 7, 8, Waves at -6dB level for 47Hz, 20Hz, 16Hz.
Conclusions about SEUL for 1 x EL34,
Notes and calculation checks for all.
SE OPT Easy Method for calculating any SE OPT.
For Basic Parameters only.
1E. Calculate RLa and PO.
2E. Calculate Afe.
3E. Calculate required Lp.
4E. Calculate Np.
5E. Calculate minimum primary wire size.
6E. Calculate minimum core window size L x H.
7E. Calculate Core T + S.
8E. Calculate µe.
9E. Check Lp, Fsat.
10E. Calculate Air gap.
High Voltage testing of transformers.
Fig 23. Schematic
for testing insulation of transformer.
OPT3 bobbin details for 25W
web pages created
OPT3 from 2006.
Fig 24. Details
testing of 8 Watt SE
Amp example, measuring Lp,
and Fsat, and setting the air gap.
TUBE AMP VOLTAGES !!!!!!!!!!!!!!!
Fig 19 resembles a very good
type of 8W audio amplifier
for an old radio where there may have been a 6V6 used to make 3
Usually the best old AM radios will have a B+ supply generated
winding with a 5Y3 rectifier giving about +280Vdc available
at up to
50mAdc at the SE OPT connection. In many old AM radio sets there
be a field coil choke on the speaker with Rw = 1k5, and then there
voltage drop across the 5Y3 anode resistance and across OPT
Electrolytic filter caps are often quite low so ripple voltage is
The ancient 1940 speaker
in such poor condition it is best to replace it.
Usually the existing OPT will need to
be replaced because of its poor bandwidth,
high winding losses, and incorrect impedance ratio. A modern
with no field coil may have lower sensitivity but have a better
Treble response in an old speaker may have only extended to 3kHz,
coned HF unit may be mounted concentrically inside the bass unit
HF to give almost hi-fi performance, with at least 9kHz BW.
To drive a modern speaker,
audio amp may be vastly improved. The 5Y3
is replaced with silicon diodes charging into a CRC filter using
replacement 220uF x 450Vdc rated caps and R may be say 235 ohms
pair of 10W rated 470 ohms in parallel. Usually 70mA may be had
existing old power transformer which may be still serviceable, and
verge of burning out with repeated continued use. Some power
will benefit from being removed from the chassis, and soaked in
varnish over night, then baked for 4 hours at 125C the next
With the reductions of series
resistance in the PSU, the Vdc at C1 at 70mA
dc draw may be may be +420Vdc, and at the OPT connection at
Such a nice high B+ allows the use of an EL34, 6L6, ( even 807 )
the chassis socket meant for 5Y3, and where the 6V6 once was,
a 12AX7, or 6J7 in triode mode. I find the 12AX7 plus EL34 in
is simplest and will give 6 watts even if using the old OPT if it
Fig 13 above shows a best practice use of OPT with SE+CFB
If there is a moderate Vdc drop
across PSU resistance and OPT primary and
across the cathode biasing RC network then often it is possible to
healthy Ea across the EL34 of +350Vdc, and up to 8 Watts of output
The OPT shown in Fig 13
has 15% CFB windings, but a simpler effective OPT
may be made with one anode winding with a CT which is used for a
to the EL34
The new OPT with larger core size
and lower winding losses will give little
B+ voltage drop but the EL34 could have Ea at 300V and and still
better power and lower THD than a 6V6, 6F6 etc or 6BQ4/EL84.
Triode operation is also generally superb compared to normal
in pentode mode, and with triode connected EL34 there is no need
NFB. Often it is found the old ancient OPT and speaker made prior
(when accountants took over design sections of radio production)
will work very well with EL34 in triode. Triode operation of EL34
with Ea at +350V to +420V, and Ia at idle should be adjusted to
most power with least THD at 400Hz. Often the Ia will be no more
was used in the original use of 6V6, so there may be no extra heat
in the power transformer. The 6L6 or 807 beam tetrodes need more
but draw less screen current, but sound excellent in triode or
SEUL or with 15% CFB.
I have recently re-built a
few ancient radios from the 1935
to 1950 period
using SEUL or triode connected EL34. I have quite a few to use,
set up with Pda = 18Watts, they will last maybe 15 years easily. I
second hand EL34 in the AM radio I designed and built in 1999, and
produces excellent sound and has never needed to be serviced.
My radio customers appreciate me putting a pair of RCA sockets
used for CD, FM tuner, whatever, for very pleasing mono sound for
lounge room or kitchen listening.
20 shows the rear view of
a completely re-engineered 1935 AM radio chassis
from July 2011. The modern 10" bass-midrange speaker with
is shown. There is a mains fuse, IEC mains cable input and
safety. There is a replacement PT on rhs, with EL34 and 12AX7
There is a new OPT in the chassis centre. There are RCA
for use with FM tuner or CD player. Sound will be mono, but very
The other tubes used to replace rare European odd ball types
6AN7 mixer, 6N8 IF amp, 12AU7 detector, 12AU7 treble control
OPT has a non-wasteless
pattern of old low grade EI
The OPT in Fig 14&15 was
calculated for use in an old floor
with Tongue = 25mm, Stack = 24mm, Window L =
ML = 189mm. Np = 3,200 turns of 0.3mm Cu dia wire.
Primary nominal RL = 6,400 ohms.
Secondary windings may be configured to give 4//81t for 4.1ohms,
3//108t for 7.3 ohms, 2//162t for 16.4 ohms, but the 7.3 ohm sec
because the speaker for this project is 8 ohms.
power operation will be used. The 10" speaker is mounted in
a large open backed cabinet which has a natural cut off at about
There is no point in making an amp which needs to produce full
below 50Hz. The amplifier bandwidth has been tailored to have a
Fsat at 0dB output level could be allowed to
be higher than for a true
hi-fi amplifier with Fsat and cut off at below 20Hz.
Notes about testing the OPT
amp performance :-
Before proceeding to test the SE audio amp the circuit needs
3 times against the schematic before turning anything on. Fuses
used in mains supply and between HT winding CT and 0V, so that you
don't ruin an old power or output transformer or a
field coil choke or anything
else. Typical HT fuse might be 0.5A fast blow, or a value which
protection if there is bias failure in the output tube when Ia may
before the output tube self destructs. The mains fuse should be
or 110V x 1A slow blow, or whatever value is needed to prevent
fuse blowings, but never so high to offer almost no protection.
Once the winding of the OPT
is complete, the UN-VARNISHED OR
OPT is temporarily placed into the
fully operational with all NFB loops
If the bobbin
varnished with cold cure polyurethane two pack
mix, the core may be assembled into the wound bobbin without the
application of applied varnish to the
completed item which is best done
by potting in a 50-50 mix of dry
clean dry sand and slow setting epoxy
The core air gap MUST be
to be adjusted before the final potting
varnishing procedure, because after potting it is impossible to
change the air gap.
The air gap used initially should
what has been calculated,
and it may or may not
The amplifier output voltage at
clipping should be measured
using a dummy
resistance load which is the load value for absolute maximum
power using a sine wave, with THD < 2%, and at 400 to
With higher output voltage levels
the sine wave should show equal clipping to
positive and negative wave
crests, ie, the wave clipping is symmetrical.
output voltage at just
under 2% THD at the onset of symmetrical
wave clipping at 1kHz with a
sine wave at 1kHz is the 0dB reference
oscilloscope must be used to
monitor all the anode currents,
and distortions at the output terminals. Sine waves with THD
be used to drive the amp from a signal generator capable of F
2Hz and 2MHz and with a flat response for all ranges which may
for level and F.
anode to OPT, as R15, Test Point 2. The schematic also shows
CFB winding as R14, giving Test Point 1 so monitoring anode signal
more convenient than having the R at a high voltage level. The R14
the screen dc current which will be about 7% of anode current.
screen signal current flow so measurements of anode current are
at TP2. But anode current waveform distortion is most easily seen
With GNFB present, the OP voltage at Sec may appear linear while
current may appear slightly distorted because the NFB is keeping
linear by applying a grid signal voltage containing an "error
to the output
tube grid. As the output signal is increased towards clipping the
grid signal and anode current signal will show increasing wave
distortions, but at
about 2dB below clipping, all wave distortions should be less than
distortion will not affect measurements and calculations.
The amplifier should be
to be optimized for unconditional stability at LF
and HF so that with GNFB connected it will not oscillate at any F
secondary load connected. If no GNFB is to be used, the EL34
as a triode. Slightly less output power is available than pentode
but there will less
distortion and measurements will be easier because the Ra
is so much lower than
with a 6V6 or 6BQ5 in tetrode or pentode. EL34 in
triode gives Ra = 1k3 approx, which may be 1/5 of RLa, so damping
with triode is high enough to not need GNFB. With 50% SEUL, Ra
about 3k0, and some GNFB is useful, but seldom is there any need
than 12dB of GNFB. I do not like ordinary pentode connection which
more GNFB. 15% CFB will reduce EL34 Ra to less than triode.
Below are photos of oscilloscope
screen using a 1983 dual trace
Hitachi oscilloscope ( CRO. )
The images below show typical
common to many tube
SE amps with Global
The amp tested to gain these images has 1 x Sovtek EL34 set up in
configuration with UL tap at 50% of anode turns.
CRO 1 shows two traces,
the secondary output voltage
with rated load
onset of symmetrical clipping at 1khz.
The negative wave crests at Ea minimum show clipping because of
current and the inability of the EL34 anode voltage to swing any
wave crests at Ea maximum are just about to reach Ia cut off.
The bottom trace shows Ia signal wave across R14 in the Fig 13
Peaks on current wave are produced by
effect of NFB trying to drive tube
into producing more current and
force Ea to swing lower.
The Ia is monitored at the cathode and the current
peaks produced by
grid current and screen current peaking because the load
no such current peaks. At this point the coupling cap driving the
is gaining a negative charge from grid current which tends to make
the tube effectively biased to conduct less idle Idc.
Sustained over drive slightly upsets DC working conditions.
The 0.0dB Reference VO signal would
be set slightly below levels
which indicate THD = 2% approximately.
The primary inductance should be
measured at the -6dB Vo level
at the lowest frequency easily measured and where THD < 3%.
2 top trace shows Vo
At this voltage level
there is no saturation and the primary load is solely the
The bottom trace shows the anode current signal flow which is seen
across R14 in Fig 13 schematic.
There is no severe distortion and EL34 load at the primary is a
inductive reactance without saturation effects.
The anode to cathode Vac is measured and in this case = 100Vrms.
This is the total Vac signal across all primary windings.
If 15% CFB windings are used, there would be 100Vrms measured
between anode and cathode, with 15Vrms between cathode and 0V,
and 85Vrms between anode and 0V, with the two voltages Va and Vk
having opposite phases. But with SEUL, there is just 100Vrms
anode and 0V, or across the whole anode winding from B+ to anode.
In this example Vac across R14 = 0.265Vrms, and so Ia =
So Lp reactance, XLp, at 25Hz = Vac / Iac = 100V / 0.0265A = 3,773
Lp = XLp / ( F x 6.28 )
where L is Henrys, F is in Hertz, and 6.28 = constant = 2 x pye.
So Lp = 3,773 / ( 25 x 6.28 ) = 24H.
Because XLp = 3.8k approx, the Vo wave shows about 2% THD because
of the loading effect of a reactance approaching 1/2 the ideal RLa
The bottom 25Hz current
waveform shows 5% THD because the
is trying to correct the voltage distortion. If you examined the
signal you might find it had 5% THD or more, because a fraction of
THD at the amp output is fed back to V1 12AX7 cathode then
so it appears at V2 EL34 grid with its phase inverted, therefore
reduce the output THD as it is produced.
The above waveforms show that the
air gap would be nearly
If there was severe THD at say -6dB at say 60Hz, it may indicate
is far too small or too large or that the OPT being tested may be
for the intended application.
The frequency response
for where THD < 2% may be plotted on a
graph at 0.0dB, -6dB, and -12dB. -6dB is where Vo = 1/2
and -12dB is where Vo =
1/4 Vreference. Graphs for F response may
be done using an exercise book
pencil and oscilloscope used to
measure relative voltages.
There are blank sheets at the bottom
of this page which you may print out
to enable graphs to be made in your
workshop with a pencil.
How to adjust the air gap
for optimum operation.
21 shows wasteless
The calculated gap is initially
used for tests if it was
There are TWO magnetic paths around the TWO core windows
together as ONE magnetic path.
The Iron magnetic path length,
ML = 2H
+ 2L + 22H/7 for any core material.
E&I lams, ML = 5.57 x T.
If the gap was not
calculated or preset, which may be the case with an old
stock OPT being trialed, then use at
least 0.05mm gapping material.
The dc flow in the OPT core will
usually be enough to draw the
Es and Is tightly together and leaving the clamping bolts loose
the E&I to
close as the gap material permits. But a newly wound
OPT may be found to be slightly forcing the stack of Es and Is
so mechanical clamping together may be necessary. This is a real
the OPT performance and MUST be checked very carefully during
Hand made bobbins using say 1mm thick cardboard are very prone
slight bulge during winding and and inexperienced DIYers may not
the bobbin size with some clearance, or may not have a winding
clamps the bobbin properly.
The OPT may tend to howl while
testing with applied
sine waves because
of some slight movement of windings and core before final
Without any DC flow in the core, the core
may be easily prized apart to
enable enough layers of paper to be carefully inserted so they
make up close to the
calculated air gap. Once the calculated paper gapping
has been inserted, the yoke bolts are
very slightly tightened and the Is tapped
tp to be tight against the Es. With C-cores, the clamps around
slightly drawn up.
Adjustments to the gap size can
only be done with the B+ turned
off so Idc
does not flow thus permitting the E&I to be prized apart to
sheets of gap material during observations of primary inductance
the air gap.
The signal voltage
at Sec should be set at the
0.0dB reference level
at 1kHz with the rated load, and measured. The load is then
and output voltage will rise so it should be then adjusted
down to equal
the reference 0.0dB level. The distortion without any load
lower than with a load.
The frequency is
then reduced slowly down and it
should be able to be
reduced to at least 100Hz without any serious increase in
But as F is reduced
below 100Hz, some THD will
visible on the CRO screen.
CRO 3 shows top trace
Vo = 0.0dB, 100Hz, With
The primary inductance is the only load, and in this example the
found to be 24H measured at the -6dB level at 25 Hz. At 100Hz,
reactance of the Lp > 15k.
The Vo wave has low THD because of the
high RLa value and because the EL34 has
slightly more gain with the
high load thus making the effective applied GNFB = 13.5dB
The bottom trace shows Ia current waveform has some distortion
because the iron is beginning to
show some non linear behaviour.
CRO 4 top trace has
reactance has begun to become a non linear reactance suffering
field collapse during part of each wave cycle. The actual Fsat
onset was at approximately 35Hz. The Vo wave has a
the high THD seen above.
The bottom trace shows the 32Hz current waveform with high
The iron has become highly non linear, and all music would be
CRO 5 top trace has Vo
The primary reactance
has become a very non linear reactance suffering substantial
collapse during each wave cycle.
The bottom 25Hz current waveform contains extreme distortion.
has become highly non linear.
So from these observations we
see Fsat = 35Hz. Now the air gap
thickness should be increased by using a sheet of paper at
re-measured by observing the frequency of onset of saturation
distortion. The inductance must also be re-measured, and may be
the Fsat onset, but at the -6dB Vo level.
I fill one exercize book each
fortnight with notes and
related to my work and to whatever else I think about. Each book
sheets, and is 6mm thick, so each sheet = 6mm / 120 thick
The use of paper is a convenient
way to adjust an air gap, but
if you have
some polyester sheeting of known thickness it is also quite
When the gap is reduced
slightly, the Lp inductance may
If Lp increases, but Fsat may occur at too high a
The Fsat should be found to be a
lower F as the air gap is
If the LP inductance reduces
with increased air gap, its value
recorded for the loading effect it may cause.
22 has three graph
curves for the SE OPT used for the
shown in Fig 13 which has EL34 for CFB operation. The SE OPT
be used for SEUL if all the primary turns are used in the anode
and there are suitable screen taps.
air gap initially
calculated and used was 0.25mm, with gap
of approx 0.12mm thick.
This gave Fsat = 35Hz, and XLp =
RLa at 43Hz. These F seem high,
but in many old radios you would find Fsat = 80Hz and XLp = RLa
The designers intended for low power levels and bass response
down to only 120Hz. Hence the inadequate walnut sized OPTs which
finally approved by the company accountant who dines with the
If the air gap is reduced to a
minimum, ie, with no gapping
one might expect the maximum iron µ to be 2,500 which
lams are fully interleaved. But with butted stacks of Es and Is,
acts as though there still is a gap, but in fact, there isn't.
change of grain
in crystalline structure and the imperfections of the butted
to much less than if all E&I were maximally interleaved
max of 2,500, and where µ max is as low as 2,500, then the
with butted Es and Is may be less than 750. The Fig 17 graph for
uncertain below where the gap < 0.15mm.
So, reducing the gap to the
minimum might increase Lp to 38H,
give XLp = RLa at 30Hz, and raise Fsat to 60Hz. Having such a
is more undesirable than having negligible loading with high Lp
If the gap is increased above
0.25mm to say 0.35mm, then Fsat
to 28Hz, and XLp = 18H, with XLp = RLa at 55Hz. This will mean
will see a partially reactive load = 0.707 x RLa, ie, 4k5 at
tube distortion at 0dB will begin at F above Fsat. This is
Fsat occurring above F where load reduces.
In other words, if Fsat is low enough at 0dB, and XLp may be
at a higher F, and it is a bonus if XLp = RLa at a lower F than
If the gap is further increased
to say 0.45mm, then Fsat moves
lower to 26Hz, while Lp becomes 15H and XLp = RLa at 70Hz which
think is too high, even for a humble AM radio OPT.
In this case, from the Fig 17
graph we may read off the ONE
where XLp = RLa AND where Fsat
0dB output level.
This is seen where the Fsat B
curve intersects the XLp C curve
at point P
which is at 40Hz.
The air gap required to achieve this F may be read vertically
air gap axis and = 0.22mm.
Adjusting the air gap so finely may not be easily possible if
sufficiently fine enough material to use.
If the air gap was left at
0.25mm then Fsat = 35Hz with XLp =
RLa at 43 Hz,
then the operation is quite OK for this application.
This example OPT for a radio OPT
will be found to sound quite
ordinary listening levels, providing the source signal has THD
4Vrms from the AM detector, or 1Vrms from CD player, or FM
Normal audio detector circuits used in 99% of old radios fail to
a true hi-fi OPT
for use with a single EL34 to make 8 Watts,
one would begin with AFe = 450 x sq.root PO.
this case max PO will be
If Pda = 18Watts then PO = 8.1Watts.
So Afe = 450 x sq.rt 8.1 = 1,280sq.mm, and the core should have
Tongue = 32mm and Stack = 40mm. This core size is obviously much
than is ever seen in most SE high-end
amps producing 8 watts from
tubes like 6550, KT88, EL34, 6L6 or 300B, but to obey all the
mentioned at this website, one must start with much more iron
amplifiers made in large batches with design work by Bean
utterly despicable character who seems to find employment to
quality everywhere he goes.
What I have described here is
acceptable for the intended use
demonstrates how to set air gaps
to maintain a high enough ohm value
for the anode load down to low enough bass frequencies and
core saturation for most music.
Within music signals, the bass
between 50Hz and 200Hz may average
CRO 6 has Vo = -6dB, F =
and with secondary RL so PO = 1.85
There is no sign of saturation or loading related
The bottom Ia trace is also very clean.
levels above 200Hz, which means that one would not want bass
the -6dB level so there is voltage headroom for
much clipping. An ancient old radio made in 1937 with large floor
may have a 10" speaker driver with sensitivity at 1kHz = of 93dB/W
and only give a minimal amount of bass. The OP tube might be a 42,
6BQ5, 6M5 and because OPT winding losses might be 25%, and B+
PO max might only be 2.5 Watts, so at the highest practical level
PO = 0.63 Watts, which is enough to fill a lounge room with sound.
But so often the old 1935 speaker has so many faults that it must
with a modern 10" driver unit which has a flatter response curve
and more pleasing bass level relative to its midrange than the
But the modern speaker sensitivity may be 90dB/W/M at 1kHz, and
expect less THD/IMD and higher levels, so it is prudent to upgrade
possible PO to 8 Watts or more so that healthy average levels can
Most modern drivers with permanent magnets and more supple cone
diaphragms will operate with far less THD or cone break up than
in 1935. HF performance of all 10" drivers is usually poor with
I recommend that a 65mm cone drive unit be used to cover the
2kHz and 8kHz which is sufficient for excellent radio sound for
In a recent example, I made a holding strap for the small driver
section of 20mm x 3mm so the smaller driver could be mounted
"inside" the larger bass unit. This meant there was no need to
timber cabinet by trying to make a 75mm hole for an additional HF
Some drivers meant for use in car audio
systems or for ceiling mounting have
a small cone midrange and dome tweeter already concentrically
These make good replacements for many 8" 1935 speakers where the
speaker was also 8" ( 200mm dia ) The speaker must be obtained
OPT is wound to ensure the OPT has the right turn ratio and
7 top trace shows Vo
with NO RL.
on the EL34 output tube is a mainly linear inductive reactance of
The reactance of 24H at 20Hz = 3k0, less than 1/2 the rated load
so hence one can see approximately 3% THD caused by the low value
inductive load at LF.
The bottom trace for Ia shows THD at about 7%. The
loop of GNFB
is not succeeding to reduce the voltage distortion very much. But
little music content occurs at 20Hz, the sound will remain OK.
8 top trace
has Vo = -6dB, F = 16Hz, with NO
The load on the
EL34 output tube has become a non linear inductive reactance of
and saturation is occurring. In fact the onset of saturation at
and half the Fsat for the 0dB Vo level. This shows shows
be a voltage
caused phenomena with no relation to the RL load used. The EL34 is
with the low value inductive load which at 16Hz = 2k4. If the Vo
at the -12dB level, and at 8Hz, RL would be 1k2, and saturation
The bottom trace for Ia shows THD at about 15%. The
loop of GNFB
is not succeeding to reduce the THD at Vo.
Conclusions for this example
for 1 x
Fsat should be found using Vo = 0dB.
This 0dB voltage level is obtained at the symmetrical clipping
with the RLa load value which gives the maximum possible SE output
The Fsat is measured at the 0dB level without any RL connected and
saturation distortion becomes visible on the CRO, ie, exceeds 2%.
For pentode and tetrode output tubes without any NFB, the LF
the amplifier without any RL connected will reduce at
per octave below where XLp = Ra. If Lp = 24H, and EL34 Ra = 12k0,
Fco = 80Hz. It should be possible to have GNFB connected to
Vo level below the unloaded Fco until saturation occurs, or where
ie, the loaded Fco, and if RLa = 6k4, loaded Fco = 42Hz.
Fsat at 0dB level should be below the loaded Fco frequency. Where
no load used, the Fsat can usually be measured at the 0dB level at
The XLp may become too low at LF if Vo is maintained at the 0dB
and THD will begin to exceed 2%. The response should be plotted
levels limited by THD = 2%, and the Fsat may be found at the lower
level than at 0dB.
Often such a situation may suggest the air gap is too small, and
increased to lower the loaded Fco while raising the Fsat, until
= Fsat and both are at satisfactory minimum frequency where the
is at 0dB.
Measurements of inductance should be made at between 20Hz and 50Hz
and Vo levels below which saturation or inductive loading
clearly visible on the
In the 8 Watt SEUL OPT example, the Lp was found to be 24.2H
At what F is Lp reactance = RLa?
RLa = 6k4, and F = RLa / ( 6.28 x Lp )
Is this acceptable? In previous calculations with example OPT4,
The design principles try to give XL = RLa at Fsat = 14Hz for a
Watt hi-fi amplifier.
But the example here is for only 1 x EL34 used in SEUL mode to
in an AM radio to give about 3 times the 2.5Watts from a 6V6
ancient and very inefficient and inferior quality OPT.
Fsat at 42Hz is OK for an old radio.
From measuring Lp, the iron µe is calculated :-
µe = 1,000,000,000 x
Lp x mL
= 1,000,000,000 x 24.2 x 189 / ( 1.26 x 3,200 x 3,200 x 24
x 25 )
Bdc is then calculated :-
Bdc = 12.6 x
x Np x Idc
= 12.6 x 590 x 3,200 x 0.051 / ( 189 x 10,000 )
Bac is then calculated at the onset of saturation Fsat :-
Fsat = 35Hz, Va = 220Vrms at 0dB at 7.6W into 6k4.
x V x
= 22.6 x 220 x 10,000 / ( 25 x 24 x
3,200 x 35 )
Total Bac + Bdc = 0.64 + 0.74 =
Is the air gap correct?
Calculate air gap using estimate of maximum permeability for the
perhaps 3,000 :-
Air gap = mL x ( µ
- µe )
= 189 x ( 3,000 - 590 ) / ( 3,000 x 590 )
= 0.26mm. Use 0.12mm gap material across the two magnetic gaps.
What are consequences if gap is
Fig 17 graph above shows
results gained with changes in air gap.
Increase air gap to lower
µe to say 300.
Lp will become 12.3 Henrys.
XLp = RLa at Fco and will become 82Hz.
The Bdc will reduce to 0.38Tesla.
Allowable Bac = 1.38 - 0.38 = 1.0 Tesla, which means higher Vac
be applied at 35Hz without saturation, but Vo could not reach 0dB
because XLp is too low.
Therefore air gap must not be increased.
Decrease air gap to obtain
µe = 750 which probably would be the maximum
possible µe with normal air gapping without partial air
Lp will become 30.8 Henrys, and XLp = RLa at Fco = 33Hz.
The Bdc will increase to 0.94 Tesla.
Allowable Bac = 1.38T - 0.94T = 0.44Tesla, which means Fsat will
at a higher F.
Fsat = 22.6 x V
= 22.6 x 220 x 10,000 / ( 25 x 24 x 3,200 x 0.44 )
This is too high and therefore air gap should not be reduced to
All things considered, the air gap for SEUL operation is about
Triode operation instead of SEUL could possibly be used instead of
max could be about 4.4Watts with EL34 Pda = 17.8W. Load line
would have to be done.
Triode operation may require higher Ea and lower Ia to gain the Ea
swing and also
higher RLa but
the Fco and Fsat and Lp must all be examined lest music be spoiled
by not optimizing the design.
this an easier
For cores with
Idc flow in windings
in one direction,
1E. Calculate RLa, maximum PO.
For SE tube or paralleled tubes the Center Value RLa is
Eadc and Iadc. The Center Value RLa is the load where maximum
SE power is possible for the given Ea and Ia. Other loads need
Pentodes, Beam Tetrodes, UL, CFB RLa = 0.9 x Ea/Ia,
Max PO = 0.45 x Ea x
Also Max PO = 0.5 x RLa x Ia squared.
Triodes, RLa = (Ea/Ia) - (2 x Ra),
Max PO = 0.5 x RLa x Ia squared.
eg, For 1 x EL34, SEUL, Ea = 350V, Ia = 50mA.
RLa = 0.9 x 350 / 0.05 = 6,300 ohms.
PO = 0.5 x 6,300 x 0.05 x 0.05 = 7.87 Watts.
2E. Calculate Afe,
x sq.rt PO.
Eg, For SEUL EL34 pentode Afe = 450 x
sq.rt 7.87 = 1,262sq.mm.
3E. Calculate required Lp so XLp = RLa at 20Hz.
Lp for hi-fi = RLa / ( 6.28 x F ) and for 20Hz,
Lp = RLa / 125
eg, For 1 x EL34, SEUL, RLa = 6k3, Lp = 6,300 / 125 =
primary turns, Np.
Np = Lp x Idc x 1,400,000
eg, For 1 x EL34 SEUL example,
Np = 51.2 x 0.05 x
1,400,000 / 1,262 = 2,839
NOTE, the above simple
derived as follows :-
Lp = 1.26 x Np x Np x
Afe x µe .........equation 1
1,000,000,000 x ML
µe = Bdc x 10,000 x ML
12.6 x Np x Idc
Bdc will remain constant.
Bac + Bdc at Fsat at 20Hz, should not be more than 1.4Tesla to
lowest grade iron which may be
available from surplus re-cycled cores.
Higher grade GOSS cores will have a magnetic headroom for
but the 1.4 Tesla design limit allows for slight variations.
Therefore Bdc for all designs should be 0.7Tesla.
µe = 0.7 x 10,000 x ML
555 x ML
12.6 x Np x Idc Np x
this µe from
eqtn 2 can be substituted into
eqtn 1 to give
Lp = 1.26
Afe x 555 x ML .........equation 1
Then Np = Lp x
Idc x 1,400,000
1,000,000,000 x ML x
Np x Idc
Np x Afe
The constant of 1,430,000 may
rounded to 1,400,000.
Idc in any part of primary winding.
Idc = anode Ia dc plus
Ia dc = Ia +
Ig2 = approximately 1.25 x Ia.
Minimum Cu wire dia = 0.8 x sq.rt
not exceed max = 2A per
eg, For 1 x EL34, Ia = 50mA,
12mA, allow Idc max = 62mA.
Cu dia minimum = 0.8 x sq.rt
0.062Amps = 0.199mm.
Therefore minimum size
= 0.20mm Cu dia
From wire table, oa wire dia =
6E. Calculate minimum winding window size area
and L x H dimensions.
H = 1.092 x sq rt Np x
eg, For 1 x EL34,
1.092 x sq rt 2,839 x 0.245mm = 14.25mm.
Therefore L = 3H = 43.5mm.
This is derived from
Most cores have window
with aspect ratio 3:1, or L = 3H,
so area of window = 3H x H =
The primary wire occupies
area = 0.28 x L x H = 0.28 x 3Hsquared,
= 0.84 x H squared.
The primary wire occupies
area = Np x oa dia squared.
So 0.84 x H squared = Np x oa
H = square root of ( Np x oa
squared / 0.84 )
= sq.rt Np x oa dia /
Calculate Core Tongue and Stack, T
The minimum window H
been calculated above.
Consider wasteless pattern
T = 2H.
eg, For 1 x EL34, from Step
6E, T =
2 x 14.25 = 28.5mm.
Select from the range of
size T for old wasteless E&I lams,
0.75" = 19mm; 1.0" =
25.4mm; 1.125" = 28.57mm; 1.25" = 31.75mm;
1.5" = 38.1mm; 1.75" =
44.45mm; 2.0" = 50.8mm; 2.5" = 63.5mm.
The standard wasteless T
28.57mm ( 1.125" ).
/ T = 1,262 / 28.57 = 44.17mm ( 1.74" ).
Consider non wasteless
E&I or C-cores.
H and L and Afe all must not
smaller than calculated so far.
Therefore choose the C-cores
standard E&I lams based on the
For E&I, measure the T
and Stack = Afe / measured T.
Some samples of non standard
wasteless E&I lams have T size less
than 2H. For example, if H =
T might be 22mm.
So if Afe = 1,262sq.mm, S =
22 = 57.4mm.
For C-cores, the S dimension
"strip width" of the wound core
material. Measure the build
thickness of one C-core,
and Strip width = Afe / ( 2 x
The build-up of a C-core is
height of the wound core material
which has been wound with
layers and glued
Double C-cores are usually used hence T = 2 x build-up.
For example, go to the Eilor website which lists C-core sizes,
Consider the window sizes available and select the cores with
window size *above* that which has been calculated so far.
eg, For 1 x EL34 SEUL, H x L = 14.4mm x 43mm, and the Eilor
has a "T32" C-core with the following dimensions :-
Window = 15.9mm x 50.8mm.
Build up = 10mm,
Strip width = 32mm.
Therefore T = 20mm, so required strip width = 1,262 / (2 x 10) =
Therefore one might use 4 x T32 cores, ie, two pairs of 00
stacked on top of each other to give an Afe section = 20mm x
The benefit of the larger window allows the wire size to be
increased to reduce all winding losses. The Eilor C-cores are
things to use, and are beautifully made, but the price is many
wasteless pattern E&I lams. A very fine 8 Watt SE OPT may be
using some old re-cycled core material which may not have cost
except the labour involved to extract the material from old
which have burnt out windings.
The secondary pattern and accurate
fitting of wire, insulation must
be worked out using the complex longer methods above.
8E. Calculate µe,
assume wasteless E&I laminations.
µe = 3,111 x T / ( Np x
eg, For 1 x EL34, the core is
28mm tongue size, Ia = 62mA
µe = 3,111 x 28 / ( 2,839 x 0.062 ) = 495.
This has been derived as follows:-
Assume always Bdc = 0.7T.
For wasteless pattern E&I, ML
= 5.6 x
µe = Bdc x
10,000 x iron ML
12.6 x Np x Idc
= 0.7 x 10,000 x 5.6 x T / ( 12.6 x Np x Idc )
If µe < 750, µe may be achieved by normal air gap
Now maximum possible Bdc
= 0.7 Tesla = 12.6 x µe x Np x Idc / ( 10,000 x ML ).
For a given RLa, and the same PO, Idc must not be changed.
If µe was calculated > 750, then to achieve
µe the Np must be
increased, or ML made longer, ie, use a larger core size, say T
with ML = 180mm. For a given Afe, the µe must not be
increasing the air gap because the Lp would then become too low.
With most E&I, even if the Es and Is are tight together
an actual gap, µe may not rise above 750 unless the
Partial air gapping might be used but it involves more patience
and error assembly and testing.
But with C-cores the µe may easily be varied from anywhere
max µ of say 10,000 to 100.
9E. Check Lp and
Lp = 1.26 x Np x Np x
Afe x µe
1,000,000,000 x ML
eg, For 1 x EL34, Np = 2,839t, Afe = 28mm x 44mm,
Lp = 1.26 x 2.839 x 2.839 x 28 x 44 x 495 / (
1,000 x 5.6 x 28 ) = 39.5Henrys.
If Lp is not as high as wanted in Step 3E, Core S may be
easily because Lp is proportional to S.
eg, For 1 x EL34, For wasteless E&I, the standard plastic
might be for 51mm stack or perhaps 57mm so S = 50mm gives Lp =
and S = 57 gives Lp = 51H which will be enough.
At Fsat, Bdc = Bac = 0.7Tesla,
Fsat = 22.6
x Va x 10,000
Np x Bac
PO = 7.87
W into 6k3, Va =
223Vrms, Bac = 0.7Tesla,
Fsat = 22.6 x 223 x 10,000 / ( 28 x 44 x 2,839 x 0.7 ) = 20.6Hz
Notice Fsat is inversely proportional to Afe, or S, so using
means lower Fsat.
10E. Calculate Air
Gap and gap
Ag = ML x (
eg, for T = 28, ML = 5.6 x T = 157mm, µe = 601 from
Assume E&I core has max possible µ = 3,000.
Ag = 157 x ( 3,000 - 601 ) / ( 3,000 x 601 ) = 0.2088mm.
Therefore the gap on two sides of magnetic path = Ag / 2 =
The OPT should
survive HV testing without the OPT
having been varnished
or waxed, and should survive the application of
+4,000 Vdc to the primary
for 1 minute without any
with all the secondaries and core well grounded.
Fig 23 shows a
schematic for applying approximately +3,920Vdc
primary of a transformer under test via 9 x 1M 2W metal film
The rectifier is built with a "ladder network" to step up the
an old surplus transformer secondary using 0.1uF caps rated for
and 1N4007 diodes.
If there is an old radio set transformer with secondary of
the available Vac max is 760Vac, or 1,070V peak available.
the ladder network would need only 4 step ups to generate
4,000Vdc, but the 0.1uF caps used should be 2,000V rated and the
diode "rung" in the ladder would need to consist of say 4 x
each with 1Meg across each to ensure the reverse bias across
is limited to less than 300Vpk, and about equal for each
may be varied by using a variac to control the
primary of the
and secondary is connected to a 10k0 x 5Watt
is taken 0V of the power supply which MUST be also grounded
green and yellow Earth wire from the wall power outlet. A Vdc
connected across the 10k0 resistance.
is turned on and voltage is raised slowly to
No damage or
smoke should occur to
anything during testing.
volt meter should not show any voltage across the 10k
does occur, there will be about +4,000Vdc across 9 Megohms and
Idc = 0.44A. This current
will flow through the 10k0 and produce a reading
of 4.4Vdc. This is the maximum
current flow, and indicates a short circuit
somewhere between primary and anything at
earth potential. If arcs occur,
they may be intermittent and pulses will be
seen on the meter. If a constant
Vdc is seen across the 10k0, it indicates
there is some DC current leakage
through what must be resistance in the insulation,
and insulation is faulty.
used for the measurement may be a normal cheap analog
Output transformer OPT3 for
a previous design
Arcs may pulse, and digital meters may not give read out numbers
in 2006 edition of the website.
OPT No3 is capable of around 25 Watts of SE
wish to analyze what tubes and operating conditions
be suitable for OPT3.
to SE OPT calc Page 2.
to SE OPT calc Page 1.