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 AM radio.

Fig 19. Schematic for 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 waveforms 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 lamination dimensions.

ALTERNATIVE METHOD to calculate max PO of any pentode

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 OPT from web pages created May 2006.

OPT3 from 2006.

Fig 24. Details of OPT3.

---------------------------------------------------------------------------------------------------------

Practical testing of 8 Watt SE Amp example, measuring Lp,

and Fsat, and setting the air gap.

GREAT CARE MUST BE TAKEN WHEN MEASURING

TUBE AMP VOLTAGES !!!!!!!!!!!!!!!

Fig 19.

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 watts.

Usually the best old AM radios will have a B+ supply generated from a

320V-0-320V HT 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

may be a field coil choke on the speaker with Rw = 1k5, and then there is a large

voltage drop across the 5Y3 anode resistance and across OPT winding resistance.

Electrolytic filter caps are often quite low so ripple voltage is high.

The ancient 1940 speaker may be 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 replacement speaker

with no field coil may have lower sensitivity but have a better bass response.

Treble response in an old speaker may have only extended to 3kHz, so a small

coned HF unit may be mounted concentrically inside the bass unit to extend

HF to give almost hi-fi performance, with at least 9kHz BW.

To drive a modern speaker, the audio amp may be vastly improved. The 5Y3

is replaced with silicon diodes charging into a CRC filter using modern

replacement 220uF x 450Vdc rated caps and R may be say 235 ohms using a

pair of 10W rated 470 ohms in parallel. Usually 70mA may be had from an

existing old power transformer which may be still serviceable, and not on the

verge of burning out with repeated continued use. Some power transformers

will benefit from being removed from the chassis, and soaked in electrical

varnish over night, then baked for 4 hours at 125C the next day.

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 C2 is +400Vdc.

Such a nice high B+ allows the use of an EL34, 6L6, ( even 807 ) to be used in

the chassis socket meant for 5Y3, and where the 6V6 once was, there may be

a 12AX7, or 6J7 in triode mode. I find the 12AX7 plus EL34 in triode mode

is simplest and will give 6 watts even if using the old OPT if it is suitable.

Fig 13 above shows a best practice use of OPT with SE+CFB configuration.

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 get a

healthy Ea across the EL34 of +350Vdc, and up to 8 Watts of output power.

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 50% UL tap

to the EL34 screen.

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 produce

better power and lower THD than a 6V6, 6F6 etc or 6BQ4/EL84.

Triode operation is also generally superb compared to normal 6V6/EL84

in pentode mode, and with triode connected EL34 there is no need for global

NFB. Often it is found the old ancient OPT and speaker made prior to 1955

(when accountants took over design sections of radio production) are OK and

will work very well with EL34 in triode. Triode operation of EL34 is best

with Ea at +350V to +420V, and Ia at idle should be adjusted to give the

most power with least THD at 400Hz. Often the Ia will be no more than what

was used in the original use of 6V6, so there may be no extra heat produced

in the power transformer. The 6L6 or 807 beam tetrodes need more grid drive

signal, but draw less screen current, but sound excellent in triode or with 50%

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, and when

set up with Pda = 18Watts, they will last maybe 15 years easily. I used a

second hand EL34 in the AM radio I designed and built in 1999, and it

produces excellent sound and has never needed to be serviced.

My radio customers appreciate me putting a pair of RCA sockets which may be

used for CD, FM tuner, whatever, for very pleasing mono sound for intimate

lounge room or kitchen listening.

Fig 20.

Fig
20 shows the rear view of
a completely re-engineered 1935 AM radio chassis

from July 2011. The modern 10" bass-midrange speaker with
concentric HF
driver

is shown. There is a mains fuse, IEC mains cable input and
grounded
chassis for

safety. There is a replacement PT on rhs, with EL34 and 12AX7
audio amp.

There is a new OPT in the chassis centre. There are RCA
terminals
provided

for use with FM tuner or CD player. Sound will be mono, but very
listenable.

The other tubes used to replace rare European odd ball types
from 1935
are

6AN7 mixer, 6N8 IF amp, 12AU7 detector, 12AU7 treble control
stage with

slight gain.

The
OPT has a non-wasteless
pattern of old low grade EI
laminations

with Tongue = 25mm, Stack = 24mm, Window L =
52mm, H
= 16.6mm.

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
is used

because the speaker for this project is 8 ohms.

where low power operation will be used. The 10" speaker is mounted in

a large open backed cabinet which has a natural cut off at about 60Hz.

There is no point in making an amp which needs to produce full power at

below 50Hz. The amplifier bandwidth has been tailored to have a cut off at

40Hz.

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 and amp performance :-

Before proceeding to test the SE audio amp the circuit needs to be checked

3 times against the schematic before turning anything on. Fuses should be

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 will offer

protection if there is bias failure in the output tube when Ia may rise to 300mA

before the output tube self destructs. The mains fuse should be 240V x 0.5A

or 110V x 1A slow blow, or whatever value is needed to prevent nuisance

fuse blowings, but never so high to offer almost no protection.

Once the winding of the OPT is complete, the UN-VARNISHED OR

UN-WAXED OPT is temporarily placed into the amplifier which is

fully operational with all NFB loops connected.

If the bobbin has been varnished with cold cure polyurethane two pack

mix, the core may be assembled into the wound bobbin without the final

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

casting resin.

The core air gap MUST be able to be adjusted before the final potting

or varnishing procedure, because after potting it is impossible to easily

change the air gap.

The air gap used initially should be what has been calculated,

and it may or may not need any adjustment.

The amplifier output voltage at
clipping should be measured
using a dummy

resistance load which is the load value for absolute maximum
possible
output

power using a sine wave, with THD < 2%, and at 400 to
1,000Hz.

With higher output voltage levels
the sine wave should show equal clipping to

positive and negative wave
crests, ie, the wave clipping is symmetrical.

The
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

voltage
level.

An
oscilloscope must be used to
monitor
all
signal
voltage
measurements

and distortions at the output terminals. Sine waves with THD
< 0.5%
must

be used to drive the amp from a signal generator capable of F
range
between

2Hz and 2MHz and with a flat response for all ranges which may
be
adjusted

for level and F.

anode to OPT, as R15, Test Point 2. The schematic also shows 10ohms below

CFB winding as R14, giving Test Point 1 so monitoring anode signal current is

more convenient than having the R at a high voltage level. The R14 also carries

the screen dc current which will be about 7% of anode current. There is some

screen signal current flow so measurements of anode current are most accurate

at TP2. But anode current waveform distortion is most easily seen at TP1.

With GNFB present, the OP voltage at Sec may appear linear while the anode

current may appear slightly distorted because the NFB is keeping the Sec voltage

linear by applying a grid signal voltage containing an "error signal" to the output

tube grid. As the output signal is increased towards clipping the output tube

grid signal and anode current signal will show increasing wave distortions, but at

about 2dB below clipping, all wave distortions should be less than 2% and thus

distortion will not affect measurements and calculations.

The amplifier should be able to be optimized for unconditional stability at LF

and HF so that with GNFB connected it will not oscillate at any F even with

no secondary load connected. If no GNFB is to be used, the EL34 should be

strapped as a triode. Slightly less output power is available than pentode mode

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 factor

with triode is high enough to not need GNFB. With 50% SEUL, Ra will be

about 3k0, and some GNFB is useful, but seldom is there any need for more

than 12dB of GNFB. I do not like ordinary pentode connection which requires

more GNFB. 15% CFB will reduce EL34 Ra to less than triode.

Below are photos of oscilloscope screen using a 1983 dual trace 15Mhz

Hitachi oscilloscope ( CRO. )

The images below show typical wave forms common to many tube

SE amps with Global Negative Feedback, GNFB.

The amp tested to gain these images has 1 x Sovtek EL34 set up in SEUL

configuration with UL tap at 50% of anode turns.

CRO 1.

CRO 1 shows two traces, top is the secondary output voltage

with rated load at onset of symmetrical clipping at 1khz.

The negative wave crests at Ea minimum show clipping because of grid

current and the inability of the EL34 anode voltage to swing any lower.

The positive 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 schematic.

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 voltage shows

no such current peaks. At this point the coupling cap driving the EL34

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 shown

which indicate THD = 2% approximately.

Measuring primary inductance.

The primary inductance should be measured at the -6dB Vo level

at the lowest frequency easily measured and where THD < 3%.

CRO 2.

CRO 2 top trace shows Vo = -6dB, 25Hz, NO RL. At this voltage level

there is no saturation and the primary load is solely the inductive reactance.

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 linear

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 between

anode and 0V, or across the whole anode winding from B+ to anode.

In this example Vac across R14 = 0.265Vrms, and so Ia = 0.0265Arms,

So Lp reactance, XLp, at 25Hz = Vac / Iac = 100V / 0.0265A = 3,773 ohms.

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 value.

The bottom 25Hz current waveform shows 5% THD because the GNFB

is trying to correct the voltage distortion. If you examined the EL34 grid

signal you might find it had 5% THD or more, because a fraction of any

THD at the amp output is fed back to V1 12AX7 cathode then amplified

so it appears at V2 EL34 grid with its phase inverted, therefore trying to

reduce the output THD as it is produced.

The above waveforms show that the air gap would be nearly optimised.

If there was severe THD at say -6dB at say 60Hz, it may indicate the gap

is far too small or too large or that the OPT being tested may be unsuitable

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 Vreference,

and -12dB is where Vo = 1/4 Vreference. Graphs for F response may

be done using an exercise book page and 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 size for optimum operation.

Fig
21.

Fig
21 shows wasteless
pattern
E&I laminations.

There are TWO magnetic paths around the TWO core windows
which
act

together as ONE magnetic path.

The Iron magnetic path length,
ML = 2H
+ 2L + 22H/7 for any core material.

For Wasteless
Pattern
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
blocks of

Es and Is tightly together and leaving the clamping bolts loose
may
allow

the E&I to
come as
close as the gap material permits. But a newly wound

OPT may be found to be slightly forcing the stack of Es and Is
apart
slightly,

so mechanical clamping together may be necessary. This is a real
danger
to

the OPT performance and MUST be checked very carefully during
tests.

Hand made bobbins using say 1mm thick cardboard are very prone
to

slight bulge during winding and and inexperienced DIYers may not
design

the bobbin size with some clearance, or may not have a winding
lathe
which

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
varnishing.

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
lay flat
and

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
the
cores are

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
remove
or add

sheets of gap material during observations of primary inductance
and
Fsat.

Adjusting
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
removed,

and output voltage will rise so it should be then adjusted
down to equal

the reference 0.0dB level. The distortion without any load
should be

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
distortion.

But as F is reduced
below 100Hz, some THD will
inevitably become

visible on the CRO screen.

CRO
3.

CRO 3 shows top trace
has
Vo = 0.0dB, 100Hz, With
NO
RL.

The primary inductance is the only load, and in this example the
Lp was

found to be 24H measured at the -6dB level at 25 Hz. At 100Hz,
Lp

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
approx.

The bottom trace shows Ia current waveform has some distortion

because the iron is beginning to
show some non linear behaviour.

CRO
4.

CRO 4 top trace has
Vo =
0.0dB,
at
32Hz,
with
NO
RL.
The primary

reactance has begun to become a non linear reactance suffering
partial

magnetic
field collapse during part of each wave cycle. The actual Fsat

onset was at approximately 35Hz. The Vo wave has a
sudden
onset of

the high THD seen above.

The bottom trace shows the 32Hz current waveform with high
distortion.

The iron has become highly non linear, and all music would be
ruined.

CRO
5..

CRO 5 top trace has Vo
=
0dB,
25Hz,
NO
RL.
The primary reactance

has become a very non linear reactance suffering substantial
magnetic
field

collapse during each wave cycle.

The bottom 25Hz current waveform contains extreme distortion.
The
iron

has become highly non linear.

So from these observations we
see Fsat = 35Hz. Now the air gap
material

thickness should be increased by using a sheet of paper at
0.05mm and
the

Fsat
re-measured by observing the frequency of onset of saturation

distortion. The inductance must also be re-measured, and may be
done at

the Fsat onset, but at the -6dB Vo level.

I fill one exercize book each
fortnight with notes and
schematic drawings

related to my work and to whatever else I think about. Each book
has
120

sheets, and is 6mm thick, so each sheet = 6mm / 120 thick
=
0.05mm.

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
suitable.

When the gap is reduced
slightly, the Lp inductance may
increase.

If Lp increases, but Fsat may occur at too high a
frequency.

The Fsat should be found to be a
lower F as the air gap is
increased.

If the LP inductance reduces
with increased air gap, its value
must be

recorded for the loading effect it may cause.

Fig
22.

Fig
22 has three graph
curves for the SE OPT used for the
example amp

shown in Fig 13 which has EL34 for CFB operation. The SE OPT
could

be used for SEUL if all the primary turns are used in the anode
circuit

and there are suitable screen taps.

The
air gap initially
calculated and used was 0.25mm, with gap
material

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
at
100Hz.

The designers intended for low power levels and bass response
extending

down to only 120Hz. Hence the inadequate walnut sized OPTs which
were

finally approved by the company accountant who dines with the
CEO.

If the air gap is reduced to a
minimum, ie, with no gapping
material,

one might expect the maximum iron µ to be 2,500 which
occurs if
the E&I

lams are fully interleaved. But with butted stacks of Es and Is,
the
core

acts as though there still is a gap, but in fact, there isn't.
The
change of grain

in crystalline structure and the imperfections of the butted
join will
limit µe

to much less than if all E&I were maximally interleaved
which might
give µ

max of 2,500, and where µ max is as low as 2,500, then the
maximum µe

with butted Es and Is may be less than 750. The Fig 17 graph for
Lp
becomes

uncertain below where the gap < 0.15mm.

So, reducing the gap to the
minimum might increase Lp to 38H,
and thus

give XLp = RLa at 30Hz, and raise Fsat to 60Hz. Having such a
high Fsat

is more undesirable than having negligible loading with high Lp
reactance

at 60Hz.

If the gap is increased above
0.25mm to say 0.35mm, then Fsat
moves down

to 28Hz, and XLp = 18H, with XLp = RLa at 55Hz. This will mean
the tube

will see a partially reactive load = 0.707 x RLa, ie, 4k5 at
55Hz, and
thus

tube distortion at 0dB will begin at F above Fsat. This is
preferable
to having

Fsat occurring above F where load reduces.

In other words, if Fsat is low enough at 0dB, and XLp may be
less than
RLa

at a higher F, and it is a bonus if XLp = RLa at a lower F than
Fsat.

If the gap is further increased
to say 0.45mm, then Fsat moves
only slightly

lower to 26Hz, while Lp becomes 15H and XLp = RLa at 70Hz which
I

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
frequency

where XLp = RLa AND where Fsat
occurs at
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
below P
on the

air gap axis and = 0.22mm.

Adjusting the air gap so finely may not be easily possible if
there is
not any

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
superb at

ordinary listening levels, providing the source signal has THD
< 1%
at say

4Vrms from the AM detector, or 1Vrms from CD player, or FM
tuner.

Normal audio detector circuits used in 99% of old radios fail to
produce

low distortion.

For
a true hi-fi OPT
for use with a single EL34 to make 8 Watts,

then
one would begin with AFe = 450 x sq.root PO.

In
this case max PO will be
0.45 x
tube pda.

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
larger

than is ever seen in most SE high-end
hi-fi
amps producing 8 watts from

tubes like 6550, KT88, EL34, 6L6 or 300B, but to obey all the
rules

mentioned at this website, one must start with much more iron
than in
most

amplifiers made in large batches with design work by Bean
Kounter, that

utterly despicable character who seems to find employment to
remove

quality everywhere he goes.

What I have described here is
acceptable for the intended use
and it

demonstrates how to set air gaps
to maintain a high enough ohm value

for the anode load down to low enough bass frequencies and
without

core saturation for most music.

CRO
6.

CRO 6 has Vo = -6dB, F =
47Hz
and with secondary RL so PO = 1.85
Watts.

There is no sign of saturation or loading related
distortions.

The bottom Ia trace is also very clean.

levels above 200Hz, which means that one would not want bass signals to exceed

the -6dB level so there is voltage headroom for midrange&treble levels without

much clipping. An ancient old radio made in 1937 with large floor standing cabinet

may have a 10" speaker driver with sensitivity at 1kHz = of 93dB/W at one metre,

and only give a minimal amount of bass. The OP tube might be a 42, 6F6, 6V6,

6BQ5, 6M5 and because OPT winding losses might be 25%, and B+ rather low,

PO max might only be 2.5 Watts, so at the highest practical level of -6dB Vo,

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 be replaced

with a modern 10" driver unit which has a flatter response curve and much higher

and more pleasing bass level relative to its midrange than the 1935 driver.

But the modern speaker sensitivity may be 90dB/W/M at 1kHz, and people

expect less THD/IMD and higher levels, so it is prudent to upgrade the maximum

possible PO to 8 Watts or more so that healthy average levels can be achieved.

Most modern drivers with permanent magnets and more supple cone suspension

diaphragms will operate with far less THD or cone break up than anything made

in 1935. HF performance of all 10" drivers is usually poor with cut off at 2kHz, and

I recommend that a 65mm cone drive unit be used to cover the response between

2kHz and 8kHz which is sufficient for excellent radio sound for most listeners.

In a recent example, I made a holding strap for the small driver using an aluminium

section of 20mm x 3mm so the smaller driver could be mounted concentrically

"inside" the larger bass unit. This meant there was no need to RUIN a beautiful old

timber cabinet by trying to make a 75mm hole for an additional HF driver.

Some drivers meant for use in car audio systems or for ceiling mounting have

a small cone midrange and dome tweeter already concentrically mounted.

These make good replacements for many 8" 1935 speakers where the original

speaker was also 8" ( 200mm dia ) The speaker must be obtained before the

OPT is wound to ensure the OPT has the right turn ratio and impedance ratio.

CRO 7.

CRO 7 top trace shows Vo = -6dB, F = 20Hz, with NO RL. The load

on the EL34 output tube is a mainly linear inductive reactance of 24H.

The reactance of 24H at 20Hz = 3k0, less than 1/2 the rated load of 6k4,

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 because

little music content occurs at 20Hz, the sound will remain OK.

CRO 8.

CRO 8 top trace has Vo = -6dB, F = 16Hz, with NO RL. The load on the

EL34 output tube has become a non linear inductive reactance of 24H

and saturation is occurring. In fact the onset of saturation at -6dB is about 17Hz,

and half the Fsat for the 0dB Vo level. This shows shows saturation to be a voltage

caused phenomena with no relation to the RL load used. The EL34 is struggling

with the low value inductive load which at 16Hz = 2k4. If the Vo was inspected

at the -12dB level, and at 8Hz, RL would be 1k2, and saturation would have just

begun.

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 EL34, SEUL.

Fsat should be found using Vo = 0dB.

This 0dB voltage level is obtained at the symmetrical clipping level at 1kHz

with the RLa load value which gives the maximum possible SE output power.

The Fsat is measured at the 0dB level without any RL connected and where

saturation distortion becomes visible on the CRO, ie, exceeds 2%.

For pentode and tetrode output tubes without any NFB, the LF response of

the amplifier without any RL connected will reduce at approximately 6dB

per octave below where XLp = Ra. If Lp = 24H, and EL34 Ra = 12k0, then

Fco = 80Hz. It should be possible to have GNFB connected to maintain 0dB

Vo level below the unloaded Fco until saturation occurs, or where XLp = RLa

ie, the loaded Fco, and if RLa = 6k4, loaded Fco = 42Hz.

Fsat at 0dB level should be below the loaded Fco frequency. Where there is

no load used, the Fsat can usually be measured at the 0dB level at the loaded

Fco.

The XLp may become too low at LF if Vo is maintained at the 0dB level,

and THD will begin to exceed 2%. The response should be plotted for Vo

levels limited by THD = 2%, and the Fsat may be found at the lower Vo

level than at 0dB.

Often such a situation may suggest the air gap is too small, and it may be

increased to lower the loaded Fco while raising the Fsat, until loaded Fco

= Fsat and both are at satisfactory minimum frequency where the level

is at 0dB.

Measurements of inductance should be made at between 20Hz and 50Hz

and Vo levels below which saturation or inductive loading distortion becomes

clearly visible on the oscilloscope.

In the 8 Watt SEUL OPT example, the Lp was found to be 24.2H at 32 Hz.

At what F is Lp reactance = RLa?

RLa = 6k4, and F = RLa / ( 6.28 x Lp )

= 42.1Hz.

Is this acceptable? In previous calculations with example OPT4,

The design principles try to give XL = RLa at Fsat = 14Hz for a good 40

Watt hi-fi amplifier.

But the example here is for only 1 x EL34 used in SEUL mode to work

in an AM radio to give about 3 times the 2.5Watts from a 6V6 coupled to

an 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

1.26 x Np squared x S x T

= 1,000,000,000 x 24.2 x 189 / ( 1.26 x 3,200 x 3,200 x 24 x 25 )

= 590.

Bdc is then calculated :-

Bdc = 12.6 x µe x Np x Idc

mL x 10,000

= 12.6 x 590 x 3,200 x 0.051 / ( 189 x 10,000 )

= 0.64Tesla.

Bac is then calculated at the onset of saturation Fsat :-

Fsat = 35Hz, Va = 220Vrms at 0dB at 7.6W into 6k4.

Bac = 22.6 x V x 10,000

S x T x Np x F

= 22.6 x 220 x 10,000 / ( 25 x 24 x 3,200 x 35 )

= 0.74Tesla.

Total Bac + Bdc = 0.64 + 0.74 = 1.38Tesla.

Is the air gap correct?

Calculate air gap using estimate of maximum permeability for the example,

perhaps 3,000 :-

Air gap = mL x ( µ - µe )

µ x 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 changed?

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 could

be applied at 35Hz without saturation, but Vo could not reach 0dB level

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 gapping.

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 occur

at a higher F.

Fsat = 22.6 x V x 10,000

S x T x Np x Bac

= 22.6 x 220 x 10,000 / ( 25 x 24 x 3,200 x 0.44 )

= 59Hz.

This is too high and therefore air gap should not be reduced to increase µe.

All things considered, the air gap for SEUL operation is about correct.

Triode operation instead of SEUL could possibly be used instead of SEUL and PO

max could be about 4.4Watts with EL34 Pda = 17.8W. Load line analysis

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.

----------------------------------------------------------------------------------------------------------

Some may find this an easier method for SE OPT......

1E. Calculate RLa, maximum PO.

For SE tube or paralleled tubes the Center Value RLa is
first
calculated from

Eadc and Iadc. The Center Value RLa is the load where maximum possible

SE power is possible for the given Ea and Ia. Other loads need not be considered.

Pentodes, Beam Tetrodes, UL, CFB RLa = 0.9 x Ea/Ia,

Max PO = 0.45 x Ea x Ia,

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.

Eadc and Iadc. The Center Value RLa is the load where maximum possible

SE power is possible for the given Ea and Ia. Other loads need not be considered.

Pentodes, Beam Tetrodes, UL, CFB RLa = 0.9 x Ea/Ia,

Max PO = 0.45 x Ea x Ia,

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,

Calculate Afe = 450 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 =
50.4H.

4E. Calculating primary turns, Np.

Np = Lp x Idc x 1,400,000

Afe

eg, For 1 x EL34 SEUL example,

Np = 51.2 x 0.05 x 1,400,000 / 1,262 = 2,839 turns.

NOTE, the above simple calculation was derived as follows :-

Lp = 1.26 x Np x Np x Afe x µe .........equation 1

1,000,000,000 x ML

4E. Calculating primary turns, Np.

Np = Lp x Idc x 1,400,000

Afe

eg, For 1 x EL34 SEUL example,

Np = 51.2 x 0.05 x 1,400,000 / 1,262 = 2,839 turns.

NOTE, the above simple calculation was 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
......................equation 2

12.6 x Np x Idc

12.6 x Np x Idc

Bdc will remain constant.

Bac + Bdc at Fsat at 20Hz, should not be more than 1.4Tesla to allow for

lowest grade iron which may be available from surplus re-cycled cores.

Higher grade GOSS cores will have a magnetic headroom for 1.6Tesla,

but the 1.4 Tesla design limit allows for slight variations.

Bac + Bdc at Fsat at 20Hz, should not be more than 1.4Tesla to allow for

lowest grade iron which may be available from surplus re-cycled cores.

Higher grade GOSS cores will have a magnetic headroom for 1.6Tesla,

but the 1.4 Tesla design limit allows for slight variations.

Therefore Bdc for all designs should be 0.7Tesla.

Therefore,

µe = 0.7 x 10,000 x ML = 555 x ML ......................equation 3

12.6 x Np x Idc Np x Idc

Therefore,

µe = 0.7 x 10,000 x ML = 555 x ML ......................equation 3

12.6 x Np x Idc Np x Idc

Therefore,
this µe from
eqtn 2 can be substituted into
eqtn 1 to give

Lp = 1.26
x
Np
x
Np x
Afe x 555 x ML .........equation 1

1,000,000,000 x ML x Np x Idc

Lp = Np x Afe

1,430,000 x Idc

Then Np = Lp x
Idc x 1,400,000 1,000,000,000 x ML x Np x Idc

Lp = Np x Afe

1,430,000 x Idc

Afe

The constant of 1,430,000 may be rounded to 1,400,000.

5E. Calculate minimum size of primary wire.

Calculate expected maximum Idc in any part of primary winding.

Idc = anode Ia dc plus screen Ia dc = Ia + Ig2 = approximately 1.25 x Ia.

Minimum Cu wire dia = 0.8 x sq.rt Idc at idle.

DC current density must not exceed max = 2A per sq.mm

eg, For 1 x EL34, Ia = 50mA, Iga = 12mA, allow Idc max = 62mA.

Cu dia minimum = 0.8 x sq.rt 0.062Amps = 0.199mm.

Therefore minimum size primary wire = 0.20mm Cu dia minimum.

From wire table, oa wire dia = 0.245mm.

6E. Calculate minimum winding window size area

and L x H dimensions.

H = 1.092 x sq rt Np x oa dia.

eg, For 1 x EL34, H = 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 dimensions with aspect ratio 3:1, or L = 3H,

so area of window = 3H x H = 3 x Hsquared.

The primary wire occupies window area = 0.28 x L x H = 0.28 x 3Hsquared,

= 0.84 x H squared.

The primary wire occupies window area = Np x oa dia squared.

So 0.84 x H squared = Np x oa dia squared.

H = square root of ( Np x oa dia squared / 0.84 )

= sq.rt Np x oa dia / 0.916

7E. Calculate Core Tongue and Stack, T and S.

The minimum window H dimension has been calculated above.

Consider wasteless pattern E&I laminations.

T = 2H.

eg, For 1 x EL34, from Step 6E, T = 2 x 14.25 = 28.5mm.

Select from the range of standard 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 could be 28.57mm ( 1.125" ).

Stack S = Afe / T = 1,262 / 28.57 = 44.17mm ( 1.74" ).

Consider non wasteless pattern E&I or C-cores.

H and L and Afe all must not be smaller than calculated so far.

Therefore choose the C-cores or non standard E&I lams based on the

window size.

For E&I, measure the T dimension and Stack = Afe / measured T.

Some samples of non standard wasteless E&I lams have T size less

than 2H. For example, if H = 14.4mm, T might be 22mm.

So if Afe = 1,262sq.mm, S = 1,262 / 22 = 57.4mm.

For C-cores, the S dimension is the "strip width" of the wound core

material. Measure the build up thickness of one C-core,

and Strip width = Afe / ( 2 x build-up ).

The build-up of a C-core is the height of the wound core material

which has been wound with many layers and glued together.

Double C-cores are usually used hence T = 2 x build-up.

For example, go to the Eilor website which lists C-core sizes,

http://www.eilor.co.il/_Uploads/102ccore.rtf

Consider the window sizes available and select the cores with nearest

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 C-core

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) = 63.1mm.

Therefore one might use 4 x T32 cores, ie, two pairs of 00 double C-cores

stacked on top of each other to give an Afe section = 20mm x 64mm.

The benefit of the larger window allows the wire size to be considerably

increased to reduce all winding losses. The Eilor C-cores are very nice

things to use, and are beautifully made, but the price is many times ordinary

wasteless pattern E&I lams. A very fine 8 Watt SE OPT may be made

using some old re-cycled core material which may not have cost anything,

except the labour involved to extract the material from old transformers

which have burnt out windings.

The secondary pattern and accurate fitting of wire, insulation must

be worked out using the complex longer methods above.

The constant of 1,430,000 may be rounded to 1,400,000.

5E. Calculate minimum size of primary wire.

Calculate expected maximum Idc in any part of primary winding.

Idc = anode Ia dc plus screen Ia dc = Ia + Ig2 = approximately 1.25 x Ia.

Minimum Cu wire dia = 0.8 x sq.rt Idc at idle.

DC current density must not exceed max = 2A per sq.mm

eg, For 1 x EL34, Ia = 50mA, Iga = 12mA, allow Idc max = 62mA.

Cu dia minimum = 0.8 x sq.rt 0.062Amps = 0.199mm.

Therefore minimum size primary wire = 0.20mm Cu dia minimum.

From wire table, oa wire dia = 0.245mm.

6E. Calculate minimum winding window size area

and L x H dimensions.

H = 1.092 x sq rt Np x oa dia.

eg, For 1 x EL34, H = 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 dimensions with aspect ratio 3:1, or L = 3H,

so area of window = 3H x H = 3 x Hsquared.

The primary wire occupies window area = 0.28 x L x H = 0.28 x 3Hsquared,

= 0.84 x H squared.

The primary wire occupies window area = Np x oa dia squared.

So 0.84 x H squared = Np x oa dia squared.

H = square root of ( Np x oa dia squared / 0.84 )

= sq.rt Np x oa dia / 0.916

7E. Calculate Core Tongue and Stack, T and S.

The minimum window H dimension has been calculated above.

Consider wasteless pattern E&I laminations.

T = 2H.

eg, For 1 x EL34, from Step 6E, T = 2 x 14.25 = 28.5mm.

Select from the range of standard 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 could be 28.57mm ( 1.125" ).

Stack S = Afe / T = 1,262 / 28.57 = 44.17mm ( 1.74" ).

Consider non wasteless pattern E&I or C-cores.

H and L and Afe all must not be smaller than calculated so far.

Therefore choose the C-cores or non standard E&I lams based on the

window size.

For E&I, measure the T dimension and Stack = Afe / measured T.

Some samples of non standard wasteless E&I lams have T size less

than 2H. For example, if H = 14.4mm, T might be 22mm.

So if Afe = 1,262sq.mm, S = 1,262 / 22 = 57.4mm.

For C-cores, the S dimension is the "strip width" of the wound core

material. Measure the build up thickness of one C-core,

and Strip width = Afe / ( 2 x build-up ).

The build-up of a C-core is the height of the wound core material

which has been wound with many layers and glued together.

Double C-cores are usually used hence T = 2 x build-up.

For example, go to the Eilor website which lists C-core sizes,

http://www.eilor.co.il/_Uploads/102ccore.rtf

Consider the window sizes available and select the cores with nearest

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 C-core

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) = 63.1mm.

Therefore one might use 4 x T32 cores, ie, two pairs of 00 double C-cores

stacked on top of each other to give an Afe section = 20mm x 64mm.

The benefit of the larger window allows the wire size to be considerably

increased to reduce all winding losses. The Eilor C-cores are very nice

things to use, and are beautifully made, but the price is many times ordinary

wasteless pattern E&I lams. A very fine 8 Watt SE OPT may be made

using some old re-cycled core material which may not have cost anything,

except the labour involved to extract the material from old transformers

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 Idc )

eg, For 1 x EL34, the core is wasteless, 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 T,

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 adjustment.

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 lower µe the Np must be

increased, or ML made longer, ie, use a larger core size, say T = 32mm

with ML = 180mm. For a given Afe, the µe must not be reduced by

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 without

an actual gap, µe may not rise above 750 unless the material is GOSS.

Partial air gapping might be used but it involves more patience and trial

and error assembly and testing.

But with C-cores the µe may easily be varied from anywhere between

max µ of say 10,000 to 100.

9E. Check Lp and Fsat calculations...

µe = 3,111 x T / ( Np x Idc )

eg, For 1 x EL34, the core is wasteless, 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 T,

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 adjustment.

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 lower µe the Np must be

increased, or ML made longer, ie, use a larger core size, say T = 32mm

with ML = 180mm. For a given Afe, the µe must not be reduced by

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 without

an actual gap, µe may not rise above 750 unless the material is GOSS.

Partial air gapping might be used but it involves more patience and trial

and error assembly and testing.

But with C-cores the µe may easily be varied from anywhere between

max µ of say 10,000 to 100.

9E. Check Lp and Fsat calculations...

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, µe = 495,

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 increased

easily because Lp is proportional to S.

eg, For 1 x EL34, For wasteless E&I, the standard plastic bobbin size available

might be for 51mm stack or perhaps 57mm so S = 50mm gives Lp = 44.8H,

and S = 57 gives Lp = 51H which will be enough.

Check Fsat.

At Fsat, Bdc = Bac = 0.7Tesla,

1,000,000,000 x ML

eg, For 1 x EL34, Np = 2,839t, Afe = 28mm x 44mm, µe = 495,

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 increased

easily because Lp is proportional to S.

eg, For 1 x EL34, For wasteless E&I, the standard plastic bobbin size available

might be for 51mm stack or perhaps 57mm so S = 50mm gives Lp = 44.8H,

and S = 57 gives Lp = 51H which will be enough.

Check Fsat.

At Fsat, Bdc = Bac = 0.7Tesla,

Fsat = 22.6
x Va x 10,000

Afe x Np x Bac

Afe x 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 OK.

Notice Fsat is inversely proportional to Afe, or S, so using higher S

means lower Fsat.

10E. Calculate Air Gap and gap material thickness.

Ag = ML x ( µ -µe )

µ x µe

eg, for T = 28, ML = 5.6 x T = 157mm, µe = 601 from Step 6E.

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 = 0.1044mm,

say 0.1mm.

----------------------------------------------------------------------------------------------------------Fsat = 22.6 x 223 x 10,000 / ( 28 x 44 x 2,839 x 0.7 ) = 20.6Hz OK.

Notice Fsat is inversely proportional to Afe, or S, so using higher S

means lower Fsat.

10E. Calculate Air Gap and gap material thickness.

Ag = ML x ( µ -µe )

µ x µe

eg, for T = 28, ML = 5.6 x T = 157mm, µe = 601 from Step 6E.

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 = 0.1044mm,

say 0.1mm.

HIGH VOLTAGE TESTING.

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 arcing with all the secondaries and core well grounded.

Fig 23.

Fig 23 shows a
schematic for applying approximately +3,920Vdc
to the

primary of a transformer under test via 9 x 1M 2W metal film
resistors.

The rectifier is built with a "ladder network" to step up the
200Vac from

an old surplus transformer secondary using 0.1uF caps rated for
630V

and 1N4007 diodes.

If there is an old radio set transformer with secondary of
380V-0-380V,

the available Vac max is 760Vac, or 1,070V peak available.
Therefore

the ladder network would need only 4 step ups to generate
approximately

4,000Vdc, but the 0.1uF caps used should be 2,000V rated and the
each

diode "rung" in the ladder would need to consist of say 4 x
1N4007 in
series

each with 1Meg across each to ensure the reverse bias across
each diode

is limited to less than 300Vpk, and about equal for each
diode.

The +HVdc
may be varied by using a variac to control the
primary of the

supply transformer.

The core
and secondary is connected to a 10k0 x 5Watt
resistance which

is taken 0V of the power supply which MUST be also grounded
directly

to the
green and yellow Earth wire from the wall power outlet. A Vdc

voltmeter is
connected across the 10k0 resistance.

When power
is turned on and voltage is raised slowly to
maximum, the

volt meter should not show any voltage across the 10k
resistance. But
if

an arc
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.

The meter
used for the measurement may be a normal cheap analog
type.

Arcs may pulse, and digital meters may not give read out numbers
which

can
be understood.

-----------------------------------------------------------------------------------------------------

detailed in 2006 edition of the website.

OPT No3 is capable of around 25 Watts of SE power output.

Fig 24.

Readers may
wish to analyze what tubes and operating conditions
may

be suitable for OPT3.

Back
to SE OPT calc Page 2.

Back
to SE OPT calc Page 1.