SE OUTPUT TRANSFORMER
CALCULATIONS.
FOR
SINGLE ENDED AMPLIFIERS WITH NET DC FLOW IN ONE DIRECTION.
For calculations
for Push Pull Output Transformers go to 'PP OPT
calculations'
This page contains :-
Brief reference to
Radiotron Designer's Handbook, 4th Ed, 1955,
Comments about SE
amplifier OPTs and Ongaku,
Fig
1. Schematic of OPT No3 for 25 watt SEUL amp.
Design
logic method, steps 1 to 51 for designing an SE OPT using OPT No3 as the
example.
No apologies for the complexities involved, go fishing
if its all too hard!
Design method contains lots of calculations and list of
Primary and Secondary interleaving configurations likely to be used with
tube audio PP OPT.
Fig 2.
OPT Secondary sub-sections for load matches with 2 and 3 secondary layers.
Fig 3. OPT Secondary sub-sections for load matches with 4
secondary layers.
Fig 4. OPT Secondary sub-sections for load matches with 5 secondary layers.
Fig
5. OPT Secondary sub-sections for load matches with 6 secondary layers.
Fig 6. OPT bobbin winding details for OPT
No3 used for an SE amp.
Calculations for shunt capacitance at the
input of the SE OPT.
More checks of final design, calculation of leakage
inductance, acB, dcB, air gap, primary inductance
and final winding
height.
PRACTICAL TESTING OF SE AMPLIFIERS AND OPT.
Adjusting the air gap and practical checking of
the gap and primary inductance.
Metric
winding wire size chart for grade 2 polyester-imide wire.
----------------------------------------------------------------------------------------------
Where does one go to start with SE output transformer
design? Well, there is a lot of good design advice in the Radiotron
Designer's Handbook, 4th Edition, 1955, but its mostly about PP OPT because
hardly anyone in 1955 thought that
real hi-fi could be had from a single
ended design.
Millions of SE transformers were designed and built and
installed into radios and radio-grams of the 1950s era which employed one 6V6 or
EL84 as a 4 watt class A pentode amplifier with rather high thd/imd but not so
high to be problem to those listening to the cricket on a sunday
afternoon.
Very occassionally some brain addled commercial maker might
use an 807 which had been bought in a large job lot as military disposals after
WW2 when millions of 807 were over produced, delighting many DIYers until about
1965 when
WW2 stocks finally ran out.
The trend in 1955 was that
wherever more than 5 watts was required, a PP circuit was employed for
99%
of occasions, and 10 watts AB1 became possible from a pair of 6BM8, and the amp
was cheaper to make than
a single 807 or a 6L6.
PP was considered deluxe,
SE considered primitive and old fashioned. Millions of TV sets had a lone
6BM8;
deluxe TV sets had a PP amp and with fullrange speakers for the one
channel.
But a few hi-fi fanatics were not concerned about the costs, and
found their ears told them them an 807 or 6L6 strapped as a triode could make
a far better sounding 6 watts than
the first 6 watts from a pair of little tubes in PP and in some ways better than
any PP amp.
In Japan after WW2 the japanese remained short of everything for
about 25 years. Their houses were small, and made with paper screens and wood,
so big loud high power PP amps were not required for most of their
needs.
Most Japanese got by with a single 6V6 or EL84, and one for each
channel when stereo came in.
But eventually PP amps became the norm in Japan
as the people became less frugal and less poor, and indeed became wealthy due to
their skills with making cameras and motor vehicles for export.
But a
determined small band of diehards persisted with the idea that SE was all that
was ever needed and discovered that a lone transmitting tube such as the 211 was
able to handle music with such outstanding preservation of the soul in music
that
there was just zero need for anything else. The lone 2A3 or 300B were also
little kings.
The Japanese also cleverly built sensitve speakers to suit the
amps so their amplifiers were always working
in the low distortion region
and required no negative FB.
The pinacle of SE amplifier performance could be
in the creations by Audio Note in Japan, and one of the most expensive
amplifiers is the 25 watt Ongaku with a single 211 output tube.
It costs a
huge sum of money. The designer Kondo San died in Jan 2006 after 30 years of
building amplifiers.
I have read some of his quoted texts and although I
don't understand all he says about why he gets such great sound from his amps I
see that he does place great significance on the use of silver winding wire in
his SE OPTs. His quoted
speaches seem to have a range of inconsistencies and
vaguities, but are food for thought. I see Kondo as the shy master craftsman who
has great natural intuitive talent with iron tubes and wire but who could only
talk about what he makes
with gracious difficulty.
Of course almost
nobody else uses silver wire in output transformers, so its easy to say silver
makes the difference
and it would be more difficult to proove silver is
better in AB tests than it is to run a crafty sales campaign extolling the
virtues
of silver and get sales. He also uses GOSS iron plus nickel content
of about 50-50% in his OPT which also is supposed to
definately but subtly
affect the sound heard.
Anyway, feel free to use silver wire and
nickel cores, or amorphous cores if you like, but to me the GOSS
is just
fine for all types of OPTs. Kondo San does not have any information about the
turn count, core sizes,
winding geometry, manner of impedance matches, all of
which would affect the sound, along with using
whatever tube topology
preceeding the output stage and not to mention the selection of type types and
whether they are
NOS or not. The sound one hears from any sound system is
the sum of the effect of its parts and the room and the status of the recording,
and imho, anyone could build an amplifier quite equal in sonic performance to
anything made by the legendary Audio Note of Japan. They will need to get their
numbers right though, and here is where I hope to be of great asistance to
all,
which is more than I can say of Audio Note.
The method for
SE OPT design is based on the theory in the RDH4 or other sources I have
collected over the last 10 years. After having wound many very fine SE output
transformers with full power bandwidth as wide as 20Hz to 70 Khz, I feel well
qualified to speak from experience.
The list of logic steps involved in
producing the best possible OPT is based on designing for low winding
losses,
core saturation at full power at 14Hz, and adequate interleaving to
extend the HF response up to at least 70kHz
by keeping both the leakage
inductance and shunt capacitances to low quantities. The end result gives a well
filled winding window with several impedance matches possible without having
wasted sections of any secondary winding
so the winding losses and response
is the same for any of the chosen load matches.
The design method is very
similar to the Push Pull design method, so the simplest way to proceed is to
base
the SE method upon that used for PP OPT but with the necessary
modifications to allow for the effects of
DC flow in one direction only and
that always the SE OPT is connected to a pure class A amplifier. This will save
the reader from trying to link back and forth to either SE or PP design
pages.
Let us examine the output transformer No3 ( which has identical windings to
OPT No1, but has different connections
and a gapped core ) . OPT No3 is
capable of around 25 watts of SE power output.
First I will display the
schematic of OPT No3 :-
Fig1
The above schematic of an OPT is typical of what I may use in an SE
circuit.
The following design logic flow could be used to construct a computer program
where one would enter the design requirements such as power, secondary load,
tube Ra, core dimensions, then with a click on a "design" button, out
would come a terrific design including all the exact wire sizes and a cross
sectional drawing of the bobbin windup details that could be understood easily
by anyone with some winding experience.
Alas, I am not a computer expert, but
i can give you the flow of logic used to get a design finalised.
I invite anyone interested to prepare a PC
program to encompass all the horrible fiddly
details of fitting the wire
that is available from
the supplies into the cores available for superb OPTs.
I will be
probably have to wait a long time before anyone converts my logic flow to a PC
program
since the brain tends to just "get things intuitively" and a PC never
will.
Output Transformer No3.
Design example for
25 watt SE OPT for 3,100 to 5 ohms ( approximate secondary load ).
1. Choose the tubes, operating conditions
and primary load for the tubes applied across the
full primary, known as the
anode load, PRL, ohms.
Careful loadline
analysis is required for accurately determining loading and power output
calculation
and this is not in this list of
steps. See my pages on load matching.
OPT3, Tubes will be 2 x 6550, Ea = 500V, Ia = 70mA each,
PRL =
3.1kohms............................................................................................3,100ohms
2. Choose the
secondary nominal speaker load value.
allow a default value, SRL,
ohms.
OPT3, SRL =5
ohms.....................................................................................................5ohms
3. Choose the maximum power at
clipping for the
PRL chosen above, PO, watts.
OPT3, PO = 24.5 watts,
...................................................................................................24.5W
4. Calculate the minimum required core cross
sectional area, Afe,
for a nearly square core centre leg cross
section.
Afe = 450 x sq.rtPO in
sq.mm
**Note. This
formula has been derived from a basic formula for
core size used for mains
transformers, Afe = sq.root power input / 4.4
where the Afe is in sq inches.
This ancient formula is based on signal ac B max being about
1 Tesla at 50Hz
but we would want B max = approx 0.33 Tesla for
an SE OPT.
After considerable trials the above formula is a good
guide for SE audio OPT.
OPT3, Afe = 450 x sq.rt 24.5 60 = 450 x 4.94 = 2,223
sq.mm..............................2,223sq.mm
5. Calculate the core tongue
dimension, T.
For a square core section, tongue dimension = stack height,
ie T = S.
T x S = Afe.
Therefore
theoretical T dimension = sq.rt AFe
= th T ......................................th T, mm
OPT3,
thT = sq.rt 2,223 = 47.15
mm...................................................................47.2mm
Choose
suitable standard T size from list of available wasteless E&I lamination
core materials.
T sizes commonly available for OPTs :-
20mm, 25mm,
32mm, 38mm, 44mm, 51mm, 63.5mm...........................................enter T.
mm
**Note. Choosing a
standard T size above thT gives lower copper winding losses, higher
weight,
and choosing T below thT gives higher losses and lower weight. Afe
will be the same for either 44mm
or 51mm chosen from above so the LF response
won't change with tongue size. HF peformance
depends entirely upon the
interleaving geometry and insulations.
OPT3, Choose core T =
44mm............................................................................44
6. Calculate theoretical stack
height, thS using the chosen T size.
thS = Afe / T, then adjust to a larger
height to suit nearest standard plastic bobbin size if available,
mm.
OPT3, S = 2,223 / 44 = 50.5mm, choose
...........................................................51mm
7.
Adjusted Afe = chosen T x chosen
S..............................................................Afe,
sq.mm
OPT3, Adjusted Afe = 44 x 51 = 2,244
sq.mm...................................................2,244sq.mm
**Note. Some constructors will be using non
wasteless E&I lams,
or C cores which do not have the same relative
dimensions as E&I Wasteless Pattern cores.
The actual sizes of the T, S,
H, & L of the core to be used must be confirmed.
8. Confirm the height of the winding
window, H, mm.
OPT3, 44T wasteless material has H =
22.............................................................22mm
9. Confirm the length of the
winding widow, L, mm.
OPT3, 44T wasteless material has L =
66..............................................................66mm
10. Calculate the theoretical
primary winding turns, thNp
Np = sq.rt( PRL x PO) x 20,000 / Afe, turns.
**Note. The formula here is derived from
more complex and complete formula taking B and F
into account for ac
operation. If we assume ac magnetic field strength B = 0.8 Tesla, and F = 14 Hz,
which is a suitably
low F for where saturation is commencing ( because there
is already about 0.8 tesla of dc magnetization, )
and express V in terms of
load and power,
we get the above short easy equation for primary turns
required.
The full formula for calculating ac B is in step 40 below. The V
factor can be expressed as
sq.root of ( Primary RL x power output ) as in
the above simplified equation.
OPT3, RL = 3,100 ohms, PO = 24.5w, Afe =
2,244 from step 7 above,
thNp = sq.rt ( 3,100 x 24.5 ) x 20,000
/ 2,244 = 2,456 turns..................................2,456
turns
11.
Calculate theoretical Primary wire dia, thPdia.
**Note 4. The Primary wire used for the
transformer will occupy a portion
of the window area = 0.28 x L x H. The
constant of 0.28 works for 99% of OPT.
Each turn of wire will occupy an area
= oa dia squared.
Overall or oa, dia is the dia including enamel
insulation.
Therefore theoretical over all
dia of P wire including enamel insulation
= sq.rt ( 0.28 x L x H
/ Np
).......................................................................thPoadia,
mm
OPT3, thoadia P wire = sq.rt ( 0.28 x 66 x 22 / 2,456
)
= sq.rt
0.1655
= 0.4068
mm....................................................................0.4068mm
12. Find nearest suitable oa wire size
from the tables, Pdia, mm
OPT3, Try oa wire size = 0.414mm, ( for
Cudia = 0.355 mm. )............................0.414mm
13. Establish bobbin winding
traverse width................................................ Bww, mm
**Note 5. Bobbin traverse width is the distance
between the cheek flanges and varies depending
on who made the bobbin, but
each flange thickness = 2mm maximum is common, but can be slightly less.
Where bobbin flanges are not used,
and insulation is simply extended to the full window length L,
the traverse
width will be the same as in the case of where bobbin does have
flanges.
Ie, the winding will traverse
a distance = L - 4mm.
OPT3, Bww = 66 - 4 = 62
mm............................................................................................62mm
14. Calculate no of theoretical P turns
per layer, thPtpl, turns.
Ptpl = 0.97 x
Bww / oa dia from step 12.
**Note. The constant 0.97 factor allows for
imperfect layer filling.
Leave out fractions of a turn.
OPT3, Ptpl =
0.97 x 62 / 0.414 =
145.26...................................................................145
turns
15.
Calculate theoretical number of primary layers, thNpl,
then round down
or up to convenient even or odd number of
layers.
Theoretical Npl = thNp /
Ptpl, then round
up/down.............................................thNpl, no
OPT3,
thNpl = 2,456 / 145 = 16.93 layers; round down to
16................................16 layers
**Note. Rounding down may reduce the Npl needed
for Fs = 14 Hz.
But the actual turns used will still allow Fs = approximately
15 Hz, which is ok.
For those wanting to maintain Fs, or have Fs marginally
lower than 14 Hz,
the Afe can be increased by increasing S from say 51 mm to
62 mm, and still use a standard
sized
bobbin, and have Fs at 12Hz, which is marginal and not going to improve
the bass much.
16.
Calculate actual Np.
Np = P layers x
Ptpl, turns
OPT3, Np = 16 x 145 = 2,320
turns..........................................................2,320
turns
17. Calculate
average turn length, TL, mm.
TL = ( 3.14 x H ) + ( 2 x S ) + ( 2 x T
), mm.
OPT3, TL = ( 3.14 x 22 ) + ( 2 x 51 ) + ( 2 x 44 ) =
259 mm............................259mm
18. Calculate primary winding resistance,
Rwp.
Rwp = ( Np x TL ) / ( 44,000 x Pdia
x Pdia )
where 44,000 is a constant, and P dia is the copper dia from
the wire tables .......Rwp, ohms
OPT3, Rwp = 2,320 x 259 / ( 44,000 x
0.355 x 0.355 ) = 108.36 ohms................109ohms
19. Calculate primary winding loss
%,
P loss % = 100 x Rwp / ( PRL + Rwp
).. .....................................................P loss,
%
OPT3, P loss = 100 x 112 / ( 3,100 + 109 ) =
3.4%..............................................3.4%
20. Is the winding loss
larger than 4%? ..................................................yes or
no.
If yes, the design calcs must be checked again, and a larger core/window
area selected.
If no, proceed to 21.
OPT3, P winding loss is less than 4%
**Note. If the P winding losses
are less than 2%, there is a possibility that the wire size could be
reduced
to increase the turns per layer, and possibly reduce the number of P
layers by say 2, but the
stack height would have to be increased to keep, Fs
low.
21. Choose the interleaving pattern
from the list below for the wattage of the transformer.
All OPT will have the secondary sections containing
only one layer of wire.
While this may be subdivided into further secondary
sub sections, there are no designs here which require
bifilar or trifilar
winding or rectangular wire.
A section of a winding is defined as a layer or group
of layers devoted solely to P or S.
The term "section" is not to be
confused with "layer". For tube OPT, most P sections will have
more than one
layer of wire.
In general, all OPT should comply with the
following P&S layer number relationships :-
Where the first and last winding on is a
primary section, then these sections should have near 1/2 the layers of the
inner sections, hence if there are 3 outer p layers in a P section, the inner
sections might be either 5, 6 or 7 p layers. When this
guide is adhered to
there is the best HF response because the leakage inductance is fairly evenly
and symetrically distributed.
When starting and finishing with an S section
all internal P sections should have the same number of p layers
but it is not
always possible and having say 2 sections of 4 p layers and 2 sections of 5 p
layers is OK.
The size of such "internal sections" should not vary more than
25%. Both the forgoing conditions
avoid unpredictable resonances in the
upper HF response.
For transformers to suit low drive devices
such as mosfets or transistors, even with an SE amp, the same amount of
interleaving is required for a a given power level. The number of p layers will
be reduced as Primary RL becomes low, and wire dia will increase.
An 8 ohm :
8 ohm OPT with very low dc voltage differences between P and S would have equal
numbers of turns for P and S and perhaps be simply interleaved so each layer of
thick wire is alternatively devoted to either P or S.
The bandwidth can then
be very easily made to go up to 500kHz. As the primary or secondary load is
reduced, the effect of shunt capacitance diminishes, so insulation thickness can
be reduced. Transformers for electrostatic speakers which step up the
amplifier voltage between 50 and 100 times need to have good insulation for
voltages involved and to lower capacitance,
and they resemble OPTs powered
"backwards" and can be designed with the method here.
But for matching tubes to normal 3 to 9 ohm
speaker loads, the interleaving list below with the number of primary layers per
section possible will give at least 70 kHz of bandwidth, and where there is a
highest number of interleavings the bandwidth
can be 300kHz. Using more
interleaving than listed leads to less available room on the bobbin for wire due
to too many layers of insulation, and poor HF due to high shunt capacitances,
and higher winding losses.
For lower Primary RL and higher amplifier power
the larger the OPT becomes and for a given number of interleavings the
HF
response becomes less due to increasing leakage inductance so the larger the OPT
becomes, the number of interleaved sections increases. So a small 15 watt OPT
may only need
3S + 2P sections for 70kHz, but a 500 watt OPT may need 6S +
6P sections.
LIST OF
PRIMARY AND SECONDARY WINDING SEQUENCE ON THE BOBBIN
:-
Up to
15W, 10 to 20 p layers.....S - 5p to 10p - S - 5p to 10p -
S
3S + 2P
sections
2p to 4p
- S - 4p to 8p - S - 4p to 8p - S - 2p to
4p 3S + 4P
15W to 35W,
12 p layers .........2p - S - 4p - S - 4p - S -
2p
3S + 4P
S - 4p - S - 4p - S - 4p -
S
4S + 3P
16 p
layers..........3p - S - 5p - S - 5p - S -
3p
3S + 4P
S - 4p - S - 4p - S - 4p - S - 4p -
S
5S +
4P
2p - S - 4p - S - 4p - S - 4p - S -
2p
4S +
5P
18 p
layers..........3p - S - 6p - S - 6p - S -
3p
3S + 4P
20 p layers.........3p - S - 7p - S - 7p - S -
3p
4S + 3P
S - 5p - S - 5p - S -
5p - S - 5p - S
5S + 4P
2p - S - 5p - S - 6p - S - 5p - S -
2p
4S +
5P
35W to 120W, 14
p layers...........S - 3p - S
- 4p - S - 4p - S - 3p - S
5S +
4P
2p
- S - 3p - S - 4p - S - 3p - S -
2p
4S + 5P
16 p layers............S - 4p - S - 4p - S - 4p - S - 4p -
S
5S +
4P
2p
- S - 4p - S - 4p - S - 4p - S -
2p
4S +
5P
18 p layers............S - 4p - S - 5p - S -
5p - S - 4p -
S
5S +
4P
2p
- S - 5p - S - 4p - S - 5p - S -
2p
4S +
5P
20 p layers............S - 5p - S - 5p - S -
5p - S - 5p -
S
5S +
4P
2p
- S - 5p - S - 6p - S - 5p - S -
2p
4S +
5P
22 p layers............S - 5p - S - 6p - S -
6p - S - 5p -
S
5S +
4P
2p
- S - 6p - S - 6p - S - 6p - S -
2p
4S +
5P
120W to
500W, 10 p layers.........2p - S - 2p - S - 2p - S - 2p - S -
2p
4S +
5P
S - 2p - S - 3p - S - 3p - S - 2p -
S
5S +
4P
1p
- S - 2p - S - 2p - S - 2p - S - 2p - S - 1p
5S + 6P
S - 2p - S - 2p - S - 2p - S - 2p - S - 2p - S
6S + 5P
12 p layers...........2p - S - 3p - S - 2p - S - 3p - S -
2p
4S +
5P
S - 3p - S - 3p - S - 3p - S - 3p -
S
5S +
4P
1p - S - 2p - S - 3p -
S - 3p - S - 2p - S - 1p
5S +
6P
S - 2p - S - 3p - S - 2p - S - 3p - S - 2p -
S
6S + 5P
S - 2p - S - 2p
- S - 4p - S - 2p - S - 2p - S
6S + 5P
14 p layers..........2p - S - 3p - S - 4p -
S - 3p - S - 2p
4S + 5P
S - 3p - S - 4p - S - 4p - S - 3p - S
5S +
4P
1p - S - 3p - S - 3p - S - 3p - S - 3p - S - 1p
5S +
6P
S - 2p - S - 3p - S - 4p - S - 3p - S - 2p - S
6S + 5P
16 p layers............2p - S - 4p - S - 4p
- S - 4p - S -
2p
4S + 5P
S - 4p - S - 4p - S - 4p - S - 4p -
S
5S +
4P
2p - S - 3p - S - 3p -
S - 3p - S - 3p - S - 2p
5S +
6P
S - 3p - S - 3p - S - 4p - S - 3p - S - 3p - S
6S + 5P
18 p layers............2p - S - 5p - S - 4p -
S - 5p - S -
2p
4S + 5P
S - 5p - S - 4p - S - 4p - S - 5p -
S
5S +
4P
2p - S - 4p - S - 3p -
S - 3p - S - 4p - S - 2p
5S +
6P
S - 3p - S - 4p - S - 4p - S - 4p - S - 3p - S
6S + 5P
20 p layers............3p - S - 5p - S - 4p - S - 5p - S -
3p
4S + 5P
S - 5p - S - 5p - S - 5p - S - 5p -
S
5S +
4P
2p - S - 4p - S - 4p -
S - 4p - S - 4p - S - 2p
5S +
6P
S - 4p - S - 4p - S - 4p - S - 4p - S - 4p -
S
6S + 5P
22 p layers.............3p - S - 5p - S - 6p - S - 5p - S -
3p
4S + 5P
S - 5p - S - 6p - S - 6p - S - 5p -
S
5S +
4P
2p - S - 5p - S - 4p -
S - 4p - S - 5p - S - 2p
5S +
6P
S - 4p - S - 6p - S - 4p - S - 6p - S - 4p -
S
6S + 5P
Record the choice of primary layers in
step 15.
choose a suitable P& S interleaving pattern from the above
list.
OPT3, 16 primary layers are used, choose ........................................................................4S + 5P.
22. Choose
insulation, i, in mm used between primary layers, mm
**Note. Usually p to p insulation for all OPT needs to only
be 0.05mm thick.
OPT3, Choose i =
0.05mm.......................................................................................0.05mm
23. Choose insulation, I,
in mm used between Primary and Secondary layers, mm
**Note. Usually, for where Ea is
above 450V, and RL above 1k, the p to S
insulation is first reckoned = 0.6mm
to keep shunt capacitance low with good enough
insulation.
OPT3, I = 0.6
mm
.........................................................................................0.6mm
24. Calculate portion of bobbin
winding height comprising primary wire layers, p to p insulation,
and p to S
insulation.
height of P+I+i = ( no of layers
x oadia P wire ) + ( no x i ) + ( no x I ) ................P+i+I ht, mm
OPT3, P = 16 x 0.414 = 6.624 mm,
i ht = 11 x 0.05 = 0.55
mm,
I ht = 8 x 0.6 = 4.8
mm,
Total ht of above = 11.974
mm............................................................................11.98mm
25. Calculate maximum total available
height of all windings on bobbin.
Max wind ht = 0.8 x H, mm.
**Note. The constant of 0.8 will
suit most OPT.
OPT3, 0.8 x 22 = 17.6
mm...................................................................................17.6mm
26. Calculate the max theoretical
oa dia of the the secondary wire in secondary layers, thSoadia.
max thSoadia = ( max wind ht - [ P+i+I ht ] ) /
no of S layers, mm.
**Note. The available height for secondary layers = maximum bobbin winding
height - ( primary + all insulations ).
OPT3, We have chosen 4 layers of secondary wires; height from step 24 =
11.98mm
thSoadia = ( 17.6 - 11.98 ) / 4 = 5.62 / 4 = 1.405
mm....................................1.406mm
27. Find nearest oa dia wire size
less than thSoadia calculated in step 26.
OPT3, Try 1.351mm oa dia, which is 1.25 mm Cu
dia ...............................................1.351mm
28.
Calculate the theoretical S turns per layer, to nearest turn, thStpl.
Get the
Bobbin winding width from step 13, Bww.
Theoretical S turns per
layer, thStpl = Bww / thSoadia from step
27, no
OPT3, thStpl = 62 / 1.351 = 45.89, choose
45..............................................45 turns per layer
**Note.
The calculated turns per layer are for the thickest wire possible but could more
turns of a smaller dia
to obtain the wanted turn ratios to to give the wanted
load matches. For a full layer of wire across
the full bobin winding
width no less than 45 turns per S layer can be used
because the increase in
wire size will give a total height of the winding which exceeds the allowable
total winding height.
29. Calculate
the nearest full S turns needed for loads of 3.5 ohms, 5 ohms and 7
ohms.
Secondary turns = primary turns /
square root of impedance ratio.
OPT3, 2,320 P turns. PRL = 3.1k ,
for 3.5 ohms want 78 turns,
for 5.0
ohms want 93 turns,
for 7.0 ohms want 110 turns.
30. Choose a pattern of Secondary
winding sections from Fig 2, 3 or 4 below to give a suitable variety
of at
least two secondary load matches of between 3 and 9 ohms to suit most modern
speakers
while the Primary load is
considered to be 3,100 ohms.
OPT3 From step 15 we caculated 16 primary layers.
In the P&S winding
list in step 21 we selected the 4S + 5P winding pattern.
Inspect the range of
secondary arrangements where there are 4 secondary layers shown
within Fig2,
Fig3, Fig4, Fig5.
Reading the charts below could be
confusing!!!
Each rectangle represents a given separate OPT . The
figure of N and its multiples are shown to give the
relationship between
numbers of turns in each winding shown as a thick line.
Consider example 2A
in Fig2..
There are two layers, each divided into 2N and N turns, which means
there could be 50 and 25 turns respectively.
Where it says "3 @ 2N" means
there are 3 parallel windings of 2N turns each
In the case of 2A, it means
there are in fact 2 windings of 2N each and the third is made up of N+N in
series.
Ns, or the secondary turn number for the transformer = 2N turns,
since paralleling any number of same turn windings
does not alter Ns .
"2
@ 3N consists of two parallel windings each consisting of 2N + N turns in
series.
In the case of all transformers the first line of impedance loads are
listed for a given number
of N as "Z = 1.0 1.7
3.0 5.0" , and the figures are starting reference impedances for
each transformer.
The next line below for an increased number of N give the
relative values of Z for that number of N
and the vertical columns of Z
values give the relative impedance relationships for the various numbers of
N
in windings.
So reading 2A, if we have 1.0 ohms as the match possible
for 2N turns, then for 3N turns
the match is for 2.3 ohms, and for 6N turns
the match is to 9 ohms.
There could also be a match where 2N = 3.0 ohms, and
reading down the figures 3N gives 6.8 ohms, 6N gives 27 ohms.
In 2B, there is 4 @ 1N, Z = 1.0, etc, but
the figure of N for 2A and 2B have no relationship.
I hope I have made
it easy for everyone to get easy valuable winding information.
Fig 2.
Fig 3.
Fig 4.
Fig 5.
Here we have 17 possible arrangements for secondary windings for many
different OPTs.
Each rectangle represents the secondaries in a given OPT.
There is no need to include the primary layers because
the number of primary
layers could vary hugely without any change to the secondary layout and sub
section divisions.
The secondaries are always going to have turns appropriate
to relatively low loads in the majority of OPT.
For the OPT No3 example we have chosen to use 4S + 5P so we can choose the
secondary subsections
from any of the 3 examples 4A, 4B, or 4C shown
in the above Fig 3. We have already calculated in step 28
that we could possibly have 45 turns per
layer.
Compare all available arrangements of secondary sub-sections and
decide which arrangement offers the most useful
range of load matches, and 2 load matches between 3 and 9
ohms.
OPT3, Examine 4A from Fig3.
There is a total of 6
windings with the numerical relationships of 3 windings of N turns, and 4
windings of 3N turns.
So we can have the top S layer divided into 3 windings
of 15 turns each and 3 other layers of 45 turns each.
This allows the connection of windings to be
as follows :-
4 parallel windings of 45 turns as calculated in step
29...........................................1.17 ohms
3 parallel windings of 60 turns consisting of 45 + 15 turns
each................................2.1 ohms
2 parallel windings of 90 turns
each consisting of 2 x 45 turns in series...................4.48 ohms
4
series windings of 45 turns each = 180turns
.....................................................18.72 ohms
**Note. This arrangement would be
fine if we wanted a match to about 5 ohms only, and relied on the
amplifier
to cope with a load of anything between 3 and 30 ohms, which is quite likely to
be OK
if all we want is a couple of watts with the remaining wattage just for
transients in a loungeroom situation.
At low power the SE class A amp will cope with any load above
1k.
But where we knew the speaker Z
was between 2.5 ohms to 5 ohms, or between 5 ohms and 10 ohms
for
the main power range between 100Hz and 1 kHz, then it would best to be able to
set the amp to suit the
speaker impedance to reduce distortions and increase
the power ceiling to a maximum optimum.
**Note. The calculated number of turns
per layer of 45 just happens to be exactly divisible by 3.
If the layer was
not exactly divisible, say it was 41, 44, 46, or 47 turns, we would be forced to
adjust the turns
to the next highest number
of turns divisible by 3, and revise our load match
calculations.
Examine 4B.
There is a total of 8 windings, 4 @ 4N,
4 @ 1N.
So we could have 4 @ 36 turns and 4 @ 9
turns.
This allows the connection of windings to be as
follows :-
5 parallel windings of 36 turns
each.....................................................................0.746
ohms
4 parallel windings of 45 turns as calculated in step
29.........................................1.17 ohms
2 parallel windings of 90 turns each consisting of 2 x 45 turns in
series.................4.48 ohms
4 series windings of 45 turns each =
180turns .....................................................18.72
ohms
Examine 4C.
There is a total of 8
windings, 4 @ 2N, 4 @ 1N.
So we could have 4 @ 45 turns and 4 @ 15
turns.
This allows the connection of windings to be as
follows :-
6 parallel windings of 30 turns
each.....................................................................0.52
ohms
4 parallel windings of 45 turns as calculated in step
29..........................................1.17 ohms
3 parallel windings of 60 turns consisting of 45 + 15 turns
each...............................2.1 ohms
2 parallel windings of 90 turns
each consisting of 2 x 45 turns in series..................4.48 ohms
4
series windings of 45 turns each = 180turns
.....................................................18.72
ohms
Choose between available options.
4A, 4B and 4C do not give two load
options between 3 and 9 ohms and thus none are
ideal.
Note the best or closest to the wanted condition of
having two load matches between 3 and 9 ohms.
**Note.
When 4A is selected, the number of secondary turns per layer
could be increased and possibly keep within the
limits
for wanted low winding losses.
Examine alternative numbers of turns per
layer.
OPT3,
48
turns give 1.3, 2.3, 5.3 ohms, NOT OK, only one match between 3 and 9
ohms.
51 turns give 1.5, 2.6, 6.0 ohms, NOT OK, only one match between 3 and
9 ohms.
54 turns give 1.7, 3.0, 6.7 ohms, OK, two matches between 3 and 9
ohms.
57 turns give 1.9, 3.3, 7.4 ohms, OK, two matches between 3 and 9
ohms.
60 turns give 2.1, 3.7, 8.2 ohms, OK, two matches between 3 and 9
ohms.
Choose final option to suit nominal 4 ohms and 8 ohm
speakers which can often be 3 or 6 ohms
The 4A secondary pattern with 54 or 57 turns per layer
seems most suitable.
31. Confirm the turn selection in
step 30 with regard to actual available wire size.
Theoretical oa wire dia = bobbin winding width /
S turns per layer.
Select wire from tables, Cu wire dia, mm.
OPT3, oa wire dia = 62mm / 57 = 1.087 mm, so from wire
tables select 1.00mm wire.
oa dia of wire = 1.093, so 57 turns = 62.3mm.
This
will be a tight squeeze but by staggering the positions of entry and exit points
on the bobbin cheeks the wire will fit.
OPT3
.............................................................................1.0mm
32. Calculate chosen secondary
option winding resistance.
List the following :-
Primary turns chosen for
Ns...................................................................................no
S wire Cu dia for chosen tpl, from wire
tables........................................................dia, mm
Turn
length is from step
17.....................................................................................TL
,mm
No of parallel sections from option chosen to make up Ns
turns..............................no
Secondary load for Ns to give the
selected primary RL, ie :-
SRL = selected RL / TR
squared............................................................................ohms
S winding resistance
= Ns x TL / ( No
of parallel S sections x 44,000 x wire dia x wire dia ),
ohms.
OPT3 Consider 57 turns per layer, arranged for 3 parallel
windings of ( 57 + 19 ) = 76 turns each.
Ns
.......................................................................................................................76
Eg,
wire Cu dia
mm...............................................................................................1.00mm
TL
from step 17,
mm.............................................................................................259mm
No
parallel S sections,
no.......................................................................................3
SRL
= 3,100ohms / 932
........................................................................................3.3ohms
(
932 = turn ratio squared, ie, the impedance ratio of the OPT. )
Swr = 76 x
259 / ( 3 x 44,000 x 1.0 x 1.0 ) = 0.153
ohms......................................0.145ohms
33. Calculate S loss % for chosen option
for chosen Ns turns.
Record SRL from step 32, ohms,
OPT3, Ns = 76
turns for 3.3
ohms...........................................................................3.3
ohms
S loss % = 100 x Swr / ( SRL + Swr
), %
OPT3, S loss = 100 x 0.145 / ( 0.145 + 3.3 ) =
4.2%.............................................4.2%
**Note. If the option where 45
turns per secondary layer was chosen the wire size is 1.25mm dia
and the
secondary winding losses would be lower, but we cannot always get minimal
winding losses
and a good load match.
34. Calculate total winding
losses, chosen option,
Total winding loss
% = P loss + S loss, ttl loss, %
Record P loss from step19,
%
Record S loss from step 33, %
OPT3 P loss from step 19,
%.............................................................................3.4%
S
loss from step 33, %
.......................................................................................4.2%
Total
loss = 3.4 + 4.4 = 7.8
%............................................................................7.6%
35. Are total winding losses
acceptable?.............................................................yes/no
Check the dc current rating for primary.
Ia = 140mA, 0.355mm Cu dia wire is rated for 296mA at 3A/sq.mm, so
actual
DC is less than 1/2 of rated amount, OK.
OPT3, Let us accept that 7.6% is
OK...........................................................................yes.
**Note. The winding losses have been calculated
on the basis of the actual load
being 3.3 ohms. For 7.6% losses, with 24.5
watts produced at the anode, 1.9 watts is lost as heat in the
OPT windings.
But in fact there may be a 4 ohm nominal value load where the speaker Z averages
4 ohms,
but may have a dip to 3 ohms and peaked Z up to 30 ohms so the
winding losses
will vary with load. An average load of 4 ohms gives winding
losses of 6.4%, 5 ohms gives 5.1%
which would be acceptable.
If the total winding losses are unacceptable the initial core selection could
be based on using a core with the next size up for tongue width, but with the
same Afe.
A core with T = 51mm x S = 44 would be slightly heavier than the T
= 44T x S = 51mm material but the window
is 76mm x 25mm which would allow
much larger wire sizes but whole design must be re-worked.
In this case the
51mm T material would probably reduce losses to under 5%, and reduce losses by
about 0.6watts
which is a tiny reduction for the extra weight and size of the
larger laminations.
36. If yes to step 35, proceed to check
winding height will actually be practical.
37. Check the final winding height and
bobbin thickness to make sure
the completed wound bobbin will fit into the
core window.
OPT3,
Primary layers, 16 x 0.414mm = 6.624mm.
Insulation, p to p
layers, 4 x 3 x 0.05mm = 0.6mm.
Secondary layers, 4 x 1.08mm = 4.32
mm.
Insulation P to S layers, 8 x 0.6mm = 4.8mm.
Insulation over top of
last on primary, 0.6mm.
Bobbin bass thickness, 2mm.
Total winding height including bobbin bass = 18.94.
Remaining clearance = window height of 22mm - 18.94mm = 3.06mm = OK
**Note. The above wind up is for plain
UL. If CFB windings are used, there will need to be some increase in
some of
the primary to primary insulations because the CFB windings will be at 0V
potential.
so instead of at least 2 x 0.05mm p-p insulation layers there may
be extra 2 x 0.6mm.
**Note. There is some room for bulge
in the windings as they are wound on.
Wire will not lay tightly as layers are
put on and will tend to spring up across the rectangular core.
The bulge will
distribute itself from start to finish and wires will not be naturally tight in
the vertical direction
hence the importance of impregnation with varnish to
make windings adhere to each other.
38. Calculate leakage inductance,
where is is considered to be
an equivalent quantity of inductance in series
with the primary load
looking into one end of the primary, with the other end
grounded.
LL =
0.417 x Np
squared x TL x { ( 2 x n x c ) + a }
1,000,000,000 x n squared x b
Where LL = leakage inductance, in
Henrys,
0.417 is a constant for all equations to work,
Np = primary
turns,
TL = average turn length around bobbin,
2 is a constant, since
there is an area at each end of a layer where leakage occurs,
n = number of
dielectrics, ie, the junctions between layers of P and S windings,
c = the
dielectric gap, ie, the distance between the copper wire surfaces in P and S
windings,
a = height of the finished winding in the bobbin,
b = the
traverse width of the winding across the bobbin.
Distances are all in mm!
OPT3,
LL = 0.417 x 2,320 x 2,320 x 267 { ( 2 x 8 x 0.7 )
+ 17.7 }
1,000,000,000 x 8 x 8 x 62
=
0.00435 Henry = 4.35 mH.
39. Is the leakage inductance
low enough?
Calculate reactance of LL at 100 kHz.
ZLL at 100kHz = L in henrys x 2 x pye x F
= L x 6.28 x 100,000Hz
.....................................................ohms
OPT3, ZLL =
0.00435 x 6.28 x 100,000 = 2,731 ohms at 100 kHz.........2,731 ohms
Is
ZLL less than PRL at 100
kHz?...................................................yes/no
OPT3, We
have PRL = 3,100 ohms, ZLL at 100 kHz is
less....................................yes.
40. If
answer to step 39 is yes, leakage inductance is low enough.
40A. Calculation of
input or OPT total shunt capacitance.
The methode of calculation has been set out in 'PP Output
Transformer Calculations'.
See step 46 on that page.
Usually if the
guidlines for P to S insulation thicknesses have been adhered to, ie, P-S
insulation
is more than 0.5mm and the insulation material dielectric constant
is less than 2.0, the shunt
capacitance should not cause undue HF attenuation
for an ideal number of interleavings.
In effect, although it may be possible
to reduce LL to half the value calculated in step 38,
the number of required
interleavings is excessive because there would be more P-S interfaces and C
would
increase too much, quite negating the effect of reducing the LL.
41. Check that calculation
of primary turns gives less than 0.8 Tesla magnetic field strength, B,
with
signal at full power and at 14 Hz.
B =
22.6 x V x
10,000
S x T x Np x F
where B is in Tesla for ac signals,
22.6 and 10,000 are constants for all
transformer equations,
V = Vrms signal voltage across the primary,
S =
core stack height,
T = core tongue width,
Np = primary turns,
F =
frequency at which B is to be measured.
All dimensions in mm!!
OPT3, For 24.5 watts into 3.1k, Va = 275Vrms.
B at 14Hz = 22.6 x
275 x 10,000 = 0.85 Tesla =
OK
51 x 44 x 2,320 x 14
42. Calculate the wanted minimum primary
inductance, Lp, Henrys.
**Note.
The core of the SE OPT will have an air gap which will needs to greatly reduce
the maximum permeability, µ,
when the laminations are fully intermeshed. This
prevents the core from becoming magnetically saturated by the dc current
flow.
The primary inductance is proportional to the effective permeability,
µe, the permeability with a gapped core.
Wanted Lp will have reactance in ohms = nominal Primary RL at no higher than
20 Hz.
Minimum Lp =
PRL
6.28 x F
Where Lp = Henrys, 6.28 = a constant of 2pye for all
equations to work and F is the frequency.
OPT3, Min Lp = 3,100 / ( 6.28 x 20 ) = 24.7H
43. Calculate the
effective permeability, µe.
µe = 1,000,000,000 x Lp x mL
1.26 x Np
squared x S x T
Where µe is effective permeability of core with
an air gap, and is just a number, no units.
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 is length of winding window H is the
height of winding window anmd 3.14 is pye,
for all equations to work. ( for
eg, L is 76mm and H = 25mm for 51mm tongue width wasteless E&I )
S =
stack height,
T = tongue width.
All
dimensions in mm !!!
OPT3, µe =
1,000,000,000 x 24.7 x 240 =
389.
1.26 x 2,320 x 2,320 x 51 x 44
44. Calculate the air
gap required to get µe in step 42.
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 permeabilty
maximum,
µe = effective permeability with air gap.
all dimensions in mm !!!
OPT3, Air gap = 240 x ( 17,000 - 389 ) / ( 17,000 x 389 ) =
0.602mm.
**Note. The air gap is the total
gap. In an E&I core which is air gapped, there are TWO
gaps
inserted in the magnetic path around each rectangle of two which form the
core.
Therefore the
dimension of the plastic material or paper used to make the gap will be
HALF that calculated above.
**NOTE. See below for practical method to check
that the gap is correct and to confirm the
primary inductance inductance is
sufficient.
45. Calculate the dc field
strength created in the core by the dc flow, Tesla.
Bdc = 12.6 x µe x Np x I
mL x 10,000
Where Bdc is in Tesla, 12.6 and 10,000 are constants for
all equations to work,
µe = effective permeability,
Np = the primary
turns,
I = dc current in AMPS,
mL is the iron magnetic path length.
OPT3, Bdc = 12.6 x 389 x 2,320 x 0.14 / ( 240 x 10,000 ) = 0.66
Tesla.
46. Calculate maximum total
magnetic field strength at 14Hz.
Maximum B = acB + dcB.
OPT3,
From step 40, ac B max = 0.85Tesla.
From step 44,
dc B = 0.66 Tesla.
Total max B = 1.5Tesla.
47. Is the total B
max below the core material maximum at saturation?
OPT No3 core material is GOSS which will saturate at approx 1.6
Tesla
Total B max = 1.5 Tesla at 14Hz, therefore design is OK.
48. Have all parameters
been
satisfied?...............................................yes/no
49. If yes to step 48, Design is OK and
sourcing materials can be undertaken.
50. Are the wire sizes available?
.......................................................yes/no
51. If no to step 42, find
out what sizes are available, and design to
suit
these
sizes without compromises!!!
52. Draw up the bobbin winding
details for the proposed OPT No 1 ready for
the guy who is going
to wind the OPT.
Fig
6.
53. Shunt capacitance of an
OPT.
There are several areas in an audio transformer where capacitance exists, and
with an OPT we are primarily interested
in the total measured capacitances
when we measure the capacitance at the anode terminals of the OPT.
There is
capacitance between primary wires in the form of the "self capacitance" of the
primary layers of wire
and between layers of wire adjacent to secondary
sections which have much lower signal voltages
and are effectively at 0V
potential.
To calculate the primary shunt capacitance in an audio transformer such as
OPT No3, refer to the above bobbin winding
layout. Neglect the self
capacitance of the primary windings; it will be such a small amount compared to
the
main shunt C between adjacent P to S interfaces.
The distance between the copper surfaces of primary and secondary layers
including the insulation thickness
of 0.6mm and the wire enamel of about
0.05mm = approx 0.7mm.
Then you must allow for the curved surface of the wire
turns so total distance = approx 0.75mm.
Capacitance between two metal
plates = ( A x K ) / ( 113.1
x d )
where
Capacitance is in pF,
A is the area
in square millimetres of the plates assumed to be of equal size,
K is the
dielectric constant of the material between the plates, air being =
1.0,
113.1 is a constant for all equations to work,
d is the distance in
millimetres between the plates and is the same for the area of the
plates.
For example, if the turn length around the wound bobbin for the first primary
layer wond on at the bottom
of the above drawing = 210mm, and winding
traverse width
= 62mm, then area = 210 x 62 = 13,020 sq.mm.
Let us say
the K for the polyester = 2.
( The C can be measured if unknown using metal
plates of known area, and using a sample of polyester
clamped tight between
the plates for the whole area. Once the C measurement has been recorded,
the
plates are set up with a very small width strips of polyester leaving the plates
the same distance apart
but with mostly air between the plates, and the C
measured again.
K = C measured with full amount of polyester / C measured
with just air. )
The d we calculated above = 0.75mm.
C in pF = 13,020 x 2 / ( 113.1 x 0.75 ) = 307pF.
The amount of capacitance in each P to S interface varies with turn length so
that at the top of the wind up
where the turn length is about 250mm, the C
would be 365pF.
Therefore to minimise the effects of shunt capacitance in an
SE transformer the anode should be connected to the
end of the primary where
the turns are shortest.
To simplify the math, let us say the average capacitance between a P layer
and an S layer is say 330pF.
The first P-S interface up from the anode
connection is below the AB secondary.
It is at a position along the primary =
14.5 layers / 16 layers, and the C at this point is transformed
to appear as
( 14.5 x 14.5 ) / ( 16 x 16 ) x C because the transformation of an impedance is
at the square of the turn ratio.
So the C between the P winding and below the
AB sec = 0.82 x 330pF = 271pF.
The C between the top of secondary AB and the
P winding is at a position of 13.5 layers / 16 layers ,
or a at a turn
ratio of 0.844, so the impedance ratio = 0.71 so the C that appears at the anode
connection
= 330 x 0.71 = 234pF.
We can move up to the secondary CD
and by similar reasoning work out the C at the anode below CD = 170pF,
and
above CD = 142pF,
C at the anode due to C below EF = 73pF, and above EF = 54
pF.
C at the anode below G-L = 8pF, and above G-L = 3pF.
The
total C seen at the anode connection is the total of all the transformed
capacitances
= 271 + 234 + 170 + 142 + 73 + 54 + 8 + 3 = 955 pF.
(
Notice that in 'PP Output Transformer Calculations the shunt C at each anode was
calculated as 455pF
with a transformer No1 which has an identical winding
geometry. So the calculation for the single anode of
the SE OPT is just twice
the case of one anode's C where the wind up details are the same ).
Cathode Feedback use further complicates the capacitance calculation but the
effect of the capacitance
on amplifier bandwidth is effectively reduced by
the NFB because the NFB reduces the Ra of the
tubes.
If the above transformer No3 is used with a pair of KT88 in beam tetrode in
SE parallel mode then if
the Ra is simply about 18,000/2 ohms at each
anode and without a load the gain of the tube will reduce
-3dB from
approximately being equal to µ of the tube at say 500Hz at where the
capacitive reactance = Ra. Since C = approx 955pF, then this -3dB pole
is found easily at a frequency = 159,000 / ( Ra x C in uF) = 159,000 / (
9,000 x 0.000955 ) = 18.5kHz.
In practice this would be about correct, and one way to measure the C shunt
with a tube
is to use a high Ra tube unloaded such as a tetrode or pentode,
and work from the observed -3dB point.
The leakage inductance will have
little effect on the unloaded response.
The capacitance and leakage inductance will react together to form a tuned
circuit and
low pass filter with an ultimate slope of more than
6dB/octave.
So rapid phase shift increase occurs as F becomes high so it is
important to minimise C and LL
to force the frequency of resonance to be as
far as possible above the audio band and where
the phase shift with loop NFB
does not cause oscillations.
Trying to establish a much more accurate
equivalent model of the complex LCR offered by such a simple OPT as No3
is
beyond my abilities and there is little point to achieve such modelling. It is
simply easier to
establish low values of C and LL by empirical methods and
then critically damp the HF gain
of the amp to achieve low overshoot on
square waves with a 0.22 uF across the output without any R load,
while
maintaining a maximal HF pole with a solely R load.
A common mistake by
would-be experts who try to wind OPT without enough know-how is to
closely
couple P and S windings so the distance between the P and S is say only
0.2mm,
and this would increase the shunt C from 955pF to 3,581pF, or even
higher if they used the space
saved for more P turns and interleavings where
neither is wanted or a
benefit.
----------------------------------------------------------------------------------------------------------------------------
PRACTICAL TESTING OF SE AMPLIFIERS AND
OPT.
Once the winding of the OPT is complete, the UN-VARNISHED OR UN-WAXED
OPT is temporarily placed into the amplifier and tests should be done to confirm
that the gap size and primary inductance is correct.
If the bobbin has been
varnished with cold cure polyurethane two pack mix, the core is assmbeled into
the
winding but without the final surface applied varnish over the completed
item.
The core gap MUST be adjusted, and final varnish coating of the core
with air drying varnish
cannot be applied until the gap size has been
confirmed.
The amplifier should be checked to make sure all is well with circuit
function, and a dummy load resistance equal to the
rated load is connected
across the output. The enthusiast should have an oscilliscope connected to
monitor
the OPT secondary output voltage and the test signal must be from a
low impedance low distortion sine wave generator with a frequency range from 2Hz
to 2MHz. If global NFB is used, it should be dis-connected, and any phase
tweaking
networks disconnected so the response is the best possible without
global NFB.
OPT with local CFB will have to be tested with their output stage
NFB left connected.
A 10 ohm resistance should be connected from ground to
the cathode circuit which may include the cathode bias
resistance and
cathode bypass capacitors. This will enable monitoring of the dc anode current
or anode + screen currents,
as well as the signal current through the anode
load and primary inductance.
The air gap can be set at first by using the calculated gap using sheets of
paper placed on both sides of the OPT core
so that if the air gap was
calculated at 0.7mm, there will be 0.35mm on each side of the core.
The
actual gap need not actually be of air, but consist of sheets of paper or other
non-magnetic or non-metalic material.
One sheet of notebook paper may be
0.07mm thick. To determine the paper thickness, measure the thickness of the
notebook less its cardboad covers and divide the total thickness by the number
of sheets in the book. A 100 sheet book may measure 7mm thick.
Once the paper
thickness is known , cut paper sheets to suit the core area to be divided by the
gap.
For example, a 50mm stack of 50mm tongue E&I wasteless pattern
laminations will need paper cut
about 55mm wide x 160mm long which will
allow some over hang when fitted into the core which will have a gap
area
right across between the whole stack of E and I which is equal to 50mm x
150mm.
The OPT will be finished with everything assembled and terminated. The
rectangular yokes should be in place with retaining bolts but not fully
tightened. I may add that the yokes for E&I
laminations on SE OPT should be made from
brass, copper or
aluminium or very thick fibreglass reinforced board so the magetic flux acting
across the gap
does not suffer interference.
When the amp is set up and running, it may be tested for full power and the
output tube conditions carefully
monitored.
The dc flow in the OPT core will usually be enough to draw the E and I
tightly together when the bolts
holding the core remain loose enough to
allow the E&I to come as close as the gap material permits
I may add that where the OPT is in a high voltage output stage where Ea
would typically be +1,000 Volts,
GREAT CARE MUST BE TAKEN
WITH SAFTY AND VOLTAGE MEASUREMENTS.
The output circuit
with the OPT should survive testing without the OPT having been varnished or
waxed,
and indeed should survive the application of +4,000 Vdc to the primary
with all the secondaries and core well grounded
and for 1 minute without any
arcing.
If this HV test is made it is from a +4,000V dc supply fed through 8
x 1 watt x 1M metal film resistors rated for
being able to take 1,000V across
the R, which is a flow of 1mA, and at a power of 1 watt.
The earth point of
all the secondaries and core should be taken via a 22k resistor to the OV of the
test power supply which must be grounded to the mains OV.
If an arc occurs
between the primary and earthy parts, 0.5mA will flow and a voltage of 11V will
occur across the
22 ohm resistor, but otherwise no damage or smoke should
occur to anything.
The meter used for the measurement should be a normal
cheap analog type.
The
windings may tend to be a little noisy while testing with signals, but this is
typical behaviour with an SE OPT which is not
damped by the final varnish or
wax treatments.
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 just slightly tightened and the Is tapped up to be tight against
the Es.
With C-cores, the clamps around the cores are slightly drawn
up.
Make sure
the wanted dc current flows at idle by careful measurements of the cathode
resistance voltages.
One should find that the 1kHz sine wave signal
should be able to be run up to a level where
the estimated maximum power is
reached without any clipping but where there is perhaps some obvious harmonic
distortion.
Usually in an SE triode amp without NFB, the THD for the load
which allows the maximum power
will be 4 to 7%, and clipping will occur on
the crests and troughs of waves about simultaneously.
( With NFB the amp may
have to be stabilised for HF by applying critical damping networks.
And LF
stability may also require attention with gain / phase reduction networks.
)
Hence minimizing NFB is a purer way to examine the OPT performance.
The
output tube grid signal response should be monitored to make sure the output
tube
input wave remains with a flat response for tests between 5Hz and
200kHz
If the air
gap was much too small, the wave form will show severe asymetrical
clipping.
But let us assume the air gap is approximately
correct, and that we can establish what the maximum
output voltage is for
clipping into the rated load value at 1 kHz.
The amp can
then be set to run at 1/4 the maximum output voltage, and the signal
generator frequency
altered to test the response without NFB of the amplifier
while the generator level kept exactly constant.
The low level response of
the OPT is of interest with a known value of source resistance.
Usually with
a triode tube, this source R is the Ra of the tube in parallel with the
connected load.
This response should be the same as when the OPT is tested
without any DC and fed directly from a sig gene
at low level with its source
resistance set up to equal the Ra of the tube with the load in parallel.
Such
a low level test only confirms the low level response without the effect of core
saturation and
high ac voltages at low frequency. The dc flow and high
voltages will not change the response from the low
level much above 50Hz
unless there is a serious error with gap or the number of
turns.
If all was
well, the low level response should be from about 5Hz to 70kHz, -3dB points, and
almost completely flat from
20Hz to 30kHz.
The
amplifer can now be tested without any load connected with output voltage set at
1/4 maximum.
( It should
remain quite stable, and if not, there is a mistake in the design, or
insufficient effort has been used to
stabilise the amplifier )
Measurements of signal currents are now made with no
load connected.
By
recording the signal voltage at 1 kHz across the 10ohm cathode current sensing
resistor, the signal current measured
should be very low, because the only
current flow possible is through the primary inductance,
and the value of the
primary inductive impedance in ohms, or reactance value as it is called may be
over 30 times the rated load value.
But with no load present, the primary
reactance = ZL = anode voltage divided by the cathode signal current,
so
that if Va = 50Vrms, and Ik = 0.5mA, then ZL = 50V / 0.0005A = 100,000 ohms.
As the frequency of operation is reduced below 100Hz, the
signal current will begin to increase due to
an increasing current flow in
the primary winding inductance because the reactance of the
inductance
reduces with reducing frequency.
The Lp reactance must be plotted for
100Hz, 70Hz, 50Hz, 35Hz, 28Hz, 20Hz, 14Hz and 10Hz
where the the signal
output level remains constant and distortion does not exceed 5%.
One may find
that there is a sudden increase in distortion at 28Hz, and this indicates
something could be wrong with the
inductance value or the gap size.
This should NOT occur while the signal level,
is 1/4 of the maximum.
The design process
calls for the the calculated minimum value of Lp to have a reactance equal to RL
at 20Hz.
So if the anode RL =
1.8k, and we measure that ZL = 1.8k at 20Hz, then primary inductance,
LP, =
1,800 / ( 6.28 x 20 ) = 14.33 Henrys.
Lp Reactance,
ZL, = L x 6.28 x F.
Lp Reactance
also = V / I.
So Lp at any
F = ( V / I ) / ( 6.28 x F )
So after carefully recording the Ik and Va
for various frequencies,
the value of inductance can be plotted and the value
we are most interested in is at 20Hz.
If the air gap is too small the inductance
will be higher than minimum calculated value, and if too large it will be
less than the minimum calculated value.
If the gap is
too small, the output signal cannot be raised to a level of 0.7 x 1kHz
maximum level without
severe assymetrical wave distortion occuring due to
saturation of the core. The gap needs to be increased
to stop saturation, but
without reducing the Lp too much.
Switch off the amp and add another layer of
paper and try the same measurements again at 20Hz.
inductance will be less
but the threshold voltage where saturation distortion occurs should be higher
with
the larger air gap.
With no
load resistance, and with correct design and correct gap and with less than 5%
thd,
the output voltage possible
should be at
least equal to the maximum 1kHz signal with a resistance
load.
If the gap is
too large, there may not be any saturation of the core but the there may
be more than 5% thd due to the
reactance of the inductive load being so low
it causes shunting tube current distortion when the output voltage is
raised
to near the maximum wanted.
During testing, it is imperative that
the distinction between iron caused saturation and tube overloading
be
clearly understood.
The In this case the Lp needs to be increased by reducing
the stack of paper gapping.
If the reduction of the gap causes no
improvement to measured thd at 20Hz at full power, some
other design mistake
has been made, but the full power cut off point may have to be accepted as being
above 20Hz.
If the gap gives more than the wanted
calculated minimum inductance, and maximum signal voltage is able to be applied
at
14Hz without saturation current spikes in the signal current wave form
then indeed the design process was correct
and you have been extremely lucky.
It is better to have a slightly overgapped
core than an undergapped, even if the minimum value of inductance calculated
cannot be attained because then the amplifier will sustain low frequency
transients better.
The full power response of the amplifier with
a resistance load connected can be measured after the gap has been established
as being correct.
It is impossible to expect that full power response
with 5% thd will be dead flat from 1kHz down to 20Hz because the
tubes
will not be able to produce the load current required for full power
voltage at 1kHz where the load is solely resistive.
At 20Hz the load
will be perhaps the R load PLUS the reactance of the Lp which may be equal in
ohms to the R load,
thus giving a load of 0.7 times the R load
value.
Where ZL = RL tube current will be 1.41 times the 1kHz level, since
the load "seen by" the tubes is the R load plus an inductance in
parallel.
The response of the amp without FB will be
-3dB when the reactance of the Lp = the load plus Ra of the tube in
parallel.
In the case of a 300B triode where RL = 5k, and the Ra = 800 ohms,
the Ra and RL in parallel = 606 ohms.
Suppose the Lp was 40H at 20Hz, then ZL
= RL at 20Hz, but the response
would be -3db where ZL = 606 ohms which would
be at 2.4Hz.
This would be easily measurable if the grid signals could be
kept level to 1Hz.
When beginning the test at the maximum 1kHz levels there
would be a considerble increase in distortion
by 20Hz without much roll off
because the load will have reduced from 5k to 3.5k.
Even at 10Hz the roll off
will be small but the core may be beginning to saturate so a combination
of
saturation distortion spikes and inductance loading is seen.
To avoid
antagonizing the tube, I allow the finished circuit to produce an output
response that is goverened by the
input circuit/driver stages so that even
with global NFB connected the output is -3dB at 10Hz
with a rapid roll off
below 5Hz. When tested from 1kHz down and at full power there should be a
smooth
roll off as F falls below 20Hz and distortion should not much increase
above 5% even at 5Hz where the response
will be down about 9dB.
The response is then plotted for 5% thd
limitation, and should show a limit rolling off at 6dB/octave below about
20Hz.
The signal current at the 10 ohm cathode sensing resistor should be
monitored during the response testing to
produce the graph. The current wave
form should show no high sudden peak for parts of the wave form which indicate
the magnetic field has collapsed due to saturation. Where saturation has
occured for part of the wave cycle
the load on the tube has reduced to only
the dc wire resistance for part of each wave cycle.
We would always wish that the load at LF
below 20Hz be one that is a pure unsaturated inductance.
The method of confirming the gap size is one
requiring natural skills in perceiving what is really happening
in a given
circuit, and many DIY enthusuiasts make terrible mistakes with SE OPTs.
Bass
performance is poor, and silly reputations are pinned onto tubes without
justification such as
" 300B have no top end or bass; just beautiful
midrange. "
This is a complete myth perpetuated by people who don't
understand what they are doing and don't have any ideas about
how to make
OPTs or amplifiers.
The method of gapping OPTs can be applied
also to chokes used for filtering ripple voltage in power supplies
or for
reactance dc supply load for driver tubes or for ISTs.
With CLC input
filters, the gap can be small where the ac signal voltage is small across the
choke, but in
LC choke input filters the gap has to be carefully established
and tested to avoid saturation wave form distortions.
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METRIC WINDING WIRE SIZE
CHART
The metric winding wire sizes were kindly given to me by a
local Sydney wire and transformer parts supplier.
The original chart
contained the same copper sizes as shown for grade 1 with less enamel
thickness
and grade 3 with more enamel thickness. I only use grade 2 which is
the only grade shown in the chart below.
Grade 2 is the only grade stocked by
my supplier because it is the industry norm for 99% of high temperature rated
winding wire for electric motors and stressful industrial applications.
The
range of sizes shown are not all obtainable off the shelf, and to get some sizes
a wait for an order is involved,
so I sometimes have to design around the
wire size available, which adds to the challenge.
Anyone not used to
measuring in millimetres better start getting used to metric because here the
diameter measurement matters more than the wire guage, and there are is AWG,
SWG, BS, all very confusing,
and I don't have conversion charts so if you
work in guages and inches and feet, provide your own solutions.
Before
winding anything, make sure you have an accurate micrometer to confirm that the
size is correct.
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