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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