转一篇E文的输出变压器计算文章！
希望看的懂的大侠能翻译一下，看有没有值得学习的地方。j版看看这东西有参考价值么？没的话请将它删除。SEOUTPUT TRANSFORMER CALCULATIONS.
FORSINGLE 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 capacitanceat 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 mightuse 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 afar 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
http://www.turneraudio.com.au/schem-opt1-se.gif
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, & Lof 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 hasL = 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 willtraverse 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 theoreticalnumber 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 WINDINGSEQUENCE 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.
http://www.turneraudio.com.au/opt-sec-sub-sections1.gif
Fig 3.
http://www.turneraudio.com.au/opt-sec-sub-sections2.gif
Fig 4.
http://www.turneraudio.com.au/opt-sec-sub-sections3.gif
Fig 5.
http://www.turneraudio.com.au/opt-sec-sub-sections4.gif
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 the3 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 of2 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 between5 ohms and10 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 of2 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 of2 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.
Theoreticaloa 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 loadfor 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 at14Hz=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
Whereµe 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, thereare 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.Ifno 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.
http://www.turneraudio.com.au/bobbin-winding-detail-se-opt-no1.gif
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 betweenlayers 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 18,000/2ohms 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 signalshould 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 outputvoltage, 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 reactancemust 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 20Hzthe 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.
---------------------------------------------------------------------------------------------------------------------------
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.
http://www.turneraudio.com.au/wire-sizes2.gif 哎看不懂呀:4sda 哇！ 这个要翻译的话， 累死人咯:Q 都是“英国老鼠”要买过翻译CAT才可以看明;P 中文的才好啊！这个看不懂的 看不懂 找不到中文版的啊！发上来看那位大大能看懂 我就第一行看懂了:L
回复 #1 happymcu 的帖子
SE 输出变压器计算。对于独身者用一个方向的净直流流量结束了放大器。
因为对于推的计算拉输出变压器去到 'PP 选择计算'
这页含有 :-
作摘要关于 Radiotron 的参考设计者的手册，第 4个 Edmund,1955,
关于 SE 放大器的注解选择和 Ongaku,
无花果树 1.概要的没有选择 25 瓦特 SEUL 安培 3 个。
设计逻辑方法, 设计 SE 的第 1 到 51 阶段选择使用没有为例选择 3。
没有被牵涉的错综度的道歉, 去钓鱼如果它的全部都太硬式了!
设计方法含有可能一起用的许多计算和最初的和第二插入配置的列表
管声音 PP 选择。
无花果树 2.选择第二的子为和 2 和 3个第二的层负载匹配区分之。
无花果树 3.选择第二的子为和 4个第二的层负载匹配区分之。
无花果树 4.选择第二的子为和 5个第二的层负载匹配区分之。
无花果树 5.选择第二的子为和 6个第二的层负载匹配区分之。
无花果树 6.因为为一个 SE 安培没有选择 3 用，选择筒管绕阻明细。
为分路电容的计算在 SE 的输入选择。
最后设计的较多的检查，漏电感， acB ， dcB ，空气隙的计算, 最初的电感
而且最后的蜿蜒画面高度。
SE 放大器的实际测试 AND 选择。
调整空气隙和间隙和最初的电感实际的检查。
公制的蜿蜒线尺寸为 2 级多元酯-醯亚胺线制成图表。
----------------------------------------------------------------------------------------------
一去从 SE 输出变压器设计开始哪里?好吧, 在 Radiotron 中有许多好设计忠告设计者的手册，第 4 版,1955, 但是它的大概有关 PP 选择因为几乎不任何一个在 1955 年被想那的人
真正的嗨-fi 可能从一个被结束的设计被有。
数以百万计 SE 变压器进入无线电和无线电之内被设计而且建造而且安装-1950 年代的克使用了 6V6 或者 EL84 的时代当一个 4 瓦特分类一个五极管放大器由于相当高 thd/imd 但是不是如此高是对在一个 sunday 午后正在听蟋蟀的人的问题。
最 occassionally 被使商业的制造人腐坏的一些脑可能使用一 807 哪一个已经在如 WW2 后的军事的处理一个大的工作总量被买当数以百万计 807 结束生产, 使许多 DIYers 高兴直到大约 1965 当
WW2 存货最后耗尽。
1955 年的趋势是无论超过 5个瓦特在何处需要, 一个 PP 电路被使用
99% 的场合 , 和 10 瓦特 AB1 从一对 6BM8 变成可能了, 和安培廉宜做超过
一 807 或 6L6.
PP 被视为高级的,SE 考虑了原语和旧的被形成。 数以百万计电视机有了孤单的 6BM8;
高级的电视机有了一个 PP 安培和和 fullrange 喇叭对于一个沟流。
但是一些嗨-fi 狂热不关心成本, 而且发现他们的耳告诉他们他们 807 或者 6L6 捆绑如一个三极管可以在 PP 和一些方法比任何好的 PP 安培使更为好的共呜成为 6个瓦特胜于来自一对小的管的最初 6个瓦特。
在 WW2 后的日本， japanese 每件事物保持了很短大约 25 年之久。他们的房子很小, 而且以纸荧屏和木制造了,如此大大声的高度能力 PP 安培为他们的需要的大部分没被需要。
被争取的大多数的日本人被由于 6V6 或 EL84, 和一对于每个沟流当立体进来了的时候。
但是当由于他们的熟练人由于为输出品制造照相机和车辆变成比较不节俭和比较不贫穷, 而且的确变成富有的时候，最后 PP 安培变成了在日本的标准。
但是曾经被需要而且发现一个孤单的发射管 , 像是 211 能够用音乐的灵魂的如此杰出保存处理音乐，一个顽强抵抗者的坚决小带继续努力主意： SE 是所有的了
有对其他事物的确实零的需要。 孤单的 2A 3 或 300B 也是小的国王。
日本人也聪明地建造 sensitve 喇叭适合安培，如此他们的放大器总是功
在低的失真区域中和必需的没有负 FB。
SE 放大器绩效的 pinacle 可能在日本的声音注意的创造，而且最贵的放大器之一用一个 211 输出管是 25 瓦特 Ongaku 。
它值极大量的钱。 设计者窃贼桑河在 30 年之后在 2006 年一月建筑放大器死了。
我已经读一些他的引述电文而且虽然我不了解全部他有关他为什么从他的安培拿如此棒的音我见到他在他的 SE 中做场所棒的重要性通银绕阻线的使用选择说。 他的引述
speaches 似乎有多种的不一致性和 vaguities, 但是可供思考的事。 我视窃贼为用铁管和线除了谁只可以说话有关他做的之外有伟大的自然直觉的才能的胆怯的主人工匠
藉由亲切的困难。
当然几乎，没有人别的在输出变压器中使用银的线, 因此它的容易说银有重要性
而且它在 AB 试验会对 proove 银更困难比它更好运行颂杨德行的一个狡诈的售卖活动
银而且得到售卖。 他也使用 GOSS 铁加上在他的大约 50-50% 的镍含量选择哪一个也应该
明确地但是敏锐地感染音听到。
无论如何, 觉得免费使用银的线和镍心 , 或无定形心如果你喜欢, 但是对我 GOSS
对所有类型是确实美丽选择。 窃贼桑河没有关于匝计数，心尺寸的任何资讯,
绕阻几何学，阻抗匹配的样子, 全部都会感染音, 向前由于使用
每个管哪一拓扑学 preceeding 输出级和不要再提到型态类型的选择而且是否他们是
的不。 健全的听到从任何的健全系统是它的零配件和房间和浮凸槽记录的状态, 和 imho 的效应的总数, 任何人可以建立一个放大器相当在音波的绩效对任何事相等被日本的传奇性声音注意创下。他们将会需要虽然确实了解他们的数目，而且这里是 , 我希望的地方为对全部的棒 asistance,
哪一个超过我能说及声音注意。
回复 #13 bg7mrj 的帖子
:L :L :L回复 #2 happymcu 的帖子
给 SE 的方法选择 , 设计以 RDH4 或我已经收集过了过去 10 年的其他的来源理论为基础。 在有创伤多数非常美丽 SE 输出变压器之后用全功率带宽像 20Hz 一样的宽对 70个千赫，我从经验说很好地感觉合格的。参与产生最好人的逻辑阶段的列表可能的选择以为低的蜿蜒损失设计为基础,
在 14Hz 的全功率的心饱和, 和适当的插入扩充达至少 70 仟赫的 HF 响应
藉由保存对低的量漏电感和分路电容。 结束结果用一些阻抗给一个很好地填充的蜿蜒窗囗相配可能的没有有浪费任何二次绕组的断面
因此蜿蜒的损失和响应是任何一个被选择的负载匹配的一样。
设计方法非常相似对推拉设计方法, 因此最简单的方法着手进行将作基础
在给 PP 的那使用者之上的 SE 方法选择但是由于考虑到效应的必需修改
一个方向的直流流量只有和总是 SE 选择被连接到一个纯粹的 A类放大器。 这将会从尝试来回地对 SE 或 PP 设计页环接把读者存档。
让我们检查输出变压器没有 3( 哪一个有同一的绕阻没有选择 1, 但是有不同的连接
而且一个被缝隙的心 ).没有选择 3 能够 SE 功率输出的大约 25个瓦特。
首先我将会显示示意图没有选择 3:-
Fig1
上述示意图一选择是我可能在一个 SE 电路使用的典型。
下列的设计逻辑流量可能被用装配一个一会进入设计需求 , 像是能力，第二次的负载，管 Ra, 心大小,然后以一个轻击通一个 " 设计 " 钮的电脑程式, 在外会来包括所有的精确线尺寸和可能用一些蜿蜒的经验容易地被任何人了解的筒管终结明细的一个十字形断面制图的一个非常的设计。
唉, 我不是一个计算机专家，但是 i 能给逻辑的流量用的你使一个设计被结束。
我邀请任何人使为一个个人计算机程序作好准备包含配合线的所有可怕 fiddly 明细感兴趣
那从供给到可用来的心是有效的极好选择。
我将会是或许在任何人变换一个长的时间之前必须等候我的逻辑流到一个个人计算机程序
因为脑倾向到正直的 " 直觉地拿事物 " 和一部个人计算机不曾决意。
输出变压器没有 3.
给 25 瓦特 SE 的设计例为 3,100 到 5个欧姆 ( 大约的第二负载 ) 选择。
1. 为管选择管, 动作条件式和最初的负载应用的横断面那
全部的最初, 已知的如阳极负载， PRL ，欧姆。
小心的 loadline 分析为正确地决定装货和功率输出计算被需要
而且这不在阶段的这一个列表中。 在负载匹配上见到我的页。
选择 3, 管将会是 2 x 6550 ， Ea=500 V ， Ia=70 妈每个,
PRL=3.1kohm............................................................................................3,100个欧姆
2.选择第二的公称喇叭负载数值。
允许一个内定的数值， SRL ，欧姆。
选择 3, SRL =5 欧姆.....................................................................................................5个欧姆
3.在截割选择最大的能力为那
被选择上方，波河，瓦特的 PRL。
选择 3, 波河 =24.5 瓦特,...................................................................................................24.5W
4. 计算最小的必需心交叉剖面积， Afe,
对于一个几乎正直的心中心走截面。
在 sq.mm 中的 Afe=450 x sq.rtPO
**注意。这一个反应式已经起源于一个基本的反应式
为输电干线变压器被用的心尺寸, Afe= sq.root power 输入 /4.4
在 Afe 在 sq 寸中的地方。 远古的反应式被建立大约的通讯号 ac B 最大的这
1 在 50Hz 的泰斯拉但是我们会想要 B 最大 = 大概给 SE 的 0.33个泰斯拉选择。
在上述的反应式是 SE 声音选择的针对的好导引的相当多的试验之后。
选择 3, Afe=450 x sq.rt 24.560=450 x 4.94=2,223 sq.mm..............................2,223sq.mm
5.计算心榫舌大小， t 。
对于一个正直的心断面, 榫舌大小 = 堆积画面高度， ie T=S。
T x S=Afe。
因此理论上的 t 大小 = sq.rt AFe= th t......................................th t, 毫米
选择 3, thT= sq.rt 2,223=47.15 毫米...................................................................47.2个毫米
选择来自有效无废料的 E&我层压心事物的列表的适当的标准 t 尺寸。
t 按规定尺寸制作普遍有效因为选择 :-
20个毫米， 25个毫米， 32个毫米， 38个毫米， 44个毫米， 51个毫米,63.5个毫米...........................................进入 t. 毫米
**注意。选择在 thT 上面的一个标准的 t 尺寸提供较低的铜绕阻损失，较高的重量,
而且在 thT 下面选择 t 提供较高的损失和较低的重量。 Afe 将会对任一 44个毫米是相同的
或被选择的 51个毫米从在上面因此 LF 响应以榫舌尺寸将不变更。 HF peformance
完全地仰赖插入几何学和绝缘。
选择 3, 选择心 t=44 毫米............................................................................44
6.计算使用被选择的 t 尺寸的理论上的堆画面高度， thS 。
thS=Afe / T, 然后对一个较大的画面高度调节适合最近的标准塑料的筒管尺寸如果有效的, 毫米。
选择 3, S=2,223/44=50.5 毫米, 选择...........................................................51个毫米
7.调节了的 Afe= 选择被选择 S 的 t x..............................................................Afe, sq.mm
选择 3, 调节了的 Afe=44 x 51=2,244 sq.mm...................................................2,244sq.mm
**注意。 一些构造函数将会使用非无废料的 E&我打,
或不有如同 E&我一般的比较大小无废料的图案心的 C语言心。
t ， S ， H 的实际尺寸,& 心的 L 被用一定被确认。
8. 确定蜿蜒窗囗， H ，毫米的画面高度。
选择 3,44个 t 无废料的事物有 H=22.............................................................22个毫米
9.确定蜿蜒遗留字串， L ，毫米的长度。
选择 3,44个 t 无废料的事物有 L=66..............................................................66个毫米
10.计算理论上的一次绕组匝， thNp
Np=sq.rt( PRL x 波河) x 20,000/Afe,转。
**注意。这里的反应式起源于带 B 和 F 的更复杂和更完全的反应式
进入解释 ac 运算之内。 如果我们承担 ac 磁场强度 B=0.8个泰斯拉, 和 F=14 赫兹是一合适地
为饱和正在开始的地方低 F( 因为已经有直流磁化的大约 0.8个泰斯拉,)
而且快递 V 根据负载和能力,
因为提名候选人的预选会变必需的，我们拿容易的反应式给上述的短。
给有心机的 ac B 的完整反应式在下面的第 40 阶段。 V 因数能被表示成
sq.root( 最初的 RL x 功率输出 ) 当做在上述的单一化反应式中。
选择 3, RL=3,100 欧姆，波河 =24.5 w, 在上面的来自第 7 阶段的 Afe=2,244,
thNp= sq.rt(3,100 x 24.5) x 20,000/2,244=2,456个匝..................................2,456个匝
11.计算理论上的最初线 dia ， thPdia 。
**注意 4. 对于变压器被用的最初的线将会占领部分
窗囗面积 =0.28 x L x H.0.28个作品的常数为 99% 的选择。
线的每个匝将会占领被一致的 area= oa dia 。
全部的或 oa, dia 是 dia 包括瓷漆绝缘。
因此理论上的超过包括瓷漆绝缘的 P 线的所有 dia
= sq.rt(0.28 x L x H/ Np).......................................................................thPoadia, 毫米
选择 3, thoadia P 线 = sq.rt(0.28 x 66 x 22/2,456)
= sq.rt 0.1655
=0.4068 毫米....................................................................0.4068个毫米
12. 来自表， Pdia 的发现最近的适当 oa 线尺寸, 毫米
选择 3, 尝试 oa 线尺寸 =0.414 毫米,( 对于 Cudia=0.355 毫米。 )............................0.414个毫米
13.建立筒管绕阻横向宽度................................................ Bww, 毫米
**注意 5. 筒管横向宽度是那距离在中模凸缘之间而且改变靠
制造了筒管的通, 但是每个凸缘厚度 =2 毫米最大是通常的, 但是可能是些微比较少的。
筒管装凸缘的地方没被用，而且绝缘只是被延长到完整的窗囗长度 L,
横向宽度将会是相同于在筒管确实有凸缘的地方情况。
Ie, 绕阻将会横向 distance= L- 4个毫米。
选择 3, Bww=66- 4=62 毫米............................................................................................62个毫米
14. 没有理论上每层， thPtpl ，匝 P 匝计算。
来自第 12 阶段的 Ptpl=0.97 x Bww/ oa dia.
**注意。 固定的 0.97个因数考虑到有缺点的层充填物。
遗漏匝的部份。
选择 3, Ptpl=0.97 x 62/0.414=145.26...................................................................145个匝
15.计算最初层， thNpl 的理论上数目,
然后把下弄圆或向上到方便的甚至或层的单号。
理论上的 Npl=thNp / Ptpl, 然后弄圆在/上面下.............................................thNpl, 没有
选择 3, thNpl=2,456 / 145 =16.93个层; 把下弄圆到 16................................16个层
**注意。 圆化下可能减少 Npl 对 Fs=14 赫兹的需要。
但是被用的实在的匝将会仍然允许 Fs= 大约 15个赫兹, 是 ok。
对于正在想要边缘地维持 Fs, 或有 Fs 的那些比 14个赫兹低,
Afe 能被藉由增加从发言权 51个毫米到 62个毫米的 S 增加, 而且仍然使用一个标准
按规定尺寸制作筒管, 而且在 12Hz 有 Fs, 是边缘的和不去改良椴树很多。
16. 计算实在的 Np 。
Np= P 分层堆积 x Ptpl, 转
选择 3, Np=16 x 145=2,320 匝..........................................................2,320个匝
17. 计算平均的匝长度， TL ，毫米。
TL=(3.14 x H)+(2 x S)+(2个 x t), 毫米。
选择 3, TL=(3.14 x 22)+(2 x 51)+(2 x 44)=259 毫米............................259个毫米
18. 计算一次绕组电阻， Rwp 。
Rwp=( Np x TL)/(44,000 x Pdia x Pdia)
哪里 44,000 是一个常数, 和 P dia 是来自线表的铜 dia.......Rwp, 欧姆
选择 3, Rwp=2,320 x 259/(44,000 x 0.355 x 0.355)=108.36个欧姆................109个欧姆
19. 计算一次绕组损失 %,
P 损失 %=100 x Rwp/.( PRL+ Rwp)。 .....................................................P 损失,%
选择 3, P 损失 =100 x 112/(3,100+109)=3.4%..............................................3.4%
20. 蜿蜒的损失比 4% 大吗? ..................................................是的或号码
如果是的，设计 calcs 一定在再一次被检查, 和一个较大的心/窗囗面积挑选的。
如果不, 前进到 21.
选择 3, P 绕阻损失是少于 4%
**注意。 如果 P 绕阻损失比 2% 更少, 有一种可能性：线尺寸可能被减少
增加每层匝, 而且可能地减少发言权 2 的 P 层的数目, 但是那
堆画面高度会必须被增加到生计, Fs 最低值。
21.选择来自列表的插入图案在对于变压器的瓦数下面。
全部选择将会有第二的断面只有含有线的一个层。
这可能被细分为比较进一步的第二潜水艇断面, 在这里没有设计哪一个需要
双线的或 trifilar 绕阻或矩形线。
绕阻的一个断面被定义为一个层或群体的层投入的独自地对 P 或 S。
期间 " 节 " 是不要再和 " 层 " 感到困扰. 因为管选择, 大多数的 P 断面将会有
线的超过一个层。
大体上, 全部选择应该遵从下列的 P&S 层编号联系 :-
在总而言之蜿蜒的通是一个最初的断面的地方,然后这些断面应该有在 1/2个内部断面的层的附近, 因此如果在 a P 断面中有 3个外部的 p 层, 内部断面可能是或 5,6 或 7p 层。 当这
指南被黏附在有最好的 HF 响应因为漏电感是公平地平均地，而且 symetrically 分配。
当以 S 断面启动而且完成所有的内部 P 断面应该有 p 层的相同数目的时候
但是它不总是可能，而且有 5个 p 层的 4个 p 层和 2个断面的发言权 2个断面是好。
如此 " 内部的节 " 的尺寸不应该改变超过 25% 。 那两个没有条件式就算了
避免上面的 HF 响应的不可预知的谐振。
因为变压器适合低的磁盘装置 , 像是 mosfets 或者电晶体, 甚至以一个 SE 安培，相同数量的插入为 a 一个给定的功率位准被需要。 p 层的数目将会被减少如最初的 RL 变成低, 和线 dia 将会增加。
一个 8 欧姆 : 8个欧姆选择由于非常低的直流 P 和 S 之间的电压差会为 P 有匝的相等数量和 S 而且也许只是被插入纸厚线的如此每个层二者择一地投入于 P 或 S。
然后带宽可能是非常容易地组成去达 500 仟赫。 当最初的或第二的负载被减少,分路电容的效应减少，因此，绝缘厚度能被减少。 为静电的喇叭变压器哪一个阶段在那上面
在 50 和 100 次之间的放大器电压需要为电压有好绝缘牵涉并且降低电容,
而且他们相似选择 " 向后 " 有力量而且能在这里与方法一起设计。
但是为匹配管给正常 3 到 9个欧姆喇叭负载, 在下面的插入列表以最初每断面层的数目可能将会至少给 70 仟赫的带宽, 和哪里有一个插入的最高数目带宽
可能是 300 仟赫。 由于太多数层的绝缘，使用插入更甚于为线列出了到在筒管上的比较不有效房间的铅, 和贫穷的 HF 由于高的分路电容, 和比较高的蜿蜒损失。
对于比较低的最初 RL 和比较高的放大器有力量那比较大的那选择变成和对于一个插入的给定数目那
HF 响应对逐渐增加的漏电感变成比较不适当因此那比较大的那选择变成, 插入纸的断面增大的数目。 因此一个小的 15个瓦特只选择可能需要
给 70 仟赫 , 但是一个 500 瓦特的 3S+2 P 断面选择可能需要 6S+6 P 断面。 翻译软件通常翻译得真差, 看样子是篇好文章，可惜是鸟语看不懂 :L看着头大啊，可还是看不了。 我就第一行看懂了 唵是中国人种地的，看不懂:lol
页:
[1]