آموزش ساخت آمپلیفایر و سیستمهای صوتی های فای

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آموزش ساخت پروژه لامپی سینگل با لامپ 6v6
نویسنده : عارف - ساعت ۱۱:٥٢ ‎ق.ظ روز شنبه ۱٢ فروردین ۱۳٩۱
 

آمپلی فایرهای لامپی با ساختار سینگل بدلیل مزیتهای اون از قبیل دیستورشن ثابت و هارمونیک های زوج بیشتر و بالانس بودن دونیم سیکل سیگنال و.. به نوعی در حالت ایده آل تقویت کنندگی قرار داره .

یک لامپ درحالت سینگل در بایاس کامل کاری قرار داره یعنی چه سیگنالی تقویت بکنه و چه نکنه لامپ در حالت کارکامل قرار داشته و از عمر اون کاسته خواهد شد . واین یکی از معایب اصلی ساختار سینگل هست . درحالیکه در ساختار پوش پول با گرفتن قدرت بیشتر لامپ بیشتر مستهلک خواهد شد.

با تمام این اوصاف سینگل ها به دلیل ویژگی های خاص خودشون علی رغم این مسائل طرفداران خودشونو دارند . و بهترین آمپلی فایرهای لامپی دنیا بیشتر براساس سینگل ساخته میشن .

تصویر زیر شماتیکی از یک آمپلی فایر لامپی سینگل براساس لامپ 6v6 هست

 

 

  دلیل انتخاب این شماتیک علی رغم کم هزینه بودن و کم بودن تعداد قطعات و سادگی مدار .استفاده از لامپ 6v6 هست که لامپی است با پاسخگویی خطی خوب در حالت پنتود و پاسخگویی خطی عالی در حالت تریود . تا حدی که  audio note از این لامپ در حالت تریود در آمپلی فایر مدل  Ankoru برای درایو دو عدد لامپ 211 استفاده کرده !!

برای پیچیدن چوکهای این آمپلی فایر از فرمولهای زیر میتونید استفاده کنید. این فرمولها کاربردی بوده و نتایج خوبی دارند. فقط باید با دقت تمام مراحل رو پیش برید . به این صورت چوک آمپلی فایر لامی شما هم کار دست خودتون بوده و از نتیجه کار بیشتر لذت خواهید برد.

 برای محاسبات چوک این آمپلی فایر توان امپ رو حدود 6 وات و امپدانس ورودی چوک رو 5000 اهم در نظر بگیرید.

در این فرمولها مراحل کامل محاسبات چوک از قبیل اندازه هسته . ضخامت سیم ها و عایق ها و الگوهای سیم پیچی کاملا توضیح داده شده.

فقط یادتون نره که چوکهارو باید با دقت تمام و بصورت ردیفی و منظم بپیچید تا نتیجه کار خب باشه.

سیم پیچی این چوکهارو حتما در زمان مناسب با حوصله و صبر و دقت کافی شروع کنید چرا که ساخت یک چوک دست ساز  چند روز زمان صرف خواهد کرد .


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.

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