AUDIO TR�NSF0$E EIGN MNUL
ROBERT G. WOLPERT
AUDIO TR�SFORE DEI6N MANUAL
ROBERT G. WOLPERT
PREFACE
This manual is intended to show how to design and manufacture audo transformers. It will ms explain things that have to the be considered and the proble that the wi ll various be encountered in ach iev ing desi red results. It wi ll show how to design to meet the requirements and give examples and test results. It will aso give methods of manufacturing to achieve the results desired, with some of the things that will prevent the design from being successfuI. This manual is wr itten with th e assum ption that the designer has expe rience in the de sign of power transfor mer s he refore, it cover s only those manufacturng techniques that are different and necessary to achieve the proper results in an audio transformer. If the designer needs more information on the construction methods, the TRANSFORMER DESIGN AND MANUFACTURING MANUAL published by the author in 1984, or some other publication, should be conslted. The appendix will have various tables and charts to assist in the design of the transformers.
Robert G. Wolpert
page 1
TABLE OF CONTENTS PART I. DESIGN CONS IDERAT IONS
1.0 1.1
GENERAL REQUIREMENTS Frequency response
1.2 1.3 1.4 15 16 1.7 1.8
Impedances Power level Total harmonic distortion Direct current in windings um reduction Longitudinal balance Insertion loss
CHAPTER
2.0 2.1 2.2 2.3 2.4
LOW FREQUENCY RESPONSE Open circuit inductance Primary voltage and current Secondary voltage and current Core size and material
CAPTER
3.0 3.1 3.2
IG FREQUENCY RESPONSE nterleaving the winding Leakage inductance
CAPER
40 4. 1 4.2 4.4
SPECIAL REQUIREMENTS otal harmonic distotion Shielding for hum reduction Longitudinal balance Inserton loss
5.0 5.1 5.2 5.3
DESIGN METODS Flux density Construction suggestions Calculating the physical parameters
CAPTER
4.3
CAPTER
page 2
PART II.
DESIGN EX AM PLES
Example 1
Voice frequency telephone transformer
Example 2 :
Audio output transfomer
Example 3:
Line to voice coil transformer
APPENDIX
Symbos used DB-Watts table Wire table Lamination table Total harmonic distortion table Wire size turns - lamination tables
page 3
PART I. DESIGN CONSIDERATIONS
page 4
PAR T I DESIGN CONSI DE RATIO NS CHAPTER 1. GE NE RAL REQUIREM E NTS
By definition, an audio transformer s designed to operate within the audio range of frequencies However, the upper and ower limits are extended beyond the audio range for many uses. For examp le , usag e in h igh fide l ity ci rcuits migh t desire a range from 10 Hz to 30000 Hz or hgher, while a telephone transformer could have a range from 30 Hz to 3000 Hz The frequency range, both high and low limits, will determine to a great extent, the design and method of manufacture. Transform ers des igned to wo rk at aud io feque ncie s can be put into th ree general categories hese are input, output and im pedance m atch ing. Actually, the only differences between these are the usage and the im pedance ratios al l can be consid ered from as i m pedanc e matching mpedance to transformers, as theyThey are used to tansmit signals one another impedance ether higher or owe o sometmes, when isoation ony is desred between equal impedances. There are several things to be considered in the design of audio transformers: a. b c d.
Frequency response Impedances Powe level TH D o r total harmonic distort ion
e. f g. h.
Value of D C. in windings, if any Hum reducton leve, f equired Longit udina balance nsertion loss
n addition, the flux density of the core material must be considered in order to not operate the core into saturation.
page 5
1. 1
Frequency Response
The frequency response of a transformer is that range of frequencies that is desired to be passed. It is desirable have the same levelwilloffalall off. frequencies withto in th is range. Th e response extre mes or of voltage th e range These are usually called out as a range of DB, such as 3 DB or 1 DB, etc. This m eans that al the vo ltages between the two extremes will not vary more than the limits shown. The variations called out will sually be referred to a certain frequency in the c enter of the resp onse . Usua l ly th is wi l l be 1 00 0 Hz for audio transforme rs. Thus, if a certain vol tage or DB lev el is caled out for 1000 Hz, then all frequencies within the range should be within the imits of ±3 DB or ±1 DB or whatever is required. The -3 at DBthe frequencie wi frequency l have a vol tage level that is 70 7°/ of the votage middle ofsthe range. The best way to measure the response is to use a meter that is calibrated in DB rather than trying to calculate the DB from the voltage levels. However, the DB can be calculated from the voltages by using the following formula: Eo DB= 20 LOG
-
ErN
The ower imit of the frequency range is controlled by the primary inductance. This will fall off 3 DB at the frequency where the inductive reactance of t he pri ma ry equals th e pri mary im pedance. It wi ll fal l off 1 DB at two ti mes th e pri mary im pedance and 0 5 DB at approximat ely 4 times the prim ary im pedance .
page 6
The fol lowing formulas are us ed :
L
Where:
-3 DB
L
L
=
0.5 DB
L Z F
=
-1 DB
primary inductance pri mary im pedane ower frequency desired
The upper frequency will be down 3 DB where the normalized impedance equals the reactance of the leakage inductance The frequency response ca n be m easured using the ci rcuit in Figure 1
Figure 1
R
the i mpedan ce of the pri mary
R
the impedance of the secondary
page 7
The voltage of E1 must be hed const ant for al l frequenci es . The output 1000 Hz and the voltage, E is then set for a middle frequency, usually deviation from this voltage is the response over the frequency range.
1.2
Impedances
The impedances of both the primary and the secondary must be known or can be calculated The i m pedan ce ratio is equal to the turns ratio squared and also equa to the voltage ratio squared 2
2 =
1.3
=
Power Leve
The oper ating power eve l is usual ly ca ll ed out Th is can be sine wave power or music power Sine wave pow er wi ll be app roximately th ree times the effective music power The transformer must be abe to handle the ful votage and current at any frequency within the operatng range for sine wave power. For music power it must handle the full votage but only one third of the current for heating purposes Thi s wil l al low the u se of smal ler magn et wire and wl result i n a sm aler unit 1.4
Total Harmonic Distortion
The total harmonc distortion is mainly a functon of the operatng flux density in the core at the lowest operating frequency. Reducing the flux density will reduce the distortion The distortion at a given frequency and flux density will vary with the type of ma gnetic materia used
page 8
15
Drect Current n the Wndings
When one or more of the wndings is requred to carry unbalanced direct current, it is necessary to design that windng fo the proper inductance th e same way you woul d design an inductor carrying di rect current. The proper air gap spacer must be put in the magnet ic path Th is wi ll result in a larger unit than one that does not carry unbalanced direct current.
1.6
Hum Reduction
When designing for low level usage, it is often necessary to keep the external flux fie ld s at a very low level . Levels of -60 DB to 80 DB requ irement s are not unu sual . his is ac co mp lish ed by enclosing the unit in a c ase o r cases using hig h 80 °/o nick el is often used for th is app lcation. A permeabilty materials. single case or a nest of cases using alternate cases of high permeability materal and co ppe r may be needed .
1.7
Longtudinal Ba ance
ongitudinal balance is a measure of a transfome's balance and abilty to prevent longitudinal signals or signals that have been induced in the power line from being transferred into the secondary of the transformer
1. 8
Inserton Loss
Insertion loss is a measure of the power available out of the tansformer versus the power induced nto the transforme.
page 9
CHAPTER
2.
LOW FREQUENCY RESPONSE
- reated and must The requirements of an audio transformer are al nter a l l be cons ide red in th e des ig n. Th e first step is to des ign for the ow frequen cy Th is wil l est a bli sh the size o f the cor e and the num ber of turns needed It is al so the easies t pa rt of the design 2.1
Open circuit inductance
Calculate the in ducta nce neces sary Assum ing a -3 D B requirement at the ow frequency end, the inductance needed wil be: L =
For
DB and
-0.5
as called out in
1.1.
1
DB requirements, the formua is changed accordingy,
It can now be seen that the higher the primary impedance, the arger the inductance needed Th is transl ates to a l arger core a nd/o r more turns It al so makes it more difficut to obtain the higher frequency limit as wil be seen later 2.2
Primary voltage and current
Calculate the pri ma ry voltage an d current Th is depe nds on the information giv en . If the powe r (wattage) is given an d the im pedance is known, Ohm's law may be used to calculate the votage and current
AND
I
If the power is given in DB or DBm, the power in watts must be either calcul ated o r ta ken from a ch art. A cha rt of DB versu s watts is in the Append ix DBm is DB for a 600 ohm im pedanc e
page 10
2.3
Secondary voltage and current
Calcu late the secondary voltage an d current h e seconda ry voltage can be calculated from the impedance rato By rearranging the formula from 1.2:
he current ratios are inversely proportional to the voltage ratios so the secondary current can be calculated by rearranging the equaton:
-
2.4
=
Core size and material
he core materia l w ll depend large y on the a ppcation If the transformer s a smal, low - evel unit, the material can be either 5 0 °/o or 80°/o nicke l These a re h igh permea bil ity materias and wi l l requ ire fewer turns than 4 °/o si licon steel If a hig her level un it is desi red, 4 °/o silicon steel wil proba bly be the bes t choice as the cos t of nicke la minatio ns wll be proh ib itive in la rger units . Al so, th e operatng flux densi ty can be higher in sil icon st eel, which wil l resut in a sma l ler un it he power requ irem ents w ll h el p choose the core size If the core size is ca l led out, th ere is no choi ce If not, then exp erence from past de sgns, or an adjustment to the information from the 60 Hz tables wl have to be used For example, a 150 watt, 60 Hz core wi ll proba bl y be a bout righ t for a watt, 20 Hz aud io transform er h is wi ll be a good start ing point. See the table in the Appendix 50
he core sze obtaned in this manner is an approximation and adjustments wi ll h ave to be mad e to get the pr oper fil l Wh en the turns are calculated in the next step and the wire sizes are chosen, the fit in the core can be caculated
page 1
When the core has been chosen, an easy and quick way to calculate the turns needed to give the required inductance for any given core sze and magnetic materia is to refer to a lamination catalog put out by the various ma nufacturers A formula is given for i nductance f o r each size of core Th is can be turned a round to find the turns The only other requi remen t is that th e permea biity of the material must be kno wn Thi s can also be obtained from data published by the manufacturers For exam ple , from a catalog , a core size o EI- 1 00 , with a square stack, ha s an inductance formula : L
= . 5289 x 10 x
AND
N
K
L x 10 5289 x
Where:
x UA c x N
K
2
x UA c
is the stackng factor UA c is the permeabilty of the material K
Th e wire size is det ermi ned by th e current The p rim ary current wl l be as calc ulated in 22 Th e seconda ry current can be caculated fr om 2 3 If the requrement is for sine wave power, the sze of the wire shoud be from 650 to 1 000 ci rcul ar-m il s per am pere Th is wil be determined b y the nsertion loss al owe d In gener a l , 800 CM /A wil be abou t rght , and wll be used as a sta rting point in this ma nual If musc power is cale d for, the size can be reduced to about 300 CM/A By reerring to The core size and wire sizes have now been determined the ta bles in the Appen dix (wire sizes an d turns versu s la mi nation szes), the fil l can be checke d before going any further In gen era l , the prima ry wire should use up half the fil and the secondary the other half This chapter should pretty wel tie down the core size turns and wre sizes for both the pri m ary and second ary It may be necessary to ma ke adjustments if interleaving is necessary to meet the hgh requency response
page 12
CHAPTER
3.
HIGH FREQUENCY RESPONSE
The limit of the high frequency response is controled by the eakage inductance and the impedances. The lea kage induct anc e i s propo rtonal to th e squa re of th e tur ns, thus, it is to reduce limit this depends value greatly byprimary reducng the turns,and butthat since thepossible lower frequency on the inductance is controlled by the turns and the type of magnetic materal used in the core, the t urns are pret ty wel l set Al so, care must be taken that the ma ximum flux density of the material s not exceeded in reducng the turns . The leak age c an be r educ ed by interleaving the prma ry and secondary windings.
31
Interleaving the winding
As mentoned before, the high requency response is determned by the leakage inductance and the windng impedances. In order to reduce the leakage inductance, it s sometimes necessary to interle ave the wind ings . That is, sp lit th e wind ings a nd wind one part pri mary, one part secondary, one par t prim ary, etc Th is can resut in 1 :2 2: 3 3 4 et c, int erle avi ng. The exam ples wi ll show desg ns with thi s interle aving used . A 3 :2 interleave would be 1/3 primary, 1/2 secondary, 1/3 primary, 1/2 seconda ry and 1/3 pri mary, a nd so on In so me cases i nterleavings of 5:4 6:5 and more are used
32
Leakage inductance
The leakage inductance of a transformer can be calculated in many ways. Some of these are ext reme ly c om pi cat ed . A good compromise for a transformer that is wound concentricaly, that is one windi ng ov er the other, is the following : L =
06 x N x MT x x S x T 5 x WL x 09
+
=
HYS
page 13
Where:
N ML H WL T s
= number of turns = mean length turn wind ing heigh t winding length = insulaton space = number of inteleaves = =
The high frequency response can now be calculated by using the values of leakage induc tance and the norma lize d i m pedanc e . The upp er frequency limit will be 2 ZT
-
x Z2 + Z1
page 14
CHAPTER 41
4.
SPECIAL REQUIREMENTS
Tota harmonic distorton
If THD is called out, the flux density in the core and the type of core materi al must be consid ered . Th is may requ ire th e co re size or turns or both be modi fied . The distortion is normaly only of concern at the lowest frequency as it falls off rapidly as the frequency increases. If you consi de r the th ree most com m only used types of materials, 80°/ nick el , 50°/ nicke l, and si lcon steel , th e flux densities wil l vary from the lowe st for the 80° / nicke l to the h ig hest for the si l icon s teel . A table in the Appendix will give some representative values for the three materials at variou s flux level s These numbers wil al low you to estmate the distortion that can be expected at the flux level and frequency of operation. 4.2
Shelding for hum reducton
Many low level transformers are requred to function within an external field without picking up that field and transmitting it into the operating circuit. The transformer can be enclosed within a case or cases to achieve the desire d result s . For exam pl e, a sing le case of steel wil l gi ve about 10 DB of sh ie ld ing A case made of 50 °/ nick el wil l giv e about 20 DB . A case made of 80°/ nick e w ill g ive 30 DB A nest of cases consisting of an 80°/ nic ke l case, a co pp er case and anothe r 80°/ nckel case wi ll g ive 60 DB of shielding. If th is is extended to th ree 80°/ nc kel cases with two copper cases in between, it wil gve 90 DB of shieding. These 80°/ nic kel cases must be prope rly ann ealed and th e coppe r cases must be soft copper to obtain the desired results. Care must be taken that the external field is not so high as to saturate the 80°/ nic kel o r th e exp ected re sults wi l l not be obtai ned . If th is occurs, a stee case on the outside may be used to reduce the field to a level that c an be tol erated . These methods wi ll also a pp ly wh en the tra nsform er is placed in a ver y sensitive circuit and the external fux of the transformer must be reduced
© 1989, ROBERT G. WOLPERT Rev.
2004
page 15
4.3
Longtudinal balance
A longitudina signal is one that is induced along the power lines and enters the tran sform er in both t he pri mary lea ds as if they were one lead . This signal must be prevented from passing through the transformer as a signal in the output. To through, the so it has the the passing same coupling romtransformer the input must eads be to balanced the output thatprevent terminations See Figure 2
Fgure This coupling is represented by the two capactors,
C 1 and C2
I c 1
equals C 2 th en th ere is no feed through and no im bala nce Of course, this is a simplification as the coupling can occur anywhere within the windings. There are several ways to reduce imbalance. shown later.
Some of these will be
Figure 3 shows a method of measuring the longitudinal balance.
,5Ra
gure 3
page 16
.5 R 1 + 5 R
= primary Z and must be match ed to with in 0 .°/ for a
40 DB balance and to be within 0 . 0 1°/ if 60 DB of b alance is re quired . R2 he core sh oud be will be equal to th e i m pedan ce of the secondary. connected to g round If a sh ie ld is used it sh ou ld be connected to the core and ground
E2 shoud be a voltmeter E2 reads zero Wh en connected to read Ei capacitance un balance.
with a DB scale.
Perfect balance will be when
measur ing E2, there should be no voltmeter Short direct wiring should be used to minimize The l evel of E should be the operating level of
the transformer, or as called out by the customer.
4.4
Insertion loss
The insertion loss of a transformer is a measure of the efficiency as it shows ho w much power is consumed by the transfo rme r. Thus, it is im portant to kee p these losses a s lo w as possib le he most obvious of these losses i s th e DC resist ance of th e windings . he larger the size of the wire that can be used, the ower the losses he losses in th e magnet ic material can also cont ri bute to the to tal loss he magnetic material losses can usually be ignored if the flux density is kept within reasonable limits. The following method can be used to determine the approximate insertion loss before bui ldi ng the unit. The total or norm al ized windi ng resstance, that is the resistance of the secondary referred to the primary, or the primary referred to1the wl normaly berox 10 imately to 20°/ 0of. 5the resistance If it is 0°/,secondary, it wi l l have a lo ss of app DBload and, if 20°/, it will be approximately 1.0 DB. By using the calculated values of the winding resistances, the caculation of the normalized resistance can be done as folows: 2
page 17
Then the percentage loss will be:
100
Where:
Ri
-
DC resistance of the primary winding
R2 = DC resistance of the secondary winding
Figure 4 shows how to measure the inserton loss
Figure 4 Where
E
Voltage across the primary
E2
-
Load voltage
Ri
=
R2
Prima ry i m peda nce Secondar y i m pedance
It should be noted that the insertion loss test circuit is the same as the frequency response circuit, except that the primary voltage is taken after the primary resistor instead of across the generator. Then the insertion loss is
E2 � Ei
20 LOG
The votages obtained from this test can be used to caculate the insertion loss at any given frequency
page 18
CHAPTER 5. DESIGN M ETHODS
Before the design of any transformer can begin, it is necessary to understand the general construction and how to calculate the turns, winding fill and DC resistances of the winding. The following will show the methods used to determine these values 5.1
Flux density
The formula for calculating the open circuit primary inductance needed to meet the low frequency response is given on page 7 . In this formula, no provision is made for the flux density. It is always neces sary to calcula te the flux density to be sure that te core will not be operating in saturation and adjustments mad e to te turns, if necessay. Ti s may result in more turns than the m ini mu m needed fo the req ui red indu ctance The resulting inductance may be higer than is necessary. The operational for the various can of be th is obtained from a flux core density manufacturer's catalo g . core or materials the pur poses gauss for 29M6 manual, the maximum flux densities used will be 17 kilo material 8 k il o- gauss for 50°/ nicke l and 5 ki lo -gauss for 8 0°/ nick el If other materials or flux densities are desirable, the literature should be consulted The formula for determining the turns for a given core is:
N = 4. 44 x A x F x B
Were:
N E
= Num ber of t urns = Voltag e appl ied
= Effective coe aea in square inches = Lowest frequency of op eratio n Flux density in ines per square inc (Gauss x 6.45 = lines per square nch) 444 is a constant A F B
=
page 19
5.2
Construction suggestions
The construction of an audio transforme differs from a regular power transformer in that the leakage inductance must be kept as low as possible The interleaving has been covered for the leakage inductance, but the advantages of interleaving can be offset by improper or soppy winding. The windings must be d irectly a bove one another The m arg ins cal led out must be ma intained and the windin g lengths must b e ful ly ut i li zed . If the turns are not sufficient to fill out the winding length, then the wire must be spiraled to fill it out The ma rgins for a l l windings mu st be the sam e If one size wire calls for a 1/4" margin, for instance, then all windings must have 1/4" margins The windings, when layer - wound, must be even and no cross-overs a ll owed . In bob bin windin gs, they cann ot be in perf ect layers, bu t they should be wound as evenly as possible. In general, good, high quality workmanship is essential for an audio transformer to meet the design goa ls . No m atte r how good the design is, sl oppy con structio n techni ques can result in a fai le d transf orm er. These methods should be called out in the construction specifcations 5.3
Calculating the physical parameters
The fol lowing design exam ple is for a si mp le power transformer. This is used in orde r to conse rve space It will d em onstrate th e principl es used for calcu lations of th e turns, winding fil l and DC resist ances o f the magnet wire the same as for an aud o transformer It is desir ed to design a transforme r to operate from a 1 1 5 volt line at 60 Hert z an d to del iver 6 . 3 vots at 1 0 am pe res AC. The ph ysica l size is not g iven
page 0
Writ e down al inform ation known
5 v
.3
60 HZ
10 A
Schematic diagram
Ep = 11 5 V 60 Hz F Es 6.3 V s = 1 0 A = =
Calculate the total VA: VA
=
Es x Is = 6 3 x 1 0
=
63
Caculate the primary current: p = (VA x 1 . 1 1 ) Ep = (63 x 1 1 1 ) 1 1 5 = 0 . 060 8 A Choose a core from the Lamination Table in the Appendix From the VA column, it is see n tha t E- 1 1/8" size wit a 1 1/8 stack hei gh t has a V A rating of 65 Thi s shou ld be a good core for th is tran sform er.
page 21
The manufacturers will gve the core losses at flux densities in kilo-gauss This can be converted to lines by multiplying by 6.45. For example, 15
KG
or 1 50 00 gau ss x 6. 45 = 96750 line s
The conversion can be done directly in the formula for primary turns by using the gauss number and adding the 6.45 factor below the lne The window of the l am ination i s 9/16" x 1 1 1/1 6" . area i s 1 164 square inch
The effective core
We wi ll choose t o try 29M 6 g rade la mination s with a flux densi ty of 95 000 l ines as a star ting point Thi s is 1 4. 72 KG. Calculate the primary turns 11 5 x 10 N =
444 x A x F x B
=
390 T
4.44 1.164 60 x 95000
Calculate the secondary tu rns Ns
Np / Ep x 1 .0 5 x E s = 390 / 11 5 x 1. 05 x 63 = 22. 4 turns
Change the turns to an even number or 22 turns Choose the wre sizes The primary wire should be .608 x 800 circular mils = 486.4.
For this we
wil l see that #23 wi re is the closes t with 5 09 5 cm See Wi re Table in the Appendix The secondary w re should be 1 0 x 80 0 cm 800 0. # 1 1 wire has 8234 cm and w il l be used It should be noted that thi s is co nservative and in practice the re is room f or adjustment up or down , if needed The only limiting factors will be temperature rise and reguation
page 22
Cal culate th e turns per la yer and number of ayer s The window length is 1 1 1/1 6" long n order to fit, the c oi l length should be 1/1 6" sho rter or 1 5 /8" long From the wire tab e, it is s een tha t the margin for #23 wi re should be 1/8" on each end The margin for # 1 1 wire sh oul d be 1/4" on ea ch end The turns per la ye r is determi ned by the winding length x the t urns per inch for tha t wi re size This is al so obtained fr om th e wi re table The va lues sh oul d be put down on the wor k sheet clearly to show the construction of the coil With a coil length of 1 5/8", the winding length for #23 wire will be 1 3/8" and a margin of 1/8" on each end he turns per layer wil be: 1 3/8" x 374 (turns per inch from tabe) = 52 turns Layers 390 / 52 = 7 . 5 layer s Use 8 . #11 wire winding length = 1/8", margins 1/4" each end Turns per laye r 1 1/8" x 1 0. 2 = 1 1 turns Layers = 22 1 1 = 2 It should be noted that for a power transformer, the margins of the wind ing s do not h ave to be the same for al l win d ings Figur e 5 sh ows the cros s secti on of a coi l with 4 la yers The spac e between the layers is, for practica purposes, the layer insulation and the im pregnating com pound The mean length turn is the distan ce a round the winding at the c enter See Figure 6. n Figure 6, the mean e ngth turn would be between layers 2 and 3.
Figure
5
Figure 6
page 23
Calculate the f il l of the window This is don e by add ing up al l the va riou s thicknesses of winding tube, wire diameter, layer insulation and wrappers. The layer insulation is determined by the thickness needed to suppor t tha t particula r wi re size This is c a ll ed out in the Wi re Table in the Appe ndix. The winding tube thickness is determined by what is needed to support the coi l . Sm al l cois with fi ne wi e need ess support than la rger coi s with heavy wi re his can var y from . 020 t o 070 or more. A coi fo the size used in this example wi generally use a winding tube thickness of . 030" to 040" he wrap per is th e insulation used between wind ings . Thi s is det ermined by the voltage isolation needed and the support needed for the next wind ing For thi s exam ple w e wil l use 0 0" thick in sulation , as this is the value n eede d to support the # 1 1 w ire and, since there are no unusual ly high voltages involved, it will be used. It is now necessary to put down on paper the various thicknesses, add them up and calculate the percentage of availabe space in the window that is needed This lam ina tio n h as a wind ow wid th of 9/16" or, in d ecima s, 5625" Figure 7 shows the sze of the lamaton (E-112) 1" 3 DIA.
HOLES
l
•
S
I
- f
�C
Figure 7
page 24
The f il l can n ow be calcula ted : Winding tube 8 layers #23 Laye r ins Wrapper 2 layers #11 Layer ins
= 0400 1920 = 0210 = 0100 = .1858 = .0100
Wrapper
= .0100
=
4688 . 5625 x 1 0 0 = 8 3 3°/
Total fill
This is an acceptable fill. The voltage d rops and resi stan ces can now be ca lcul ated In order to obtain the voltage drops in each winding, it is first necessary to picture the buld up of the coil as calculate d ab ove i n the fil l . This b uild -up is accomp lished in the f ol lowing or der: 1
Winding tube
2 3 4 5
Pri ma ry wi e sep arat ed by the laye r insulation Wrapper between windings Secondary wire separated by the layer insulation Fin a l ly, t he out side wra pper
The mean length turn can be determined by taking the build-up and adding up the va rious se ctio ns. Figure 8 s h ows a view o f th e tube up on wh ic h the w ire is wound .
. 040"
1 1/ 8"
040
1 1/8,,
Figure 8 page 25
n order to simplify the calculations it is advisable to reduce the winding tube to a squa re if it is not a l ready one Th is is done by ta king the t ota l distance a round and dividing it by 4. Thi s wi l l gi ve an equiva lent dimension of one side only. For example, a winding tube that is 1 1/2" x 1 3/4 would be: 1 1/2" x 2 + 1 3/4 x 2 square.
3 + 3.5 = 65 4
=
1625 equivalent
n the example used the winding tube is already a square so the 1 1/8" dimension will be the starting point. Starting with the size of the lamination and adding the winding tube thickness to each side, the actual dimension of the winding form will be obtained. The wi re and insula tion is ad ded on top of th is . Lamination
= . 12 50
Tube x 2wir e 8 #23 Insulation
=
Total
=
(1 1/8 ")
0800 1920 0210 14180
Thi s gi ves the build-up in one di rection of the p rim a ry wind ing . When th s number is mutiplied by 4, it will give the length of one turn in the center of the winding, or the mean length t urn of the pri ma ry wire Thus, 3 . 75 Ohm s. The 1 .696 6 is the 1.4180 x 4 390 x 1.6966 1000 resistance of this size wire per 1000 inches.
t should be noted here that this mean length turn value is used in the calculation o f the winding lea kage inductance of an au dio transf ormer The value, 14180, is the buid-up to the center of the primary winding, so the primary values must be added in again to get to the start of the secondary winding . The enti re buil d-up i s now repeate d to clearly show the calculations.
page 26
Lam. Tube 8 - #23 Insul
=
=
11250 .0800 .1920 0210 1 .41 80 x 4 x 390 x 1 .69 66 1 00 0
=
3 . 75 x .608
=
2.28
v
=
8 - #23 Insul Wrap 2 #11 Insul
=
=
1920 .0210 .0100 .1858 .0100 1 .8368 x 4 x 22 x . 1 05 0 1 00 0
=
.0169 x 10
=
.169
v
These values can be used to determine the output voltage under loaded conditions This i s done by subtracting the p rim a ry voltage d rop from the input voltage and, from the turns ratio, obtain the secondary voltage. The secondary voltage drop is then subtracted from this value to obtain the loaded voltage. 1 12.72 115 - 2.28 =
v
This is the effective input votage. rom the t urns ratio, 1 12 . 72 / 390 x 22
=
6358 V
Subtracting the se conda ry voltag e drop, 6. 35 8 . 169
=
6. 189 V
This is lower tha n the 6. 3 V desire d so adjustmen ts must be ma de. This can be done by adjusting either the primary turns down or the secondary turns upis to Thshow is w i how ll nottobe carrie dthe anyfif urthe r aresistances s the purpoofsethe of thi s example caculate and the windings. The weight of the wire can be obtained by using the DC resistances. Referring to the wi re table, the weights are given in Oh ms per pound. For example, #23 wire is 12.88 Ohms per pound Then the weight is 3 75 / 12 .88
=
.291 pounds.
page 27
As indicated previously, ths example was used only to show the methods of calculating the winding layers, turns per layer, fill and DC resistance. It can also be used as a guide for designing power transformers. The complete design and construction of power transformer and inductors can be found in the TRANSFORMER DESIGN AND MANUFACTURING MANUAL p ubli shed by the auth or i n 1984. The methods described above will be used in the folowing examples of audio transformer s.
page 28
PART II. DESIGN EXAMPLES
page 29
PART II . DESI GN EXAMPLES
In the a ctua l design of an aud io tra nsforme r it s nec essa ry to co nsider al l of the requirem ents one at a ti m e However, m any o f these requiremen ts interact and this will affect the results so they must all be kept in mind while doin g the design. The exam ples th at fol low wil l cover a broad range of au d io tr ansformers. Some of them wi ll be relativey ea sy to design and so me wi ll be more difficult and time consuming. An attempt will be made to go through each design, step-by-step, explaining the thi nking as the design progress es The calculations of the various parameters wil be given and then the actual tran sfo rmer wil l be buil t and tested The results wi l be com pare d with the calculated values In or de r to demonstrat e the results of i nterleavng , a desig n for a 6 00 watt transformer has been made and built for a 4 Ohm to 200 volt line transformer The com plet e design wi ll not be shown as the f olowing designs wil thorough ly de monst rate th e design and construction of enough types of audio transformers to sufice. The lamination size is El- 2 1/8" wth a 2 1/8" st ac k. The fi rst design was not int erleaved It was bui lt with the p rim ary first and then the secondary The h ig h frequ ency response was c a culated to be down 3 The measured response was down 3
DB
DB
at 9600 Hz.
at 1 1000 Hz.
The u nit wa s then redesigned and b uil t with a 2: 1 interle aving . The secondary was split in half and the primary was put in the center. The calculated 3 DB down-point was 26400 Hz and the measured frequency was down 3 DB at 30000 Hz. The low frequency response was not afected. The following designs have been chosen to demonstrate the three most representative types of audio transformers.
page 30
EXAMPLE 1: VOICE FREQUENCY TELEPHONE TRANSFORMER This example will be for an audio transformer used in a telephone circuit that has a voice frequency response requirement 5 60 OHMS
600 S
Crcuit Dagram
The pola rity dot s indicate the instantaneous pol a rty o the windin gs. This is important in many audio transformers he specifications call for the folowing Impedances are 600 Ohms to 600 Ohms Frequency response is ± 1 0 DB from 300 Hz to 35 00 H z Insertion os s is 1 0 D B max imum Longitud inal ba lanc e is 60 DB m inimum f rom 2 0 Hz to 1 00 0 Hz an d 40 DB at 400 0 Hz Operating level is + 10 DBM H D at 300 Hz = 0 5° maxim um Prima ry DC c urrent is . 090 a m peres Physica l size is g iven as EI- 3/8" l am ination with a squa re stac k to be wound on a printed circuit bobbn Primary DC resistance 5 0 Ohms max imum Secondary DC resistance = 65 Ohms maximum From page 7, the inductance needed is: z
L=
F
© 1989, ROBERT G. WOLPERT, Rev.
2004
=
600 x 300
= 636 HY
page 31
The design for turns on the primary is done the same as for an inductance th at ca rries di rect cu rrent Any m eth od for obtai ning the proper result is s atisfact ory In this case Hannas curves were used to calculate the turns and air gap needed to obtain the proper inductance. This tran sfor mer ca n be constructed using et her 29 M6, 50°/ nckel or 80 °/ nicke l. 29M6 is th e preerred choice, i it wil resu t in th e proper inductance and the wire sizes wil meet the specification for resistance, because it is the least expensive From Hannas curves 29M6 will be chosen to start the turns calculations. for 29M6 core ma terial, th e tur ns wi ll be 1 0 0 0 and the g a p spacer needed is 005. Since the gap spacer is put across both legs and the center E, the spacer is di vided by 2 for a thi ckne ss of 0 025 ". A spacer of 00 3 w il l be us ed to sta rt as this is a stand ard th ickn ess of ins u lating pa per. have to be adjusted when the unit is tested.
This v a lu e may
The fux densit y in the co re mu st be checked . + 1 0 DBM ( 1 0 DB n 60 0 Ohms ) i s the power level . By checking the DB Expr essed in W atts Ta ble in the Appe ndix , the power level is seen to b e 0 1 watts The volt age is : E
x 600
=
=
2.45 V
The flux dens ity i s ca lcuate d : B
=
245 x 10 444 x .1336
=
1376 lines
213 gauss
x 300 x 1000
This is a low f ux density In general , when using H anna's curv es, it is not necessary to calculate the flux density as they are designed to keep it within the proper ra nge The curves in the Total Harmonc Dstorton Table in the Appendix show the e xpected THD fo r th is m ateria l and f l ux densty. They sh ow that the THD wi ll be less than . 03 °/ . This is low er than the re qu i red 0 . 5°/
page 32
The current s 01 watts divided by 2.45 V, which is .004 amperes The 090 ADC n the primary and the required resistance wll determine the wre s zes to be used : .09 x 0. 7 = 063 ( 0 7 is 700 C M/ A) From the Wire Table in the Appendx, #32 AWG can be used, however, the customer has called out the DC resistance and to meet ths requirement, the EI375 Laminaton turns table shows that #33 AWG will fit and sh ou ld m eet th e resstance requ irement #34 AWG w ll b e used for the secondary. It can be seen from these calculations that 29M6 material wil meet the inductance and OCR requirements. The frequency response wil not be dfficut to meet, but the longitudinal bal ance mu st be consi de red This wil l cal f or sp l ittin g the prim ary in two parts as a 2: 1 nterleav e Thi s wi ll be wound on a b obb in as r equ ired by the customer. The followng calculatons are explained in Section 53
the
The fil wl l not be calc ul ated : The window of th e a mination is 5/16 x 3/4" The winding en gth of the bob bin is .673 " (from ma nufacturer 's cata log ) #33 wire turns per layer = 75 layers
=
500 / 75 = 666 (use 7)
#34 wire turns per layer = 84 layers = 1000 / 84 Bobbin 7- #33 Wrap 12 #3 4 Wrap 7- #33 Wrap
=
11. 9 (use 12)
= 0300 = 0553 = .0060 .0840 = 0060 = .0553 = .0060 =
.2426 3 125 = 776 x 10 0
=
78° fll
page 33
Ca lc ula ting the DC resistances . .3750 .0600 .0553 4903 x 4 x 500 / 1000 x 17.2416 0553 0060 0840 6356 x 4 x 10 00 / 10 00 x 2 1 . 7416 0840 .0060 .0553 .7809 x 4 x 500 I 10 00 x 1 7.2416
16.9 Ohms
=
55.27 Ohms
26.92 Ohms
This is 16.9 + 26.92 43 .82 Oh ms t ota l for th e prm ary, 55. 27 Oh ms for the seconda ry . These a re in conform ity with th e requ irement s =
The custo mer ca led for a 1: 1 t urns ratio so the tu rns cannot be adj usted. The insertion loss is calculated using the formula from page 17: 2
Since the turns are equal, the first value becomes 1
page 34
Then: R = R2 + Ri = 55.27 + 4382 IL =
RT
=
9909
=
99.09
X 10 0 = 165°/
600
z
f thi s is int erpol ated it wi l l be 75 DB for 15 °/ , so it is a pproximately 82 DB for 16 5 °/ . The requi remen t of 1 . 0 DB wi ll be met See page 1 7. The next requ irement to be consid ered i s the long itud inal ba lance. This is a small transformer with impedances of 600 Ohms for both windings. The size and the low impedance makes it easier to meet the requirements. There is no easy way to calculate the ongitudina balance so past experience mu st be ca l led on A 2: 1 interleave ha s been chosen The high frequency response is only 3500 Hz so that should be no problem and can be met without interleaving but the balance will require that voltage gradients be considered. The 2: 1 interleave wi l sp li t one winding so th at the voltag e g rad ient on each side of the center winding is small
.s ov
ov
I V
I
f the start of the primary winding is 0 volt and the finish is 1 volt and the secondary will be the same since they have the same number of turns then the center of the primary will be 0.5 volt to 0 volts on the start of the secondary and aso 05 volt to 1 volt to the finish of the secondary. This wil result in a difference of 0.5 volt from the primary to both the start and finish of the se cond a ry. This wi ll provide a fai rly equal voltage g rad ient Of cou rse th ere a re other paths th at can up set a perfect balance for example from windings to core and the dressing of the leads page 35
Another way to increase the balance is to put shields in between the windings and co nnect these to g round . A furt her inc rease can be accompished by putting in additional shields and using box shields that compl etey enclose both wi nd ings . These metho ds are used in i nstrumen t transformers, where maximum isolation is necessary. This interleaving wil result in a frequency response much higher than requi red for thi s transf orm er The fr eque ncy response can now be ca lcuated The leakage inductance is calculated using the formula from page 13 Where: N MLT s
T H WL
-
= =
= -
1000 2.54" 2 2692" 006 673
Assi gnin g th e proper val ues:
LL -
10.6 x 1000 x 2.54 x (2 x 2 x .006 + 69)
= .00293
2 9 2 x 673 x 10
page 36
The high freq uency li mit is calc ula ted :
zT
_
[
2 x 600
+ 600 = 1 2 0 0
=
65215 H Z
2 x .00293
The transformer manufacturing specifications can now be written up and the unit built and tested. The frequency response test results were plotted on the curve shown on page 43 These were ru n without DC on the p rim ary and no air gap an d with D C and the nece ssar y ai r ga p. They comp are f avor aby with the calculated results. The inser tion l oss me as urements were tested to be 0 . 78 DB. The calcua ted val ue was a ppr oxi mately 0 .8 2 DB. The total ha rm onic di stortio n from the curves i s 0 0 3 °/ The mea sured distor tion was ap proximately 0 . 03°/ . This com pared t o the re quiremen t of 0 5°/ maximum The longitudinal balance test was measured as - 60 DB requirement was for a minimum of
- 75 DB at 300
Hz.
The
This same transformer was constructed wth shields between the primary ha lves and th e seconda ry t o show th e i mprovement tha t can be ob tained This resulte d in a me asu rem ent of - 86 DB, an increase of 11 DB
page 37
ROBERT G. WOLPERT
TRANSFORMER DESIGN SERVICES WINDNG SHEET
PAGE
OF
-
PAGES
SPEC NO. EGINEER.
DATE
5/16" x 3/4"
WINDOW
78 /o
COIL BUILD
NET GROS S
3/8" X 3/8" BOBBIN
TUBE (NET GROSS) OVE R TUBE
218 GAUSS
DENSITY FREQUENCY
300 HZ
AREA
0129 +10 DBM
AT
VOLTS
TERMINALS
1
4 -
r .673"COI 1
A A 2
3
A
B
CONN EC A'S OGE THE R AND BLIND WIDIG WIRE SIZE
#33
#34
#33
TOTAL TURS
500
1000
500 673
TAPS
673
.673
MARGI
BO BB IN WI ND
NO MA RG IN S
TURNS PE R LAYER
RANDOM WIND IN
LA YER AS CLOSE
WIDING LEGTH
AS POSSIBLE
% IL NO OF YERS LAYER ISULATIO WRAPPER
2L MYAR
TERM COI
1A
TAPE 3-4
A2
START AT
page 8
ROBERT G. WOLPERT TRANSFORMER DESIGN SERVICES MATERIAL S HE ET PAGE
2
OF PAGES
SPEC NO.
CORE
PART NO. EI- 3 / 8 " 29M6
COPPER #33 MAGNET WIRE 3 MAGNET WIRE
AM .110#
O PRICE
TO PRICE
O PRICE
.03 .026#
CAN LID- LID-B
ERMINALS BOBBIN
x
1
3/8" x 3/8"
1
TERM BOARD LUG PANEL BK
LEADS
HORIZ. FRAME #22 SLW x
7"
LONG
#1
BLACK
1
#2 #3 #4
BROWN RED GREEN
1 1 1
NOTES:
page 39
ROBERT G. WOLPERT
TRANSFORMER DESIGN SERVICES FNSHING 3
PAGE
OF _ PAGES
SPEC NO.
COLOR
LEADS SIZE
LENGTH OUT OF COIL
LEAD#
#22 SLW
BLACK
6"
1
#22 SLW
BROWN
6"
2
#22 SLW
RED
6"
3
#22 SLW
GREEN
6"
LUGS OR LUG P AN EL PART#
LEAD#
SPEC IAL I NSTRUCT IO NS:
page 40
ROBERT G. WOLPERT
TRANSFORMER DESIGN SERVICES STACKING & ASSEMBLY PAGE .
4
OF
PAGES
TELEPHONE
AM NATON EI- 3/8" 29M6
SIZE GRADE STACK H EGH
3/8"
NTERE AVE
1x1
KEEPERS CUT OFF E'S GAP SPACER
.003K
BRUSERS SZE SHELD U NSU LATORS SZE BRACKETS -
Q
1
3/8" x 3/ 8" HORIZON TAL FRAME
HARDWARE Q: Q Q TO BE REMOVED
NO
SPEC AL NSTRUC TON S BU STACK WITH .003" GAP SPACER VACUUM VARNISH AFTER ADJUS TING FOR PROPER INDUCT ANCE - LEADS O UT BO TTOM
page 41
ROBERT G WOLPERT
TRANSFORMER DESIGN SERVICES TEST INSTRUCTIONS PAGE
OF
5
PAGES
SPEC NO. PROCEDURE
2,
lST TEST 2ND ES
s, 2, ,
3RD TEST {AFER VARNISH) FNAL ES
1.
NO LOAD VOLTAGE RATIO APPLY
HZ TO TERM
v
MAX.
lex
V T ERM
READ
V TERM. V TERM
2.
IN DUC ANC E TEST APPY
10
READ "L
3
1000
v
.636 H Y
HZ TO TERM .
1-2
&
0090
ADC
MIN
INDUCED VOLTAGE TEST APPY
v
MUST MEG
5
H I POT
HZ TO TERM
FOR
MEGOHMS MIN
VOLTS D.C
LEAD NO.
O
VOTS
1
3
100
CORE CASE
100
3
6
CONTINUIY
7
SPECIAL TESTS
SEC
R 600 ohms RL = 600 oh ms Frequency response =
page 42
EXAMPLE 2: AUDIO OUTPUT TRANSFORMER
Audio amplifers using vacuum tubes are again in favor after a period where transistors were used exclusively. Vacuum tubes have a high plate-to-plate impedance while transistor mpedances a re usua ly qu ite low This ma kes the design of tra nsformer s for use wi th vacuu m tu bes muc h mo re d ifficul t. The im peda nces can be 10000 ohms, platetoplate and above and are therefore a more difficut design in order to obtain the high frequency response The output transf orm er chosen for this exam pl e s a 1 0 0 wat t, sine wave pow er, unit with a prim ary plate-to-p late im peda nce of 12 50 oh ms . The secondary must del iver pow er to 16, 8, and 4 oh m im pedances The f requency re sponse is to be ±1 DB from 20 Hz to 2000 0 H z Th e prim ary is to h ave s creen taps for ultr a l in ea r o pera tio n. These taps a re plate at thevoltage 60°/ is point o f th p rim a ry from the cente r ta p . Th at is, 60 °/ of the applied toethe screens. See the circuit diagram
PLATE 16 OHM
SCREEN
8 OHM 4 OH
R
COM.
PT
Cicuit Diagram
page 44
First it is necessary to choose a core size, if it is not called out by the cus tomer As e xp a ine d in page 1 1, the co re size wil l be a bout the s ame as for a 3 0 0 watt, 60 z t ransformer. Since the windings must b e interleaved, it wi ll nee d to have a sl ig htly la rge r core An EI 1 75 lamination with a square stack will be tried Area 1. 75 x 1 7 5 x 92
2.81
The inductance needed will be L -
1250
= 19.89 HY
x 20
We will desi gn for 20 HY. Refering to a manufacturer's catalog, the formula for inductance for this la mina tion is : L
=
9283 x 1 0
-8
x
K
x U x N
2
Then using a permeability of 4000 for 29M6 material:
20 9283 x 10
-8
765
x . 92 x 40 00
A permeabi lity of 40 00 is about right for 29M6 la minations. volt age for 1 0 0 wa tts , 125 0 oh ms wil be :
The prim ary
2
R E
100 x 120
33 V
page 45
The tur ns for a m axi m um f lux den sity o f 1 7 KG at 20 Hz wi ll be :
N
353 x 10
=
= 1290 T
4.44 x 281 x 20 x 17000 x 645
1300 turns wil be used In this case the turns needed for the inductance are less than those necessary for the flux density so the 1300 turns will have to be used The B+ votag e wil l be ap pl ied to the cent er of the win ding 60 ° of the votage will be at 1300 / 2 = 650 x 60 = 390 turns from the center tap This wil l be 650 - 39 0 = 260 tur ns from the p late en d of th e windings. Secondary voltages wil be 16 OH M E =
8 OHM
E
4 OHM E
10 0 x 1 6
100 x 8
=
100 x 4
40 V
2828 V
20 V
The secondary turns will be: 1300 353
x 1 0 5 x 40 = 15 5 T x 2828 = 1 10 T x 20 = 78 T
The 1 05 fact or is to com pen sate for th e osses.
page 46
Prim ary curre nt :
I =
w E
Ip =
100
- .283 A
353 Secondary currents
16 OHM -
8 OHM =
4 OHM -
100 40
= 2 .5 A
100 = 353 A 28 . 78
100 20
=
5. 0 A
The config uration m ust be ch osen. Th is is for th e most p a rt, a guessing ga me, using past experience. A 4 to 5 interleave wi ll be tied for th is design since the primary can natually be divided into 4 sections in series and the se conda ry wi ll be wound in 5 se ctio ns that wil be put in paral lel . The lamination size, the number of turns and the currents ae known and the configuration has been decided upon, the wire sizes can be chosen The primary current is 283 amperes Using 750 CM/A, 283 x .750 will be chosen
=
212, so #27 wire with 201 circular mils
page 47
The second a ry is d ivid ed into 5 part s The l argest cu rrent for any one wnding wil be for the 4 ohm winding, 5 amperes dvided by 5 or 1 A per section Then, 1 x 75 0 = . 75 0 So #21 wire with 81 0 cir cul ary mls wil be used .
The fol owi ng cir cuit diagr am wi ll s how how the win d ing wi l l be done . The circled numbers indicate the winding sequence
Wnding Dagram The lead numbers shown wil be connected so that lke numbered leads are c onnected together and brough t out as o ne lea d For the pr im ary leads nu m bered 1 and 5 wi ll be the plates Leads numbered 2 and 4 will be the screen taps. Leads numbered 3 wil be the center ta p for the B+ voltag e.
page 48
For the secondary, leads numbered 6 will be for the 16 ohm winding. Leads nu m bered 7 wil l be for the 8 oh m winding Lead s nu m bered 8 will be for the 4 ohm winding and the number 9 leads will be the common point lead The fil l can now be calcu lated The windo w for the EI 1 75 lamination is The coi l len gth wi l be 1/16" less than the 7/8" wide by 2 5/8" long win dow length or 2 9/16 " Using the methods shown
in the e xam ple in Sec tion 5 . 3 :
The fill wil be calculated using the winding oder as shown in the circuit diagram on page 48
page 49
The total tu rns for the secon da ry are 1 5 5 . This wi l l be on each se condary winding as they are t o be put i n pa ral l e l The tur ns for the pri ma ry are spl it as s hown on page 47 . The cir cuit diag ra m sh ows h ow this is done. The winding length is 2 9/16". 1 2 3 4 5 6 7 8 9
#21 wire turns per layer = 52 #27 wire turns per ayer = 130 #2 1 wi re tu rns per la ye r = 52 #27 wire turns per laye = 130 #21 wire tur ns per laye 52 #27 wire turns per laye = 130 #2 1 wi re tu rns per la ye r = 52 #27 wire turns per layer = 130 #21 wire turns per layer = 52
Layer s = 1 5 5 / 52 Layers = 260 / 13 0 Layers = 1 5 5 / 52 Laye rs = 3 9 0 / 1 30 Layer s = 1 55 / 52 Laye rs = 3 9 0 / 13 0 Layer s = 1 5 5 / 52 Layers 260 / 1 3 0 Layers = 1 5 5 / 52
= 3 =2 = 3 3 = 3 =3 = 3 =2 = 3
The winding tube will be made up of .040" thick materia. Winding tube 3 layers #21 Layer ins. Wrapper 2 layers #27 Layer ins. Wrapper 3 ayers #21 Layer ns. Wrapper 3 ayers #27 Layer ns. Wrapper 3 layers #21 Laye r ins . Wrapper
= = = = = = =
.0400 .0903 . 0 100 (2 layer s o .0 05 " Kra paper) .0050 .0308 .0020 .0050 .0903 .0100 .0050 .0462 .0040 .0050 .0903 .0100 .0050
ins#27 . 3Layer ayers Layer ns. Wrapper 3 ayers #21 Layer in s. Wrapper 2 layers #27 Layer ins Wrapper 3 layers #21 Layer in s. Wrapper Total
= =
.0100 .0462 .0040 .0050 .0903 .0100 .0050 .0308 .0020 .0050 .0903 .0100 .0100 . 7675 I .875 0 x 100
86. 5°/ fi ll
page 50
This fil l is h ig h er tha n the 8 5° idea l , but it wi l fit wit h a good winding job An aud io tr ansformer that is layer wou nd m ust be wo und care ful ly. The margins at the ends of the windings must be maintained even if it is This is so that the windings are di rectly necessary to spiral the windings. above on another a nd no t stag gered Stag gering wi ll greatly incr ease the leaka ge ind uct ance and th row of f the ca lcu lations . For this transformer the #21 wire at 52 turns per layer on the secondaries wi ll take 52 x . 0 30 1 = 1 565" o f winding space . Then 1 565 / 2 . 3 12 5 = 67° fi l l . So the s econda ry windings will have to be spiraled, however each secondary will have 2 taps brought out and that wil l ta ke u p some space. A l ittl e experiment ation on the fi rst winding will result in a properly filed winding space The prima ry windings are 1 30 tun s per laye of #27 wire. 130 x .0 1 54 = 2 0 02" I 2 3 125
=
86 . 5° fi ll
Thi� is about right and thee should be no problem in holding the margins. The DC resistance of the windings can now be calculated example on page 27
See the
1 . 7500 0800 .0903 .0100 1 .930 3 x 4 x 1 55 x 1 .06 66 / 1 00 0 = 1 27 6 oh ms 0100 0903 0050 .0308 .0020 2.0684
page 51
DC resistance calculations, continued 2 0684 x 4 x 260 x 42 891 / 1 0 00
=
9226 ohms
0020 0308 0050 0903 0100 22065 x 4 x 1 5 5 x 1 .066 6 I 10 00 = 1 459 ohm s 0100 0903 0050 .0462 .0040 2 36 20 x 4 x 39 0 x 42891 / 1000
= 15 .8 0 ohms
.0040 0462 0050 0903 .0100 2 5 175 x 4 x 15 5 x 1 .066 6 / 1 00 0 = 1 .664 oh ms 0100 .0903 0050 0462 0040 267 30 x 4 x 390 x 42891 / 10 00
=
1 788 oh ms
0040 0462 0050 .0903 .0100 28285
page 52
DC resistance calculations, continued 2.8285 x 4 x 1 5 5 x 1 .06 66 10 00 = 1 87 ohm s .0100 0903 0050 0308 0020 2.9666 x 4 x 260 x 42891 1000
=
13 .23 ohms
.0020 .0308 .0050 .0903 .0100 3.1047 x 4 x 155 x 10666 / 1000
=
2053 ohms
The total primary resistance is: 9226 + 15.80 + 1788 + 1323
=
56 . 13 ohms
The volt age dr op in the pr ima y is 56 14 x .2 83 = 1 5 88 V. The s econda ies a re a ll in par al lel so that the su Rs
m of the windings will
1 =
1 1 .27 6
1 1 459
1 1 .6 64
1 1. 87
1 20 53
be
= . 32 3 OHMS
This will be fo r th e 16 o hm windi ng Calcul ate the output voltage
:
(353 15.88) 1300 x 155 = 4 0.195
- .809 = 39.38
v
In order to increase the output to the desied 40 volts, the secondary turns shoul d be in crease d to 1 57 Then the outp ut wi ll be 39 9 V
page 53
The other taps can be calculated rapidly by taking a percentage of the total winding. For the 8 ohm wnding it wil l be 1 10 I 1 57 x 323 226 ohms. Then .226 x 3 53 = 798 V d rop (353 15 8 8) / 1 30 0 x 110 = 28 5 2 .7 98 2772 v Adjus ting for 28 .28 V = 1 12 tu rns =
=
For the 4 oh m windin g it w il l be 78 / 1 57 x 323 = . 160 ohm s Then, 160 x 5 .802 V d rop. (353 - 15 .88) I 13 00 x 78 = 2022 802 = 19 42 v . Adjusting for 20 V 8 0 turns. =
=
These adjustments will result in a small change in the DC resistances, but it won't be necessar y to go back and m a ke a l l the ne w calc uations Calculate the leakage inductance usng the formula from page 13 1300
N MLT WL
-
s T
H
L
=
2 5 1 75 x 4 = 1 0 07 23125 4 (nterleaves) 005 7675
(from the center winding )
1 0.6 x 13 00 x 1 0. 07 x x 4 x 005 4 x 2.3125 x 10
+
.
=
00393 HY
For the hi gh frequency resp onse : 2 1300 157
7
x 16 + 1250 = 2347
2347 x 00393
= 95095
page 54
This caculates to be 0. K. The design is now ready to write up the manufacturing specifications and bui ld and t est the unit The graph on page 61 shows the frequency response test results
page 55
ROBERT G. WOLPERT
TRANSFORMER DESGN SERVICES WINDING SH EE T
S P E C NO ENGINEER.
TO
TPE
100
7 KG H Z IN VOLTS
353
r29/16'�
_s
8
4
7 x 5 6 /o NET GROSS 3 x 3 x
WINDOW COIL BUILD TUBE (NET GROSS) OVER TUBE DENSITY FREQUENCY AREA AT
6
DATE �
RGW
-
1- 1
COIL 1
4 2 2
-----
1
-----
3 3
CONNECT LI KE NU MB ERED LEADS TOGETH ER; WHEN WINDING # 2 WIRE, HOLD TO /8 " MARGINS EVEN IF YOU HAV E TO SPIRAL WIND HOLD ALL MARGINS T O /8 " WINDNG WIRE SIZE TO TA L TU RN S TAPS WINDING LENGTH MARGN T RN S E R YER % FL NO . O F YERS YER INSUTON WRAPPER TE RM CO STAR AT
#21 157
#27 260
#21 157
2 5/16"
2 5/16"
2 5/16"
#27 390
130 53 53 70 87°o 70°o 3 2 3 .005K 002K .005K .005K 005K 005K 6, 7, 1-2 l -
l -
l -
l -
6, 7,
2 5/16 "
#21 157
#27 390
#21 157
#27 260
#21 157
2 5/16"
2 5/16"
2 5/16"
2 5/16
2 5/16"
130 130 53 130 87% 70% 87°o 70°o 87 70% 3 3 3 3 2 3 002K 005K 002K 005K .002K .005K .005K 005K 005K 005K .005K .005K 2-3 6, 7, 9 4-5 9 53
53
l -
l -
6, 7,
3-4
l -
l
l
6, 7,
page 56
ROBERT G. WOLPERT TRANSFORMER DESIGN SERVICES
MATERIAL SH EET PAGE
2
PAGES
OF
SPEC NO.
PAR NO.
AM
O PRCE
TO PRICE
O PRICE
EI- 1
CORE
29M6
87#
COPPER #21 MAGNET WIRE
1.63#
#27 MAGNET W IRE
0.68#
CAN LID- LIDB
ERMNALS UBE
1 3/ 4" x 1 3/4 x . 04 0
1
x 2 9/16 LONG TERM BOARD LUG PANEL BK
1
HORIZ. "L"
BOLS
4
#10
4
NUS
#1 0 STEEL
WASHERS
4
#10 x 2"
8
#10 FIBER #20 SLW x 10" ONG
4
#1
BLACK
1
#2
BROWN
1
#3
RED
1
#4
YELLOW
1
#5
GREEN
1
#6
BLUE
1
#7
ORANGE
1
#8
WHITE
1
#9
VIOLET
1
WASHERS LEADS
NOES:
page 57
ROBERT G. WOLPERT
TRANSFORMER DESIGN SERVICES FNISHNG 3
PAGE
OF - PAGES
SPEC NO.
COLOR
LEADS SZE
LENGTH OU OF COIL
LEAD# 1
#20 SLW
BLACK
8"
#20 SLW
BROWN
8"
2
#20 SLW
RED
8"
3
#20 SLW
YEL LOW
8"
4
#20 SLW
GREEN
8"
5 6
#20 SLW
BLUE
8"
#20 SLW
ORANGE
8"
7
#20 SLW
WHITE
8"
8
#20 SLW
VIOLET
8"
9
LUGS OR LUG PANE: PAR#
EAD#
SPECI AL N SRUCIO NS : FINISH ALL LIKE NUMBERED LEADS TOGETHER.
page 58
ROBERT G. WOLPERT
TRANSFORMER DESIGN SERVICES STACKING & ASSEMBLY PAGE
4
OF
PAGES
LAMINAION: EI- 1 3/4 " 29M6
SIZE: GRADE: SACK H EIGH:
1 3/4"
INERLEAVE:
lxl
KEEPERS:
2
CU OFF E'S: GAP SPACER: BRUISERS: SIZE: SHIELD: U INSULAORS: SIZE: BRACKES -
Q
4
1 3/4" HORIZONAL "L"
Q Q
4
BOS #0-32
4
NUS #10-32
HARDWARE:
Q Q
WASHER S # 10 STEEL 4
O BE REMOVED
WASHERS #10 FIBER NO
SPECIA L INSTRUC IONS : VACUUM VARNISH
page 59
ROBERT G. WOLPERT TRANSFORMER DESIGN SERVICES TEST INSTRUCTIONS PAGE
PAGES
OF
5
SPEC NO -
PROCEDURE
6
1ST TEST 2ND TEST 3RD TEST FINAL TEST (AFTER VARNISH) 1
7 -
s, 6
NO OAD VOTAGE RATO APPLY
v
HZ TO TERM
READ
lex
MAX.
V T ERM V T ERM V T ER M
2
ND UCTANC E TEST APPLY
v
HZ TO TERM
READ "L
&
ADC
MN
N DUCED VOL TAG E TES T
3
APPY
4.
MUST MEG
5
H POT
v
HZ TO T ERM
OR
MEGOHMS MIN
EAD NO. 6
TO
SEC VOTS DC.
6
VOTS 000
CORE
000
CASE
6
CONTNUTY
7
SPECAL TEST S :
s
=
5 ohm
6 ohm e i made acro the entre econday e reult on the foowng page =
page 60
> u z w O w a I
.
•
f
10
-
�. t
7
"!' -· r· • .
! !
.
.
0
0 'I
:;' :..
. �
0 N
.
.
�
.
.. ·=
.
-·
:
..
. . . ..
EXAMPLE 3: LINE-TO-VOICE COIL TRANSFORMER
Linetovoice coil transformers are widely used in the transmission of mu sic an d for pub ic ad dres s systems. A desig n for th is type tra nsformer is one of the m ost dif ficult to obtain The fol lowi ng exam pe wil l sh ow the design of a typical line-tovoice coi transformer. The r equ iremen ts fo r this un it are: 70 . 7 volt lin e to 4, 8 a nd 16 ohm s outpu t wi th pow er r atings of 8, 16 a nd 32 watts. The freque ncy respo nse i s ± D B from 30 z to 15 000 z. Inse rtion loss o f 0 5 D B maxim um .
8 OHMS
Circut Dagram
Connections to the appropriate leads will give the desired wattage and output impedance Sinc e this unit wil be about 3 DB down at 15 z (1 DB at 30 z), it will require a lamination that can support about 32 watts x 4 12 8 VA at 60 Hz. =
By divid ing 60 z by 15 z, a facto r of 4 i s obt ai ned for power requirements of the lamination for this transformer. From the Lamination Tabe in the Appendix, E l- 1 1/4" x 1 1/4" is good for 90 VA A 1 1/2" stack of this size amination shoud be about right as a starting point for 128 VA.
page 62
The primary impedances can be calculated from Ohm's law.
Z =
E
w
-
70.7 32
=
156
=
707 16 707 8
312
=
625
Then calc u la te the prim a ry ind uctan ce necessary, from page 7 For the 32 watt prima ry This windi ng is u sed as it wi ll ha ve the least nu mber of turns and, as shown later, the maximum flux density must be calculated for this wi ndi ng . z
L
156 30
7 30
= 1 .65 HY
From the ma nu facturer s cat a log fo r this l am ination : L
=
665 x 0
8
x K x UA x N
For a 1 1/2" stack it will be changed
This formula is for a square stack. by a ra tio of 1 . 5 / 1 . 25 = 2. The f ormula wil l then be: L
665 x 10
-8
x 1 2 x K x UA x N
Using a per me ab il ity of 4000 f or M6 l am ination N
665 665 x 1 . 2 x .9 2 x 4000 x
=
237
page 63
These are the turns needed for the 32 watt winding to obtain the proper inductance. The a rea of this co re i s 1 1/4 x 1 1/2 x .92 = 1 7 25 sq in A reasonable flux density is about 15 KG or 96750 lines Using this flux density and calculating the turns needed from the formula on page 19: 8 707 x 10 4 44 x 1 72 5 x 30 x 96750
N =
= 8
It can be seen that 237 turns as calculated for the inductance wil result in too high a flux density, so the 31S turns must be used for the 32 watt windin g As me ntioned be fore, this wind ing has the ow est nu mber of tu rns an d 7 0 7 volts wil be applied to all the prim ary wind ing s as th ey are used By taking 31S turns for the 32 watt winding, since the turns ratio is the square root of the impedances ratio, the turns for the other taps on the primar y will be : 16
w
x 31S = 450
SW=
The turns for the sec on da ries can now be calc ula ted By ta kn g the 31 S tur ns for 3 2 watts , t he tu rns for the secondaries wil be :
4 H M =
S O HM =
1 6 HM =
x 31 S = 51
x 31 S = 72
x 31S = 02
page 64
he configuration as shown later will result in a change of turns in both the prima ry an d sec ondary. The configuration must now be considered. t is advisable to split the prima ry so that a 2 : 1 interleave is achi eved . By splitting the primary in the center of the low impedance winding (32 watts) and adjusting the winding conf igu ration of the p rim ary and sec onda ry the mi ni mu m volt age gradient can be obt ai ned . If the 31 8 turns c a n be sp lit into 15 9 + 1 59 an d then pu t the 4 watt wind ing in be tween The 4 wat t wi nd in g is 5 2 turn s his is spl it into 26 + 2 6. So we now have 159 T primary 26 T secondary primary
26 secon dar y ; 159
We need 318 turns more on the primary and 52 more turns on the secondary In the o rdeadditional r to prese rve the vo ltage nts anand d split the windings evenly, primary turns shouldgradie be divided added a t each en d. he addition al s econ dary tu rns sh ou ld be put in the center of the secondary. he n the win din g conf igu ration wil l be 1 59 + 15 9 ; 26 T + 2 6 + 2 6 + 2 6 ; 1 59 + 1 59 T t rema ins to
his will give the total primary turns and secondary turns. number these windings so they will be connected properly.
159 + 159;
26
2 -3
7-B
3-A
+ 26 + 26 +
5-6
6 -7
26; 159 B-8
A4
+
159 12
©0
page 65
This cir cuit diag ra m wi show how the win din gs are int erle aved circed numbers in the diagram show the order in which the windings will be wound .
The
Circuit Diagram
The primary will be 3-A, A-4 = 318 turns for 3 2 watts 2-3-A-4 = 477 turns for 1 6 watts 23A4
=
636 turns for 8 watts
be wound as one winding of Since 23 3A are together they can turns tapped at 59 turns an d not be se para te windin gs.
38
The secondary will be: 56 + 67 +7B + B8 67 + 7B + B-8 7B + B8
=
=
04 turns for 6 ohms
78 tu rns for 8 ohms
= 5 2 turns for 4 ohms
This will result in the correct turns for the 8 watt and 3 2 watt windings, but the 6 watt will be 477 turns This is a de viation of 6°/o which is acc eptabe . Also, the second ary turns have al so bee n adjust ed and will result in a deviation for the 8 ohm w ind ing of a pproxi mately 6/o .
page 66
The next step is t o determine the wire si
zes and cal cul ate the f il l .
The highest cur rent in the primar y wil be in the I =
w -
E
32 0
= .4 52A 32 W
16 0
= 226A
8 0
= . 1 13A 8 W
3 2 watt winding.
16 w
he highest current in the secondary winding wll be the and at 3 2 watts.
I=
�
4 ohm wnding
= 2 . 8 2 A 4 HM
= 2.0 A B OHM
= 141 A 16 OHM
#26 AWG wie will be
254 / 452 = 562 CM/A
#1 6 AW G wire will be
1624 / 2.82 = 575 CM/A
for the primary. for the seconda ry.
These sizes are smaller than suggested previously, but will be suicient, as they are worst case for 32 watts and 4 ohms.
page 67
CALCULATE THE FILL The window for ths size la mi nation is should be 1/32" shorter than the window Coil length
5/8" x 1 7 /8"
The coil lengt h
1 27 /32"
#26 winding length 1 19/32" , the ma rgins will be 1/8" each en d Turn s per layer 80 (fro m the Wir e Table in the Appen dix) Layers 31 8 80 4
1 19/32", the margins will be 1/8" each end. # 16 winding length Turns per layer 26 Layers 26 / 26 1
#16, 26 turns 1 layer # 16, 26 turn s 1 layer #16, 26 turns 1 layer #26, 80 turns per ayer #26, 80 turns per ayer
159 80 159 80
2 layers 2 layers
The fi l can now be calcu lated using the conf igu ration on page 67 Wnding tube 4 L #26 wre Layer ins. Wrap 1 L #16 wre Wrap 1 L #16 wire Wrap 1 L #16 wire Wrap 2 L #26 wre Lay e ns Wrap 2 L #26 wire Layer ins. Wrap Tota
= = =
.0400 .0684 .0090 (3 layers of .003 " Kaf t paper) .0100 K .0527 .0100 K .0527 .0100 K .0527 0100 K .0342 .0060 K .0100 K .0342 .0060 K .0100 K .4265 I . 625 x 100
70°/ fil
This fill is 0 K
page 68
The wndng resstances can now be calculated (see page 26). From the fll figures and the lamnaton sze: The equvalent of a square tube wl be 125 + 1.5 Core
2 75 / 2
1375.
1.3750 .0800 0684 0900 15324 x 4 x 318 1000 x 34005
6.628 ohms
.0090 0684 0100 .0527 1.6725 x 4 x 26 1000 x 3346
=
.0582 ohms
.0612 ohms
=
.0652 ohms
=
.0693 ohms
.0527 .0100 .0527 17609 x 4 x 26 1000 x .3346 .0527 .0100 0527 1.8763 x 4 x 2 6 1000 x 3346 .0527 .0100 .0527 1.9917 x 4 x 26 1000 x .3346 .0527 .0100 0342 .0030 2.0916 x 4 x 159 1000 x 3.4008
4.523 ohms
=
4. 706 ohms
.0030 .0342 .0100 .0342 .0030 21760 x 4 x 159 I 1000 x 3.4008
page 69
The mean length turn will be in the center of the windings which will be This will be used in calculating t he leaka ge 1 . 8 763 x 4 7 5" . inductance This winding configuration has all of the secondaries in between the primaries, so a 2 to 1 interleave will be used. From page 13, the formula for leakage inductance, using the values for this transf ormer i s:
10.6
x
(636)
x
7.5
x
(2
x
2
x
.010
x
10
+
.4265) =
2
x
1.593
.00235
Where: 5
2
W H MLT
-
N
1 19/32 010 .4265 75 636 (the total pri ma ry turns)
The hi gh frequency re sponse c an be c al culated us ing this va lue of leakage inductance
+
Z
F
104
16
1223 2
7
625
1223
82895 HZ
x .00235
page 70
It should be noted that this has been calculated using the entire primary an d second ary. Wh en taps are u sed f or diff eren t wattages and outputs, the high frequency response will not be as good. This is due to the windings that are not used causing the effective leakag e ind uctance to be d ifferent. Thi s is dificu lt to cacu late an d, when the upper end is beyond the requirements, it will not be attempted However, it will be tested and the results put on the frequency response graph to show the difference. he inserti on loss requi rement is for 5 DB maxi m u m From pa ge 64, the total p rima ry turns are 636 . The total sec on da ry tu rns are 04. By adding the resistances, as calculated on page 69, the tota primary resistance is 6.28 + 4523 + 4 76 155 ohms =
The total secondary resistance is: 582
R
+
=
612 + .652
t IL
R Z1
R
+
+
X 00
693
R
=
=
2539.
x 254
=
04
25 625
x
00
=
+
15.5
=
25
4°
Referring t o pa ge 7 , this b ein g ess tha n 0°, it wi ll m eet the 5 DB maximum he inserti on l oss will be df ferent for the dif ferent tap s This m akes it nece ssary to cal cul ate what pr oba by is the worst case Thi s is for 32 watts and 4 ohms as they are the windings that will carry the highest current
page 71
Using the resist a nce of the 32 wat t wind ing , 3A, A4: 3-A will be 6628 / 38 x 59 = 3314 A-4 is 4523 The total will be 3314
+
4 . 523 = 7. 84 ohms . h is is for 3 1 8 turns.
The 4 ohm winding is 7-B, B8. 7B = 0582 B8 = .0693 0582 + 0693
. 275 ohms
his is for 52 turns.
he im pedance for 32 wa tt is 1 56 ohm s
hen: 2 RT
IL
x .1275
12.60 =
+
7.84 = 12.60
52
156
x 100
8 08°
his is a so less tha n 1 0 ° fo r the 0. 5 DB m axi m um . The design is now complete and the manufacturing specifications can be written u p, the un it bu il t an d tested .
page 72
ROBERT G. WOLPRT
TRANSFORMER DESIGN SERVICES WINDING SHEET PAGE
1
PAGES
OF
SPEC NO. ENGINEER
RGW
DATE
AUDIO W TO
PE
16 OHMS
5/8" x 1 7/8"
WINDOW
70 /o
COIL BUILD
NET GROSS
1 1/4" X 1 1/2 " X 040 K
TUBE (NET GROSS) OVER TUBE DENSITY
15 KG
FREQUENCY
30 HZ
AREA
1725
IN
707
AT
VOLS
TERMINALS
1- 1
1 27/32
7
1
7
2 2 3 A A
8 5
-
r
COIL 1
�
B B
-
-- -
6
4 A
6
B
HOLD AL L MARGINS T O 1/8 " A's, Bs, 2's & 7s CONNECT TOGETHER WNDNG WRE SIZE TOTA TURNS TAPS WNDNG ENGTH MARGN TURNS ER YER % FL NO. O AYERS YER NSULATON WRAPPER TERM COI
#2 6
# 16
# 16
#1 6
# 16
#2 6
#2 6
3 18
26
26
6
26
159
159
159
--
1 19 / 2"
1 19 /2 "
1 19 / 2"
1 19 / 2"
1 19 / "
1 19 /2 "
1 19 / 2"
80 86%
6 86°/o
26 86°/o
26 86%
26 86%
80 86 °/o
80 86%
4
.003K l
OlOK 23A
1
l
OlOK 7-B
1
l -
OlOK 5-6
1
1
l -
1L -
.OlOK 67
OlOK B 8
2
2
003K
003K
ll OlOK
OlOK
A4
1 2
l -
START A
page 73
ROBERT G. WOLPERT TRANSFORMER DESIGN SERVICES
MATERIAL SHEET PAGE
2
OF _ PAGES
SPEC NO . �
PART NO CORE
EI - 1 29M6
COPPER #26 MAGNET WIRE #16 MAGNET WIRE CAN
AM
TO PRICE
O PRCE
TO PRCE
3.75#
0.3# 0.5#
LD- LDB TERNALS UBE
1 1/4 " x 1 1/2 " x 040 x 1 27 LONG
1
1 1/4" HORIZ. "L" #832 X 2" LONG #8-32 #8 STEEL #8 FIBER
4 4 4 8 4
TER BOARD LUG PANE BK BOLS NUS WASHERS WASHERS LEADS #1 #2 #3 #4
LEADS #5
#6 #7 #8
#20 SLW 10 LONG BLACK BROWN RED ORANGE # 16 SLW x 10" LONG YE LL OW GREEN BLUE VIOLET
1 1 1 1 1 1 1 1
NOES :
page 74
ROBERT G. WOLPERT TRANSFORMER DESIGN SERVICES
FINISHING 3
PAGE
OF - PAGES
S P E C NO . � COLOR LEADS SIZE
LENGH OU OF COIL
LE A D #
#20 SLW
BLACK
8"
1
#20 SLW
BROWN
8"
2
#20 SLW
RED
8"
3
#20 SLW
ORANGE
8"
4
#1 6 SLW
YELL OW
5
#16 SLW
GREEN
6
#16 SLW
BLUE
7
# 1 6 SLW
VIO LET
8
LUGS OR LUG PANEL PA R #
LEA D #
SPECIAL INSTRUCIONS: CONNECT ALL LIKE NU MB ERED AN D LET TERED LEADS T OGETHER. BL IND A's & B's
page 75
ROBERT G. WOLPERT TRANSFORMER DESIGN SERVICES STACKING & ASSEMBLY PAGE
4
O F � PAGES
SPEC NO. LAMNATON
EI- 1 1/ 4" 29M6
SE GRADE SACK EG
1 1/2" 1x1 2
NERLEAVE KEEPERS CU OFF E'S GAP SPACER BRUSERS SE SED: U NSUAORS SZE BRACKES -
Q
4
1 1/4" HOR IZON TAL "L"
Q Q
4 4 8
BOLTS 832 x 2 NUTS 832 #8 WASHERS, STEEL #8 WAS ERS FIBE R
HARDWARE
Q Q
4
TO BE REMOVED
NO
SPECAL NSRUC ION S:
VACUU M VARNISH
page 76
ROBERT G. WOLPERT TRANSFORMER DESIGN SERVICES
TEST INSTRUCTIONS
PAGE
5
PAGES
OF
SP EC NO PROCEDURE
6
1ST TEST 2ND TES 3RD TES FINA ES (AFTER VARNISH)
7 --
6, 5
NO LOAD VOLTAGE RATIO
1.
APPLY
v
HZ TO TERM
lex
MAX
V TERM
READ
V TERM V TERM
2
IN DUCTANCE TEST APPLY
HZ TO TERM.
v
READ "L
&
ADC
MIN
IN DUCE D VO LTAGE TEST
3
APPLY 4
MUST MEG
5
H IPOT
v
HZ TO TER M
FOR
MEGOHMS MIN
LEAD NO.
SEC VOLTS DC.
TO 5
VOLS 500
CORE
500
CASE
6
CONTINUITY
7
SPECIAL TESTS :
s
=
625 ohms
L = 1 ohms Test across the tota pimary and secondary Test resuts on te foowng page
ge 77
r
> u z w : O w a u
- � t._ -�
-.!-
·
f
;
a Q
_,
0
0 NI
-
APPENDIX
page 79
SYMBOLS U SED A B
-
-
E
E2
-
F F F2 H
-
-
1
12
-
K
L L
Efectve area of core n sq. nches. Flux densty n Gauss Prmary voltage Secondary voltage Frequency Low frequency lmt Hgh frequency mt Wndng buld-up Prmary current n amperes Secondary current n amperes Stackng factor Prmary open crcut nductance n Henrys Leakage nductance n Henrys
ML=
Mean ength turn n nches
N
Prmary turns
N2 n Ri
-
-
-
R2
Secondary turns Number of ayers DC resstance of prmary DC resstance of secondary
-
RT
s
otal or normalzed resstance Number of sectons or nterleaves hckness of nsuaton n nches
U Ac =
Incremental permeablty of core materal
w
Power n Watts
-
WL =
Wndng length n nches
Z
Prmary mpedance
Z2 ZT
-
-
Secondary mpedance otal or normalzed mpedance
page 80
DB EXPRESSED N WATS
DB
MICRO-WATS
80 70 -60
000001 00001 0.001
-so 40 - 30
0.01 0.1
-20 - 19 -18 -17 -16
10 10.0 12.26 lS.9 20.0 2S.1
ls 14 -13 12
31.6 39.8 S0. 1
-11 -10 - 9 - 8 - 7 - 6
79.4 100.0
s - 4 - 3 2 - 1 0
631
1260 1590 2000 2S10
DB
WATS
00 10 20
0001 0.0013 0.0016
30 40 s.o 60 70 80 90 100 11.0 12.0 130 140
0.002 0.002S 0.0032 0004 ooos 0.0063 0.0079 0.01 0.0126 0.01S9 0.02
lSO 160 170 180 19.0 20.0
0.0316 0.0398 ooso 0.063 0.079 0.10 1.0
002S
3160 3980 SOlO
300 400
631.0
60.0
7940 1000.0
70.0
10000.0
80.0
100000.0
so.a
100 100.0 10000
page 81
WIRE TABLE
SIZE 42 41 40 39 38 37
36 35
34 33 32 31 30 29
28 27 26 25 24 23 22 21 20 19 18 17 16 15 14 13 12 11 10 9 8
OHMS/ 1000" 139.00 110.25 87.42 69.32 5497 43.59 34.56 27.42 2174 17.24 1367 10.84 860 6.82 5.41 4.29 3.40 2.70 2.14 1.70 1.345 1.067 .8458 6709 .5320 .4220 .3346 .2653 .2104 1669 1323 .1050 .0833 .0660 0524
.0007 .0007 .0007
1/16 1/16 1/16
CM AREA 625 7.8 99
.0007 .001 .001 .001 .001 .001 .0015 .0015 .0015 .0015 0015 0015
1/16 1/16 3/32 3/32 3/32 3/32 3/32 3/32 3/32 3/32 1/8 1/8
125 15.7 198 250 3152 39.75 5013 632 79.7 100.5 126.7 1598
.002 .002 .002 002 003 .003 .003 005 005 007 007 010 010 .010 .010 .010 010 .010 010 010
1/8 1/8 1/8 1/8 1/8 1/8 1/8 1/8 1/8 1/8 1/8 1/4 1/4 1/4 1/4 1/4 1/4 1/4 1/4 1/4
I N SUL
MARGIN
201.3 254.1 320.4 4040 5095 6424 8101 10220 12880 16240 20450 25330 3257.0 4107.0 5778.0 6530.0 82340 103800 130900 165100
TURNS/ IN 304.0 267.0 239.0
84648.0 54045.0 35610.0
.0040 .0045 .0051 0056 0062 0070 0079 .0089 .0099 .0110 .0123 0137
215.0 1923 170.4 155.5 142.0 125.6 110.4 99.0 88.9 810 72.5 63.0
22047.0 12887.0 8077.0 5248.0 3375.0 2106.0 1305.0 8100 5300 333.4 204.6 132.5
.0154 0171 0192 0215 .0240 0268 0301 0336 .0376 .0421 0471 .0527 .0590 0661 0741 .0829 .0929 .1042 1168 1310
57.6 520 45.9 41.9 37.4 336 303 26.7 24.3 217 193 17.6 15.9 14.2 12.8 11.5 102 9.5 8.2 72
DIA 0028 .0032 .0036
OS/LB
82.5 527 32.7 207 12.9 822 512 323 2.04 1.29 805 .5101 .3193 .2016 .1258 0794 .0500 0315 .0198 .0125
page 82
LAMINATION TABLE
VA
AREA
WINDOW
SIZE
STACK HT
EI-187
3/16
05
035
3/16
10 3.0
0625 1406
1/4
EI-3/8
1/4 3/8
EI-5/8
5/8
7.0
390
EI3/4
3/4
14.0
.5625
3/8
EI-3/4
1.0
19.0
EI-7/8
7/8
300
EI-7/8
10
320
EI - 1
10
EI-1 EI -1 1/8
EE-2425
x
7/16
x
1/2
WEIGHT LBS . 015
3/4
034 108
15/16
.392
1 1/8
678
. 75
3/8 x 1 1/8
904
.765
7/16 x 1 5/16
105
.875
7/16
x
1 5/16
120
450
1.00
1/2
x
1 1/2
155
1 1/4 1 1/8
500 650
125 1.265
1/2 x 1 1/2 1 11/16
1.94 2.24
EI- 1 1/4
1 1/4
900
1.5625
1 7/8
308
EI1 3/8
1 3/8
1250
1.89
11/16
2 1/16
4.17
EI1 1/2
1 1/2
1600
2.25
3/4
EI- 1 3/8
1 3/4
160.0
2.40
11/16
EI-1 1/2
2 1/2
3000
3.75
3/4
x
2 1/4
8.91
EI1 3/4
1 3/4
3400
306
7/8
x
2 5/8
861
EI1 3/4
2.0
4000
3.50
7/8
x
2 5/8
9.84
EI-1 5/8
2.0
4500
325
1 1/4
5/16 5/16
9/16 5/8
x
x x
x x
x x
2 1/4 2 1/16
x
x
2 5/8
5.35 5.31
7.78
Note: Values shown for area and weight must be modified by the stacking factor K
page 83
TOTAL HARMONIC DISTORTION TABLES
The fo lo wi ng ta bl es a re the results of test in g actua transformers. They sho uld onl y be used as gu idel ines for estimati ng the di stort ion that can be expec ted usi ng th ese three material s at dif ferent flux levels . The hysteresis curves of these materials show that the distortion increases as the flux level approaches the knee of the curve.
FLUX (KG)
S O /o NICKEL (/o)
8 0 /o NICKEL ( /o )
29M6 ( /o )
1
.01
.012
.03
2
.01
.03
.03
3
066
.06
03
4
.10
.18
03
5 6
.12 .14
.4 5
1.45
.04 .05
7
.16
1.6
.10
8
.46
2.3
15
.75
.25
10
.30
11
.36
12
.41
13 14
66 98 .
15
125
16
165
17
2.10
18
2.55
page 84
The following tables show the turns of each size of wire that can be fitted nto dif ferent s izes o f lam inations, both layer w oun d an d bo bbin woun d The se tables w l be a he lp i n dete rmin in g the possibi lty of fit in a des ign after the turns and wre szes are chosen and before spending a lot of time calculating the f ill . For point, ex am page ple, if 23 th e de sign shown in of 5. 3 pages 20 and 2 was s toppe the where 390 turns #23 wire was determned for the d at p rimary and the EI-2 Lamnation Table in the Appendix is consulted, it can be shown that it will fit, 390 / 848 x 100 46° Th s s le ss th an /2 of the aval ab le spa ce and s a bout r gh t for the p rim ary wn din g in a nor mal design. =
These tables can save a lot of design time that might be necessary in jugglng between turns, wire sizes and laminaton sizes However, where ext ens ve interleaving is used, 46° fl l fo r the pri ma ry win din g mi ght resu lt in a tg ht ft.
page 85
EE LAMINATION MAXI MU M TURNS FOR L AYER WOUN D COILS COIL LENGTH = / WINDOW WIDTH / 0 . " MEAN LENGTH TURN =
Wire Gage
Min Layer Insulation
Turns/ Layer
Max. Layers
Max. Turns
Resistance square stack
22
1 - 003K
6
3
18
0237
23
1 - 003K
7
3
21
0345
24
1 - 002K
7
4
28
0587
25
1 - 002K
8
4
32
086
26
1 - 002K
9
5
45
150
27
1 - 002K
11
5
55
231
28
1 - 002K
12
6
72
38
29 30
1 0 015 K - 0 015 K
13 15
7 8
91 120
608 101
31
1 - 001K
17
8
136
1445
32
1 - 00K
19
9
171
2292
33
1 - 0 01 K
2
0
210
3548
34
1 - 0 01 K
24
11
264
5625
35
1 - 001K
27
13
351
9431
36
1 - 001 K
30
14
420
14228
37
1 - 0 01 K
34
16
544
2324
38
39
1 - 001K 1 - 00075K
43
8 20
666 860
3587 5842
40
1 - 00075K
48
22
1 056
9047
41
1 - 0005K
53
26
1378
14889
42
1 - 0005K
59
28
1652
22572
43
1 - 0 005K
67
31
2077
35688
44
1 - 0005K
77
35
2695
58810
45
1 - 0005K
85
38
3230
88310
46
1 - 0 005K
94
41
3854
132410
37
page 86
EI- LAMINA ION MAX IM UM U RNS FOR L AYER WOUN D COILS COIL L ENGH /4" WINDOW WIDH = / MEAN LENGH TURN = " =
Wire Gage
Min Layer Insulation
Turns/ Layer
Max. Layers
Max urn s
Resistance sq ua re stack
22
1 - 003K
5
20
037
23
1 - 003K
4. 5
5
22
0512
2
1 002K
5
6
30
0892
25
1 002K
6
6
36
135
26
1 - 002K
65
7
5
212
27
1 - 002K
7
8
56
33
28
1 - 0015 K
8
9
72
51
29 30
1 - 0015 K 1 - 001K
9 10
10 12
90 120
853 135
31
1 - 001K
11
13
13
2155
32
1 - 001K
125
1
175
3326
33
1 - 001K
1
16
22
5368
3
1 001K
16
18
288
871
35
1 - 0 01K
18
20
360
1372
36
1 - 001 K
20
22
0
211
37
1 001 K
225
2
50
3272
38 39
1 - 001K 1 - 0 0075K
25 29
27 31
675 899
51 57 8662
0
1 - 0 0075K
32
34
1088
13220
1
1 0005K
35
39
1365
20920
2
1 - 0005K
39
2
1638
3170
43
1 - 0005K
5
7
2115
5150
1 - 0005K
51
53
2703
83660
5
1 0 005K
56
57
3192
123790
6
1 - 0005K
63
62
3906
19030
page 87
EI LA MINAT ION MA XIMU M TURNS FOR L AYER WOUN D COILS COIL LENGTH = /" WIN DOW WIDT H = /" MEAN LENGTH TURN ." =
Wire Gage
Min Laye Insulation
Turns/ Layer
Max Layers
Max Turns
Resistance square stack
22
1 - 0 025K
9
5
45
0841
23
1 - 0 03K
10
5
50
1165
24
1 - 0 02K
15
6
69
2052
25
1 - 0 02K
13
6
78
2925
26
1 - 0 02K
145
7
101
4774
27
1 - 0 02K
16
8
128
7631
28
- 005K
18
9
162
1218
- 0015K
20
10
200
1 89 5
29 30
1 - 0 01K
23
12
276
330
31
1 - 0 0K
255
13
331
500
32
1 - 001 K
28
14
392
745
33
1 - 001 K
315
16
504
1208
34
1 - 001 K
36
18
648
1958
35
1 - 001 K
405
20
810
3087
36
1 - 001 K
45
2
945
4541
37
1 - 0 01 K
505
24
38
39
1 - 0 0075K 1 - 0 0075K
56 65
27 31
2015
11550 19420
40
1 - 0 0075K
73
34
2482
301 60
41
1 - 0 0075K
80
39
3120
47800
42
1 - 0 0075K
88
42
3696
71630
43
1 - 0 0075K
101
47
4747
1 1 5700
44
1 - 0 005K
115
53
6095
188650
45
1 - 0 005K
127
57
7239
280740
46
1 - 0 005K
41
62
8742
426000
122
7344
1512
page 88
EE LAMIN ATIO N MAXIM UM TURN S FOR L AYER WOUN D COIL S COIL LE NGTH /" WIND OW WID TH /" MEAN LENGTH TURN . =
=
=
Wie Gage
Min Layer Insulation
Turns/ Layer
Max Layers
Max Turns
Resistance sq ua re stack
22
1 - 0 03K
11
6
66
1660
23
1 - 0 03K
13
7
91
2887
24
1 - 0 02K
14
8
112
4480
25
1 0 02K
16
8
128
6422
26
1 - 0 02K
18
9
162
103
27
1 - 0 02K
20
10
200
1 602
28
1 - 0015 K
22
12
264
267
29 30
1 - 0015 K 1 - 001K
25 28
13 15
325 420
4143 6753
31
1 - 0 01 K
31
17
527
1068
32
1 - 001K
34
18
612
1564
33
1 001 K
39
20
780
2514
34
1 - 0 01 K
44
23
1012
4114
35
1 - 001 K
50
25
1250
6410
36
1 - 001 K
55
28
1540
9952
37
1 - 001 K
62
30
1 860
15160
38 39
1 - 0 0075K 1 - 0 0075K
69 80
35
39
2415 3120
24820 40430
40
1 - 0 0075K
89
43
3827
62550
41
1 - 0 0075K
98
50
4900
101000
42
1 - 00 05K
108
54
5832
152000
43
1 - 0005K
123
61
7503
246000
44
1 - 0 005K
140
68
9520
396300
45
1 - 0005K
155
73
11315
590200
46
1 - 0 005K
172
80
13760
901900
page 89
EE- LAMINATION MA XIMU M TUR NS FO R LAYER WOU N D COILS COIL LENGTH = /" WINDOW WIDTH = /" MEAN LENGTH TURN = ." Wire Gage
Min Layer Insulation
Turns/ Layer
Max. Layers
Max Turns
Resistance square stack
22
1 - 0 0 3 K
16
6
96
3125
23
1 - 0 0 3 K
18
6
108
4435
24
1 - 0 0 2 K
20
7
140
7250
25
1 - 0 0 2 K
23
8
184
1201
26
1 - 0 0 2K
26
9
234
1 92 6
27
1 - 0 0 2K
29
10
290
301
28
1 - 0 015 K
33
12
396
5184
29 30
1 - 0 015 K 1 - 0 0 1K
36 40
13 16
468 640
7724 1247
31
1 - 0 0 1K
45
16
720
1889
32
1 - 0 0 1 K
50
18
900
2980
33
1 0 0 1 K
56
20
1120
4674
34
1 - 0 0 1 K
64
22
1408
5876
35
1 - 001K
72
25
1800
11945
36
1 - 0 0 1K
80
27
2160
18070
37
1 - 001 K
90
30
2700
28490
38 39
1 00 1 K 1 - 0 0 0 7 5K
100 1 16
34
38
3400 4408
45230 73960
40
1 - 0 0 0 7 5 K
129
42
5418
109300
41
1 - 0 0 0 5 K
142
49
6958
185700
42
1 - 0 0 0 5 K
157
53
8321
280800
43
1 0 0 0 5 K
180
59
10620
44
1 - 0 00 5 K
205
66
13530
730000
45
1 - 0005K
226
71
16046
1083600
46
1 - 0 0 0 5 K
251
78
19578
1661300
450700
page 90
EI LAMI NATIO N MAXIMU M TURNS FOR L AYER WOUN D COILS COIL LENGTH = /2" WINDOW WIDTH = /" MEAN LENGTH TURN = 2. Wire Gage
Min Layer Insulation
" Turns/ Layer
Max Layers
Max. Turns
Resistance square stack
0
1 - 0 05 K
13
6
78
1
1 - 0 05K
15
7
105
3079
1 0 03K
16
8
18
4734
3
1 - 0 03K
18
9
16
7558
4
1 - 0 0K
1
10
10
1 3 5
5
1 - 0 0K
3
11
53
1 876 7
6
1 - 0 0K
6
1
31
917
7 8
1 - 0 0K 1 - 0015 K
9 31
15 16
435 496
5130 7376
9
1 - 0015K
36
17
61
11476
30
1 - 0 01 K
40
0
800
189
31
1 - 0 01 K
44
968
886
3
1 - 0 01K
49
4
1176
444
33
1 - 0 01K
55
7
1485
7041
34
1 - 0 01 K
63
30
1890
1130
35
1 - 0 01 K
71
34
414
180
36 37
1 001K 1 0 01K
78 85
40
886 3400
7433 40758
38
1 - 0 01K
96
44
44
63848
39
1 - 00075 K
107
51
5457
10401
40
1 - 0 0075K
119
56
6664
1600
41
1 - 0005K
133
66
8778
661 37
4
1 - 0 005K
15
74
1810
37
1148
43000
page 91
EI- LAMI NATION MAXIMUM TURNS FOR LAYER WOUND COILS COIL LENGTH = /" WINDOW WIDTH = /" MEAN LENGTH TURN = " Wire Gage
Min Layer Insulation
Turns/ Layer
Max Layers
Max Turns
20
1 - 003K
15
6
21
1 - .0 03K
17
7
119
.396
22
1 - .0 03K
20
8
160
.6714
23
1 .0 03K
22
9
198
1 .048
24
1 002K
24
10
240
1 .60 2
25
1 - 002K
27
11
297
2.50
26
1 - 002K
31
12
372
3947
27 28
1 - 002K 1 - .0015K
34 39
13 15
442 585
5.915 9.87
29
1 .0015 K
43
17
731
15.55
30
1 - 0 01K
48
19
912
24.47
31
1 - .001 K
54
21
1134
38.36
32
1 - .001 K
59
23
1357
5810
33
1 - .001 K
67
26
1742
93.71
34
1 - 001 K
76
29
2204
118.60
35
1 - 001K
86
32
2752
235.40
36 37
1 - 001K 1 - 001K
107
39
35
3325 4173
35860 56760
38
1 - 00075K
119
45
5805
995.50
39
1 - 00075K
127
50
6350
1373.30
40
1 - 00075K
154
55
8470
2310.00
41
1 - 00075K
168
64
0752
3698.50
42
1 - .0 005K
187
70
1 3090
5694.00
43
1 - 0005K
214
78
16692
9131.00
44
1 - .0 005K
243
87
21141
14687.00
90
Resistance sq ua re stack .2375
95
page 92
EI LAMI NATION MAXIM UM TURNS FOR L AYER WOU N D COILS COI LENGTH = /" WINDOW WIDTH = /" MEAN ENGT H TUR N ." =
Wire Gage
Min ayer Insulation
Turns/ Layer
Max ayers
Max Turns
Resistance squ are stack
18
1 - 0 03K
15
5
75
145
19
1 - 003K
16
5
80
194
20
1 - 003K
18
6
108
330
21
1 - 0 03K
20
7
140
541
22
1 - 003K
23
8
184
896
23
1 - 003K
25
9
225
1 38 2
24
1 - 002K
31
10
310
240
25 26
1 - 0 02K 1 - 0 02K
35 39
11 12
385 468
376 576
27
1 - 0 02K
43
13
559
868
28
1 - 0 015 K
49
15
735
1439
29
1 - 0 015 K
54
17
918
2266
30
1 - 001 K
61
19
1159
3610
31
1 0 01K
68
21
1428
5610
32
1 - 0 01 K
75
23
1725
8540
33
1 - 00 1K
85
26
2210
13795
35
1 - 0 01 K 1 0 01 K
96 108
29 32
2784 3456
21910 34300
36
1 - 001K
120
35
4200
52560
37
1 - 0 01K
135
39
5265
831 00
38
1 - 00075K
150
45
6750
134300
39
1 - 00075K
174
50
8700
218300
40
1 - 00075K
194
55
1 0670
337700
41
1 - 0005K
213
64
1 3632
5441 00
42
1 - 0005K
236
70
16520
833800
34
page 93
EI LAMIN ATION MAXIM UM TURNS FOR LAYER WOU N D COILS / COIL LENGTH WIN DOW WIDTH /" MEAN LENGTH TURN ." =
=
=
Wire Gage
Min. Layer Insulation
Turns/ Layer
Max. Layers
Max. Turns
Resistance squ are stack
18
1 003K
18
6
108
249
9
1 - 003K
20
7
140
513
20
003K
23
7
161
590
21
1 - 003K
26
8
208
960
22
1 - 003K
28
9
252
1 467
23
1 - 003K
32
10
320
231
24
1 - 002K
36
12
432
400
25 26
1 - 0 02K 1 - 0 02K
40 46
13 15
520 690
607 1016
27
1 - 002K
51
16
816
1515
28
1 0015K
57
18
1026
2400
29
1 - 0015 K
63
20
1260
3720
30
1 001K
71
24
1704
6345
3
1 - 0 01K
79
26
2054
9641
32
1 - 001K
87
29
2523
14938
33
1 - 001 K
99
32
3168
23650
35
1 001K 1 - 001 K
112 126
39
35
3920 4914
36900 58330
36
1 - 001 K
140
43
6020
901 00
37
1 - 001K
157
47
7379
139600
38
1 - 0 0075K
175
54
9450
224900
39
1 - 0 0075K
203
61
12383
371600
40
1 00075K
227
67
15209
575600
41
1 - 0005K
248
77
1 9096
91 1 500
42
1 - 0005K
275
85
23375
34
1411000
page 94
EI LAMINATION MA XIM UM URNS FOR LAYER WOU N D COILS COIL LENGH = / WIN DOW WIDH = / 6 " M E AN LENGTH TURN = 0 Wire Gage
Min. Layer Insulation
Turns/ Layer
Max. Layers
Max. urns
18
1 - 00 3 K
22
7
154
38
19
1 - 003K
25
8
200
73
20
1 0 0 3 K
28
9
252
21
1 - 0 0 3K
31
10
310
1
22
1 003K
35
11
385
1 9 7
23
1 - 0 0 3 K
39
12
48
39
24
1 - 002K
44
14
1
4
25 2
1 - 002K 1 00 2 K
49 55
1 17
784 935
10 103
27
1 00 2 K
1
1 1 59
25 0
28
1 - 0 015K
9
19 22
29
1 - 0 015 K
77
24
1848
351
30
1 0 0 1 K
8
28
2408
10440
31
1 - 0 0 1K
95
31
2945
1090
32
1 0 0 1 K
0
34
304
24840
33
1 - 0 0 1 K
20
38
450
3930
34
1 - 0 0 1 K
13
42
5712
1500
35
1 - 001 K
53
47
7191
99370
3
1 001 K
170
51
870
37
1 - 001 K
191
57
10887
239200
38
1 - 0 0 0 7 5K
213
5
13845
383550
39
1 - 0 0 0 7 5K
24
74
18204
3000
40
1 - 0 0 0 7 5K
275
81
22275
981400
41
1 - 0 0 0 5 K
302
93
2808
15000
42
1 0 0 0 5 K
334
102
3408
2393900
1518
Resistance squ are stack
1074
4138
151050
page 95
EI AM INATION MAXIM UM TU RNS FOR AYER WOU N D COILS COIL ENGTH / " WIN DOW WIDTH /" MEAN LENGTH TURN " =
=
=
Wire Gage
Min Layer Insulation
Turns/ Layer
Max. ayers
Max Turns
Resistance square stack
16
1 - 005K
21
6
1 26
17
1 - 005K
23
7
161
18
1 - 003K
26
8
208
19
1 - 003K
29
9
261
1 0 0
20
1 - 003K
33
10
330
1 60
21
1 - 003K
37
11
407
248
22
1 - 003K
41
13
533
410
24
23
1 - 003K 1 - 0 02K
46 52
14 16
644 832
624 1016
25
1 - 002K
58
18
1 044
1608
26
1 - 002K
65
20
1300
2524
27
1 - 001K
72
22
1584
3880
28
1 - 0015 K
81
25
2025
6253
29
1 - 0015 K
90
27
2430
9461
30
1 - 001K
102
32
3264
16027
31
1 - 001K
113
35
3955
24480
32 33
1 - 0 01K 1 - 0 01K
125 141
43
38
4750 6063
37090 59700
34
1 - 001 K
159
48
7632
94740
35
1 001 K
180
53
9540
149340
36
1 001 K
200
58
11600
228940
37
1 - 0 01 K
225
64
14400
358400
38
1 - 00075K
250
73
18250
572800
39
1 - 0 0075K
290
83
24070
952600
40
1 - 0 0075K
324
91
29484
1471600
241 388 632
page 96
EI LAMINATION MA XIMU M TURN S FO R LAYER WOU N D COIL S 1 / COIL LENGH WINDOW WI DTH / MEAN LENGH URN = " =
=
Wire Gage
Min. Layer Insulation
uns/ Layer
Max Layers
16
- 0 05K
24
7
168
371
17
- 0 05K
27
8
216
602
8
1 - 0 03K
30
10
300
984
9
1 - 0 03K
34
11
374
1 656
20
1 - 0 03K
38
12
456
255
21
- 0 03K
425
13
552
389
22
1 - 0 03K
475
15
712
632
23
1 - 0 03K
53
16
848
938
24
1 0 02K
59
18
1062
500
25
1 0 02K
67
20
1340
2386
26
1 - 0 02K
75
23
725
3872
27
1 - 0 02K
83
25
2075
5874
28
- 0 01 5K
94
29
2726
9730
29
- 001 5K
1 04
32
3328
14978
30
1 - 001K
1 17
37
4329
24570
31
1 - 001K
131
41
5371
38430
32 33
1 - 001K 1 - 0 0 1K
144 162
44
6336 7938
57190 90330
34
1 - 0 01 K
1 84
55
10120
145220
35
1 - 001 K
208
61
12688
229600
36
1 - 001 K
230
67
15410
351570
37
1 001 K
258
74
19092
549300
38
- 0 0075K
288
85
24480
888100
39
1 - 0 0075K
333
96
31968
1462500
40
1 0 0075K
372
105
39060
2253500
49
Max urns
Resistance square stack
page 97
EI LAMINATION MAX IM U M TU RNS FOR LAYER WOU N D COILS COIL LENGTH /" WINDOW WIDTH / MEAN LENGTH TURN = " =
=
Wre Gage
Mn Layer Insulaton
Turns/ Layer
Max. Layers
Max. Turns
Resistance square stack
16
1 - 0 05K
27
8
216
521
17
1 - 005K
30
9
270
822
18
1 - 0 03K
34
10
340
1 30 4
19
1 - 0 03K
38
11
418
2022
20
1 - 0 03K
43
13
559
341
21
1 - 0 03K
48
14
672
517
22
1 - 0 03K
54
16
864
838
23
1 - 0 03K
60
18
1080
1321
24
1 - 002K
67
20
1340
2067
25
1 - 0 02K
75
22
1650
3210
26
1 0 02K
85
25
2125
5202
27
1 - 0 02K
94
27
2538
7848
28
1 - 0015 K
106
31
3286
12810
29
1 0015 K
117
34
3978
19320
30
1 - 001K
132
40
5280
32740
31
1 001K
147
44
6468
50560
32
1 - 001K
162
48
7776
76660
33
1 - 001K
183
54
9882
122840
34
1 - 001K
207
60
12420
194680
35
1 - 001K
234
67
15678
309900
36
1 - 001 K
260
73
18980
473000
37
1 - 001 K
292
80
23360
7341 00
38
1 - 0 0075K
325
92
29900
1184900
39
1 - 0 0075K
376
104
39104
1954200
40
1 00 075K
421
114
47994
3024800
page 98
EI- LAMINATION MAXIM UM TUR NS FOR L AYER WOU N D COILS COIL LENGTH = " WINDOW WIDTH /" MEAN LENGTH TURN = " =
Wire Gage
Min Layer Insulation
Turns/ Layer
Max Layers
Max Turns
Resistance sq ua re stack
13
1 - .0 10 K
19
6
114
.146
14
1 0 10 K
21
7
147
.24
15
1 - 0 10 K
24
8
192
39
16
1 - .0 07K
29
8
232
60
17
1 - 0 07K
33
10
330
1 .07
18
1 - .0 07K
37
11
407
1 .66
19
1 - .0 05K
43
12
516
265
20 21
1 - 0 05K 1 - 005K
48 54
14 15
672 810
4.35 660
22
1 - 005K
62
18
1116
11.50
23
1 - 005K
69
20
1380
17.90
24
1 - 0 03
78
23
1 794
29.50
25
1 - 003
88
26
2288
47.20
26
1 .003
98
29
2842
7400
27
1 003
110
32
3520
1 16.0 0
28
1 .0 015
123
37
4551
189.00
29 30
1 - 0 015 1 - .001 K
136 153
44
40
5440 6732
28700 445.00
31
1 - .001 K
172
49
8428
700.00
32
1 - 001 K
191
53
10213
1060.00
33
1 - 001K
213
58
12354
1630.00
34
1 - 001K
241
69
1 6629
276500
page 99
EI- LAMINAION MAX IM UM URNS FOR L AYER WOU ND COIL S COIL LEN GTH = / " WINDOW WIDTH / MEAN LENGH TURN = " =
Wie Gage
Min. Layer Insulation
urns/ Layer
Max. Layers
Max. urns
Resistance square stack
13
1 - 0 0 K
2
7
147
21
14
1 - 001K
24
8
1 92
34
15
1 - 001K
27
8
26
48
16
- 007K
33
9
297
84
17
- 0 06K
37
11
407
1 44
18
- 007K
4
12
492
220
19
1 - 005K
47
13
611
345
20 21
1 - 005K 1 - 005K
53
15 17
795
59
1003
560 900
22
1 - 0 03K
69
20
1380
1560
23
1 0 03K
77
22
1694
2400
24
1 - 0 02K
86
26
2236
4100
25
1 - 0 02K
97
28
2716
6350
26
1 - 002K
109
32
3488
10000
27
1 002K
121
35
4235
15300
28
1 - 0015 K
136
40
5440
24700
29 30
1 0015 K 1 - 00K
151 169
44 48
6644 8112
38100 58300
31
1 - 001K
190
54
1 0260
93300
32
1 - 001K
215
58
12528
143500
page 100
EE-- LAMINATION MAX IM U M TURNS FOR BOBBIN WOU ND COIL S ( °o FIL L) M EAN LENGT H T U RN
=
"
Wire Gage
Turns/ Layer
20 21
7 8
2 2
14 16
.0121 .0174
22
9
3
27
0373
23
10
3
30
0520
24
11
3
33
0723
25
12
4
48
1325
26
14
4
56
.196
27
15
5
75
325
28
17
5
85
.469
29 30
19 21
6 7
1 14 147
955 1 .28 6
31
24
8
192
2.117
32
26
8
208
2.841
33
30
10
300
5.20
34
33
11
363
5.69
35
37
12
444
1235
36
41
14
574
2080
37
46
15
690
2973
38 39
51 58
17 20
867 1 1 60
47.00 82.80
40
65
22
1430
111.60
41
73
25
1825
164.80
42
80
28
2240
312.70
43
90
31
2790
450.00
44
102
35
3570
780.00
45
118
41
4838
1 36 9.0 0
46
123
43
5289
1 87 80 0
Max Layers
Max Turns
Resistance squa re st ack
page 10 1
EI- LAMINATION MAXIM U M TU RN S FOR BOBBIN WOU N D COILS ( °o FILL) MEAN LENGTH TURN
=
"
Wire Gage
Turns/ Layer
Max Layers
Max Turns
Resistance square stack
20 21
5 6
4 4
20 24
026 039
22
6
30
062
23
7
5 5
35
091
24
8
6
158
25
7
26
9 10
48 63
7
70
.369
27
11
8
88
572
28
13
9
117
1 42
29 30
14 16
10 12
140 192
209 2.875
31
18
13
234
393
32
20
15
300
6.25
33
22
17
374
9.89
34
25
19
475
1591
35
28
21
588
2500
36
31
23
713
3840
37
35
26
910
5970
38 39
38 44
29 34
1 10 2 1496
91 .4 0 16210
40
49
38
1862
25800
41
55
43
2365
402.90
42
61
47
2867
609.00
43
68
53
3604
990.00
44
77
60
4620
1540.00
.262
page 102
EI- LAMINATION MAX IM U M TU RN S FOR BOBBI N WOU N D COILS ( °o FILL) M EAN LENG H T U RN
=
" Max urns
Resistance square stack
3 4
30 44
0376 070
2
4
48
0963
23
4
5
70
.176
24
5
5
75
.2382
25
7
6
102
408
26
9
7
1 33
.675
27
2
8
168
1 0 7
28
24
9
216
1 7 2
29 30
27 30
10 11
270 330
2.72 4.24
3
34
12
408
6.58
32
37
14
518
1040
33
42
16
672
1672
34
47
17
799
2580
35
53
20
060
43.35
36
58
22
276
6540
37
65
25
1 62 5
1 02 90
38 39
72 83
27 32
2656 2944
5590 278.00
40
92
35
3220
429.00
41
103
40
4120
67350
42
114
44
5016
102900
43
127
50
6350
168300
44
145
57
8265
241300
45
167
66
Wire Gage
urns/ Layer
20 21
0
22
Max Layers
11022
464000
page 103
EI- LAMINATION MA XIM U M TU RS FO R BOBBIN WOU N D COILS (8 °o FILL) MEAN LENGTH URN = " Max Layers
Max urns
Resistance square stack
Wire Gage
Turns/ Layer
20 21
1 20
6 7
1 0 140
25 423
22
22
1 76
.646
23
25
9
225
1 04 3
24
2
10
20
1 64 2
25
31
12
372
2.1
26
35
13
455
4.4
27
3
15
570
66
2
43
17
731
106
29 30
4 53
1 21
64 1113
1600 26.40
31
60
24
1440
43.59
32
66
26
1716
61.30
33
75
29
2175
10210
34
4
33
2772
165.00
35
94
37
347
26200
36
103
41
4223
3900
37
116
47
5452
63600
3 39
12 147
51 59
652 673
96500 1672.00
40
163
66
1 07 5
2663.00
41
12
74
1 346
466000
42
202
3
1 67 66
633000
43
226
92
20792
10160.00
44
257
105
2695
1590000
45
29
122
36356
25230.00
page 104
E E- LAMINA TION MAX IM U M TUR NS FOR BOBBIN WOU ND COI LS ( °o FILL) MEAN LENGTH TURN
=
16" Max. Layers
Max Turns
Resistance square stack
Wire Gage
Turns/ Layer
20 2
2
5
55 72
094 8
22
4
84
2308
23
5
7
05
324
24
7
8
3
25
9
9
7
954
2
22
0
220
54
27
24
24
235
28
27
3
35
394
29 30
30 34
4
420 544
58 93
3
37
8
500
32
4
20
820
2270
33
47
23
08
379
34
52
25
300
5785
35
59
28
52
9320
3
4
32
2048
4450
37
72
35
2520
22300
38 39
80 9
39 4
320 48
34400 0400
40
02
5
5202
9500
4
4
57
498
57300
42
2
3
7938
22 4 00
43
4
7
00
3300
44
0
8
290
587000
45
8
94
7484
97000
page 105
EI- LA MINA I ON MAXIMU M URN S FOR BOBB IN WOU N D COILS ( °o FILL) MEAN LENGTH URN
=
"
Wire Gage
uns/ Layer
20 21
19 21
22
Max Layers
Max urns
Resistance squae stack
6 7
114 147
.319 5345
24
8
92
856
23
27
9
243
136
24
30
10
300
2123
25
34
11
374
334
26
38
14
532
560
27
42
14
588
8.49
28
47
16
752
1350
29 30
52 58
18 20
936 1 160
2075 3322
3
66
23
1518
5553
32
72
25
1800
8600
33
82
28
2296
13000
34
91
32
2912
210.00
35
03
36
3708
32800
36
2
40
4480
51250
37
26
45
5670
80000
38 39
140 160
50 58
7000 9280
1250.00 283000
40
178
64
1 139 2
3385.00
4
199
72
14328
524000
42
220
80
7600
812800
43
247
90
22230
1 3130.0 0
44
281
102
28662
20430.00
45
325
118
38350
3220000
page 106
EI- LAMINATION MAXIM U M TUR NS FOR BOBB IN WOU N D COILS ( °o FIL) MEAN LENGTH TURN
=
"
Wire Gage
Turns/ Layer
17 18 19 20 21 22 23 24 25
16 18 20 21 23 26 30 33
26 27 28 29 30 31 32 33
42 46 52 58 65
34
35 36 37
38 39 40 41 42
37
73
81 91 101 115 128 143 159 183 204 227 255
Max. Layers
Max Turns
5 5 6 6 7 8 9 10 11
80 90 120 126 161 208 270 330 407
13
546 644 832 1044 1300 1606 2025 2548 3131
14 16 18 20 22 25 28 31 35 39 44
49 56 63 71
79
4025 4992 6292 7791 1 0248 1 28 52 16117 20145
Resistance square stack 1189 168 284 4135 656 1085 1773 2743 4265 725 1052 1750 2740 4360 687 10650 16930 26450 43050 67000 104000 163000 283000 448000 67000 1080000
page 107
EI- LAMINATION MAX I M U M TURN S FOR BOBBIN WO U N D COILS ( °o FILL) MEAN LE NG TH T UR N
=
" Max Layers
Max Turns
Resistance square stack
Wire Gage
Turns/ Layer
16 17
17 20
6 6
102 120
.1435 .213
18
22
7
154
. 34 9
19
25
8
200
.564
20
27
8
216
.8425
21
30
9
270
137
22
34
10
340
2157
23
38
11
418
327
24
43
12
516
5.10
25 26
48 54
14 16
672 864
836 14.60
27
60
17
1 02 0
2020
28
67
20
1 340
33.60
29
75
22
1650
51.60
30
84
25
2100
84.00
31
95
28
2660
135.80
32
1 04
31
3224
201.50
33
118
35
4130
327.00
34 35
1 32 148
39
44
5148 6512
51650 83000
36
165
49
8085
129000
37
1 84
54
9936
1959.00
38
205
61
12505
3121.00
39
236
70
16520
5370.00
40
263
78
20514
8510.00
page 108
EI- LAMINATION MA XIM U M TURN S FOR BO BBIN WOU N D COIL S ( °o FILL ) MEAN LENGTH TURN
=
" Max Layers
Max Turns
Resistance square stack
Wire Gage
Turns/ Layer
16 17
20 23
6 7
120 161
202 .342
18
25
8
200
.536
19
28
9
252
852
20
32
10
320
136
21
36
11
396
213
22
40
12
480
3.25
23
45
14
630
5.39
24
50
16
800
863
25 26
56 63
17 20
952 1260
1294 2159
27
70
22
1540
3329
28
79
25
1975
5383
29
88
27
2376
8165
30
98
31
3088
3168
31
109
34
3706
20250
32
122
38
4626
319.40
33
37
43
5891
51200
34 35
155 175
48 55
7440 9625
81520 132960
36
193
61
11773
20 51 00
37
212
67
14204
3120.00
38
222
76
16872
467350
39
271
85
23035
80460
40
301
95
28595
1259700
page 109
EI-00 LAMINATION MAXIMU M URN S FOR BOBBI N WOU N D COILS ( °o FILL) MEAN LENGTH URN
=
" Max Layers
Max Turns
Resistance square stack
Wire Gage
urns/ Layer
15 16
22 25
7 7
1 54 175
227 .326
17
28
8
224
.526
18
31
9
279
826
19
35
11
385
1.44
2
39
12
468
22
21
44
13
572
34
22
49
15
735
5.5
23
55
17
935
883
24 25
61 69
19 21
1159 1449
1381 2177
26
77
24
1 848
35.
27
86
26
2236
53.42
28
96
3
288
86.75
29
17
33
3531
1341
3
12
37
444
2127
31
133
42
5586
3373
32
148
46
688
5186
33
34
167 189
52 59
8684 11151
834. 135.
35
213
66
1485
2146.8
36
236
74
1 7464
33624
37
26
81
216
51 13.
38
294
92
2748
8281
39
331
13
3493
13163.
4
368
15
4232
266.
page 110
EI- LAMINA TION MAXI MU M TUR NS FOR BO BBIN WO U N D COIL S ( °o FILL) MAN LENGTH TURN
=
."
Wire Gage
Turns/ Layer
Max. Layers
13 14
19 22
6 7
114 154
126 214
15
24
7
168
294
16
27
8
216
477
17
31
10
310
863
18
34
11
374
1 32
19
38
12
456
202
20
43
14
602
336
2
48
5
720
507
22 23
54 61
17 19
918 1159
815 1297
24
68
21
428
2016
25
76
24
1824
3247
26
85
27
2295
5151
27
95
30
2850
8067
28
1 06
34
3604
12860
29
119
38
4522
20350
30
33
42
5586
31700
31 32
147 164
47 52
6909 8528
49440 76970
33
185
59
10915
124200
34
209
67
14003
200900
35
235
75
17625
318900
Max Turns
Resistance square stack
page 111
EI- 5 LAMINA TION MAX IM U M TURN S FO R BOBBIN WOU N D COIS (5 °/o FILL) MEAN LENGTH TURN
=
"
Wire Gage
Turns Layer
12 13
19 21
6 7
114 147
109 177
14
24
8
192
291
1
27
9
243
46
16
30
10
300
723
17
34
11
374
1138
18
38
12
46
1 7
19
43
14
602
291
20
48
1
720
439
21 22
3 60
17 19
901 1 14 0
693 1100
23
67
21
1407
1721
24
7
24
1800
2776
2
84
27
2268
4410
26
94
30
2820
691
27
1 0
34
370
11040
28
118
38
4484
17484
29
131
42
02
2700
30 31
147 163
47 53
6909 8639
42840 6730
32
182
59
1 07 38
1 0 860
33
20
66
1 3 30
1 682 00
34
231
7
1732
27180
3
261
84
21924
433380
36
289
94
27166
677000
Max Layers
Max. Turns
Resstance sq ua re s tack
page 112
EI- AMINAION MAX I M U M U RN S FOR BOB BIN WOU N D COIS ( °o FIL) MEAN EN GTH URN
=
."
Wire Gage
urns/ ayer
Max Layers
1 2
7 19
6 7
1 02 1 33
082 135
3
21
7
47
188
4
24
8
92
310
15
27
9
243
494
16
30
1
330
846
17
34
2
408
132
18
38
13
494
201
19
42
15
630
324
20 2
47 53
17 19
799 1007
517 823
22
59
21
1239
1276
23
66
24
584
2058
24
74
27
2268
3716
25
83
30
2490
5 4 5
26
93
34
362
8237
27
03
37
38 1
12520
28
6
42
4872
2080
29 30
1 30 145
47 53
6110 7685
31925 50625
31
161
58
9338
77550
32
179
65
11635
121880
33
202
73
14746
194750
34
228
83
18924
35160
35
258
94
24252
509320
36
285
104
29640
784800
Max Turns
Resistance square stack
page 113
EI- LAMINAION MAXIM UM U RNS FOR BOBBIN WO UN D COILS ( °o FILL) MEAN LENGH TURN
=
"
Wire Gage
Turns/ Layer
Max Layes
10 11
18 20
6 6
108 120
.076 .106
12
23
7
161
179
13
26
8
208
292
14
29
9
261
461
15
33
10
330
.735
16
37
12
444
1.25
17
41
13
533
1.89
18
46
15
690
308
19 20
51 58
17 19
867 1102
489 7.82
21
64
21
1344
1204
22
72
23
1656
18.70
23
81
26
2106
30.00
24
96
29
2784
5000
25
101
33
3333
7552
26
114
37
4218
12049
27
126
41
5166
186.10
28 29
142 158
46 52
6532 8216
296.70 47060
30
177
58
10266
741 60
31
197
64
12608
1 148. 20
32
219
71
15549
178600
Max Tuns
Resistance square stack
page 114
