TANSFOEIR
16N
ND MAUACR UAL
ROBERT G. WOLPERT
© 198, ROBET G. WOLR Re
20
TR�SFORE DIGN AND MANUFACTUING MANUAL
ROBERT G. WOLPERT
© 1984, ROBERTG. WOLPERT Rev.2004
A PRACTICAL APPROACH TO TE DESIGN AND MANUFACTURE OF ELECTRONIC TRANSFORMERS AND INDUCTORS
B . W
PREFACE
he first part of this manual is intended to serve as a starting point in learning a method of designing transformers and other wire wound magnetic co mpon ents he second par t is incl uded to show the new designer, and others, how the transformer is manufactured The design guides, if followed carefully, will result in a design that will work as intend ed her e will be very little atte m pt to expl ai n the theor ies or ju stify the method or form ul as use d I have used these method s successfully for many years in both the designing of transformers and in the t rai ni ng of ne w desig ners I suggest that additional study into the theory of magnetism and transformer operation be conducted by the serious engineer he second part concerns the actual manufacturing of the transformer after it is years desig ned he methods ofshown he re have extensivel y for many in the manufacture transformers andbeen otherused wire wound mag netic com pone nts hey hav e prov en very useful as train in g aids for new employees and as manufacturing procedures for standard shop practices he manufacturing methods are also an aid to the new designer in showing how the transformer is actualy put together and a help in choosin g the pro per material s for const uction
Robert G. Wolpert
page 1
TABLE OF CONTENTS PART I . TRANSFORM ER DESIGN PROCE DU RES
CHAPTER
10
TRANS FOR ME R OPE RATO N T EO RY
CHAPTER
20 21 22 23 24 2.5
SNGLE PHASE POWER TRANSFORMER DESIGN Design procedure Design example Sample manufacturng specifications Lam ination table Wire table
CHAPTER
30 31 311 32 322 33
THRE E PASE POW E R TR ANSFO RM ER DESI GN The Wye configuration Wy e ex amp le The Delta configuration Delta example Three phase design example
33.1 332 3 .4
Temperature rise Regulation nterconnections
CHAPTER
4.0 41 42
AUTO TRANSFORMERS Design procedure De sign example
CHAPTER
50 POWER TRANSFORMERS SNG CAPACTIVE FITERS 51 Types of rectif ie r circ u its Fu ll wave ce nt er- ta ppe d circuit 511 512 Example 513 Full wave bridge circuit 5.14 Fu ll wav e bridge cent er- ta ppe d circuit 5.2 Correcting the efficiency 5.21 Example
CHAPTER
60 61 6 2 63 64 6.41
CONVERTER TR ANSFOR ME RS The saturating transformer Co re mater ia l Control winding voltage Design crit eri a Design recap
6. 5
The nonsatu rating tra nsf ormer
.
page 2
TABLE OF CONTENTS, continued
CHAPTER
70 71 711 72
SH E LDNG N POWER TRANS FORM ERS El ectrostatic shieds Box shieds Electro-magnetic shieds
CHAPTER
80 8 82 83
IRO N CORE FLTER CHOKES Chokes that carry a direct current Chokes with no direct current Conf igu rations o ther tha n E I la mi nations
CHAPER
90 9.1 92
AIR CORE INDUCTORS AND SOLENODS The singe layer solenoid Mutiple layer solenoid
page 3
PART II . MAN UFAC TURING PROCESSES
CHAPTER
10.0 101 102
LAYER WIND ING O N A SI NGE COI FORM Electros tat ic shie ld Wire table
CHAPTER
110
BOBB IN WINDING
CHAPTER
120 121 122
LEAD FINISHING Stranded lead wire terminatons Solder l ug te rmination s
CHAPTER
130 131 132 133
ASSE M BY AND STACK IN G O F MAGN ET IC CORES Stacking of laminated cores Assembly and bracketing Assem bly w th a f l ux shel d
CHAPTER
140
IMPREGNATION
CHAPTER
15.0
TES TIN G THE TR ANSF ORM ER
CHAPTER
160
I NS ULA TIO N MATE RIA S
APPEN DIX TABLES FOR TUR NS AND WIRE SIZ ES VER SUS LAMINATION SIZES
page 4
PART I.
TRANSFORMER DESIGN PROCEDURES
page 5
CHAPTER 1.
TRANSFORMER OPE RATION T H EORY
An deal transformer, (one in which there are no losses), will transform AC voltage directly proportonal to the turns ratio and will transform current inversely proportional to the turns ratio Thus, the votage of the secondary dvided by the voltage of the prmary is equal to the turns of the secondary dvided by the turns of the primary and the current of the secondary divded by the current of the primary is equal to the turns of the prmary divded by the turns of the secondary. Es/Ep
=
Ns/Np and Is/Ip
=
Np/N
Where : Es Ep
-
Secondary voltage Prmary voltage
Is Ip
=
Secondary current Primary current
=
Ns Np -
Secondary turns Prmary turns
The total VA of the prmary will be equal to the total VA of the secondary. Thus, the voltage of the primary multpled by the current of the prmary s equal to the voltage of the secondary multiplied by the current of the secondary Ep x Ip
Es x Is
Ep
Es
Ip
Is
Figure 1
page 6
For a transformer wth more than one secondary, the total VA of the primary wil be equal to the sum of the VA's of the secondares.
Ep x Ip
- Es x Is + Es x Is, etc
The turns ratio wll be: Es Ep
=
N s N p an d Es E p
Ns /Np, etc
=
In other words, the votage of second ary # 1 div ide d by the voltage of th e primary is e qua l to the turns of secon da ry # 1 div ided by the turns of the primary and so on for all secondaries The pola rity dots of F ig u re 2 show in sta ntane ous poa rity of the votages . •
3 1
•
l1 Ep
4
5
Ip
Es
2
Is 6
Figure 2
If arbtr ary va ues a re assgned to the wn Let
Ep Np Es Is Es Is
-
di ngs
1 15 vo ts 230 turn s 6 5 vo lt s 5. 0 am per es 45 volts 1 . 0 ampe res
page 7
Ns/Np to Np/Ep Ns/Es, we will have a formula By transposing Es/Ep for turns per volt Thus , the turns of the prima ry divided by the voltage of the prim ary is eq ua l to the turns of the second ary divided by the voltage of the secondary By filling in the assigned values, the secondary turns can be calculated
230 115
Ns 65
Solvng for secondary tu rns : Ns And Ns
(230 x 45) 115
(230 x 65) / 11 5
90 turns
The total VA of the transformer is
6.5 V x 5 0 A 45 V x 1 0 A Total
Then the prm ary cu rrent, Ip
=
- 1 3 turns
775 / 1 1 5
- 325 450 77.5 VA
067 ampees
These calculations neglect any losses in the windings and the core Polarity dots shown by each winding in the schematic diagram indicate the polarity at any given instant of time and will dictate the winding direction Therefore , if it is de sired to determine the pol a rity or winding direction by measurement, connect the windings of Figure 2 as shown below i n Fgu re 3 . This then becomes an auto t ransformer 1
•
2 3 4
6
Figure 3
page 8
f 1 1 5 volts is app le d to # 1 an d # 2, then you wil l read From # 1 to #4, 1 1 5 / 230 x (23 0 13) 12 1. 5 volts From #1 t o #6, 1 15 / 230 x (230 + 13 + 90) 16 6 5 volts
If the pola rity is r eversed on #3 a nd # 4, then you wil l read : From #1 to #4, 1 1 5 / 230 x (2 30 From # 1 to #6, 1 1 5 / 23 0 x (230
- 13) 10 8. 5 vo lts - 13 + 90) 15 3. 5 volts =
=
A center tapped winding can be treated like two separate windings for determnng the turns ratios and polarities These methods shown here will apply to all transformers that are being tested wth out a oa d . If loaded voltages an d cu rrents a re needed, it is necessary to through a much and more sophisticated method determining the go losses in the windings lamination and calculate the of voltage drops his is expla ined f u ll y in the fol lowi ng cha pters.
page 9
CHAPTE R 2. SING LE PHASE P OWER TRANSFORM ER DESIGN This chapter will give a stepbystep design procedure for a single-phase power transformer of the type generally used in OEM electonic equipment The u ser 's specifications will ca ll out wha t is desired Sometimes this wi ll only be the input voltage and frequency and the required output voltage an d curr ent. If thi s is the case, the de sig ner c a n ch oose the size of the core that wi ll best fit In other cases, the si ze is als o given al ong with temperatur e ris e and regu latio n . This is mo re re st ric tiv e and wi ll requi re more time and calculations to fit the requirements. The procedure that follows will enable the designer to arrive at a design that will fi t the req ui rement s if the ste ps a re f ol lowed care ful ly . Providing, of course, that the specifications are not so restrictive that they are impossible to meet
2 1
Des ig n pr oc edu re
This is a stepbystep procedure for the design of a power transformer for use in elect rical a nd el ectr on ic equ ip ment If the proc edu re is fo lowe d carefuly, it will result in a transformer that will function as desired. The design procedure is first given and then an example is shown using the pr oce du re #1 Assemble and put down o n paper all to be designed
the inf ormat ion avail able on the unit
#2 Calculate the total VA and the primary current VA = Es x Is Where:
Ep Es Ip Is
I p VAX 111 Ep
-
-
primary voltage secondary voltage prima ry curren t secondary current
1 1 1 is used as a f a ctor to cover losses f or a 90 °/ efficie nt tra nsform er
page 10
#3 Cho ose a cor e size from the V A colu mn of the la mi nation ta ble a t the end of ths cha pter Record the wn dow size and core a rea of the la m nation chosen. Calculate the effective area of the core Aeff = Ac x K Where:
Ac = a rea of the co re (Ton gu e width x stack heig ht) K = the sta cki ng facto r (Use 92 for 1 x 1 in tereave an d 95 for a butt stack )
#4 Calcul ate the prim ary tur ns for th is core and voltag e desire d . Ep x 10 8 Np = 444 x B x A x F Where:
444 is a constant for sne wave operation E p - prim ary voltage from the power s ou rce - flu x densit y in in es per squ are inch . (This B value wi ll depend on th e grade o f ste el used. ) - effective area of the core A - li ne fre q uency F
#5 Calculate the turns for the secondary or secondaries Ns = J x 1.05 x Es Ep Where: Np - primary turns Ns - secondary turns 105 is a factor used to adjust the turns to compensate for the losses This will vary with the size of the transformer and the desired regulation
page 1 1
#6 Choos e the wire sizes Use 800 cir cu a r mil s per a m pere for a sta rti ng point Refer to the wire tab e to fin d the sizes needed Fo r exam ple, if a cu rrent of 1 ampere is required, then AWG #21 would be chosen as it has a volume of 81 0 circuar mis #7 Determine coil ength, winding length, margins, turns per layer and nu mber of ayer s (See wire table on page 36) #8 Caculate fil l This is done by adding up the various element s of the wind in g. These are the win din g tube thic kness, dia meter of the wire multiplied by the number of layers, layer insulation thickness and interwin din g ins u latio n th ickness An acc epta ble f l is f rom 80 to 90°/ . If fil l is not a cce ptabl e, adju stments must b e mad e This can be done in severa different ways, either by more or less core stack, a change in core size turns adjustment, which wil change the flux density or wire sizes. In any case, if adjustments are made, you must be careful that all other paramete rs a re consi dere d to keep f rom exceeding an y a l lowabe l im its #9 Calculate wire res sta nces an d vo ltage dr ops in each wind in g. These may ead to further adjustments to bring the secondary voltage to the desired values #10 Calculate the copper an d lam ination we ights. #11 Calculate the osses
page 12
Step #12 Cac ulate the a pp rox ima te temperature rise. emperature ri se i n degree s centigrade:
TOTAL LOSSES 01
[
TOTAL WEIGHT 1.073
J
2/3
he total losses are the combined copper losses and the lamination or iron losses The wei g ht is the weight of the copper wi re plus the w eight of the la mi nation pu s 1 5°/ for insu latio n, wi nd in g tube, b rackets , etc
#13 Calcu late the r egul ation he regu lation is the secon da ry fu ll load vol tage subtracted from the no oad voltage and that result divided by the full oad volt age 01 0
Reguation -
No loa d full lo ad Ful oad
x 100
Now proceed to 22 and the exampe, folowing the above steps.
page 13
2.2
Design exam ple #1
It is desred t o desi gn a tra nsformer to operat e from a 1 1 5 Volt line at 60 Hertz an d to de live r 6 . 3 Volts at 10 a m peres AC, cent er tapped The physical size is not give n . Record a l l inf ormato n.
115
6 .3 V ct. @ 1 0 0 A
v
6 0 Hz .
Schematc dagram
Ep = F = Es = s =
115 V. 60 Hz 6 3 V c.t 10 A
#2 Calculate total VA VA = Es x Is - 63 x 10
- 63
Caculate prmary current p
=
VA x 111 Ep
63 x 1.11 115
0. 608 A
page 14
#3 Choose a core from the lamination table From th e VA colu mn it is seen that E 1 1/8 size with a 1 1/8 s tack height has a VA rati ng of 65 This should be a good core for this transformer The effective area is 1 1/8 x 1 1/8
1.265 sq. in. x K.
Assuming a 1 x 1 interleave, it will be 1265 x 92
=
1 16 4 sq. in
At this time it becomes necessary to explain the different types and grades of lamnations available The standard EI lamination is called out by the thickness and grade of steel used Thus, 29 M6 is a lam ination of 29 gaug e ( 0 14" thick) an d made of a grade M6 grain-oriented steel. There are several grades and thickness of laminations available, but, in most electronic transformers, and in this treatse, we will consider only th ree These a re M6, M 19 an d M 22 grades and the thi ckn esses used are 29 gauge ( 04"), 2 6 gauge (. 0 18" ) and 2 4 gauge (. 025") Only the M6 grade is grain - oriented and thus can be used at a higher flux density and wil l result in a sm al ler un it. Thi s grade o f steel a lso c osts mor e per pound, so a compromise is sometimes necessary between size and cost The core loss table will show the flux densities that can be safely used and the losses per pound that will result with each of the different grades of laminaton More extensive data can be obtained from the various man ufacturer of i nes the la nations data given he re is onl y inten ded to serve as gu sidel formithe begin niThe ng des igner. For our purposes, only two levels of flux density will be shown Sat ura tio n wi ll oc cur a t ap proximat ely 18 kiloga uss (1 16 , 10 0 lines ). This should be con sider ed the li miting val ue
page 15
#3, con tinu ed The manufacturers wil give the core losses at flux densities in kilo-gauss. This can be converted to lines by multiplying by 645. For examp le, 15 K G or 15000 Gauss
x 6.45 = 96750 lines
This, of course, can be converted directly in the formula for primary turns by using the Gauss figure and adding the 645 factor below the line The window of the am ination i s 9/1 6" x 1 1 1/ 16" The effective core area is 1 1 64 sq i n . We will choos e to tr y 2 6M 19 lam inations with a fl ux density o f 85000 li nes as a sta rting point This is 13. 17 KG.
#4 Calcuate the primary turns. Np =
Ep
x
444 x B x A x F
x .
.
x
x
x
turns
#5 Calculate the secondary turns. Ns = Np / Ep x 1 .05 x E s = 4 36 / 115 x 105
x 6 3
=
25 1 turns
Since the secondary must have a center tap, we should change the turns to an even num ber or 26 turns Actual ly, a ha lf-tu rn can be obtai ned in a transformer using this lamination by bringing out the lead on the opposite side of the core. This is usua ll y not done un less absol utely nece ssary. In order to have a full turn, the wire must pass through both sides of the window A hal fturn wi ll pa ss th roug h on y one side If the cust omer cals out t he ead position, the deci sion will be made for you .
page 16
#6 Choo se the wi re sizes The pr ima ry wir e should be 608 A x 8 00 circula r mil s
=
486.4
For this we will see that #23 wire is closest with 509.5 cm (see the wire tabe) The secondar y wir e should be 10 A x 800 cm . 8000 cm # 1 1 wi e has 8234 cm and will be chosen for this It sho uld be noted that th is is conservative and in practice thee s room for adjustment up or down if needed . The only l im iting fa ctors will be tempe ratur e rise and re g ulatio n.
#7 Calcu late the turns per layer a nd n um ber of layers he win dow length s 1 1 1/1 6" ong . In order to fit, the coil len gth shoul d be 1/ 16" sho ter or 1 5/8 " on g From the wi re tabl e, it is seen that the m argi n for #23 wi re sho ul d be 1/8" on each end The margi n for # 1 1 wire sho ul d be 1/4" on each end The turns per layer i s determin ed by the wind in g length times the turns per inch for that wi re siz e. This is also obtai ned f rom the wire ta bl e. The val ues sho uld be put down on the work sh eet clearly to show the con struc tion of the coil Figure 4 shows the size of the laminaion and the way the coil is constructed (E-1 1/8) 0
1
-la
V
0
0
4
Figure
page 17
Wndow = 9/1 6" x 1 1 1/1 6" Coil length = 1 5/8" #23 wre wndng length = 1 3/8" Mar gn s = 1/8" each end Tur ns pe r layer = 1 3/8" x 37 .4 (turns per nch r om ta ble) = 52 t urns Layers = 436 / 52 = 84 layers (use 9) #11 wre wndng length = 1 1/8" Margns = 1/4" each end Turns per layer = 1 1/8" x 10 . 2 = 11 turns ayers = 26 / 11 = 236 layers (use 3) #8 Calculate t he f l l of the wn dow by add ng up all the various thcknes se s of the wndng tube, wre dameter, layer nsulatons and wrappers (See Fgure 4) The layer nsulation s determned by the thckness needed to support that pa rtcul ar w re sze . Ths s ca led out n the w re tabl e The wnding tube thickness s determned by what s needed to support the col Sm al l cols wth f ne wre nee d less sup por t than la rger cols wt h hea vy wre Ths can vary from 020" to .0 70" or more. A co l of the sze used in this example wll generally use a wndng tube thckness of 030" to 040" The wr a pper s the ins ula ton used be twe en w nd n gs Ths is det ermned by the voltage solaton needed and the support needed for the next wnding For thsofe#xampl e w e wll use .0 10" thck nsulaton, as t va lue for. support 1 1 wre.
hs s t he
It s now necessary to put down on paper the varous thcknesses, add them up and calculate the percentage o avalable space n the wndow that s us ed Ths l am n aton has a wnd ow wdth o 9/1 6" or, n decma ls , .5625".
page 18
Winding tube= 9#23 wire Layer insu l. Wrapper 3# 1 1 wir e Layer insul. Wrapper
-
0400 2 160 (9 x diamet er of wir e) 0240 (8 layers x 003" paper from table) 0100 2787 0200 (2 x .010 paper) (outside wrapper)
Total
6037
-
The available space is .5625" 603 7 56 25 x 1 00 = 10 7° his is obviousl y too mu ch So some adjustments wi ll have to be made. As mentioned pr eviously, ther e are seve ral choi ces A larger la mi natio n can be chosen, a better grade of lamination can be used, or a larger stack of the sa me la mi nation ca n be us ed . The choice is not al ways le ft to the designer If the cust ome r has cal led out this size , then the choice wi ll be a higher gr ade o f la min atio n his wi ll al low a h ig her fl ux densit y to be used an d there fore fewer turn s. We wi ll ch oose to use a h ig her grade o f lamination of the same size and return to Step #4 to modify
Recalculate the primary turns using a new fux density of 95000 lines This i s 1 4. 7 KG and is an a rbitrar y choice. It coul d go as high as 17 KG and still be acceptable. Np
4.44
x
x
x
x
= 390 turns
#5 Ns = 390 / 11 5 x 1 0 5 x 6. 3
=
22 4 t urn s (u se 22)
#6 Wire sizes will not change for this modification; #23 and #11 will still be used
page 19
#7 Coil engths and margins will be the same. remain th e sa me #23 wire wi have 390 / 52 #11 wire will have 22 / 11
= =
Also, the turns per layer will
75 layers (use 8)
2 layers
#8 Reca cu ate the fil Tube 8# 23 w ire Insu Wrap 2#11 wire Insu. Wrap Total
-
0400 1920 0210 0100 .1858 0100
-
0150 4738
-
4738 5626 x 100
=
84°
This is an acceptable fill #9 This step is to cacuate the voltage drops caused by the DC resistance of the windings and adjust the turns, if necessary, to obtain the proper voltage under load. n order to get the voltage drop in each winding, it is first necessary to pic tur e the bu il d- up o f the col a s calcu late d in Ste p #8 Ths bui ld- up s accomplished in the following order: 1 Windi ng tube 2 . Prima ry wire separa ted by the layer insu lation 3 Wrapper between wind ings 4 Secon da ry wire sepa rated by the layer ns u lation 5 Fna ly, the outsde wrapper
page 20
If we take the buldup and add up the varous sectons, we can arrve at a mean length turn for each wndng Figur e 5 sh ow s a vew of the tube on whch the w re s wou nd
.040"
1
1 -
8
1 040"
8
Figure 5
In order to smpfy the calculatons, t s advisable to reduce the windng tube t o a squa re, f it s not already one. Ths s d one by t a kng the t ota l dsta nce around an d d ivdn g by 4 Ths wll g ve you an equ valent dmenson of one sde For example, a wndng tube that s 1 1/2" x 1 3/4" would be: 1 . 625" eq uvalent sq uare. 1 1 /2 x 2 + 1 3/4 x 2 3 + 3. 5 or 6.5 / 4 =
=
In our ex am ple, th e tube s al ready square s o w e w ll t a ke the 1 1/8" dmenso n a nd bul d- up f rom ther e. Sta rtn g with the sze of the la mnaton an d a dd ing t he w nd n g tu be thckness to each side, the actua dmenson of the wndng wll be ob taned The wire an d i nsu laton s added on t op of th s. Lamnaton Tube x 2 8-# 23 wr e Insul Total
1.1250 .0800 1920 .0210
( 1 1/8")
1.4180
page 2 1
This give s the buil d up in one direc ti on of the pr im ary win di ng. When t his numb er is multiplied by 4, it wil give the length of one turn in the center of the wi nd in g, or the mean l ength turn of the prim ary wre . Thus, 1 4 180 x 4 x 390 turns wil l gi ve the engt h of the wire in in ches If this is multipied by the resistance per 1000 inches from the wire table and divided by 1000, it will result in the DC resistance of the winding. 1 41 80 x 4 x 390 x 1 6966 1 000
375 Ohms
This vaue, 14180, is the buildup to the center of the primary winding, so the primary v a lu es m ust be added in again to get to the star t of the se conda ry wi nd i ng The entire buil d up is n ow re pea ted to clea rly show the calculations. Lam. Tube 8-#23 Insu. Total
8-#23 Insul Wrap 2#11 Insul Tota l
-
=
-
11250 0800 1920 0210 .
x
x
x
I
. x .
.
v
drop
1920 0210 0100 1858 0100 1 8368 x 4 x 22 x 10 50 10 00
0 169 x 10
169 V dr op
These votage drops can now be used to determine the output voltage under o aded conditio ns This i s do ne by su btr act in g the prma ry drop from the input voltage and, from the turns ratio, obtain the secondary volt ag e The sec on da ry voltage d ro p is then su btra cted f rom th is va lue to obtain the loaded v oltage. Th us 11 5 - 228
112 72 V . This is the e ffect ive i np ut voltage
Fro the turns ra ti o : 1 1 2 72 / 390 x 22
6358 V
Sub trac ting the se condar y dr op 6 358 169
6189 V
This is lower than the 63 volts desired so adjustments must be made.
page 2 2
Since the secondary must be center-tapped and an even number of turns is desred, it wil l be bett er to adjust the prim ary turns This is done by di viding the calculated votage by the desired voltage and mutplying this by the prima ry turns 689 6.3
982 x 390
Recalculating the 11272 / 383 x 22
383 turns
loaded volt ag e: =
64 74 169
6. 305 volt s
It should be noted that this also changes the flux density by a factor of 390 383 This wil result in a flux density of: 95000 x 390 / 383
96736 lines or 15 KG
Ths is well within the mitations of M6 material. The secon dary turns c a n al so be cha ng ed by this method . It is some times necessary to adjust both the primary and secondary turns when there is more than one secondary with a small number of turns This can result i n a jug gl in g of turns bac k and forth to get the desir ed results. Ths small change in turns will not change the winding layers or configuration #10 The wire an d la mi na tion weights can now b e ca lcu lated The wir e weight s obtained by using the calculated resistance and, referring to the wire table for #23 wire, it is seen that it has 12.88 Ohms per pound By dividing the calculated primary resistance by this value, we will get the weight o f the wire The same is done f or the # 1 1 wi re of the second ary For the #2 3 prim ary wir e : 3 75 / 1 2 88 For the #11 secondary wire: 0169 / 0500
.291 pou nds .338 pounds =
page 23
he weight of the lam in ation is obta ined f rom the lam in ation tabe or a manu facturer's cataog . For a sq ua re stack of EI- 1 1/8" la mi nati on, the ta be sh ows a weigh t of 2 24 poun ds . his must b e modified by the stacki ng factor This is 9 2 for 1 x 1 interleavin g 224 x 92
=
2 06 pounds o f lam ination
#11 The osses can n ow be cal cua ted The core loss is obtaine d from the man ufact urer s cat a ogs . Some of these are listed at the bot tom of the wire tabl e It can be s een th at M6 am in ation at 1 5 KG is 66 watts per 1 36 watts po und hen the os s wil l be 2 06 x 66
he wind ing or copper os ses are obtai ned by mu lt iplyi ng the voltage dr op by the load current. Primary
Secondary
228 x .608
1 69 x 10
1.39 watts 1 . 69 wat ts
Addin g the loss es Core loss Pri loss Sec. loss Tota
-
1. 36 139 1.69 444 watts
The total weight is: Core #23 wire # 1 1 wi re Total
2060 0291 0338 2689 x 1.15
3.09 pounds
The 1 1 5 is the 1 5° adde d for in su lat ion , brac kets, etc.
page 24
#12 The temperature rise can now be calculated. 4 44 0 1
(
309 1 . 073
)
2 /3
- 21 . 93 de grees C ise
This is very conservative No rma l tran sfo rmer ma terial ca n stand a total tem perature of 10 5C. This incudes the tempe ratur e of the surroundings, or ambient temperature, plus the temperatue rise of the tran sformer For exa m ple, if the am bient is 50 de gees, then the rise can be 55 deg rees M ost specif icati on s wi ll c all o ut eith er the am bient or the maximum temperature rise acceptable, or both. #13 Calcuat e the re g ulation d. by This secondary loadnovolta mi nus the fu ll load voltage, divide thei sfuthe ll load voltag e.n oThe loa dgevoltage has not been ca cu lated previous ly so it m ust be do ne now This is done by calculating the turns ratio without any voltage drops: 1 15 I 383 x 22
6605 V no oad
Then 66 05 - 6.305
30 / 6305
047 x 100
=
4. 7° regu latio n
A normal transformer of this size would probably be desired to be within the 5 to 0° ra nge of regulatio n. This wil l al so nor ma lly be c a lled out in the specification if it is critica I. This example is one of the simplest designs, but it clearly shows the design method Mo re comp li cated desig ns ca n readi ly be accompl ished by fol lowing the st eps an d ad apting them to fi t the requi rements
page 25
2. 2 1 Bobbin construc tion met hod of desg n
The design e xamp le wa s sho wn as be ing a layer wound coil . This could aso have been designed on a bobbin. A bobbin-wound unit can be smaller than a ayer wound unit because there is no need for end margins as the bobbin flanges protect the windings from the lamination, and also, bobbins are usually random wound whic h eli mi nates the n eed for layer in su lation . Wrap pers between windings are stil necessary for electrical isolation. Some extra preca utions m ust be taken in man ufac tu ring This wi ll be discusse d in Chapter 11 on bobbin winding To demonstrate the differences, let us return to the design example that resulte d in a f il l of 1 07°/ The bobbin for this size lam in ation wil l ha ve a winding length inside the flanges of 1545 inches (see the manufacturer's catalog). Tur ns pe r layer for # 2 3 wir e wil l be 374 x 1 545 Layers 436 I 58 8 =
Turns per ayer for #11 wire will be 10.2 Layers 26 I 15 2
5 8 turns
x
1.545
=
15 tur ns
Recalcu lat ing the fill Winding tube 8#23 Wrapper
=
=
-
and eli min ating the laye r insulation : 0400 (bobbin) 1920 0100
2#11 Wrapper
1858 .0150
Total
.4428 I . 5625 x 100
=
79°/ fill
As you can see, the use of a bobbin in this transformer would result in being able to use a lower grade of lamination, since the flux density can be lower
page 26
A more dramatic change can be obtained by reducing the core one size and increasin g the fl ux densit y This might r esul t in u sin g sma l ler sizes of wire an d the refore hi g her copper losses The tem perature rise woul d have to be calcu lated to see if this is ac cept ab le . Care must be taken to conform to the customer's requirements and to not exceed good design practices 2.3
Frequencies other than 60 Hertz
The la mi nation tabl e gives the approx imate sizes t o be used a s sta rting points for a pa rticul ar VA ratin g . This ta ble is f or 60 He rtz operat ion If the frequency is other than 60 Hertz, adjustments must be made From the tr a nsformer turns f ormul a it wou ld seem that the size woul d be inversely proportion al to the freq uen cy. A loo k at the man ufacturer 's core loss curves show that this is not true As a rule of thu m b, a 40 0 Her tz un it will be app roximately one-ha f the si ze of a 60 He rtz u nit , an d a 50 Hertz u n it wil l be a pproxi mately 1 0°/ larger than the 60 H ertz un it Thes e are only a pproximations and a re to be used for choosing a lamination size as a starting point in the design 2 4
Exci tati on cu rre nt
The excitation current, or the current needed to drive the core to the proper flux den sity, has not been previousl y d iscussed Som e customers wil l put a maxi mu m exci tati on curre nt li mitation in the ir speci fica tions If this is the case, this current must be caculated. The manufacturer's catalogs will give some values or have curves that show the values to be expected for various flux densites. Unfortunately, these curves usually are test resuts of the core materia under ideal condition s. The val ues obtai ned from these cu rves must be modfi ed to accommodat e the act ua l conditions.
page 27
Consd er ng the thre e pre vously mento ned g rades of la m natons, and a n nterleave of 1 x 1, the followng multplers should be used: For 29M6 matera at 15 KG, use 3 tmes the value shown. For 18 KG, use 1 0 tmes th e value s For 26M 19 an d 24M22 mater als at 10 shown
KG, use
1 5 tmes the values
For 14 KG, use 3 tmes the values. For values of flux densites n between those gven. t wll be necessary to interpolate There are many thngs that wll cause the exctng current to vary other than the fux densty and the grade of the co re materal . For exa mp e, if the am naton s are bent or not stacke d tg htly t o reduce the gaps, the excitng current wll ncrease drastcally. Unfortunately, the gu de nes a bove a re a pproxmat ons Experence wi ll re sul t n a better feel for what to expect under actual condtons. The followng pages contan the wre and lamnaton tables and a complete manufacturng specfcaton for the layer wound transformer that w as desgned
page 28
WINDING SHEET EXPLANATIONS
()
The tube siz e is giv en as 5 " x 1 . 15" x . 040". This size is 025" larger than the core to allow the laminations to be inserted without dig gin g into t he tub e. The . 040" is the thic kness of the tube
(2)
The leads a re shown com in g out of the c oil Noti ce that the primary winding is 8 layers, therefore, the start and the finish will both come out on th e sa me end The secon da ry win din g also has an even n u mber of ayers , 2, bu t the t a p is at the end of th e first layer Therefore, the start, #3 and the finish, #5, will come out on the sam e end, b ut #4 wil l come o ut the op pos ite en d This is the r esult of the tap coming at the end of the first layer A look at the wind ing method o f Chapte r 10 wi m ake this ce ar. The finishing in structions on sh eet # 3 of thi s specif ication cal ls for the #4 l ead to be finished across the coil in order for all the secondary leads to come out on the same end when finished . Note the li nes dr awn i n to re pre sen t the la mi nations . This is done to indicate to the winders the areas that must be free of leads
(3 )
It is cust oma ry to put an out side wra p on t he c oi wh en it is woun d and also an additional wrap of gummed paper or tape is put on after fi nis hi ng the eads
(4)
This ind icat es at which end of the coil the win di ng is star ted The A end is at the left and the B" end is at the right.
(5)
This dimension is to be fied-in if a maximum coil size is caled out on the f inished coil
page 2 9
ROBERT G. WOLPERT
WINDING SHE ET 1
PAGE
PAGES
SA MPLE DESIGN
SPEC NO. ENGINEER.
RGW POWER
TYPE
WINDW CIL BUILD UBE VER UBE DENSY FREQUENCY AREA A ERMINALS
OF
x 84 f NET RSS .5" x 5" x 040" None 96750 lies 60 HZ .64 IN2 5 VLS -2
2
3
� 5/8" - 1-<5) I COIL
(2)
4 1
5
A 1
B
3
115v 60 Hz.
V ct @ 100 A
6.3
4
2
5
See page 29 for explanation.
2 # 22
#23 383
WINDNG NO WIRE SIZE
TOTAL TURNS TAPS WINDING ENGTH
MARGIN TURNS PER LAYER % FI
NO. O ERS LAYER INSULATON
WRAPPER
TERM CO IL START AT
48 84°/ 8 1L .007K -layer .0 OK -2 A
+
9°/ 2 1L .0 K layer .0 OK layer .0 05 GK {by 3-4-5 A
pg 30
MATERIAL SHEET PAGE
PAR NO.
EI- 1 1 /8 " 2 9M6
CORE
COPPER
#23 SINGLE COATED #1 1 HEAVY COATED
AMT
2
OF
TO PRICE
PAGES
O PRICE
TO PRCE
2 .06 # 0 .2 9# 0 . 33 8#
CAN ID-T ID-B TERMNALS
TUBE
1.15 x 1.15 x 1 5/8 "
x
. 0 40
1
ERM BOARD LUG PANEL BKT
1
HORIZ. "L
4
EADS
Back Green Yelow BOLTS NUTS
#20 AWG PVC 8 LON G #12 AWG PVC 8 LON G #12 AWG PVC 8 LON G
6 - 32 6 - 32
x
1
2 2 1 4 4
WASHERS WASHERS
# #6 6 stee fiber
4 4
NOTES:
page 3 1
FINISHING
PAGE
3
OF
PAGES
SPEC NO. COLOR
LENGH OU OF COIL
#20 AWG PV C #1 2 AWG PVC
BLACK GREEN
6" 6"
#1 2 AWG PVC
YELLOW
6"
LEADS SIZE
EAD# 1 &2
3&5 4
UGS OR UG PANE: PART#
LEAD#
SPECIA INSRUCIONS #4,
insh as shown Yelow lead,
is finished across the coil
BLCK 2
3 GREEN
page
2
STACKING & ASSEMBLY
PAGE
OF
PAGES
SPEC NO. � AMINAION:
EI
SIZE GRADE STA CK HEIGH
1 1 /8 " 29M6 1 1/8 " -
1x 1 2
INEREAVE KEEPERS CU OFF E'S GAP SPAC ER: BRUISERS: SIZE
No
SHIED U INSULATORS SIZE BRACKETS:
HARDWARE
QT: 4
Q T: QT QT QT
4 4 4 4
1 1/8 " H orizontal "L "
6
-
32 x 1 1/2 " bolts 6 32 nuts #6 stee washers #6 iber washers -
SPECIA INSRUCI ONS
Assemble as shown . Fiber washers to go under bot heads core Vacuum varnish completed unit
Mark on top of
w = 3 3/8" D= 2 3/8" H = 2 13/16" MH = 13/64" X 3/8" slot MW= 2 13/16" MD= 2 1/8"
page 33
TEST INSTRUCTIONS
PAGE - OF
PAGES
5
PROCEDURE 1ST EST
1
2ND ES 3RD EST FINAL EST
1
5, 6
NO LOAD VOLTAGE RATIO APPLY
115
v
60
H TO TERM
BKBLK
V TERM.
READ
lex _ MAX
GREEN-YELLOWGREEN
V TERM.
V TERM
2
INDUCANCE ES APPL
NA
V
READ "L 3
5
&
A D. C
MIN
INDCED VOLTAGE TEST APPL
4.
H TO TERM
V
NA
MS MEG
NA
HZ O ERM
FOR
MEGOHMS MIN
SEC VOLS D. C .
HI POT EAD NO.
TO
VOS
BLACK
GR EE N
1500
CORE
1500
CASE
6
CONI NUIY
7.
SPECIAL ESS
page 34
LAMINATION TABLE VA
AREA
EI-187
05
035
EE2425
10
0625
STACK HT.
SIZE
WINDOW
WEIGH
015
034
30
1406
108
70
390
392
140
5625
1
678
190
75
1
904
300
765
X 1
105
10
320
.875
X 1
120
EI1
10
450
100
1
155
EI1
1
500
125
1
194
EI1
1
650
1265
EI1
1
900
15625
EI1
1
1250
189
EI1
1
1600
2.25
EI1
1
1600
240
EI1
2
300.0
375
2
891
EI1
1
3400
3.06
X 2
861
EI1
20
4000
3.50
X 2
984
EI-1
20
4500
325
1.0
1
224
1
308
2
417
2
5.35
x 2
1
531
X 2
778
Note: Values shown for area and weight must be modified by the stacking factor K REPRESENTATIVE CORE LOSS
CORE LOSS IN WAS PER LB GRADE
10 KG
15 KG
29 M6
066
26 M19
0.83
200
24 M22
110
260
Note
Core losses shown are from manufacturer's catalogs and are typica . The catalog s sh oul d be consu lted for mo re detailed dat a page 3 5
WIRE TABLE SIZE
INSUL MARGIN
CM AREA
625
DIA
3040
846480
0032
2670
540450
.0036
239.0
35610.0
125
0040
2150
220470
001
157
0045
19.8
0051
1923 1704
128870
001
3456
001
25.0
0056
1555
52480
0062 0070
1420
33750
42
13900
0007
41
1102 5
0007
78
40
8742
0007
99
39
6932
.0007
38 37
5497 4359
36
0028
B
80770
35
2742
001
3152
34
21 74
001
3975
33
17 24
0015
5013
0079
32
1367
0015
63.2
0089
31
1084
0015
797
0099
889
5300
30
860
0015
1005
.0110
810
3334
29
682
0015
1267
0123
725
2046
28 27
541
0015
1598
0137
63.0
132.5
429
002
2013
0154
576
825
26 25
340 270
002 002
2541 320.4
.0171 0192
520 459
527 327
24
214
002
4040
0215
419
207
23
170
003
5095
0240
374
129
22
1345
003
6424
0268
336
8.22
21
1.067
.003
8101
0301
30.3
52
1256
21060
1104
13050
99.0
8100
20
8458
005
1022 0
0336
267
19
670 9
005
12880
.0376
243
204 1 29
323
18
.5320
.007
16240
0421
21.7
17
4220
007
2045.0
.0471
19 3
805
16
3346
010
25330
176
5101
15
2653
010
3257.0
.0590
159
3193
14 13
2104 1669
010 010
41070 57780
0661 0741
142 12.8
2016 1258
12
1323
010
65300
0829
11.5
0794
11
1050
010
8234.0
0929
102
0500
1042
0527
10
0833
010
103800
95
0315
9
0660
010
130900
1168
82
0198
8
0524
165100
1310
7.2
0125
010
page 36
CHAPTER 3 THREE PHASE TRANSFORMER DESIGN Three phase transformers can be designed using a similar method to the si ng e phase transformer des ign in Ch apter 2 There a re differences in the m ethods of ca cul atin g the voltages and currents n ge neral, a t h ree phase unit can be considered as three single phase units connected together Three phase transformers can be connected in several configurations. They are Delta Deta, Delta - Wye, Wye Del ta an d Wye - Wye The Wye is also called a Star configuration Before the transformer can be designed, it must be determined what the configuration will be, and also the various relationships between voltages and currents 3.1
The Wye conf ig uration
The i ne vota ge is usu al ly given for the pr im ary an d secon da ry To obtain the phase voltage, divide the line voltage by the square root of 3, which is 1 73 2 t can be shown that this is de rived from the f act that the three vo tages are 12 0 de grees out of phase Let
E<
phase voltage
E I<
ine volta ge phase current
line current
Then E<
=
E I 1. 732
This is the voltage that must be us for a given core
ed in ca lcu lating the turns necessa ry
The phase current is equa to the line current I< The pha se power is eq ua l to E< x I< The total power is equa to the phase power x 3 or
=
E<
x
I< x 3
To demonstrate this, refer to Figure 6
page 37
FIGURE 6
The ine to ine voltage will be the voltage applied between 1 2, 1 & 3 and 2 3 Thes e voltages are 120 de gre es a part The phase voltage wil l be the volt age bet we en each l ine and the neutr al ( N ) The neut ra l is no t al ways u sed but s shown here for cl arty. It ca n now be see n th at the turns in each phase will be calculated using the phase voltage and the wir e size must be ch osen using the phase curr ent The neutr al l in e wil l car ry no current if the l ines a re bala nced 3.11
Example
A Wye secondary has a line voltage of 200 V. and a power rating of 120 VA. To fin d the ph ase voltage : E< / .7 5.4 v
The phase current s equal to the line current The total power is 12 0 VA which i s equ al to Solving for the phase current
x E x 3 .
/ 54
x
.47 A
f this transformer also has a Wye primary with a 208 volt line voltage the prma ry ph ase voltage wi l b e : E< / 7 v. =
The primary total power wi ll be : x . The prima ry cu rre nt wil l be : . / x
. VA. .7 A.
The per phase power for the primary for calculating the size of the core needed will be
/
4.
page 38
This can be considered as three transformers each capable of handling 40 VA in choosin g the core size In this case, from the lam ination t a ble, Chapter 2.4, it is seen that a 40 VA single phase transformer wil need about 1 s qu are inch of core. Th is would b e a 1" stack of EI-1 la mi nation However, the window width is larger in the three phase than in the single phas e a mi natio n, so a sm al le r size wil proba bly suf fice Ther efore, a 7/8"_ lamination with a 7 /8" stack wil l be used as a startin g point A manufacturer's catalog should be consulted for available sizes The design guide for single phase transformers can now be followed with the difference being only 1/2 of the window width can be used for each coil . Figu re 7 shows an outl ine drawi ng of a typica l three phase transformer Figure 8 shows a three phase lamination of the size used in this design
4
3 19 2 31/2 11 - 13 2
1.
-
1
3 32
DIA
4
325 FIGURE 7
-
-
� 35 l 64
64
3 1/2
FIGURE 8
page
39
(4)
3.2
The Delta configuration
The Delta configuration differs only in the way the windings are co nnect ed This cha nges the r elations hip s between the voltages an d currents. See Fig u re 9 .
1
2 3 FIGURE 9
In the Deta connections the phase voltage is equal to the line voltage. The phase current is equal to the line current divided by the squae root of three If you examine Figures 6 and 9 it can be seen that the line voltage is applied directly across the per phase turns in the Delta configuration, whie it is no t in the Wy e . Als o the lin e cu rrent must flow through the turns direc ty in the Wye and is divide d in the Deta 32.2
Delta example
If the same specifications used in the Wye example are used for the Delt a Line voltage 200 V Power 120 VA
The phase volt ag e is eq ua l to the i ne volt age , 20 0 V The ph ase current is equal to the l in e cu rren t divi ded by . 73 2. The l in e cu rrent 02 I 1732
120 / ( 200
x
3)
=
0.2 and the phase cu rrent will be:
0115 .
The turns ar e ca lcu lated using 20 0 V. an d th e wire size f rom the 0 5 A.
page 40
3.3
Three phase design example
The pr ev ious exam ples are us ed for a c o mp et e d esig n . This will be a tran sformer with a Wye pri ma ry an d a Delta s econda ry . See Fig ure 10 .
1 1
ST
2
ST F
ST
200 v L-L @ 120 VA
208 V L-L @ 60 Hz
ST
F ST
2
F 3
ST
3
FIGURE 10
As has been shown previously, the secondary current per phase will be 1 15 A The p rimar y VA = 12 0 x 1 1 1 = 13 3 2 Th e pr imary line vo lt age is 208 V, then the per phase voltage will be 120 V and the per phase current 037 A A starting point for the core sze will be EI-7 /8 with a 7 /8" stack effective core area is 7 /8" x 7 /8" x 92 = 0 704 . C the primary tu rns for 29 M6 l ami naton and 60 ) 15alculating Kilogauss (96750 lines 120
Np =
4.44
x
60
x
x
The
Hz at
10 8
. 704
x
96750
=
661 turns
Calcuat in g th e secondary turns: 661 / 120 x 105 x 200 = 1157 turns The wire size needed for the primary current of 0.37 A will be #25 Th e w ire s ize fo r the sec ondary cu rre nt of 0 1 1 5 A will be # 3 0 .
page 4 1
Calculating t he f il l Window 1 3/32" x 2 13/32" Coil Length 2 1/4" #25 winding length 2"; margins 1/8" Turns per layer 95 (this is more than the chart shows but will fit with a
wind ing lay er fil l of 9 1 ° ) 7 ayers 661 9 5 #30 winding lengt h 2"; margins 1/8" Turns per ayer 1 66 ( a wi ndi ng fill of 9 1 °) ayers 1 157 166 7
Winding tube 7-#25 Insulation Wrapper 7-#30 Insulation Wrapper
.0300 1344 0120 (6 layers of 002" paper) 0100 0770 0090 (6 ayers of 0015 paper) 0150
Total
2874 x 2
=
5748 1.094 x 100
52°
This is too sm al l a fil l . An adjustment can be made by using les s expens ive l am in ation or a sma ler size The next sm al le r size is 5/8", which is probably to o sma ll so 26M 19 lam ina tion wil l be tri ed using a f lux den sit y of 10 KG (645 00 li nes) The primary turns wil be increased by the ratio of 15 KG divided by 10
KG : 661 x 15 / 10 992 turns Seco ndar y t urns 992 / 120 x 1 . 05 x 200
1736
The same wi re sizes and turns per layer wil l be used #25 ayers 992 95 11 #30 ayers 1736 / 166 11
page 42
Recacuating the fi Tube 0300 2117 11#25 0200 Insul 0100 Wrap 1210 11-#30 .0150 nsul Wrap 0150 -
4227 x 2
Tota
8454 1 094 x 100
78°
This is suicient fill for this transformer Calcuating the votage drops (from the single phase exampe): 8750 Core 0600 Tube x 2 11-#25 2117 nsu 0200 1.1667
x
4
x
992
x
4
x
1736
x
2.697/1000
12.48 Ohm s
x
.37
8.6 / 1000
=
92.23 Ohms
x
115
=
4.62
.0200 2117 0100 1210 0150
Wrap 11-#30 Insul
1 . 5444
x
=
10.6
Ca lculating the l oade d voltage per phase 120 - 462
11538 992 x 1736
201.91
- 106
19131
v.
This is ower th an the 2 0 0 V needed. Therefo re, i t wi ll be nece ssa ry to adjust the secondary turns by the ratio of 200 / 191.3 This i s 1 045 x 1 736
1815
page 43
Recaculating the voltage drop in the secondary wth this new number of turns 92. 23 Ohms x 18 15 / 1736 11538 I 992 x 1815
=
96 43 O hms x 1 1 5
2110
110 9
=
20001
1 1 09 V dr op
v.
Th is is the c omp lete d esign Primary 992 turns of #25 wr e Secondary 1815 turns of #30 wire per coil for three identcal coils =
3.31
Temperature rise
The temperature rise of a small three phase transformer is calculated n the same manner as the single phase transformer except that the total weight and total losses are both divided by three before the caculations. The weght of the core (from the manufacturer's catalog) s 405 pounds The act ual weight is 4. 05 x 9 2 3 73 pounds
.
The weight of the w ire is #25 #30
= =
1248 I 32 69 9223 / 3334
= =
381 pounds per coil x 3 277 pounds per coil x 3
1.145 pounds .831 pounds
The total weght Core 373 1145 #25 831 #30 Total
5 706 x 1 15
The core loss Primary loss Secondary oss
=
6 56 poun ds (a dding 1 5° )
3 73 x 83 (from chart page 24) 462 x 37 1 . 7094 per coil x 3 11 09 x 1 15 1 .28 per coil x 3 =
Total losses Dvding the weight by 3: Dviding the l osses by 3 :
6.56 / 3 12.04 / 3
=
3.09 5.128 3826 12.04
219 pounds 4 01 5 wa tts
page 44
Using the temperature rse formula from Chapter 2: 01
4.015 22 1.073
(
) 2/3
=
2487 degrees rise
33 2 Re gu lat ion
The no -loa d voltage = 12 0 / 992 x 1 81 5 = 2 1 9 55 V The full load voltage 20001 V Regula ton (2 19 55 - 200.0 1) / 20 0 01 x 100 977° 3.4
Interconnections
The manufacturing specifications for a three phase transformer will be s mil ar to the sng le ph ase s pecif ica tons The three phase u nit wi ll nee d thre e coils woun d. The inte rconn ectons be tween the coils wil l be as show n in the Fi g u re 1 0 di agram . Al l fin ishes on the Wye conf igu ration will con nect together a nd g o to the neutr al , f desir ed The s tarts connect to the finishes i n th e Delta conf ig uration . The A, B, C col s are cal led out to keep the windings straight When taps are requred on three phase transformers, t is simple to put taps n the Wye conf ig ura ton . Ths is d one ju st li ke the single phase un its Ta ps n the Delta conf ig uration must be ha ndle d dierently. If a tap is put in the Delta confguration without changing the connections, t wl l resul t n c ircul ati ng curre nts n o rder to over come this, an open Delta is use d. See Figur e 1 1 1
3
3
1 1
3 c
FIGURE 11
page 4 5
Referring t o Fig ure 1 1 , for norma l oper ation , con nect: # 1 of coil A t o # 3 of coil B # 1 of coil B to # 3 of coil C #1 of coil C to #3 of coil A When the taps are used, the #s are left disconnected and the #2s are co nn ected to the # 3 s in stead of the # 1 's. The mater ial list m ust sh ow th e wire weight fo r all th ree coil s.
page 46
CH AP TER 4 AUTO TRANSFORM ERS
An auto transformer can be used when primary to secondary isolation is not nece ssa ry. It c an be either a step-u p or a st ep-down voltage ratio A distinct sze advantage s gained if the rato of voltages s not too large t be shown oformer an auto E) E ( E vary formula: will appcan roximately fr omthat a re gthe ul arsize t ransf by thetransformer Where
E E
=
hig her vo lt age lower votage
For example, if E 230 V and E 1 1 5 V , then the size wi ll be ap proxim ately 1/2 that of a re g ul ar tra nsf orm er The size advantage i s lost if the ratio becom es more tha n 3/4, wh ich is a voltage rati o of 4 to 1 . =
4. 1
Design pro cedu e
In the design of an auto transformer consder Figure 12. I 3
1
5
2
I FIGURE
12
f the line vo tage s a pped to 1 a nd 2 an d the load is ap pli ed to 3 an d 5, then the load current flow in 3 and 5 will be directly drawn through the i n put ines, 1 a n d 2, an d a tran sform er wil l not be need ed. Howev er, if a load is pl aced a cro ss 3 a nd 4, then the load cu rre nt flow wi ll be from 1 to 3, through the load resistance to 4 and back through a portion of the tu rns to 2 . See Fg ure 13 . The arrow s repre sent the cu rre nt flo w directon at any given instant.
page 47
If the diagram of Figure 13 is followed, it is readily seen that the turns from 4 to 2 is the only portion of the transformer that must carry the load cu rrent The b aa n ce of the turns will carry the dif ference be tween the primary curre nt an d the load cu rrent The turns from 1 to 2 must carry the exciting current plus the difference of the two rrent s. Thus theThe turnexciting s from cu 4 to themuc loadh pusother the cu exciting current rre2ntmu is st usucaalrry ly so cu rrent smaler than the load current and also is not directly in phase with the load so that it usu all y can be dis reg arde d in cal cul ating the w ire s ize s
1
3
RL E
Ei
l
FIGURE 13
4. 2
Desg n ex ample
An auto transformer is desired to change 120 volts to 50 volts at 1.0 am pere s This is 5 0 x 1 = 5 0 VA From the lamination table of 2.4 it is seen that a regular transformer will be about the same sizethis as the ple0 of- Chapter 2, =which is 1im1/8 " 60° For an auto transf ormer will ebe xam ( 12 5 0 ) / 12 0 ap p rox ately of the size or 38 VA From the table this wi ll be a cor e size of EI 7 /8" x 1" for a starting point Using 29M6 grade laminations: 120
The turns =
444
x
60
x
x
1 08
.05
x
96750
=
57
The secondary turns = 578 / 120 x 1.08 x 50 = 260 The curre nt for th e in put windi ng wil l be: 50 x 1 1 1 / 120 = 462 A Then the current for 3 - 4 wil be 1 - .462 = . 532 A
page 48
The wire size f o r this part of the windi ng wi ll be # 2 5 The load curre nt of 1 0 am peres wi l re qu ire #2 2 wire It sho uld be not ed here th at this is 642 circuar mils per ampere and is smaller than indicated in Chapter 2. It wi l l be seen l ater that i t wi l l be suf ficient fo r a reasona bl e te m pera ture rise The turns for the # 25 wir e wi be 578 260
318
Using a bobb in for this design, check the table f or EI87 am ination i n the Appendix to see if the turns wil fit in this core size The maxi mu m turns for #2 5 wir e are 952 turns, then 318 / 952 34 x 100 34°
The maximum turns for #22 wire are 480 turns, then 260 / 480 5 3 x 100 53 ° The total is 34 + 53
87°
This should fit as the tables are consevative Calcu lat ing th e fil l : Core wi ndow Winding en gth #2 5 turns per layer Layers #22 turns per layer Layers Winding tube 6 #2 5 Wrapper 7 #2 2 Wrapper
7/16" x 1 5/16" 1 3/16" 54
3 18 54 5. 8 or 6 layer s 34 260 34 6. 6 or 7 layer s .0400 (bobbin) 11 52 0600 (Myla r tape ) 1876 0060
-
Total fil Cacuating Core Tube 6 #2 5
3547 / 4375 the -
X
100
8 1 °
voltage drops: .9375 0800 1152 11327
x4
318
x
26975 / 1000
4
260
1 345 / 10 00
=
3.88 Ohms
1152 0060
7 #2 2
.1876 14414
2.01 Ohms
page 49
The voltage drop f or 3 4 2 01 x 532 1 0 7 vo ts The voltage d rop for 4 2 wil l be 3. 88 x 1 3 88 vol ts The tota l vot age drop acro ss the n put wi l be 1 07 + 3 88
12 0 4 95
1 15 05 I 57 8
.199 x 260
51.75 10 7
This is close enough to the desired 50 volts for this example votage is desred, the turns can be adjusted to 50 / 5068 x 260
4. 95 volts.
5068
v.
f a closer
257 turns
This wll not change the other calculations enough to make a significant difference
4.2.1 Temperature rise Core weight #2 5 weight # 22 weight
-
120 x 92 3 88 I 32.7 20 1 I 822
1104 012 .244
Total
- 136
x 115
1564 pounds
Losses: Core #25 #22 -
1104 x .66 388 x 10 1 07 x 532 Tota
Tem perature r se :
- 0638 - 3.88 - 0.569
-
5 087 Watts
5087 ) 0 (.56 3956 d 073 422 Regulation 2/3
C
Noload volt age 120 / 578 x 260 53.97 V (5397 5068) / 5068 0649 x 100 6 49° reg u la ti on
=
This is a very usab le de sg n t m ust b e noted , h owev er, that the size formul a s ap pro x mate an d is on ly a star tin g poi nt. Adjustmen ts may have to be made just as in the regular transformer designs
page 50
CHAP TER 5 POWE R TR ANSFORM E RS USIN G CAPA CITO R FI LT E RS
When a tra nsfor me r is us ed in a rect ifie r ci rcu it, it is som etime s necessa ry for the tr a nsform er desig ner to calcu late the RM S voltag es an d curre nts needed whe n the DC va lu es are given . Th is ch apter wi ll show a metho d of ca lcul atin g these v a lues In the past, in du ctive fi lter s were gen eral ly used There is mu ch wri tten on methods used to calculate this type of filter and will not be covered The m ost com mo nl y used filter now is the ca pac itan ce in pu t filter This is the ty pe of filter circui t that wil l be considered The complete calculations for the capacitor filter transformer are comp l icat ed a nd wi ll not be sh own The metho d used her e is a close a pproximat ion a n d the resu lts a re very satis fact ory in most cases. When the requirements are given for a DC load, it must also be known what type of rectifier circuit will be used and also the value(s ) of the ca pacitor f ilter 5.1
Typ es of rectifier ci rcu its
There are s evera l types of recti fier ci rcu its These are u ll wave center tapped, full wave bridge, full wave bridge center-tapped, half wave, etc. In this design guide only the first three circuits will be considered 5 1 1 Full wave ce nte r-t app ed
If the circuit of Fig u re 1 4 is exam i ned, it can be seen that the se condary of the transformer will supply power first in one direction and then in the other This the cau wi sesrethe conda one hal time. Thi s al lows sizeseto be ry cuttoi nfurnis hal f haspower the duty cyclef at is a 50°/ However, the turns must be doubled as the votage delivered is from only half of the winding ; this m ust be ke pt in mind when designin g the un it
c
-
FGURE 14
page 5 1
When the output DC voltages and currents are given, it is necessary to calculate the RMS votages and currents in order to design the trans former Aga n , ther e wil l be no atte mpt to expl ai n why certa in th ng s are d on e The theory ca n be s tudie d by th e desig ner at a later date, if desired The RMS voltage needed is equal to the DC voltage divided by 44 and a n um ber r epresent i ng the e fficiency of the entire s ystem . number is added the voltage drop in the rectifiers
To th is
Erms = V / ( 4 4 x eff) + E d Where:
Erms V 4 4 eff Ed
= =
the sec ondary voltage of th e transf ormer the D C volta ge gi ven a constant the efficency of the system the volt age drop in the diodes Ths value wil l be con sidere d to be 0 . 7 volts per diode u nl ess otherwise specifed
As a sta rtin g poin t i n the desig n a n efficienc y of 85° w il l be assumed Thus, 85 wi l be the valu e used i n the abov e formu la . n this circuit ther e are t wo dodes but only on e per t ran sforme r leg That is, ther e s o ny one dode in use at a ti me, a nd a voltage drop of O. 7 volts will be used The RMS current is equal to the DC current multplied
by 6
A x 6 Whe re
I RM S cu rren t A DC cu rrent 1 6 = a cons tant
page 52
5. 1 2 E xample
Assume the following parameters DC vot ag e DC current
=
40 V 1 0 A
he RMS voltage that must be delivered by the transformer will be: E 40 / (1 41 4 x .85 ) 3328 + 07 he RMS curr ent wi ll be I 1 0 x 16 he VA will be 34 x .6 54.4 VA
=
3398 or 34 volts 1 . 6 Ampe res
The transformer will now be designed to deliver 34 x 2 68 volts center Since the secondary will deliver the ta pped at a current o f 1 . 6 Amperes total current only half of the time, the wire in the secondary will be chosen for one h af of 1 6 or 0 8 Ampere s. he prm ay wil l be de signed using the 544 VA as this wil be the total VA delivered by the
transformer If the above calculations are considered, it will be seen that the secondary voltage drops and reguation will be larger than for a unit with a resistive oad of the same VA, as the wire size is for a current of one hal f the am oun t del ivere d he tem perature r ise wil l be cal cua ted usin g the ent ire s econ dary an d one half of the current he de sig n meth od of Chapter 2 can now be used in designing this type transformer. 5 1 3 Ful l wa ve bridge cicuit
+
c
-
FIGURE 15
page 53
This circuit utiizes the entire secondary of the transformer for calculating the volt ages hus , there will be one h alf of the turns of the pre vious exam ple , however, it wi ha ve to de liver the ful l c u rrent 1 0 0°/ of th e time t wi ll a lso be notic ed that there a re 4 diodes, 2 that a re use d at one time This wi ll result in a diode drop of 2 x 0 7 1. 4 Volts . =
E
= =
V (1 .414 x 85) + 1.4 1 6 x A
f the v a ues fr om the p re vious ex a mpl e a re used : E VA
= = =
40 (1414 x 85) + 1.4 16 x 1 1 . 6 Ampe re s 3468 x 1.6 54 49 VA
=
3468 Volts
=
=
From this point on, the transformer will be designed the same as a reguar transformer having these requirements 5. 1 4 Fu ll wav e bridge cent er- tapped
he ful wave bridge center-tapped circuit is the same as the ful wave bridge circuit except it has a center tap and will supply a plus and a mi nus v otage a t the oad This re q ui res tw o cap acito rs f the loads an d capacitors are the same, it is calcuated as if there were no center tap and the load was across the entire output M a n y times the l oads a re di fferent. If thi s is the case, the t ota l VA is cal cul ated as i f the re were two wi nd in gs with sepa rate loads Howeve r, each half of the winding will have to supply first one load and then the othe r This may al low a smal ler wi re siz e if one load is m uch less. h e wire size can be considered as supplying the larger load plus the smaller load divided by 2 This is a gui de on y and the other re q ui rement s, such as r egu lation a nd tem pera ture ri se, will govern . her e ar e tw o diodes per leg as in the fu wave brid ge circuit. See Figu re 1 6
FIGURE 16 page 54
5. 2
Cor recti ng the eficiency
The above examples show how to design a transformer using an estmated efficiency, without considering the value of the filter capacitor or the drving sou rce impedan ces The foll owing is a metho d tha t wi l give a more accurate resut by taking these values nto consideration A transf orme r s first des igne d us ng the methods prev o usly show n This will resu lt in a c rcu it wit h 85°/ eic enc y. The v al ues of ths t ran sforme r can then be used to determine the actual efficency, and corrections can be made, if necessary, that will result n obtaining voltages closer to the desir ed val ues 5.2 .1 Ex ample
As an ex am pl e take the va lues o f gure 1 An ac tua l design was made for this t ra nsformer The desgn will not be shown , but the va lues obtaned wil be used . The core use d is EI 1 " wi th a 1 " stack on a bobbin . The values ar e : Np Ns Rp Rs -
495 turns 161 turns 6 94 Ohms 914 Ohms
From igure 15, the capactor is 5000 ufd R
-
The load res stance Ths can be calcu lated b y d vd ng the load voltage by the oad current
R
-
40 1
=
40 Ohms
From these values, the driving source resistance, Rt, the equivalent resistance, Req, and the equivalent capacitance, Ceq, can be determned The driv n g so urce resist a nce, Rt, con sists of th e tra nsfor mer resi stance referre d to th e secon da ry pl us the dio de res sta nc e. The dio de re sistance s usuay in the ord er of 0 05 O hms Req i s the load res sta nce divide d by Rt
page SS
To determine Rt, divide the secondary turns, Ns, by the primary turns, Np and squ are that valu e This i s th en mu ltiplied b y the pr im ary resista nce , Rp, and added to the secondary resistance, RS, and the diode resistance, Rd. Rt Req
(Ns/Np) x Rp R Rt
+
Rs
+
Rd
The equivalent capacitance is equal to the driving source resistance mu tiplied by the capacitance and this valu e m u ltipl ied by f / 60. Wher e 11 f is the power i ne freque ncy. This valu e is 1 if the f req uency i s 60 Hertz. Ceq
Rt x C x f 60
Subst ituting the actual Rt Req Ceq
=
values :
(16 1 495 ) 1 06 x 6.94 40 1 7 5 2285 5000 x 1 . 75 x 1 = 8750 ufd
734 + 914
+
2(05)
1 . 75 Ohms
With these values the actual efficiency can be obtained with the use of the curves of Figure 17. The va ue fa lls betwee n the 80 ° a n d 86° li nes . By interpolating, it can be see n that it wil l be a bout 84 5 ° If this valu e is used : 34 68 - 14
33 28 x 1 414 x 8 45
39 76 VDC
This is probaby close enough to the desired 40 V
If it is desired to
adjust the voltage, the following procedure is used Going back to the srcinal calculations for the RMS voltage of the transformer secondary and using the new efficiency number, the value is: E
40 ( 1 41 4 x 8 45 ) + 1 4
3488 Volts
he turns can then be adjusted to obtain this voltage instead of the 3468 Volts previousy ob tai ned Resu lts a s close as this wi ll not requ ire an y adj ustment other than the turns, however, if there is a large difference, it may be necessary to change the desgn of the transformer and recalculate all the values and refer back to the curves for the proper resuts.
page 56
It sho ud be noted tha t thi s is a very efficie nt tran sform er If the physica l size of the transformer is smal or the regulation is large, it can resut in efficiencies of 60 to 70 °/ . When th is occurs the u se o f the curve s and recal cu ation s wi ll b e necessary to obt a in voltag es desi red Experience with this type of design will result in using an efficiency closer to the actual value in the srcinal calculations
page 57
1
%
93.5
I 0
%
s
-
% 8% 7% 6%
4% '
$
,
79
"
1
a
r e
t
,
C (Micfrds)
,
CHAPTER 6 CONVERTER TRANSFOR M ERS
Before a converter transformer can be designed, it is necessary to un ders tand how i t is to be used. On ly the basics wil l be covered. The transformer designer must study and explore the many different circuits and rel ationsh ips on his own, if fur the r knowledge is desir ed A converter transformer is used in a circuit to change a DC voltage into an AC voltage, which can be rectified into another level of DC voltage, if desired The frequ ency of operati on is dete rmi ned by the design of the trans former in the case of the saturat ing tra nsfo rmer circuit If the transistors are saturated, then the transformer transfers the voltages and currents in a normal manner similar to the standard power transformer. 6. 1
The satur ating tr ansf ormer
A sim ple f orm of the Royer circuit is shown i n Fig u re 18 This cir cuit use s PN P tra nsisto rs, however , N PN tr a nsist ors can be use d by ch an gi ng the circuitry with n o effect on the transformer S i n ce we are con cerned wit h the transformer only, this example will suffice.
1/2 FB
OUTPU T
1/2 FB
FIGURE 18
page 5 9
Looking at the circuit, it can be seen that a D.C. voltage source is conn ected to the center ta p of the p rim ary wi nd in g The tran sist ors, when conducting, wil allow current to flow in the winding from the DC sour ce It is desired that c u rrent wil fo w in o ne ha f of the prim ary at a time an d a lternate be tween ha lves. This will set up a fl ux in the core sim il ar to an A C votage in a norm al power tra nsformer e xcept that this wil be a squ are wave inst ead o f a sine wave The feed-bac k win din g which is con nect ed to the bases of the transisto rs, cont rols th e swit chi ng of the transistors, and alows the D.C current to flow in an alternating fashio n This is a ver y si mp ifi ed expla nation, but is al l that is ne ces sary for the tr a nsform er to be desig ned . The transformer must be designed so that the core will saturate at the desired voltage evel and oper ati ng freque ncy If the stand ard form ul a for primary turns is examined, it can be seen that the flux density, core area, frequency and votage all interact to give the number of turns It must be noted that in this type of transformer the voltage is applied to ony one half of the tu rns in t he prima ry wind ing at one time an d thus th e turns must be calculated with this in mind. E x 10 8 40 x F x A x B The 444 constant for sine waves is changed to 40 for square wave operation If the voltage is fixed and the core area and saturating flux have been chosen, then the ony variables left to be determined are the turns and the f reque ncy, o r convers ely, if the f requ ency a nd voltage i s given, then the c ore a nd the turns must be chosen to gve the proper r esults The frequency of operation and losses will determine the type of core mat erial to b e used . It wi l now be advisa ble t o explore some of the materials used in these designs.
page 60
6. 2
Cor e materials
he ideal material will have a very narrow, square hystersis loop, as this wil l de termine the sw itch ing loss es an d eficiency All m ag neti c materia ls wil l satur ate at s om e val ue of fl ux density at a given frequen cy . he c ore losses at these values, along with the copper losses, will determine the efficienc y rers of the cir gs cuitand . l ues a retousu ly given the man ufactu catalo ahese re too va extensive be alshown herein The cat alogs liste d at the e nd of this chapter sho uld be studied ca refu ll y. Tape wound toroidal cores have been used extensively for this purpose. The type and thickness of the material are chosen, depending on the desired freque ncy Orig in al ly the freq uen cy of operation was fai rly low , 1 KHz to 5 KHz, because of the unavailability of high frequency switching tra nsis tors . This is no long er a proble m and swit ch ing frequencie s of 20-50 KH z an d abov e are now common wo (2 ) and f ou r (4 mil thick ta pe cores a re suf ficient f or frequencies below 5 K Hz . Thi nne r ma terials must be used for higher frequencies and square oop ferrite cores are now used extensiv ely fo r freq ue ncies above 20 KHz M a n ufacturer's catalogs and other available literature will assist in choosng the proper core and material Many times the operating frequency range will be determined by the physica l size req ui rem ents or by the cust ome r his wil l rem ove one option from the transformer designer 6.3
Control win din g vo tag e
he control winding voltage will be determined by the amount of voltage necessa ry to switch the tra nsistors Th is wil l vary dep end ing on the tra nsistor s used . his voltag e wil l be eithe r given by the cust om er or wi ll have to be determined from the requirements of the transistors used his can be obtained from the published data on the transistors. 64
Design criteria
The actual design of the transformer will proceed according to the guidelines shown previously in the power transformer design method of Chapte r 2 he primary turns, f eed back turns, an d the output windi ngs wil l all be ca lc ul ated he construc tion wi ll be different as it is nec essa ry to hav e clos e cou pl in g betw een hal ves of the prim ary windi ngs
page 6 1
hs s acc om pl ished b y win din g the p rmary ha lve s in paral lel Also , coup lin g betwe en the prma ry and f eed back win din gs is im portant his can be accomplished by winding one half of the feedback winding frst, then the prima ry an d then the other half o f the feedback win din g The output, or sec ond ary win dngs, ca n be wou nd on top of these windings If the step up voltage ratios are large and the frequences are hgh, t s sometimes necessary to intereave the output wth the primary and feedbac k wi nd in gs In the case of la rge step -u p ra tios , it may b e advsable to use two transformers, one for the converter crcuit with a smal step-up rato and a second transformer to further step-up the voltage In ths case t he step up tr ansfo rmer is des gned l i ke a typica l non-saturating unit for square wave voltages. In the case of high power converters, it is sometimes advisable to use two transformers, one small transformer usng square loop material to switch the transstors and a arger transformer using less expensive mater ia l for the power ha nd li ng The design s of these transfo rmers wil fol sa me gu de li nes amust s pre be viously cover by edthe The cir cu tsfor to ths be used andlow thethe parameters required furnished customer type of design he osse s of the transformers mu st be kept low i f a hi gh eiciency system is to be obtain ed The losses of the core material has been covered The copper osses mu st al so be k ept l ow his is a ccomp ished by using a rger szes of wire tha n would ordin arily be r eq u ired As a rule of thumb, the wire size should be a minmum of 1000 crcular mils per am pere of current n al wn di ng s If sze is li mited, the prim ary wind ing wire can be adjusted to take nto account the fact that t is conducting curre nt only o ne ha f of t he time.
page 62
6.4.1 Design recap
Reviewing the important points: 1
The transformer is designed using the regular turns formula with the exception of using 40 instead of the 444 constant.
2.
Wire sizes should be 1000 circular mils per ampere minimum for copper losses.
3.
Flux de nsi ty used wil l be the sat ura ting flu x den sity of t he mat eri al
4
he primary windings should be wound bifilar to reduce coupling losses That is, the two halves of the windi ng a re woun d in paral le l he feedback winding halves are interleaved with the primary wind in g The pr im ary an d feedback win din gs ar e connect ed as shown in Figure 17 so they are in series
5
he s econda ry, o r output wind ing , can be put on over t he primary and feedback windings or itconvenient. can be wound first before the other windings, whichever is more
65
The non- satura ti ng tra nsformer
The non-saturating transformer needs little coverage as it is similar to a regular power transformer operating on a square wave voltage and at a freque ncy specified by the cust ome r Care m ust be taken i n choosing the core size and mater ial to accommodate the r equ irements. here have been many artices pubished on converter circuits and their design, therefore the methods shown here to design the transformer is all that needs to be covered
page 63
Manufacturers' catalogs covering the nformation necessary for choosing cores and core materials can be obtained from the followng companies:
MAGN ET IC ETA10 LS COM PANY C AMDEN NJ M 081 MAGNETICS BUTLER PA 16001 ARNOD ENGINEERING COMPANY 300 WEST STREET MARE NGO , I L 60 15 2 FERROX CUB E CORPORA TIO N SAUGERES, NY 12477
page 64
CH APTE R 7 SH IE LDING IN P OWER TRANSFORM ERS
There a re two types of sh ie ld s used in power tran sformers These are elect rostatic shi el ds an d electr o- mag netic shi eld s They a re both used to redu ce inter ference an d noise fro m bein g either ge nerated or t ra nsm itted 7. 1
Elec trostat c s elds
Electrostatic shields are used to reduce noise or interference in the power source from passing through the transformer and into the load circuit and also to reduce noise and interference from passing from the load circuit back through the t ransf ormer and into th e power i nes The most used type of electrostatic shield is placed between the primary or prima ries an d the s econd aries It is usua l y a 2 or 5 mil thick c opper sheet, however, a close wound layer of fine wire can be used, but will be less eective The shield must be well insulated from both the primary and secondary windings and must not short out on itself, otherwise it will result in a short ed turn A lead wi re of some typ e will be sol dered t o the shield a nd brought out as a ead or be c onnecte d to the c ore as a groun d. 71 1 Box s hields
Box shields are electrostatic shieds that completely enclose the winding. These are used when extremely high isoaton of interference is desired. Sometimes more than one shield is used over the primaries or secondaries or bo th . The shie d lead and the coil leads are brough t out thr ough shielded cable to keep from pickin g up no ise on the ead wir es Box shields can be constructed using copper sheet or strip or aluminum she et o r st rip. Aluminum is easier to use as it is softer and conforms to the contour of the winding easier, however, it is more difficult to attach the lead t o the al um in um Care must be taken to insulate the shields from the windings and each other Sh orted tu rns a re difficu lt to prevent in this type of shi ed and great care must be taken
page 65
7 .2
Elec tro-ma gnetic shiel ds
Electro-magnetic shields are used to reduce the external flux from the core from rad iat ing out an d in terfering with other ci rcui ts. his type of shied can be anything from a simple end cover, whch is only marginally effective, to a copper strap around the core and coil, which is the norma type of shie d Also, a sheet of magnetic materia can be put around the periphery of the core to re duce th e extern al fl ux he maxim um s hie di ng ca n be obtai ned by enclo sing the trans former in a nest of shie ld ca ns This wil l consist of several layers of encosures made of high permeability material interleaved with copper enclosures. The amount of shieding necessary will dictate the type of shield that is to be used
page 66
C H APTER 8 IRON CORE FILTER CH OKES
Iron co re filte r ch okes us ua ll y fit into two categ ories direct current in their windings and those that do not
Those that carry
Much has been written on the design of these units and magnetic materia manufacturer's catalogs give very adequate design methods for their various material s For this reason, it is only necessary to cover some o f the im porta nt poi nts a nd some of the pitfal s to avoid 8.1
Chokes that carry a direct current
A filter choke is usually designed to have a certain inductance with a particu ar val ue of di rect cu rren t flowing in the win di ng . her e a lso i s usually a limitation on the resistance of the windings, either imposed by the specification, or limited by the allowable temperature rise he core some madedirect f ancurrent ai r ga pwill is i saturate ntroduce the d into t heifma gneticadjustments path, the e are ffectnot of the di rect cu rrent fu x is r edu ced By adjustn g the length of this gap w th regard to the total length of the magnetic path, the optimum permeability for this particuar design can be obtaned. C R. Hanna published in 1927 a simple method of designing this type of choke It is stil l a very popul ar and easy method to use woul d suggest a copy of this paper be obtained A sta rti ng poin t for the physical
size c a n be obtain ed by the formu la :
VA = 188 L Where VA
The VA of an equivalent 60 ertz transformer - The inductance desired - he direct current flowing in the winding
The lamination table in Chapter 2 can be used to find a core sie that wll han dl e this VA. Us ng this c ore size for a st arting point and Han na's curves or other method, a design can be derived
page 67
It should be pointed out that EI laminatons have two paralle magnetic paths around both egs of the E and so will also have two gaps if a gap spac er is put in betwee n the E's and I's. hi s is with a butt st ack Fgur e 35 in Ch apte r 1 3 on am nating methods 8.2
See
Chokes with no direct current
Chokes that do not carry drect current can be designed using the standard formula for inductance: L=
Where
L N A u 7
-
-
0. 4 x
x A x N2 x u x10 7
Inductance in henries Nu mber of t urns Net core area Core permeability Magnetc path length of the core 3416
Al of the above vaues can readily be obtained from the szes listed in the ma n ufac turer s catalo gs except for the cor e permea bi ity The perme a b ty wi ll v ar y w ith the typ e of cor e m ater ia l u sed . Ths can be det ermned fro m the pub lshed data f or that par ticul ar material. his can va ry fr om 14 o r ess f or po wdered ma terial up to 40, 00 0 o r mo re for 80°/ nickel an d other materials. Some manufac turers cat a logs give an ndu cta nce fo rmul a for each sze of am inatio n . Ths f ormul a, a long wit h the mat er al perme a bil it y, will result in an easy desgn It should be noted that many szes of cores can be used for any partcular However, t is necessar y to ca lc ul ate the f l ux den si ty t o ndu cta nce value. determine that the core is not saturatng and also to determine that the resstance of the winding is not too hgh to meet the requirements.
page 68
8.3
Config uration s oth er than EI lam inations
There are many core configurations that can be used for chokes, both with an d without direc t current flowin g in their w nd n gs These incu de such cores as toroids, pot cores, H cores, and many others. These will not be covered as the manufacturer's catalogs and published desgn dat a m ake their design exc eedin gy sm ple
page 69
CHAPTER 9. AIR CORE IN DUCTOR S AN D SOLENOI DS The previous cha pters ha ve dealt with ir on cor e com ponents. t is som etimes desir ous to have an ind uct or designed usin g no mag net ic c ore These are referred to as air core inductors The formua formagnetic inductance applies to this except for the the fact basic that there is no path for the flux,type and unit, for this reason flux path wil radia te out i nto space aroun d the c oi l This makes usage of the standard formula more difficult
the
This chapt er will deal with two co nfigu rations of a ir core coils and an eas y way to design them The se are the sing le layer solenoid an d th e mul tiple layer solenoid
9.1
he single ayer solenod
A sing le ayer soenoid is wound using on ly one layer of mag net wire sim ila r to a sc reen door spring It is usual ly wou nd on a coil f orm and held in pace with tape or cement. See Figu re 1 9
FIGURE 19 Many times a high resistance carbon resistor is used as a coil form for sma ll un its This works out v ery wel if the r esis tan ce is sufficiently hig h as to not affect the results La rger co il form s can be obtained o r made from various m aterials a nd w il work f i ne if they a re no nm ag netic an d non-conducting t is important that the length to diameter ratio not be too large or the ind uctance wil l not increa se as desir ed w ith additiona l turns . f possibe the length to diameter ratio should be no more than 3 to 1 for accurate resuts
page 7 0
t should be noted that in al coils the inductance will increase as the turns squ ared For exam ple, if the turns are dou bled, the nd ucta nce will increa se by a ratio of 2 sq ua red or 4 times f th is formul a is exam ine d, it can be see n that it is neces sary to ha ve the size o f the coi a nd the desired v a lu e of in du cta nce bef ore star tn g L x ((9 x a) a
N =
Where: L N a I
+
(1 0 x I))
Inductance desired (in microhenries) Nu mber of turns Radiu s of the col Len gth of the coi
For exampe, if vaues are assigned to the variables Let
L a I
20 mcro-henries .5" 1
-
The n the re q ui red n u mber o f turns will be
N
x ((9 x 5)
+
(10 x 1))
= tuns
.5
So 34 turns w ll be nee ded In order to ma ke a cose woun d coil wi th 34 turns on a 1" wnding, t is necessary to choose a wire size that will have 34 turns per nc h. f the wire table of Ch apte r 2 is consu lted it can be seen that #2 2 AWG wi re wil l fil l the req ui remen t f th s wire size s la rge enoug h to carr y the cu rren t, then th e design is comp lete Sometimes it wl l be nece ssary to inc rease the size an d reca lcu late. In creasing the dia meter of the winding will give a larger inductance with less turns so a larger wi re can be used .
page 7 1
9.2
Mu ltiple lay er solenoid
The multipe ayer solenoid has one more dimension to be considered This is the bui d u p o f the wire. See Figu re 20
1
c
FIGURE 20
he ind uct an ce fo rmu la for this coil is : L =
Where
L N a I
c
-
-
-
08 a N Ga + 9 + 10c nductance desired (in micro-henries) N um ber o f turns need ed Rad iu s of the col Length of the coil Bui ld of the windi ng
n this case the value of "a" is measur ed to the ce nte r of the wire build inst ea d o f th e radi us of the co il
up
From here on it is a matter of calculating the desired number of turns and det ermin in g the wi re size that wi ll fi t into the a l lott ed space adjustments need to be made, the new sizes can be used to recalculate the turns or the inductance can be calculated to determine how close it is
If
to the desired value
page 7 2
9.2
Summary
As can be seen, it is quite possible that more than one try will be needed to arriv e at the exact va lu es desi red H owev er, the formul as do work an d are v ery ha ndl e in the design of these ty pes of in du ctors. In the construction of the solenoids, the wire can be wound on a bobbin with end f a nges or can be layer wound usin g layer insul ation n eit her case the actual winding ength of the wire must be used for calculating purposes
page 7 3
PART II.
MANU FACTURIN G PROCES SES
The following standard processes and procedures are designed to show shop personnel and new engineers how to build and assemble the ma gnetic parts These methods, if care ful ly fo lowed , wi ll result in a component that wi l m eet the cust ome r's requi rements he indiv idua l specfcations, as called out by the designer, must be followed in all cases, where other than standard procedures are needed he materia to be used for insulation, the type of magnet wire, and other materias must be caed out in the manufacturing specifications.
page 74
CHAPTER 10.
LAYER WINDIN G ON A SING LE COIL FORM
Winding a single coil on a winding tube without flanges requires the wire to be wound in ayers with individual turns side by side and with no cros sove rs In divi du al layers mu st be sepa rated with insu lation hea vy enough to support the coil wire without it pulling down between the wires of the pre vious ayer A margin m ust be ma intai ned at each end of the coil wide en ou gh to preve nt t he turn s from fal li ng off the en ds. he insulation thickness and minimum margins needed for each size of wi re is cal led out in t he wi re tabl e in th is cha pter Fig ure 2 1 shows a cross se ction of a coi l woun d on a wi ndi ng tube a nd layer insu lation .
MARGINS LAYER INS ULATION
LE N GTH
FIGURE 21
A wrapper is put on over the completed winding, sufficiently thick to suppor t the next wi nd in g The thickness and ma teri a l of this wrapper is determ ine by the electrical requirements of the design and also the size of the wire This wi ll be c a l led ou t on th e wi nd in g spe cificati on by the designer Before starting the winding it is necessary to correctly anchor down the start of the coil wire The coil wire is anc ho red with a s tri p of el ectrical ta pe, ad hesive side u p The start of the coi l wi re is pla ced on the tube a nd a nchored down a distance from the end equal to the margin, as called out on the speci fica tio n . This s tri p is fol ded over an d held down with an addi tional str ip of ta pe he end of the co wi re is lef t long enoug h to acc omplish Us ual ly 3 or 4 inches is the fini shi ng p roc ess as descr ibed later sufficient Figure 22 shows this process.
page 7 5
FIGURE 22
The co wire is then wound on evenly and closely without bunching or crossing over mai ntaini ng the proper margins at each e nd . At the end of ea ch ayer the nsu ati on is pu t on a nd the wnd ing cont n ued back in the ot her di rection The insu lation will be the same widt h as the coi l form a nd the thickness as cal led out on the specif ica tion. he insulation is taped down under the last turn of the ayer, wrapped around the coi a nd ta ped down a ga in his bec om es the co il fo rm for the next layer Fg u re 23 shows this proc ess
FIGURE 23
page 76
This winding and insulation process is continued until the proper number of turns have been wo und . If a ta p is r eq uired within th e winding, it is brou ght out at the proper num ber of turns, as show n in igu re 24
ANCHOR TAPE
OVER TAP E
UNDER INSUL.
FIGURE 24
he procedure is as follows: Pace two anchor tapes, adhesive side up, a few turns (2 or 3) before the tap is to be brought out. Separate them enough to accommodate the loop for the tap Continue t he wi nding ove r this ta pe u ntil the prop er n um ber of tur ns has been reached Fold the first tape over to hold the wire in pace, pull a loop in the wire long enough for the lead and then bring the wire back to the second anchor tape and fod the tape over A piece of insulation is placed under the oop and a piece of tape is placed over the loop to hold it in place and also protect it from the wire as the winding is continued If the wire is too heavy f or the anc hor tapes al one to ho ld i t in place, it may be necessary to place an additiona piece o tape over the anchors after they are folded back After the proper number of turns have been put on, the end of the winding is agai n an chored down in a simi lar ma nne r to the ta p . A piece of tape is placed under the wire, adhesive side up, a few turns from the end page 77
The winding is continued until the proper number of turns has been reached and then the t ape is f old ed ov er an d tap ed down . See F ig ure 25. The wrapper is then p ut on over this win din g and taped down . It is now ready for the next wind in g to be put o n.
FIGURE 25 0.1 Electrostatic shield
Some transformers require an electrostatic shield be placed between winding s. This is usua lly a 1, 2 or 5 mil thick copper foil wi ndin g This is 0 00 1 0 0 02 or 0 005 inches thick The shield is made t he same width length . That is the tube length mi nus the margin s The length of the shield is made ong enough to wrap around the coil with a minimum overlap of 1/8" The shield lead wil usual ly be #2 0 or #2 2 AWG tinned copp er. If oth er types of leads are required, it wil be caled out on the specification. Figure 26 shows a t ypical elect ostatic shi eld before it is pu t on the c oi . The method of put ting o n the s hield is as f ol lows : The start of the shied is taped down with the lead coming out at the proper place on the coi l, accor di ng to th e speci fica tion A strip of insulation wide enough to cover the overlap is placed over the stat end of the shield so that the finish of the shield will not touch the start
page 78
If the start and finish touch, a shorted turn will result and the coil wil hav e to be rewound or d sca rde d . A wrapper is then put on over the shield and the coil is ready for the next winding
f Winding width
Sode r SHI ELD CONSTRUCTIO N
_ Lead wire
FIGURE 26
Figure 27 shows this process
Shiel
FIGURE 27
p 7 9
10.2 Wire table
Wire Size
Minimum Insulaton
Mnimum Margin
Turns/Inch
00007 00007
116 116
304 267
36 35 34 33 32 31 30 29 28 27 26 25 24 23 22 21 20 19 18
00007 0.0007 0.001 0.001 0.001 0.001 0.001 0.00 5 0.0015 0.0015 0.0015 00015 00015 0002 0002 0.002 0.002 0003 0003 0003 0.005 0005 0.007
116 116 116 332 332 332 332 332 332 332 332 18 18 18 18 18 18 18 18 18 18 18 18
239 215 193 171 156 142 126 111
17 16 15 14 13 12 11 10 9 8
0007 0.010 0.010 0010 0010 0010 0010 0010 0010 0010
18 14 1 4 14 14 14 14 14 14 14
19 18 16 14 13 11 10 9 8 7
42 41 40 39 38 37
99
89 81 73 63 58 52 46 42 38 34
31 27 25 22
The turns per in ch a re gi ven f o r a ° fil l fac tor
page 80
CHAPT ER 1 1 . BOBBIN WINDING
A bobbin is a winding form that has flanges on each end to contain the winding turns without a mar gi n When winding on a bobbin it is not necessary to be as careful as with the layer winding Most bobbins are ra ndom woun d. This me a ns that the No layer w i re is spread eveny on the bobbin, but not in exact layers ins uation is us ed un less ca ll ed for in the specif ication If layer insu lation is called for, the procedure is the same as described for layer winding
FIGURE 28
Figu re 28 shows th e method of an choring d own the sta rt lead This i s sim iar to the layer win din g an chor ex cep t that no ma rgi n is r eq uire d . The wire is placed as close to the flange as possible A strip of insu latio n is placed over the coil wire as it is brought up the side of the flange to protect it from the turns as the wire is wound. When using fine wire it is also good practce to insulate under this lead wir e to keep it from breaking w hen bent over the edge of the f la nge taps and eads are insulated up the sides of the bobbin in this manner as they are brought out.
All
page 8 1
he wn di ng then pro ceeds. he wir e s woun d as evenly as po ssible without attemptin g to layer wind Any taps, if r equ red, a re broug ht out using the same procedure as described for the layer wnding Int erwnd ing nsu ation or wrapper s a re used Bobbin wndi ng usu a lly uses a flexible insulatng material to better conform to the contour of the wind ng Ths insu ation should compete y cov er the win din g so that the subsequent windings cannot pull down at the ends and contact the turns of ano ther windi ng It is somet i mes desira ble t o use a th ree flang e bobbi n . his is a bobbin with two wi nd in g areas In this case the pr im ary(s) a re usua ll y put in one space an d the secon da ry(s ) are put i n the o ther. his el imi nate s the need for insulation between the primares and the secondaries since the cente r flan ge pro vides the insu lation Also, an electro static shiel d is usua ll y not needed w ith this type of bobbin The wi ndi ng procedu re wil l be the sam e as wi th the stand ard bobbin Figure 29 show s an end and side vew of a three flange bobbin
FIGURE 29
page 8 2
CHAPTER 12.
12. 1
LEAD FINISHING
Stranded ead wire termination.
After a co is wound it is necessary to have terminations that are usable by the customer So metimes in the case of heavy wir e of #26 AWG or larger , addition al leads ar e not neces sary. This is cal led out in th e spe cif ica tion Howeve r man y tmes it is des ra be to attach additonal lead wir e to the coil ter m nations This can be ns ul ated, s tran ded lead • wir e or a ength of tinn ed copper wir e. The c olors, sizes and lengths are caled in the specification Coils that are layer wound on a former without flanges are finished in the following manner: A pec e of insulaton
(tab) is placed on the end o
f the coil un der the c oil
wires This of cantape be tao na owmodate strip omo f tape 30 orare a wider pece accrr om re thas an shown on e leaind . Figu Therewires then brought out on top of these tabs and a piece of tape put over them for protection and to hold them in pace
SADDLE
FIGURE 30
page 83
A saddle of suitable material, such as kraft paper, fish paper, or tape is put on the end of the co and the coil wire is taped in place on top of this sad dl e. Care m ust be taken t o keep the lead s a pa rt so no sh ort circut can result. See Figu re 30 The eads are the n sold ered to the col wire
he coi wire an d lead wi re
are twisted together to make a mechancal joint before soldering. Another pece of tape is then put over the eads to hold them in place, eavng approx m atey 1/2 inch expos ed below th s ta pe. he lead wire is then bent up over the tape and another piece of tape is put over the ent ire end of the coil to over the solder joints This is shown in Figure 1 .
SADDLE
FIGURE 31
Care must be taken during this process to leave a ittle slack in the coil wre so that t won't end up n tension and cause a breakage during handling. The bendng over of the ead wres is to keep them from pulling out when handling durng the rest of the manufacturing process and by the customer
page 84
If there are too many leads for the size of the coil to space them safely, without danger of touching, it will be necessary to alternate the taping pr oce ss and accom pli sh this in more than one s tep. This a so ca n be done by put ting tape over and under leads when th ey are bent up If leads from another winding are to be placed on the same surface, another saddle should be put on over the first set of leads and the same procedure followed. The final wrapper is then placed around the entire coil This same procedure is folowed for coils wound on bobbins, except that the t abs wi ll not be necessar y 12 2 Solder lug te rminati ons A solder lug is a piece of metal, usually copper or brass with a tin coating, that is a tt ached to the coi l a nd the coil wire, then sol dered t o it Th es e lugs top strips and then this is ta pe dare t o usually the c oil attached This s tri is cal leof d ainsulating l ug pa nelpaper . These lugs can be one of several different configurations and sizes such as soder lugs, fasten lugs, etc Figure 32 shows a loose solder lug and also a lug panel that has been made up using the oose lugs and a strip of insulating paper.
FIGURE 32
page 85
Before attaching the lug to the coil a saddle is placed on the coil in a sim ila r man ner prevous ly desc ribed . The lug pane l is then t aped i n p lace with the swagged sid e of the lug s away f rom the coi . A ta b is pla ced under the lead and it s then brought out and wrapped around the lug to make a good mecha nical join t, an d then sol dered. See Fig ure 3 3.
HLD DWN P E LUG PNEL COI L IN SU LATION ( SADDLE)
FIGURE 33
In both of the above processes nothng has been said about the soldering proces s. There a re m any d ifferent types of coating used on mag net wire. Some are of a material that wi l sode r without any str ip pin g . This type is the most commonly used because the soldering process is accomplished with a minimum of work Other types of coating require that it be stripped o ff before it ca n be so lde red . Thes e types of coati ng are usu al ly high temperature materals and are normally used in special circumstances. The strippin g process is an add ition al opera ton that should be avoided unless t s absolutely necessary
page 86
CHAPTER 13 . ASS EM BLY AND S TACKING OF MAGN ETI C COR ES
13 1 Stacking of a m inated cor es The lamination type core is made by laminating thin layers of electrical g rade s tee that has been st a mped into various sha pes These incl ude EI, EE, DU , F an d others. The exam ples g iven he re sho w the EI type stam pin g This is the most comm onl y used configu ratio n. The other configurations can easily be understood from these examples Interleaved construction is demonstrated in Figure 34
FIGURE 34
This shows a n interle aving of 1 x 1 The lam in ations are insert ed in the coil one at a time and are alternated with E's and I's from one end and then f rom th e other en d This c a n be done 2 x 2 3 x 3 or othe r, as ca l led out in the specifica tion Care must b e ta ken th at the la mi nations d o not cut into the coi former and short out the turns
page 87
The laminating process must result in a structure that is tightly stacked with a m in im um of gappin g betw een the ends of the E's a nd I's. Some specif ications c a l l for butt stackin g inductors that carry direct current
This is usua ly us ed for
Butt stacking is when the entire quantity of Es are assembled from one end of the coil a nd t he I's from the oth er A gap spa cer may be cal led f or between the E's an d I's
Figu re 35 shows this method o f stacking
GAP SPACERS
FIGURE 35
The gap thickness and material to be used will be caled out in the specification.
page 88
132 Assembly and bracketing Severa l methods of ho ld in g the lami nations toget her ca n be u sed Mo st of the methods are also the mounting method of the competed com ponent Whe n bol ts are used it is sometimes desire d to insu lat e the bolt from the la mi nations Thi s can be accompl ished by putt ing an insulating sleeve over the bolt or a fiber shoulder washer under the head of the bolt If the bolts sho rt out the various leaves of the la mi nation stack, it can result in the exciting current increasing and the bolts will get very hot The following are several of the mounting and assembling methods comm on ly used Fig ure 36 shows a hor izontal an d ver tica chann el f rame construction
FIGURE 36
Figure 37 shows horizontal and vertical L" brackets.
FIGURE 37
page 8 9
Fig ure 38 sh ows an en d bel l cons truction . These en d bel ls completel y enclose the co except for lead hoes out of the mounting surface
FIGURE 38
Fgu re 39 show s a ha lf shell construc ton This c a n be one or two half shes It s no rmal y mou nted on the ch assis by the l am ination bolt s, wth the coil protrudng through the chassis.
�
MO
MW
I
FIGURE 39
page 90
13 3 Assembly w ith a fl ux shield It is often des ired to reduce the ex terna l fu x of a m ag netic device This is especially true when the transformer is operated near a sensitive devi ce or circuit This cou ld be a cathode r ay tube or a high ga in amplifier One way to acc om plis h this is to u se a mag neti c fl ux shie l d This consi sts of a heavy cop per stra p wrapped arou nd the core an d co il The copper strap thickn ess wi l vary from 0 0 1 0" to 0. 060 . This wi ll be caled out in Figu re 40 shows this cons tru ction the specifications
FIGURE 40
The width of the strap should be the same as the length of the coil. The strap is formed to closely folow the contour of the core and coil Insulation is placed under the shield to prevent shorting to the termina lugs or ea ds. The ends of the st rap are solde red togeth er with a good solid solder joi nt Pre -tinning of the ends is customary and they are sweated together with a heavy soldering iron The strap is usualy put on before the impregnating process.
page 9 1
CHAPTER 14.
IMPREGNATION
The completed component, in order to be moisture resistant and last for many hours of operation, is usually impregnated with a moisture resistant material. The imp regnation c a n be accomp lish ed by di ppi ng or vacuu m im pregnatin g. The la rge units, using l arge size wire, can be di ppe d with good results, however, when fine wire, #26 or smaller, is used, it is better to vac uu m im pregnate . Th is wi ll result in bet ter pene trat ion throughout th e win ding s Before the impregnation process is done, the unit should be baked in an oven to drive out the moist u re in the insu lation This is an im portant proc ess. If the un it is im pregnated with moisture seale d in it can cause corro sion an d fai lure over a period o f time The un it sh oul d be bake d at a temperatur e at or a bove 10 0° ( 2 12 °F) to boil out the moist ure . The length of ingthe w ilentire l va ryunit withtothe sizethe of baking the un temperature. it The ti me mu st be long enough to bak allow reach There are man y types of im pregnating material s The most often used is an ele ctri cal grade of varnis h Mi litar y and oth er special a ppl icatio ns can req uire tha t an epoxy im pregnate b e used There are many brands o f varnish an d epoxy on the ma rket that are suita ble for this used It would be im practical to list them here. Some are what is know n as ai r dry, or room temperature curing, which means that the unit will cure without baking Other s are lis ted as a baki ng ty pe that means that t he unit wil l never cure without it being raised to a temperature that cures the im pregnate These un its ma y skin over on the outside an d ap pear to be cured, but inside they will remain wet and will eventually cause failures by attacking the magnet wire coverng. The metho ds of im pre g nating an d curing also a re n u merous Most manufacturers furnish the properties and impregnating and curing conditions for their materials, so it is not necessary to include them here. These instructions must be followed carefully if satisfactory results are to be obtai ned . Materials used and the impregnating and baking processes will dictate the methods needed and the results obtained
page 92
CHAPTER 15 . TE ST ING THE TRANSFORMER
Transformers must be tested after they are manufactured in order to be sure the results meet the desired objectives The first unit of any design should be tested completely by a test technician (or the designer) under loaded conditions to assure that the design meets the requ ireme nts of the cust om er's specific atio n . h is sh ou ld be accom pl ished befo re production is started . his first un it can usually be tested before impregnation so that any adjustments, if needed can be made without d iscardi ng the enti re un it However in the case of high voltage units it is sometimes necessary to impregnate the transformer in order to pass all tests Sheet #5 of the manufacturing specification example of Chapter 2 calls out the elect rica l tests needed Sh eet #4 is used to chec k the physical requirements Referring to Sheet #5: 1.
The no-load voltage ratio tests are made to check that the proper nu mber of turns have been put on e ach win di ng . f a maxim um exciting current is caled out it can aso be checked at the same time by putting a current meter in the primary circuit.
2.
In du ctance tests a re not norma ll y req ui red for power transform ers. If they are required an inductance or impedance bridge is usually used
3.
The induced voltage test and the hipot tests are usually used to be sure there are no electrical shorts between the separate windings an d the wi nd in gs and the core. The y wi ll a lso de termi ne if the prop er size and type of ins ul ation has been used Some specifications call for induced votage tests but usually the hipot tests a re used .
4
Mego hm tests are sometimes ca l ed for. hese a re done to measure leakage current through the insuation and in the margins and fini sh ing operation his test is sometimes used in mi litar y and medic a l eq ui pmen t. A megohm mete r is used f or this t est.
page 93
5.
The va lu e of the hi pot tests a re often cal led out b y the customer's specification or other specifications referred to on the specification These could be Underwriters Laboratories, Military specifications or others The tes t voltage on most power transformer s wi ll b e 10 00 to 15 00 volts bet we en the prim ary and sec on daries an d to th e c or e . Low voltage windings can be tested to each other and the core at 500 volts. Un less othe rwi se sp ecifie d, the se te sts wi ll run at 60 Hz
6
If other than norma operati on is requ ired, suc h as some un its that may operate at a high voltage above ground, the test voltage sho uld be 13 0°/ of the working voltage pl us 1 000 volts. For example, if a unit has a working voltage of 5000 volts, the hipot test will be 13 x 5000 6500 + 1000 7500 volts =
7
Special tests must be called out showing a test circuit, if needed, in the even t other than normal test s are need ed .
Production testing wil normally be accomplished before impregnation on al but the hi po t an d fina l con tin u ity tests This is done so that any production errors can be caught and corrected without discarding al the un its If the un it ha s been imp regn at ed it is a mos t im possible to salvage
page 94
CHAPT ER 16 . INS ULA ION MATE RIAL S
In choosing the proper materials for insulation within a transformer or ind uctor, it is necess ary to consider both the thick ness for suppo rt of the windings as well as the dielectric strength needed to withstand the voltage stresses. The wire table of page 37 gives the mnmum thickness needed for su pport of each s ize wire These va lue s a re fo r Kraft paper, but al so may be used for mos t sta ndard insu lating mat eri a ls . The thickness needed to withstand the test voltages must be determined, cons ider ing the type of insulati ng m ater ial chosen The tests referred to are the breakdown tests between windings and between winding s and the core and not within a wind ing A normal transformer will not have voltages within a winding that will require special insu lation In the case of very hi gh voltages req ui ring m any turns an d layers, it may be necessary to consider the layer to layer voltages in choosing the insulation thickness. If a dielectric or high potential test is not called out in the customer's requirement, the designer must choose a test voltage that will insure a reliab le compone nt The stan dard is usu all y 1 times the work ng volt age plus 100 0 vo lts a t 60 H z. For example: A 1 1 5 volt tra nsformer would be
tested at 1 1 5 x 1 . 3 + 10 00 volts
=
1 149 . 5 or 12 0 0 volts A com mo nl y used test for this ty pe of power transformer is 1500 volts at 60 Hz
page 95
The following is a patial list of mateals commonly used in powe tan sfomes This l ist gives the diel ectic stength in volts/mi l . (volts p e .001 inches) Mateial Kaft Kraft Fish Fish
Dielectc
volts/mil 110 230
varnshed vanished vanished
6110 00 10 00 230 600 - 1000 200 300 4000
The insulating popeties of the many types of electical tapes ae too extensive to be in clu ded hee They can ead il y be fou nd in the manufacture's liteatue These vaues ae the punctue values and should be modified fo actual use For exampe , Kaft pape , the m ost commo nly use d insu lation, would normal y be use d at 15 0 volts pe mi when vanish t eated Thus 10 mils of insulation would be the minimum fo a 1500 volt test. It is possible that pin hoes could occu in the insulation, theefore, I would suggest that two ayes of 5 mil mateial be used athe than one layer of 10 mil to educe the chance that a beakdown could occu due to pin holes.
page 96
APPENDIX
be ftted The fo lowin g ta ble s sh ow the turns of each size of wire that can into different szes of laminations, both layer wound and bobbin wound These tabes wil l be a h el p in determin in g the pos si bi lity of fit in a desig n aer the turns and wire sizes are chosen and before spending a lot of time calcul atng the fil l . For example, if the design procedure shown in 2.2 was #4 where 436 turns of #23 wre was stopped at the point after determned for the primary and the EI-112 amnation table was consulted, it can be shown that it will be a tight fit and changes need to be made: 43 6 848 x 1 0 0
=
5 1 .4°
This is more than 1/2 of the available space and is more than can be used for the prim ary wn d ng in a norm al desg n A rev iew of the design is necessary at this point to either reduce the turns or wire size as shown in the exampe. Another al ter nat ive is to u se a bobb i n The EI- 10 0 Lam ination shows that 1 1 59 turns of # 24 wie wil l fit. This is : 436 1159 x 100
=
table
3 7 . 6° ,
so a bobbi n cou ld be used for this desi g n . These tabl es ca n save a lot of design time that might be necessary in jugglng between turns, wire sizes and lamination sizes Another use for these tables is in determining the turns, wire size and core size needed in designing chokes. The standard inductance formula on page 68 can be used for caculating the turns for an AC in d uctor Then, by us ing the t a bl e for the s ize lamination chosen, it can readly be seen what size wire can be used and if it is the right size and if the resistance is correct for what is needed
page 9 7
A quick way to estimate the size for a choke carrying DC is to use the formula on page 68 an d the ta ble for that la mi na tion size . Experie nce has shown that the permeability for this type of choke will usually be in - 200 A perme abi lit y of 200 can be put i n the the range of 190 ind ucta nce form ua to fin d the turns needed. The table for that lamination will show what wire size can be used and what the resistance wil l be f these val ues agre e with what is needed the n the design c a n be continued or a quotation can be made without going through a compete design In using Hanna's curves for chokes carrying DC, it is easy to determine if the proper size wire will fit and what the resistance will be after determining the turns and before spending a lot of time calculating the fill
page 98
EE9 LAMINATION MAX IM UM TURNS FOR L AYER WOU N D COIL S COIL LEN GT H 19/" WIN DOW WIDTH 1/" MEAN ENGTH TURN 09"
=
Wire Gage
Mi n. La ye r Insulation
Turns/ ayer
Max Layers
Max Turn s
Resistance sq ua re stac k
22
- 0 0 3 K
6
3
8
0237
23
7
3
2
0345
24
- 0 0 3 K 0 0 2 K
7
4
28
0587
25
- 0 0 2 K
8
4
32
086
26
- 0 0 2 K
5
45
50
27
- 0 0 2 K
5
55
23
28 2
- 0 05 K - 0 05 K
2 3
6 7
72
38 608
30
- 0 0 K
5
8
20
0
3
0 0 K
7
8
36
445
32
- 0 0 K
7
222
33
- 0 0 K
2
0
20
3548
34
- 0 0 K
24
264
5625
35
0 0 K
27
3
35
43
36
- 0 0 K
30
4
420
4228
37 38
- 0 0 K - 0 0 K
37
6 8
544 666
2324 3587
3
- 00075K
43
20
860
5842
40
00075K
48
22
056
047
4
- 0 0 0 5 K
53
26
378
488
42
- 0 0 0 5 K
59
28
652
22572
43
0 0 0 5 K
67
3
2077
35688
44
- 0 0 0 5 K
77
35
265
5880
45
- 0 0 0 5 K
85
38
3230
8830
46
- 0005K
4
4
3854
3240
34
page 9 9
E- AMNATON MAX I MU M TURNS FOR AYER WOUN D COIL S COL LE NGTH /" WN DOW WIDT H /" MEAN ENGTH TURN 9" =
=
Wire Gage
Min. Layer Insulation
Turns/ Layer
Max ayers
Max. Turns
Resistance sq uare stac k
22
1 - 0 0 3 K
4
5
20
0374
23
1 - 00 3 K
45
5
22
0512
24
1 - 0 02 K
5
6
30
082
25
1 - 0 0 2 K
6
6
36
135
26
1 - 0 02 K
65
7
45
212
27
1 - 0 0 2 K
7
8
56
334
28
1 - 0 0 15K
8
72
541
2 30
1 - 0 0 15K 1 - 001 K
10
10 12
0 1 20
853 1 43 5
31 32
1 - 0 0 1 K 1 0 0 1K
11 125
13 14
143 1 75
2155 332 6
33
1 - 0 0 1K
14
16
224
5368
34
1 - 0 0 1K
16
18
288
871
35
1 - 0 0 1K
18
20
360
1372
36
1 - 0 0 1K
20
22
440
2114
37
1 - 0 0 1 K
225
24
540
3272
38 3
1 - 0 0 1K 1 - 0 0 0 7 5 K
25 2
27 31
675 8
5157 8662
40
1 - 0 0 0 7 5 K
32
34
1088
13220
41
1 - 0 0 0 5 K
35
3
1365
2020
42
1 - 0 0 0 5 K
39
42
1638
31740
43
1 - 0 0 0 5 K
45
47
2115
51540
44
1 - 0 0 0 5 K
51
53
2703
83660
45
1 - 0 0 0 5 K
56
57
312
12370
46
1 - 0 0 0 5 K
63
62
306
10340
page 100
E AMNA TON MAX M UM TU RNS FOR AYER WOU N D COS CO LENGTH /" WINDOW WDTH /" MEAN LENGTH TURN " =
=
Wire Gage
Mi n a ye r nsulation
Turns/ Layer
Max. ayers
Max Turns
Resistance sq ua re sta ck
22
1 - 0 0 2 5 K
9
5
45
0841
23 24
1 - 0 0 3 K
10
5
50
1165
115
6
69
2052
25
1 - 002K 1 - 0 0 2K
13
26
1 0 0 2 K
78 1 01
2925 4774
27
1 - 0 0 2K
145 16
6 7 8
128
7631
28
1 - 0 015 K
18
9
162
1218
29 30 31 32
1 - 0 0 15 K 1 - 0 0 1 K
20 23
10 12
200 276
1 89 5 330
1 - 0 0 1K 1 - 0 0 1K
255 28
13 14
331 392
500 745
33
1 - 0 0 1K
315
16
504
1208
34
1 0 0 1 K
36
18
648
1958
35
1 - 0 0 1 K
405
20
810
3087
36
1 - 0 0 1 K
45
21
945
4541
37
1 - 0 0 1K
505
24
1212
7344
38 39
1 - 0 0 0 7 5 K 1 0 0 0 7 5 K
56 65
27 31
1512 2015
11550 19420
40
1 - 00075K
73
34
2482
30160
41
1 - 0 0 0 7 5 K
80
39
3120
47800
42
1 0 0 0 7 5 K
88
42
3696
71630
43
1 - 0 0 07 5 K
101
47
4747
1 1 570 0
44
1 - 0005K
115
53
6095
188650
45
1 - 0 0 0 5 K
127
57
7239
280740
46
1 - 0005K
141
62
8742
426000
page 101
EE-- LAMINATION MA XI MU M TURNS FOR L AYER WOU ND COIL S COIL LENGTH /" WINDOW WI DTH /" MEAN LENGTH TURN ."
=
=
Wire Gage
Mi n La ye r Insulation
Turns/ Layer
Max Layers
Max. Turn s 66 91
Resistance squa re stac k 1660
22
1 - 0 0 3 K
11
23
1 - 003K
13
6 7
24
1 - 0 02 K
14
8
112
4480
25
1 - 0 02K
16
8
128
6422
26
1 - 0 02 K
18
9
162
1 0 3
27
1 - 0 02K
20
10
200
1 6 02
28
1 - 0 0 15K
22
12
264
267
29 30
1 - 0 0 15K 1 - 0 0 1K
25 28
13 15
325 420
4143 6753
31 32
1 - 0 0 1K 1 - 0 0 1K
31 34
17 18
527 612
1068 1564
33
1 - 0 0 1K
39
20
780
2514
34
1 - 0 0 1K
44
23
1012
4114
35
1 - 0 0 1 K
50
25
1250
6410
36
1 - 0 0 1 K
55
28
1540
9952
37
1 - 0 0 1 K
62
30
1860
15160
38
1 - 0 0 0 7 5 K 1 - 0 00 7 5 K
69 80
35
39
39
2415 3120
24820 40430
40
1 0 0 0 7 5 K
89
43
3827
62550
41
1 - 0 007 5K
98
50
4900
101000
42
1 - 0005K
108
54
5832
152000
43
1 - 0005K
123
61
7503
246000
44
1 - 0 0 0 5 K
140
68
396300
45
1 0 0 0 5 K
155
73 .
9520 1 1315
590200
46
1 - 0 0 0 5 K
172
80
1 37 60
901900
2887
page 102
EE LAMNATION MAXM UM TUR NS F OR L AYER WOU N D COILS COL LENGTH /" WNDOW WIDTH /4" MEAN LENGTH TURN 4" =
=
Wre Gage
Min Layer nsulaton
Turns/ Layer
Max. Layers
Max Turns
Resistance squ are sta ck
22
1 - 00 3 K
16
6
96
3125
23
1 - 0 0 3 K
18
6
108
35
2
1 - 00 2 K
20
7
10
725
25
1 - 002K
23
8
18
1201
26
- 0 0 2 K
26
9
23
1 92 6
27
1 0 0 2 K
29
10
290
301
28
1 - 0 0 5 K
33
2
396
518
29 30
1 - 0 0 5K 1 - 0 0 1K
36 0
13 16
68 60
772 127
31 32
1 - 0 0 K
45
16
720
1889
- 0 01K
50
18
900
2980
33
- 0 0 1K
56
20
1 120
67
34
- 0 0 1K
6
22
108
5876
35
- 0 0 1K
72
25
1 800
1 1 9 5
36
1 - 0 0 K
80
27
2160
18070
37
1 - 0 0 1K
90
30
2700
2890
38 39
1 - 0 0 1K 1 - 00075K
100 1 16
3 38
300 08
5230 73960
0
1 - 0 0 0 7 5 K
1 29
2
518
109300
1 - 0 0 0 5 K
1 2
49
6958
185700
2
1 - 0 0 0 5 K
1 57
53
8321
280800
43
- 0 0 0 5 K
80
59
10620
50700
44
- 0 0 0 5 K
205
66
13530
730000
5
- 0 0 0 5 K
226
71
1606
1083600
6
- 0 0 0 5 K
251
78
19578
1661300
page 103
EI AMINATION MAXIMUM TURNS FOR LAYER WOUND COILS COIL ENGTH /" W I N DOW WIDTH /" MEAN LENGTH TURN "
=
Wire Gage
M n. La ye r Insulation
Turns/ ayer
Max ayers
Max. Turns 90
Resistance squ are stack 2375
20
0 0 3 K
5
6
21
1 - 0 0 3 K
7
7
19
396
22
1 - 0 0 3 K
20
8
60
6714
23
1 - 0 0 3 K
22
9
198
04 8
24
- 0 0 2 K
24
10
240
1 60 2
25
1 - 0 02 K
27
11
297
250
26
1 - 0 0 2K
31
12
372
3947
27 28
1 - 0 0 2 K 1 - 0 0 15K
34 39
13 15
442 585
595 987
29
1 - 0 0 15K
17
731
30
1 - 0 0 1K
43 48
19
912
1 55 5 2447
3
1 - 0 0 1K
54
21
134
3836
32
1 - 0 0 1K
59
23
357
5810
33
1 - 0 0 1K
67
26
1 742
9371
34
1 - 0 0 1K
76
29
2204
1860
35
1 - 0 0 1K
86
32
2752
23540
36 37
1 - 0 0 1K 1 - 0 0 1K
107
39
35
3325 4173
35860 56760
38
1 0 0 0 7 5 K
19
45
5805
99550
39
1 - 0 0 0 7 5 K
127
50
6350
137330
40
1 0 0 0 7 5 K
154
55
8470
231000
41
- 0 0 0 7 5 K
168
64
10752
369850
42
1 - 0 0 0 5 K
87
70
13090
569400
43
1 - 0 0 0 5 K
214
78
16692
913100
44
1 - 0 0 0 5 K
243
87
21141
1468700
95
page 104
E- LAMINATON MA XM UM TURNS FOR AYER WOU N D COILS CO LENGTH /" WINDOW WIDTH /" MEAN LENGTH TURN " =
=
Wire Gage
Min Layer nsulation
Turns/ Layer
Max Layers
Max. Turn s
Resistance squ a re sta ck
18
1 - 0 0 3 K
15
5
75
145
19
1 - 0 0 3 K
16
5
80
194
20
1 - 0 0 3 K
18
6
108
330
21
1 0 0 3 K
20
7
140
541
22
23
896
25
8 9
184
23
1 - 003K 1 - 003K
225
1 38 2
24
1 00 2 K
31
10
310
240
25 26
1 - 0 0 2 K 1 - 002K
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 0 15K
54
17
918
2266
30
1 001 K
61
19
1159
3610
31
1 - 0 0 1 K
68
21
428
5610
32
1 - 0 0 1 K
75
23
1725
8540
33
1 0 01K
85
26
2210
13795
34 35
1 - 0 0 1K 1 001 K
96 108
29 32
2784 3456
21910 34300
36
1 - 0 0 1 K
120
35
4200
52560
37
1 - 0 0 1 K
135
39
5265
83100
38
1 - 0 0 0 7 5 K
150
45
6750
134300
39
1 - 0 0 0 7 5 K
174
50
8700
218300
40
1 - 0 0 0 7 5 K
194
55
10670
337700
41
1 - 0 0 0 5 K
213
64
13632
544100
42
1 - 0 0 0 5 K
236
70
16520
833800
page 105
E AMINATON MAX M U M TURNS FOR L AYER WOU N D COLS COL ENGTH / WNDOW W D TH / MEAN LENGTH TURN ." =
Wire Gage
Min ayer nsulation
Turns/ Layer
Max Layers
Max Turns
Resistance squ are stack
18
1 - 00 3 K
18
6
108
249
19
1 - 00 3 K
20
7
140
513
20
1 - 0 0 3 K
23
7
161
590
21
1 - 003K
26
8
208
960
22
1 - 0 0 3 K
28
9
252
1 46 7
23
1 - 0 0 3 K
32
10
320
231
24
1 - 0 0 2K
36
12
432
400
25 26
1 - 0 0 2 K 1 - 0 02K
40 46
13 15
520 690
607 1016
27 28
1 - 002K 1 - 0 0 15K
51
16
57
8
816 1026
1515 2400
29
1 - 0 015 K
63
20
1260
3720
30
1 - 0 0 1 K
71
24
1704
6345
31
1 - 0 0 1 K
79
26
2054
9641
32
1 - 0 0 1K
87
29
2523
14938
33
1 - 0 0 1K
99
32
3168
23650
34 35
1 - 0 0 1K 1 - 0 0 1K
112 126
35 39
3920 4914
36900 58330
36
1 - 0 0 1 K
140
43
6020
90100
37
1 - 0 0 1 K
157
47
7379
39600
38
1 - 0 0 0 7 5K
175
54
9450
224900
39
1 - 00075K
203
61
12383
371600
40
1 - 0007 5K
227
67
15209
575600
41
1 - 0 0 0 5 K
248
77
19096
911500
42
1 0005K
275
85
23375
1411000
page 106
EI- LAMINATION MAXMU M TURN S FOR L AYER WOU N D COILS CO ENGTH /" WIN DOW WI DTH /" MEAN LENGTH TURN 50" =
=
=
Wire Gage
Min ayer Insulation
Turns/ Layer
Max ayers
Max. Turns
Resistance sq ua re stac k
18
1 - 3K
22
7
14
386
19
1 - 3K
2
8
2
6763
2
1 - 3K
28
9
22
1 7 4
21 22
1 - 3 K
31
1
31
1 66 6
1 - 3 K
35
11
38
1 97
23
1 - 3 K
39
12
468
396
24
1 - 2K
44
14
616
664
2 26
1 2 K 1 - 2 K
49
16 17
784 93
166 163
27
1 - 2 K
61
19
1 1 9
2 6
28
1 - 1K
69
22
118
4138
29
1 - 1 K
77
24
1848
631
3
1 - 1 K
86
28
248
144
31
1 - 1 K
9
31
294
169
32
1 - 1K
16
34
364
2484
33
1 - 1K
12
38
46
3963
34
1 1K 1 - 1K
136 13
42
35
47
712 7191
61 9937
36
1 - 1K
17
1
867
37
1 - 1K
191
57
1887
2392
38
1 - 7 K
213
6
1 384
383
39
1 - 7 K
246
74
1824
636
4
1 - 7 K
27
81
2227
9814
41
1 - K
32
93
2886
166
42
1 - K
334
12
3468
23939
11
page 107
E 0 0 AMINA TION MA XM UM TURNS FO R LAYER WOU N D COL S CO LENGTH / " WNDOW WDTH /" MEAN LENGTH TURN " =
=
Wire Gage
Min Layer nsulation
Turns/ Layer
Max Layers
Max Turn s
Resistance squ are stac k
16
1 - . 00 5 K
21
6
126
241
17
1 - 0 0 5 K
23
7
161
388
18
1 - 0 0 3 K
26
8
208
632
19
1 - 0 0 3 K
29
9
261
1 0 0
20
1 - 0 0 3 K
33
10
330
1 6 0
21
1 003K
37
11
407
2.48
22
1 - 003K
41
13
533
410
23 24
1 - 0 0 3 K 1 - 0 0 2 K
46 52
14 16
644 832
6.24 10.16
25
58
18
65
20
1 044 1300
16.08
26
1 - 002K 1 - 0 0 2 K
25.24
27
1 . 00 2 K
72
22
1548
38.80
28
1 - . 0 0 15 K
81
25
2025
6253
29
1 - 0 0 15 K
90
27
2430
94.61
30
1 - 0 0 1K
102
32
3264
16027
31
1 - . 0 0 1K
113
35
3955
244.80
32 33
1 - 0 0 1 K 1 - .0 01 K
125 141
38 43
4750 6063
37090 597.00
34
1 - . 0 0 1 K
159
48
7632
947.40
35
1 - . 0 0 1 K
180
53
9540
1493.40
36
1 - 0 0 1 K
200
58
11600
2289.40
37
1 - . 0 0 1 K
225
64
14400
3584.00
38
1 - 00075K
250
73
18250
5728.00
39
1 - 00075K
290
83
24070
9526.00
40
1 - 0 00 7 5K
324
91
29484
14716.00
page 108
EI LAMINATON MAXIM U M URNS FOR LAYER WOUN D COL S COIL LENGTH /" WINDOW WIDH /" MEAN LENGH TURN . =
=
=
Wire Gage
Min Layer nsulation
Turns/ Layer
Max Layers
Max. Turns
Resistance squ ar e stack
16
1 - 00 5 K
24
7
168
371
17
1 - 0 0 5 K
27
8
216
602
18
1 - 0 0 3 K
30
0
300
984
19
1 00 3 K
34
11
374
1656
20
1 - 0 03K
38
12
456
255
21
1 - 0 03K
425
13
552
389
22
1 - 003K
475
15
712
632
23 24
1 - 0 0 3 K 1 - 0 02K
53 59
16 18
848 1 06 2
938 1500
25
1 - 0 0 2 K
67
20
1 340
2386
26
1 - 0 0 2 K
75
23
1725
3872
27
1 - 0 0 2 K
83
25
2075
5874
28
1 - 0 01 5K
94
29
2726
9730
29
1 - 0 01 5K
104
32
3328
14978
30
1 0 0 1K
117
37
4329
24570
31
1 - 0 0 1K
131
41
5371
38430
32 33
1 - 0 0 1K 1 - 0 0 1K
144 162
44 49
6336 7938
57 1 90 90330
34
1 - 0 0 1K
1 84
55
10120
145220
35
1 - 0 0 1K
208
61
12688
229600
36
1 - 0 0 1K
230
67
15410
351570
37
1 - 0 01K
258
74
19092
549300
38
1 - 0 0 0 7 5 K
288
85
24480
88 81 00
39
1 0 0 0 7 5 K
333
96
31968
1462500
40
1 - 0 0 0 7 5 K
372
105
39060
2253500
page 109
EI LAMINA TION MAXIMU M TUR NS FO R LAYER WOU N D COILS COIL LENGTH /" WNDOW WIDTH /" MEAN LENGTH TURN " =
=
=
Wire Gage
Min. Layer nsulation
Turns/ Layer
Max Layers
Max. Turns
Resistance squ ar e stack
16
1 - 0 0 5 K
27
8
216
521
17
1 - 0 05K
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 0 3 K
43
13
341
21
1 - 0 0 3 K
48
14
559 672
22
1 - 0 03K
54
16
864
838
23 24
1 0 03K 1 - 002K
60 67
18 20
1080 1 340
1321 2067
25 26
1 - 0 02 K 1 - 0 02K
75
22
1650
3210
85
25
2125
5202
27
1 - 0 02 K
94
27
2538
7848
28
1 - 0 0 15K
106
31
3286
12810
29
1 - 0 0 15K
117
34
3978
19320
30
1 - 0 0 1K
132
40
5280
32740
31
1 - 0 0 1K
147
44
6468
50560
32 33
1 - 0 0 1K 1 - 001K
162 183
48 54
7776 9882
76660 122840
34
1 - 001K
207
60
12420
194680
35
1 - 001K
234
67
15678
309900
36
1 - 0 0 1K
260
73
18980
473000
37
1 - 0 0 1K
292
80
23360
73 41 00
38
1 - 0 0 0 7 5K
325
92
29900
1184900
39
1 - 0 0 0 7 5 K
376
104
39104
1954200
40
1 - 0 0 0 7 5 K
421
114
47994
3024800
517
page 110
E- AMN ATION MAXMUM TURNS FOR AYER WOUND COILS CO LENGTH " WNDOW WIDTH /" MEAN LENGTH TURN " =
=
=
Wire Gage
Min. ayer nsulation
Turns/ Layer
Max Layers
Max Turns
Resistance squa re stac k
13
1 - 0 10 K
19
6
1 14
146
14
1 - 0 10 K
21
7
147
24
15
1 - 0 10 K
24
8
192
.39
16
1 - 0 0 7 K
29
8
232
60
17
1 - 0 0 7 K
33
10
330
107
18
1 - 0 0 7 K
37
11
407
1 66
19
1 0 0 5 K
43
12
516
265
20 21
1 - 0 0 5 K 1 - 0 0 5 K
48 54
14 15
672 810
435 660
22
1 - 005K
62
18
1116
1 1 5 0
23
1 - 005K
69
20
1380
1790
24
1 - 0 0 3
78
23
1794
2950
25
1 - 0 0 3
88
26
2288
4720
26
1 0 0 3
98
29
2842
7400
27
1 - 0 0 3
110
32
3520
11600
28
1 - 0 015
123
37
4551
18900
29 30
1 - 0 0 15 1 - 0 0 1 K
136 153
44
40
5440 6732
28700 44500
31
1 - 0 0 1 K
172
49
8428
70000
32
1 - 0 0 1 K
191
53
10213
106000
33
1 0 0 1 K
213
58
12354
163000
34
1 - 0 0 1 K
241
69
16629
276500
page 111
E 0 AMINA TION MAXIMU M TURN S FOR LAYER WOUN D COIS COIL LENGTH /" WNDOW WDTH /" MEAN LENGTH TURN =
=
Wire Gage
Turns/ Layer
Max. ayers
21 24
7 8
147 192
21
14
1 - 01 0K 1 - 0 0 1 K
15
1 - 0 0 1K
27
8
216
48
16
1 0 07 K
33
9
297
84
17
1 - 0 0 6 K
37
11
407
1 4 4
18
1 - 0 0 7 K
41
12
492
220
19
1 - 0 0 5 K
47
13
611
345
20 21
1 - 0 0 5 K 1 0 0 5 K
53 59
15 17
795 1003
560 900
22 23
1 - 0 0 3 K 1 - 0 0 3 K
69 77
20 22
1380 1694
1560 2400
24
1 - 002K
86
26
2236
4100
25
1 002K
97
28
2716
6350
26
1 - 0 02K
109
32
3488
10000
27
1 - 002K
121
35
4235
15300
28
1 - 0 015 K
136
40
5440
24700
29 30
1 - 0 015 K 1 - 00 1K
151 169
44 48
6644 8112
38100 58300
31
1 - 0 0 1K
190
54
10260
93300
32
1 - 0 0 1K
25
58
12528
143500
13
Min. ayer Insulation
Max. Turns
Resstance sq ua re stack 34
page 112
EE-- AMINATION MAXIM U M URNS FOR BOBBI N WOU N D COL S ( °o FL L) MEAN LENGTH TURN Wire Gage
=
."
Turns/ Layer
Max Layers
Max Turns
Resistance square stack
20 21
7 8
2 2
14 16
0121 0174
22
9
3
27
0373
23
10
3
30
052
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
114 147
955 1 28 6
31
24
8
192
2117
32
26
8
208
2841
33
30
10
300
520
34
33
11
363
569
35
37
12
444
1235
36
41
14
574
2080
37
46
15
690
2973
38
51
17
867
4700
39
58
20
1160
40
65
22
1430
11160
41
73
25
1825
16480
42
80
28
2240
31270
43
90
31
2790
45000
44
1 02
35
3570
78000
45
118
41
4838
136900
46
123
43
5289
1 8780 0
8280
page 113
EI- LAMINAION MAXIM U M U RN S FO R BOBBIN WOU N D COI S ( °o FIL) MEAN LENGTH TURN Wire Gage
=
"
Turns/ ayer
Max. Layers
Max. urns
Resistance square stack
20 21
5 6
4 4
20 24
026 039
22
6
5
30
062
23
7
5
35
091
24
8
6
48
158
25
9
7
63
262
26
10
7
70
369
27
11
8
88
572
28
13
9
117
142
29 30
14 16
10 12
140 1 92
209 2875
31
18
13
234
393
32
20
15
300
625
33
22
17
374
989
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
1 102 1496
9140 16210
40
49
38
1 86 2
25800
41
55
43
2365
40290
42
61
47
2867
60900
43
68
53
3604
99000
44
77
60
4620
154000
page 114
E- LAMINATION MAXI M U M TU RN S FO R BOB BN WOU N D COILS ( °o FLL) MEAN LENGTH TURN Wire Gage
=
1"
Turns/ Layer
Max. ayers
Max Turns
Resistance square stack
20 2
0
3 4
30 44
0376 070
22
2
4
48
0963
23
4
5
70
76
24
5
5
75
2382
25
7
6
02
408
26
9
7
33
675
27
2
8
68
07
28
24
9
26
72
29 30
27 30
0
270 330
272 424
3
34
2
408
658
32
37
4
58
040
33
42
6
672
672
34
47
7
799
2580
35
53
20
06 0
4335
36
58
22
276
6540
37
65
25
625
0290
38 39
72 83
27 32
2656 2994
5590 27800
40
92
35
3220
42900
4
03
40
420
67350
42
4
44
506
02900
43
27
50
6350
68300
44
45
57
8265
24300
45
67
66
022
464000
page 15
EI- AMINATION MAXIM U M TU RN S FOR BOBBIN WO U N D COILS ( °o FIL) MEAN ENGTH TURN Wire Gage
=
."
Turns/ Layer
Max. Layers
Max. Turns
Resistance square stack
20 2
18 20
6 7
108 140
25 423
22
22
8
76
646
23
25
9
225
1 04 3
24
28
10
280
1 642
25
31
12
372
218
26
35
13
455
448
27
38
15
570
668
28
43
17
731
1086
29 30
48
864 1113
1600 2640
31
60
18 21 24
1440
4359
32
66
26
1716
6130
33
75
29
2175
0210
34
84
33
2772
16500
35
94
37
3478
26200
36
103
41
4223
39800
37
116
47
5452
63600
38 39
128 147
51 59
6528 8673
96500 167200
40
163
66
1 07 58
266300
41
182
74
13468
466000
42
202
16766
633000
43
226
83 92
20792
1016000
44
257
105
26985
1590000
45
298
122
36356
2523000
53
page 116
EE- - LAMINA TION MAXIM U M TURN S FOR BOB BIN WO U N D COILS (° o FILL) M EAN ENGT H TURN Wire Gage
=
"
Turns/ ayer
Max Layers
Max Turns
Resisance square stack
20 21
11 12
5 6
55 72
.0946 .1618
22
14
6
84
2308
23
15
7
105
3624
24
17
8
136
616
25
19
9
171
954
26
22
10
220
1 5 4
27
24
11
264
2.315
28
27
13
351
3.94
29 30
30 34
14 16
420 544
581 963
31
37
18
666
1500
32
41
20
820
22.70
33
47
23
1081
37.9
34
52
25
1 300
57.85
35
59
28
1652
93.20
36
64
32
2048
144.50
37
72
35
2520
223.00
38
80
39
3120
344.00
39
91
46
4186
60400
40
1 02
51
5202
956.00
41
1 14
57
6498
1573.00
42
126
63
7938
22 41 00
43
141
71
10011
366300
44
160
81
12960
5870.00
45
186
94
1 7484
9170.00
page 117
E-1 AMNATION MAXM U M TURN S FOR BOBBN WOU N D COI S (8 °o FI) MEAN ENGTH TURN Wire Gage
=
.1"
Turns/ Layer
Max Layers
Max Turns
Resistance square stack
20 2
9 2
6 7
4 47
39 5345
22
24
92
56
23
27
9
243
36
24
30
0
300
223
25
34
374
334
26
3
4
532
560
27
42
4
5
49
2
47
6
752
350
29 30
52 5
20
936 60
2075 3322
3
66
23
5
5553
32
72
25
00
600
33
2
2
2296
3000
34
9
32
292
2000
35
03
36
370
3200
36
2
40
440
5250
37
26
45
5670
0000
3 39
40 60
50 5
7000 920
25000 23000
40
7
64
392
33500
4
99
72
432
524000
42
220
0
7600
200
43
247
90
22230
33000
44
2
02
2662
2043000
45
325
3350
3220000
page 18
E- LAMNATION MAXIM U M TURNS FOR BOBBN WOU N D COL S ( °o FL L) MEAN LENGTH TURN Wre Gage
=
"
Turns/ Layer
Max Layers
Max. Turns
Resistance square stack
7
16
5
80
1189
8 19
18
5
168
20
20 21
6 6
90 120
2 22
23 26
7 8
161 208
656 1 08 5
23 24
30 33
9
1773 2743
25 26
37 42
0 11
270 330 407
4265
13
546
725
27
46
14
1052
28
52
16
644 832
29
58
18
1 044
2740
30
65
20
1300
4360
31 32
73
22
687
81
25
1606 2025
10650
33
91
2548
16930
3131
26450
126
284 41 35
1750
34
01
28 31
35
15
35
4025
43050
36
128
39
4992
67000
37
143
44
6292
104000
38
159
49
7791
63000
39 40
183 204
56 63
10248
283000
12852
448000
4
227
71
6117
675000
42
255
79
20145
1080000
page 19
E- AMNA T ON MAXM U M TU RN S FOR BOBBIN WOU N D COILS (° o FLL ) MEAN LENGTH TURN Wire Gage
=
."
Turns/ Layer
Max. Layers
Max Turns
Resstance square stack
16 7
17 20
6 6
1 02 20
1435 213
8
22
7
54
.3 49
9
25
8
200
564
20
27
8
26
8425
21
30
9
270
1 3 7
22
34
10
340
2157
23
38
11
418
327
24
43
2
516
510
25 26
48 54
14 16
672 864
836 460
27
60
17
1020
2020
28
67
20
1340
3360
29
75
22
1650
5160
30
84
25
2100
8400
31
95
28
2660
3580
32
04
31
3224
20150
33
118
35
4130
32700
34
132
39
548
51650
35
148
44
652
83000
36
165
49
8085
129000
37
1 84
54
9936
195900
38
205
6
12505
32100
39
236
70
16520
537000
40
263
78
2054
851000
page 120
EI- LAMI NA I ON MAXIM U M URN S FOR BOBBI N WOU N D COILS ( °o FILL) MEAN LENGTH TURN Wire Gage
=
"
urns/ Layer
Max. Layers
Max Turns
Resistance squ are sta ck
16 17
20 23
6 7
1 20 161
202 342
18
25
200
536
19
28
8 9
252
852
20
32
10
320
1 3 6
21
36
11
396
213
22
40
12
480
325
23
45
14
630
539
24
50
16
800
863
25 26
56 63
17 20
952 1260
1294 21 59
27
70
22
1540
3329
28
79
25
1975
5383
29
88
27
2376
81 6 5
30
98
31
3088
13168
31
109
34
3706
20250
32
122
38
4626
31940
33
137
43
5891
51200
34 35
155 175
48 55
7440 9625
81520 1 32960
36
193
61
1 1 77 3
205100
37
212
67
14204
312000
38
222
76
16872
467350
39
271
85
23035
804660
40
301
95
28595
1259700
page 121
EI- 0 0 LAMINA TION MAXIM U M TU RNS FO R BOB BIN WO U N D COIS (° o FIL) MEAN ENGTH TURN Wre Gage
=
"
Turns/ Layer
Max Layers
Max Turns
Resistance square sack .227
15
22
7
16
25
7
154 175
17
28
8
224
526
18
31
9
279
826
19
35
11
385
1 .4 4
20
39
12
468
2.20
2
44
13
572
340
22
49
15
735
550
23
55
17
935
883
24
61
19
1159
13.81
25
69
1449
26
77
21 24
1 848
21 7 7 35.00
27
26
2236
53.42
28 29
86 96
30
2880
86.75
1 07
33
3531
13410
30
120
37
4440
21270
31
133
42
5586
337.30
32
148
46
6808
518.60
33
167
52
8684
834.00
34
189
59
11151
1350.00
35
213
66
14085
2146.80
36
236
74
17464
336240
37
260
81
21060
51 1300
38
294
92
27048
8281.00
39
331
103
34093
1316300
40
368
115
42320
2060600
.326
page 122
E- LAMNATION MAXM UM TURN S FOR BOBBIN WOU ND C OIS (° o F LL ) M EA N LENG TH TURN Wire Gage
=
."
urns/ ayer
Max ayers
Max Turns
Resistance square stack
13 14
19 22
6 7
114 154
126 214
15 16
24 27
7
168
294
216
477
17
31
8 10
310
863
18
34
11
374
1 3 2
9
38
12
456
202
20
43
14
602
336
21
48
15
720
507
22 23
54
61
17 19
815 1297
24
68
21
918 1159 1428
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
133
42
5586
31700
31 32
147 164
47 52
6909 8528
49440 76970
33
85
59
10915
24200
34
209
67
14003
200900
35
235
75
17625
318900
page 123
EI- L AMINA ION MA XIM U M UR NS FO R BOBBIN WOU N D COIS ( °o FI ) M EAN E NG TH URN Wire Gage
=
."
urns/ ayer
Max. ayers
Max Turns
Resistance square stack
12 13
19 21
6 7
1 14 147
109 177
14
24
8
1 92
291
15
27
9
243
465
16
30
10
300
723
17
34
11
374
1138
18
38
12
456
1 7 5
19
43
14
602
291
20
48
15
720
439
21 22
53
60
17 19
901 1 140
693 1 1 0 0
23
67
21
1407
1721
24
75
24
1800
2776
25
84
27
2268
4410
26
94
30
2820
6915
27
105
34
3570
11040
28
118
38
4484
17484
29
131
42
5502
27050
30 31
147
47
42840
163
53
6909 8639
32
182
59
10738
105860
33
205
66
13530
168200
34
231
75
17325
271580
35
261
84
21924
433380
36
289
94
27166
677000
67530
page 124
EI- AMINATION MAXIM U M T U RN S FOR BOBBIN WOU N D COILS ( °o FIL) MEAN ENGTH TURN Wire Gage
=
"
Turns/ Layer
Max. ayers
Max. Turns
Resistance square stack
11 12
17 19
6 7
102 133
082 135
3
21
7
147
188
14
24
8
192
310
5
27
9
243
494
6
30
11
330
846
7
34
12
408
32
18
38
13
494
201
9
42
15
630
324
20 21
47 53
17 19
799 1007
517 823
22
59
21
1239
1276
23
66
24
1 584
2058
24
74
27
2268
3716
25
83
30
2490
51 4 5
26
93
34
3162
8237
27
103
37
381 1
12520
28
116
42
4872
201 8 0
29 30
130 145
47 53
6110 7685
31925 50625
31
161
58
9338
77550
32
79
65
1 6 35
12880
33
202
73
14746
194750
34
228
83
1 892 4
315160
35
258
94
24252
509320
36
285
104
29640
784800
page 125
E- LAMINATION MAX M U M TU RN S FOR BOBBN WOU N D COLS ( °o FILL) MEAN LENGTH TURN Wire Gage
=
4"
Turns/ Layer
Max. Layers
Max. Turns
Resistance square stack
10 11
18 20
6 6
108 120
076 106
12
23
7
161
179
13
26
8
208
292
1
29
9
261
61
15
33
10
330
735
16
37
12
125
17
1
13
533
189
18
6
15
690
308
19 20
51 58
17 19
867 1 1 02
89 782
21
6
21
1 3
120
22
72
23
1656
1870
23
81
26
2106
3000
2
96
29
278
5000
25
101
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