Transformer Design and Manufacturing Manual - Robert G. Wolpert (2004)
July 7, 2023 | Author: Anonymous | Category: N/A
Short Description
Download Transformer Design and Manufacturing Manual - Robert G. Wolpert (2004)...
Description
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-
View more...
Comments
