Design of Voltage Regulator
A transformer is a device with two or more stationary electrical ckt that are conductively disjointed but magnetically coupled by a common time-varying magnetic Field. Transformer are basically passive device for transforming voltage and current one of the windings, generally termed as secondary winding, transformer energy through the principal of mutual induction and drivers power to the wad. The voltage level at primary and secondary windings are usually different and any increase or decrease of the secondary winding voltage is accompanied by corresponding decrease or increase in current.
Transformer are among the most efficient machines 951. efficiency being common in lower capacity ranges. While efficiency of the ordered of 99 % is achievable in high capacity range. Theoretically there is no upper limit to the power handing capacity, transports constraints, handling facilities etc. Being the limiting factors, the lower limit is governed by the allowable no-load loss.
The physical basis of transformer is mutual induction between two ckt. Linked by a common magnetic field. The primary ckt carrying a current has associated with it as a manifestation of the electrical phenomenon, of current flow, a magnetic field at any pt in the surrounding medium will vary in Both magnitude and direction in accordance with change of current with time. The secondary ckt being in the vicinity of primary ckt will link some of the magnetic flux produced by primary.
With an alternating primary current and therefore flux, the changing linkages will produced in the secondary winding an e.m.f.
Transformer works on mutual inductance. It has got two winding, namely primary and secondary winding wound on laminated core of silicon steel, it is used for step-up or step down the voltage level in a ckt.
i.e. A transformer is a static device piece of apparatus used for transformer power from one ckt to another without changes in frequency.
When sinusoidal voltage i.e. A.C. voltage applied to primary winding (NL turns) a Flux is produced in the iron core. This Flux link with secondary winding, so an e.m.f. is induced in secondary winding. According to Faraday?s law of electromagnetic induction. If load is connected to this secondary winding a current start flowing in secondary winding in such way that it opposes the flux produced by primary winding.
E. M. F. EQUATION :-
? = ? m sin ?t = ? ? m sin ?t f
Due to sinusoidal current flowing through in a primary winding a flux almost sinusoidal is produced in the iron whole of waveform as shown above.
If supply frequency is ?F? then the frequency of supply will also ?F?
? since time to complete F cycle = 1 sec.
So time to complete 1 cycle = 1/F sec.
So time to complete Quarter Evolutions = 0.25 x 1/F
= 1/4F second=d
(This is the time in which Flux Rises from 0 to ?m.)
? So average voltage induced in each turn coil = change in Flux
time interval for change in Flux
= ?m ? 0 = ?m = 4.F. ?m
1/4F -0 1/4F
We know that,
Form factor = R.M.S. = 1.11 for W Av.
So R.M.S. value of voltage induced in each turn coil = 1.11 x 4 x F x ?m
If primary turns = N 1
? R.M.S. voltage induced in primary i.e.
If secondary turns = N2
?R.M.S. voltage induced in secondary i.e.
VOLTAGE TRANSFORMATION RATIO (K)
K = Secondary voltage
= Vs = Ns ------- (1)
Also For Xmer
Primary volt. amp = Sec. Volt. amp.
E1 . I1 = E2 . I2 ??(2)
From equation (1) and (2)
Ep = Np = Is
Es Ns Ip
Also for Xmer
Primary Flux Linkage = Secondary Flux Linkage.
Np Ip = Ns .Is
K is known as constant voltage transformation ratio.
PHASOR DIAGRAM ON NO ? LOAD
No load current of Xmer has two components.
i) Magnetizing components (Im OR I?)
ii) Core use components (Ic)
And no ? load primary current (Io) is phasor sum of I? and Ic
OR Io = I? + Ic
(1) Magnetizing component :-
I? is responsible for production of working flux in Xmer core. This current has behind the applied voltage by 900.
(II) Core loss componets is presents
(1) Hysterisis loss (Ph)
(2) Gady current loss (Pe) in the core.
This current is resistive in nature so, It is in phase. With applied voltage
?m Io = I? + Ic
ON LOAD POWER FACTOR OF XMER :-
(1) LAGGING LOAD :- (INDUCTIVE LOAD).
VARIOUS LOSSES IN TRANSFORMER
Transformer losses are divided in to two categories.
(I) CONSTANT LOSSES :-
In which mainly Iron Loss take place. These losses are independent of load current, these are of two tapes.
(1) Hysterisis Loss
(2) Eddy current loss.
(1) Hysterisis Loss :- When alternating current flows through, primary winding, cyclic magnetization and demagntisation of Xmer Core take place. So Heat is produced. Hysterisis loss is proportional to valve of Flux-density and frequency.
Ph ? F. B Max1.6
To minimize hysterisis was CRCTO ? cold Rolled grain oriented steel or silicon steel is used. The permability of (RGO is very high more than 10.000.
II) EDDY CURRENT LOSSES :-
In Xmer primary and secondary windings alternating current Flows as per Faradays Law of electromagnetic induction and lene`s law opposite current are set-up in the Iron core due to which I2R losses takes place in the Iron core. Since this loss is due to Eddy current. It is called as Eddy current loss.
Eddy current in Core
It is clear that to Reduce eddy current loss. Eddy current should be loss. So electrical resistance of iron core in the path of eddy current should be more. To increase the electrical resistance of iron core the core is made-up of, thin laminated strips insulated to each other and assembled by nut-bolts.
The thickness of laminated is around 0.3 mm (1/3), also silicon increases the electrical resistance of Xmer core so silicon steel is used. But more is the silicon steel becomes brittle so silicon up to 4 % is added.
Eddy current loss is also dependant upon. Pe ? F2. Bm2
(III) VARIABLE LOSSES :-
This losses depends upon primary and secondary winding resistance and currents flowing through them. RF primary currents Ip, primary Resistance R1 and secondary current R2 and secondary winding Resistance R2. Then total variable loss = Ip2R1 + Ib2R2
This is called variable loss because it is dependant upon load current.
To minimize copper losses primary and secondary winding Resistance should be as small as possible.
When Xmer is loaded with a constant primary voltage, then the secondary terminal voltage drops (assuming lagging power factor ); It will increase if power factor is leading because of its internal resistance and leakage reactance.
V2 = Secondary terminal voltage at no-load
V2! = Secondary terminal voltage at load
? % R = V2 - V2! x 100
Then % Regulation of a loaded Xmer at any power factor is given as
= ( R Cos? + X sin? ) + ( XCos ? - R sin? )
Where R = % of Resistive drop
X = % of reactive drop
COS? = Lagging or leading P.F.
Efficiency of Xmer are
(I) ? = O/P
= out put
out put + losses
O/P + const.loss(Pc) + variable loss (Pcu)
II) % ? = I/P ? losses
= Input _ Losses
% ? = 1 ? losses
Generally the efficiency of Xmer is very high around 96 % to 99 %.
EQUIVALENT CKT OF Xmer :-
Refer to Secondary
re2 = r2 + r1 ( N2 )2
xe2 = x2 + x1 ( N2 )2
r1 = Primary winding resistance.
r2 = Secondary winding Resistance.
x1 = Primary winding Reactance.
x2 = Secondary winding Reactance.
RC = Resistance Representing core loss.
X? = Mag Reactance.
N1 and N2 = No of turns primary and secondary windings.
CONDITION FOR MAXM EFFICIENCY OF XMER
We know Iron Loss = Cu loss
EFFICIENCY = Output = Output
I1p Output + Losses
O/P + Iron loss + copper loss
? = V2 .I2 cos ?2 ????(1)
V2 .I2 cos ?2 + Pi + I22 re2
For maximum efficiency dx = 0
so differentiate equation (1)
dx = d V2.I2.cos ?2
dI2 dI2 V2 .I2 cos ?2 + Pi + I22 re2
dx = (V2 .I2 cos ?2 + Pi + I22 re2 )V2cos ?2 - V2cos ?2 (V2 . cos ?2 + 0 + 2I2. re2)
dI2 (V2 I2 . cos ?2 + 0 + 2 I2. re2 )
? (V2 .I2 cos ?2 + Pi + I22 re2 )V2cos ?2 - V2 I2cos ?2 (V2 . cos ?2 + 2I2. re2) = 0
? (V2 .I2 cos ?2 + Pi + I22 re2 ) V2cos ?2 = V2 I2 cos ?2 (V2 . cos ?2 + 2I2. re2)
? V2 .I2 cos ?2 + Pi + I22 re2 = V2 I2 cos ?2 + 2I22. re2
Pi + I22 re2 = 2I22. re2
Pi = I22. re2
? i.e. For maximum efficiency Iron loss is equal to copper loss.
TYPES OF XMER
There are mainly two type XMer are used.
(1) CORE TYPE
(2) SHELL TYPE
The difference between these two XMer are as
CORE TYPE SHELL TYPE
(1) Winding surrounds the core (1) Core surrounds the winding
(2) Winding have poor mechanical (2) Higher Mechanical strength
(3) More leakage reactance (3) Less leakage reactance
(4) Repairs easy (4) Repair difficult
(5) Better cooling of winding (5) Better cooling of core.
Due to easy repairs and better cooling core XMer are mostery more used than shell type XMer.
Kinds of transformations
The ratio of primary & Secondary Voltages (V1 & V2 ) is equal to ratio of no of turn in primary & Secondary winding
V1 = N1 N1 = N2
V2 = N2 V1 = V2
V2 = N2 V1
Similarly current transformation
I2 = N1 ? I1 N1 = I2 N2
Similarly Impedance transformation.
V1 = N1 I2 = N1
V2 = N2 I1 = N2
V1 I2 = N1 2
V2 I1 N2
From Ohm?s law secondary winding load res. RL is V2 /I2
? V2 /I2 kinds reflection of RL
RL` = N1 2
RL = N2
RL` = RL N1 2
V2 = N2
V2 = N2 V1
Volt ampear :-
As in any transformer the sec volt ampear or VA must equal the V.A.(J/P) as stated
V1 = I2
TWO BASIC DESIGN EQUATION
First is voltage equation & second is power capability equation.
P = 0.707 J f WaB x 10-8
V = Applied Voltage
F = form factor
f = frequency
a = Core crass sectional area
N = No of turn on considered wndg
B = flux density per unit area.
W = area of core window in cm2
N = 108
V 4 F fa B
N = 1
V 4 F f a B x 10-8
Some time T is made to represent N/V
T = 108
4 F f a B
It is assumed that Equation will be used with. Sinusoidal 9/P & So F is immediately assigned the value of 1.11 this value. Combine with 4 gives 4.44.
In place of 4 F
N = 108
V 4 .44 fa B
This Equation usually move practical for design purpose however to add conservation to them that enable a expressed in inches. & B in gauss.
This conservation is accomplished by including the factor 6.45 in bottom line. (There are 6.45 Sq Cm to Square inch )
? N = 108 = 108
V 4 .44 x 6.45 fa B 28.64 fa B
Or dividing N = 3.49 x 106
V f a B
Apply the factor 6.45 to equation
N = 108
V 25.8 F fa B
Finally let up consider following variant of Equation.
V = 4 F f N ? x 10-8
It this is compared to basic Equation.
? is now being used to replace a x B
B = ?
B = flux density
? = total flux
a = cross sectional area of core
? ? = V x 108
4 F f N
Which express total flux in core.
Autotransformer losses and ratios
for this case the turn voltage ratio equation is
N1 = Vin - Vd1
N2 Vo + Vd1 - Vd2
& for stepdown case.
N1 = Vin - Vd2 - Vd1
N2 Vo + Vd1
Where the voltage are at shown in fig.
Vin V2 N2
Core Selection :-
All this discussion about alloy, lamination, toroids, cut cores, metal powder, ceramic & many chart is fine, interesting even later the moment of trust must arrive ? a choice must be made from wetter of possibilities.
It you want to design power XMer . This statement immediately narrow the search because power XMer is a high flux application that is generally accepted that the core material for power XMer should be run at high flux density at possible in order to keep XMer small & less costly.
Volt ampear ratings :-
It the load is resistive in nature then they have power factor 1.0. It P.F. less than 1.0 then such load require XMer of larger volt. Ampear capacity than that indicated by load expressed in watt motors for example have power factor less than 1.0.
Other type of XMer require a greater voltampear capacity because the current in wndg are not sinusoidal & develop greater heating in the copper.
Output of Transformer :-
Let. ?m = main flux ; Bm = max. flux density wb/m2
= current density A/ m2 ; Agi = gross core area m2
Ai = net core Area = Stacking factor x gross core area m2
Ac = Area of copper in the window m2
Aw = Window Area m2
D = Distance between core centeres in
d = Diameter of circumscribing circle m
kw = Window space factor f = frequency Hz
Ei = Emf per turn Volt.
Tp, Ts = No. of turn in primary and secondary of Win
Tp, Is = Curen in primary and Secondary or withal
Vp, Vs= terminal voltage of primary & secondary of winding
ap, as = Area of conductors of primary & sec. Win m2
li = mean length of flux path in iron m
Lmt = length of mean turn of XMer Wind (M)
Gi = Wt of active iron Kg. Gl = wt of cuppe kg.
gi = Wt per m3 iron Kg, Ic = wt per m2 01.Cu kg
Pi = iron loss per Kg Pc = Cu loss per Kg.
As Tp.Ip = Ts.Is = At if we neglect magnetizing emf.
i) Single Phase Transformer
The voltage induced in a transformer winding with T turns and excited by a source having a frequency f Hz given by
Voltage per turn Et = E = 4.44 f ? m ?????(i)
The window in a single phase transformer contains one primary and one secondary windings.
Total Cu area in window
Ac = Cu area of primary winding + Cu are of secondary winding
= pri. turns x area of pri. conductor + sec. turn x area of sec. turn
= Tp . a p + Ts .as
taking the current density ? to be the same in both primary and secondary winding
a p = Ip /? and as = Is /?
Total conductor area in window
Ac = Tp.Ip/? + B.Is/?
= ( Tp.Ip +Ts.Is) / ?
= 2AT / ?
As Tp.Ip = Ts.Is = AT if we neglect magnetizing emf.
The window space factor Kw is defined as the ratio of copper area in window to total area of window.
iii) ?????? Kw = conductor area in window = AC
total area of window AW
? conductor area in window AC = kW .AW
from equation (i) and (iii)
2AT / ? = KW.AW
? AT = KW. AW . ? ?????..(iv)
Rating of 1? Transformer
Rating of 1? Transformer in KVA
Q = Vp.Ip.10-3 ( Vp ? ED )
= Ep .Ip.10-3
= Et .Tp.Ip.10-3 = AT .ET . 10-3
= Et KW.AW.? 10-3 = 4.44 f ? m KW.AW .? x 10-3
= 2.22 f ? m KW.AW .? x 10-3
But ? m = max. flux density x net. Area of core
= Bm.Ai ?????..(iv)
? Q = 2.22 f Bm ? KW AW Ai x 10-3 . KVA ???.(v)
O/P Equation Voltage per turn
Considering the output of one phase KVA rating of one phase.
? Q = Ip.Vp.10-3 = Ip x 4.44 t ? m Tp x 10-3
= 4.44 t ? m AT x 10-3 ????. (vii)
The ratio ? m /AT is constant for transformer of given type service and method of construction .
Let ? m/At = r where r = constant
From equ. (vii)
Q = 4.44 ? m f AT x10-3 = 4.44 ? m f ? m x10-3
Q = 4.44 ? m2 f/r x 10-3
? m = r. 10-3 x ? ?
? 4.44 f
Voltage per turn
Et = 4.44 f ? m = 4.44 f r. 10+3 . ? ?
= ? 4.44 f . r 103 . ? ? = k ? ?
K = ? 4.44 f . r. 103
K = 4.44 f ?.m x 103 ?
As the ratio of ?m/AT depends upon type of transformer and therefore K is also a constant. Whole value depends upon type service condition and method of construction.
Sr.No. Type K
5. Single phase shell type
Single phase core type
Three phase shell type
Three phase core type (distrib)
Three phase core type (Power) 1.0 to 1.2
0.75 to 0.85
0.6 to 0.7
Ratio of iron loss to copper loss
Copper loss per m3 = p. ?2
? Pc = 2.1 x 10-3 ?2 = 2.36 x10-3. ?2 W/kg
8.9 x 103
Pi = Pi.Gi
Ratio of iron to copper loss
Pi = Pi. Gi
Pc Pc. Gc
Transformer may designed to make one of the following quantities minimum.
i) Total Volume ii) Total weight iii) Total cost iv) Total losses
In general these requirements are contact directly and it is normally to satisfied to only one of them. All these quantity varies with ratio r = ?m / AT. If we chose the high value of r , then flux will become larger and consequently large core cross section is needed which result in higher volume, weight and cost of iron and gives higher iron loss.
On the other hand owing to decrease in the value of AT the volume and cost of cu required decrease and also the I2R loss decrease. Thus we conclude that the value of r if a controlling factor for the above mentioned quantities.
Design for minimum loss or Max. efficiency
Total loss at full load = Pi + Pc
At any Fraction x of full load, the total loss are Pi + x2 Pc
If Q is the o/p of full load, the output of fraction load is Qx
? Efficiency at O/p xQ,
i.e. ?X = x Q
xQ + Pi + x2Pc
This efficiency max. when d ? x = 0
Differentiating ?x we have
d? x = ( xQ + Pi + x2Pc ) Q ? xQ. ( Q + 2 x Pc)
dx ( xQ + Pi + x2Pc )2
for max. eff. ( xQ + Pi + r2Pc ) ? xQ ( Q + 2xPc ) = 0
? Pi = x2 .Pc
So that the mar. efficiency it obtained when the variable losses are equal to the constant losses.
Design of windings
No. of turns in primary winding.
Tp = Voltage of Primary winding = Vp
Voltage per turn Et
? N. of turn in secondary winding
Ts = Voltage of Primary winding = Vs
Voltage per turn Et
The number of turn of an low voltage winding is usually determined in a preliminary design by adjusting the voltage / turn to get the number of l.v windg. Turn per phase Al an integer.
Tl.v = Tl.v = an integer
The number of h.v. wndg per turn per phase is
Th.v = Vh.v .Tl.v
Note :- If the tapping are located in the middle part of an h.v. wndg, the no. of wndg. turn must be even to ensure the symmetry of winding. For a wndg. with tapping it is necessary to have a proper turn ratio or voltage ratio not only on the principal tapping but on the other taps us well. Therefore turns should be selected individually current in primary end.
Therefore turns should be selected indiously current in primary winding
? Ip = KVA per Phase x 103
? Is = Ip Vp
The core is made up of any of the following combination of stampings
i) E & I ii) T & V iii) E used in pairs.
The below table gives information about std. Stamping manufacture by precision pressing Division of M/s Guest Keen, Keen Williams for Small transformer and chokes.
? Single Phase Transformer?
The design of small low voltage transformer of rating 10 to 1000 VA is given. The saturation points of small transformer for design is the choice of turn per volt.
?Turns per Volt?
Sr.No VA Turn per Volt Sr.No. VA Turn per Volt
1. 10 23.3 9 200 3.5
2. 15 17.5 10 250 2.8
3. 20 14.0 11 300 2.8
4. 25 11.7 12 400 2.3
5. 50 7.0 13 500 2.0
6. 75 5.6 14 750 1.7
7. 100 4.6 15 1000 1.6
8 175 4.0 16
Now, E = 4.44 f. ?m T.
? T = T/E = ? . 44 f . ?m
flux in cox ?m = ? . 44. f. Te.
The frequency of Xmer is specified and the value of turms per volt Te is taken from above table. ? ?m is known.
Net area of core Ai = ?m/Bm
Bm = max. flux density = 1 wb/m2 (Assum)
Gross area of core Agi = Ai / 0.9
? (stacting factor = 0.9).
A shell type of construction is normally used for small transformer. The core is made up of any of the following combination of stampings.
Winding Design :-
Current in primary winding (Ip) = VA/?
The efficiency of small transformer varies from 80 to 96 percent.
Area of Primary winding conductor
Ap = Ip/? 8 mm2 when ?p is the current density in primary winding conductor in A/mm2. A value of 2.3 A/mm2 may be used.
Enammd round conductor are used for the windings of ? small transformer.
Sr.No. SWG Dim. mm Area mm2 Nominal cond. d/m Overall diam. Max.
Normal cover Thick covering
Turns in primary winding = TP = VP. Te
Current in Secondary winding cond. = Is = VA/v
Area of sec. Winding cond. as = Is/?s mm2.
When calculating the number of secondary winding turns an allowance of 5 % extra turns is made to compesule for the voltage drop in the winding
Ts = 1.05 Vs.Te.
Material use in Transformer
1. Cu Conductor :- Copper conductor having 24 SWG is used for winding coated with insulation.
2. Stampings :- For making transformer E-I stamping is use for reducing the air gap
3. Rotating Switch :- It has 6 ? junction including Zero position. It is use for turn changing of transformer.
4. Two way switch :- This switch is use for measuring voltage in both I/P and O/P side.
5. One way switch :- This switch is use for O/P Supply.
6. 5 amp. Socket :- Here Socket is use for connecting the load terminal.
7. Neaon Indicator :- Neaon indicator is use for indicating the continuity of supply.
8. Fuse :- Fuse is peace of metal conductor which when melt excessive current how through it.
9. Voltmeter :- It is an instrument use for measuring the potential diff of ckt.
10. Connection wire : These are Cu conductor with PVC coating use for completing the ckt.
11. Insulation Oil :- It is use for providing insulation.
Apart from active materials like copper and cold rolled grain oriented silicon steel, a number of ferrous, non-ferrous and insulating material are employed for building up a transformer.
1) Electrical Machine Design
- A.K. Sawhney
2) Electrical Installation System
- M.P. Vader
4) Machine and Transformer
5) Practical Transformer Design Handbook
- BPB Publication by Eric Lowdon
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