machine design 3

7/31/2019 Machine Design 3

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In our generation, power supply is very important in our everyday living in the way like our

life will not be complete without it. Thats why they invented the transformer for power

transmission.

Being an electrical engineering student, we have known everything about power transmissionthrough the use of transformer. But will it be proper to know everything about transformer. Not

just the theoretical knowledge about transformer but how to construct a generating transformer.

But first we must know what a transformer is.

A transformer is a device which uses the phenomenon of mutual induction to change thevalues of alternating voltages and currents. In fact, one of the main advantages of a.c.

transmission and distribution is the ease with which an alternating voltage can be increased or

decreased by transformers.

Transformers range in size from the miniature units used in electronic applications to the

large power transformers used in power stations. The principle of operation is the same for each.

Now that we discuss what a transformer is, we will now tackle what will be the paper all

about.

In this paper we will be using laminated core or EI core power transformer. This is the mostcommon type of transformer, widely used in appliances to convert mains voltage to low voltage

to power electronics.

This paper will discuss about power transformers which converts AC to DC supply, and step

down the supply to 19-volts 5-amperes output. It will tackle how the transformers function and

its application, and how to compute the specification to construct such transformer. And tells

what are the needed apparatus and materials for the construction and what will be the step-by-step procedure on constructing this kind of transformer.

Although the transformer is not an energy conversion device, it is an indispensable

component in many energy conversion systems. As one of the principal reasons for the

widespread use of ac power systems, it makes possible electric generation at the most

economical generator voltage, power transfer at the most economical transmission voltage,

and power utilization at the most suitable voltage for the particular utilization device. Thetransformer is also widely used in low-power low-current electronic and control circuits for

performing such functions as matching the impedances of a source and its load for maximum

power transfer, insulating one circuit from another, or isolating direct current while

maintaining ac continuity between two circuits.


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Transformers are needed in electronic apparatus to provide the different values of plate,

filament, and bias voltage required for proper tube operation, and to maintain or modify wave

shape and frequency response at different potentials.

A transformer can convert AC (Alternating Current) to DC (Direct Current) supply

through the use of a rectifier.

Another function of the transformer is to allow AC voltage to be readily converted from

one voltage level to another. This way, the power can be generated at a relatively low

voltage. A transformer is used to step the voltage up to a transmission level close to the usepoint of the power, the voltage is stepped back down, and then routed to the transformer on

the utility pole behind the house or close to the industrial plant. The final step down

transformer changes the voltage to a utilization level. This kind of transformer is called the

power transformer.

Though, this paper will focus on the power transformer which converts AC to DC supply,and for electronic devices which uses low-power low-current supply, this function of

transformer for a high-power distribution is just included for better understanding of thefunction of a transformer.

Transformers are needed in electronic apparatus to provide the different values of plate,

filament, and bias voltage required for proper tube operation, and to maintain or modify wave

shape and frequency response at different potentials.

A power transformer with 19-volts and 2.1-amperes output with 60 hertz operating

frequency are usually applied in the charger of some electronic devices or apparatus.

One of its applications is on laptop or portable computer. Most of the laptop with this

output specification is Netbook, usually small in size and only use for searching and typing..

Another application is on television. Usually apply in wide screen flat televisions which

most of them are using an adaptor to connect to a source.

Any electronic devices or gadgets and appliances with this output specification will be

suitable to use for supplying power to operate.

Occasionally someone asks why electronic transformers cannot be designed according to

curves or charts showing the relation between volts, turns, wire size, and power rating. Suchcurves have appeared in magazines and have been used for small control transformers. The


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idea is that by means of such charts any engineer can design his own transformer. However,

this idea has not been found practicable for the following reason.

(a)Regulation. It regulates 220 Volts to 19 Volts, 2.1 Amperes and 60 Hertz output.This property is rarely negligible in electronic circuits. It often requires care and

thought to use the most advantageous winding arrangement in order to obtain the

proper IX and IR voltage drops. Sometimes the size is dictated by such

considerations.

(b)Frequency Range. It frequency is 60 Hertz fixed. The low frequency end of atransformer operating range in a given circuit is determined by the transformer open-circuit inductance. The high frequency end is governed by the leakage inductance and

distributed capacitance. Juggling the various factors, such as core size, number of

turns, interleaving, and insulation, in order to obtain the optimum design constitutes a

technical problem of the first magnitude.

(c)Voltage. It voltage is 19 Volts. It would be exceedingly difficult, if not impossible, toreduce to chart form the use of high voltages in the restricted space of a transformer.Circuit considerations are very important here, and the transformer designer must be

thoroughly familiar with the functioning of the transformer to insure reliable

operation, low cost, and small dimensions.

(d)Size. Its size is small, the same size with the adaptor of a netbook and notebooklaptop. Much electronic equipment is cramped for space and, since transformers often

constitute the largest items in the equipment, it is imperative that they, too, be smallsize. An open-minded attitude toward this condition and the use of good judgement

may make it possible to meet the requirements which otherwise might not be fulfilled.

The use of new materials, too, can be instrumental in reducing size in some

instances down to a small fraction of former size.

(e)Rectification. A rectifier is an electrical device use to converts alternating current(AC), which periodically reverses direction, to direct current (DC), which flows inonly one direction.

(f) Filter. A device that is designed to physically block certain objects or substanceswhile letting others through. This circuit reduces ripple voltage to become smooth DC

voltage.

Some advantages properties of a power transformer using EI core will be enumerated

below.

Widely available in power ratings ranging from mW to MW Insulated lamination minimizes eddy current losses Small appliance and electronic transformers may use a split bobbin , giving a high

level of insulation between the windings

Rectangular core


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Core laminate stampings are usually in EI shape pairs. Other shape pairs aresometimes used

Mu-metal shields can be fitted to reduce EMI (electromagnetic interference) A screen winding is occasionally used between the 2 power windings Small appliance and electronics transformers may have a thermal cut out built in

Occasionally seen in low profile format for use in restricted spaces Laminated core made with silicon steel with high permeability

Magnetic Circuit is made up of one or more closed loop paths containing a magneticflux.

Power factor is defined as the ratio of the real power flowing to the load to theapparent power in the circuit.

Split bobbin is the method of winding a transformer whereby the primary andsecondary are wound side by- side on a bobbin with an insulation barrier between the

two wingdings.

Mu-metal is a nickel-iron alloy that is notable for its high magnetic permeability. Eddy current loss is the induction of eddy currents within the core causes a resistive

loss.

Electromagnetic interference (EMI) is disturbance that affects an electrical circuitdue to either electromagnetic induction or electromagnetic radiation emitted from an

external source.

Permeability is the degree of magnetization of a material in response to a magneticfield.EI core is a sheets of suitable iron stamped out in shapes like the letters "E" and "I",

are stacked with the "I" against the open end of the "E" to form a 3-legged structure.

Biasing is the method of establishing predetermined voltages or currents at variouspoints of an electronic circuit to set an appropriate operating point.

Filament is a thin heating element. Alternating Current (AC) the movement of electric charge periodically reverses

direction.

Direct Current (DC) is the unidirectional flow of electric charge. Impedance is the complex ratio of the voltage to the current in an alternating current

(AC) caircuit.

RMS Value, In mathematics, the root mean square (abbreviated RMS or rms), alsoknown as the quadratic mean, is a statistical measure of the magnitude of a varying

quantity. It is especially useful when variates are positive and negative, e.g.,

sinusoids.


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AWG (American Wire Gauge) also known as the Brown & Sharpe wire gauge is astandardized wire gauge system used since 1857.

Interpolation means to estimate a value of (a function or series) between two knownvalues; to make insertions or additions.

Transformer is a device used to transfer electric energy from one circuit to another,especially a pair of multiply wound, inductively coupled wire coils that affect such a

transfer with a change in voltage, current, phase, or other electric characteristic.

Voltage Regulation is the change in voltage magnitude that occurs when the load (ata specified power factor) is reduced from the rated or nominal value to zero, with no

intentional manual readjustment of any voltage control, expressed in percent of

nominal full-load voltage.

Wiring is the material, as wire or rope, wound or coiled about anything, or a singleround or turn of the material; as (Elec.), a series winding, or one in which the

armature coil, the field-magnet coil, and the external circuit form a continuous

conductor; a shunt winding, or one of such a character that the armature current isdivided, a portion of the current being led around the field-magnet coils.


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Specification Needed:

Primary Voltage: 220 Vrms

Secondary Voltage, Current: 19 Volts dc, 2.1 Amperes

Operating Frequency = 60 Hz

Step 1:

Compute the total voltage across the secondary:

Esn = 2.35 Edc

Esn = 44.65 Volts

Where:Edc => Secondary Voltage in Volts

2.35 => twice the ratio of the RMS to average value plus

5% regulation

Step 2:

Compute the secondary currents from:

Is = k Idc

Is = 2.226 Amperes

Where:

Is => secondary current in Amperes

k => k factor (see Table 1)

Step 3:

Compute the output power in Watts.

Pout = Es Is

Pout = 99.3909 Watts

Where:

Es => Secondary Voltage in Volts

Is => Secondary current in Amperes


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Step 4:

Compute the required input power in Volt-Amperes.

VA =

out

(n)VA = 128.4568 Volt-Amperes

Where:

Pout => Required Output Power computed previously

n => efficiency of the transformer

0.9 => power factor

VA => Power input in Volt-Amperes

Step 5:

Assume a square core area for convenience and compute for the

required area and the size of the EI lamination.

A =

A = 2.0312 in2

Where:

A => cross section area of the core in square inches (n2).

Compute for the tongue width ( tw ) and the stacking height ( g ).

g = tw = g = tw = 1.4252 in

Then use Table 3 to choose the size of the EI lamination required.

g = tw = 1.5 in

The size of the EI lamination also gives the winding length ( wl ), which should

not exceed the tongue width ( tw ) and the allowable winding build-up which

should not be greater than the EI laminations window The details are shown in

Figure 6.


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Step 6:

Compute for the number of the primary and secondary turns Np

and Ns.

Np = Ep 1

f

Np = 473.9506 turns 474 turns

Ns = 1.05 Np ()Ns = 101.0105 turns 102 turns

Where:

Np => number of turns in the primaryNs => number of turns in the secondary

Ep => primary voltage in RMS (22CV)

Es => secondary voltage in RMS

f => frequency of operation in Hz (60Hz)

A => tongue area in square inches

B => flux density in Gauss (Table 3)

Step 7:

Compute for the size of the magnetic wires to be used for the coils.

Ip =

Ip = 0.5839 Ampere

Where:

Ip => current in the primary

Compute for the diameter of the wires for the primary and the

secondary coils.

d = 1.13 * I- login+

1

dp = 0.0246 inch diameter

ds = 0.0479 inch diameter


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Where:

d => diameter of the wire in inches

I => current in the wire in Amperes

Pin => required power in Volt-Amperes

The commercially available wire gauges based on the diameter of

the wire are shown in Table 4.

Primary wire size:

# 22 AWG, dp = 0.0253 inches

Secondary wire size:

# 16 AWG, ds = 0.0508 inches

Step 8:

Compute for the number of turns per layer. Initially compute for

the allowable winding length.

wl = tl2(margin)2(bobbin allowance)

wl = 1.936 inches

Where:

wl => winding length

tl => tongue length

margin => 0.125 inchesbobbin allowance => 0.032 inches

The number of turns per layer:

turns

layer=

wl

diameter of wire with insulation

Primary:

turns

r= 72.7820 turns/layer 73 turns/r

Secondary:

turns

r= 36.9466 turns/r 37 turns/r


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Step 9:

Compute the winding build up of the primary and secondary

winding.

WBtot = 1.1 (WBp + WBs + bobbin thickness + inter

winding insulation)

WBtot = 0.5086 inch

Where:

WBtot => total winding build up

WBp => winding build up in the primary

WBs => winding build-up in the secondary

Bobbin thickness => 0.095

Inter winding insulation => 0.002 number of winding

The total winding build up should not exceed 90% of the EI

laminations window If the computed value does exceed that

amount, choose a larger EI lamination and repeat steps 6, 8 and 9.

Step 10:

Compute for the total length of wires needed for the transformer.

Initially compute for the mean length of turns ( MLT ) of each

winding.

MLTp = 2 (tw + g + 4b) + WBp

MLTp = 7.3889 inches

MLTs = 2 (tw + g + 4b) + (2WBp + WBs)MLTs = 8.5306 inches

Where:

MLTp => mean length of turns in the primary

MLTs => mean length of turns in the secondary

g => stack height

b => bobbin thickness => 0.095

WB => winding build up


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The actual length of wires is computed from:

L = MLT N

Primary:

Lp = 3,502.3386 inches

Secondary:

Ls = 870.1212 inches

Where:

L => length of the wire

Step 11:

Since magnet wires are sold by the weight rather by length, thetotal weight of the wires are computed using

W(lbs) =

1

f

1

Primary:

Wp = 0.5677 lbs

Secondary:

Ws = 0.5669 lbs

Where:

W => weight of the wire in lbs


f => conversion factor (see Table 4)

Step 12:

Determine the percent efficiency ( ) and voltage regulation ( r )

from:

=out 1

outcore loss(copper loss) = 92.71%


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Vr = Is[ *+

]Vr = 0.024 or 2.4%


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Specification Needed:

Primary Voltage: 220 Vrms

Secondary Voltage, Current: 19 Volts dc, 2.1 Amperes

Operating Frequency = 60 Hz

Step 1:

Compute the total voltage across the secondary:

Esn = 2.35 Edc

Esn = 2.35 19V

Esn = 44.65 Volts

Where:

Edc => Secondary Voltage in Volts

2.35 => twice the ratio of the RMS to average value plus

5% regulation

Step 2:

Compute the secondary currents from:

Is = k Idc

Table 1: Factors K and K for Single- Phase Rectifier Supplies

Capacitor input K K

Full-wave

Half-wave

0.707

1.4

1.06

2.2

Is = 1.06 2.1A

Is = 2.226 Amperes

Where:

Is => secondary current in Amperes

k => k factor (see Table 1)


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Step 3:

Compute the output power in Watts.

Pout = Es Is

Pout = 44.65V 2.226A

Pout = 99.3909 Watts

Where:

Es => Secondary Voltage in Volts

Is => Secondary current in Amperes

Step 4:

Compute the required input power in Volt-Amperes.

VA =out

(n)

Table 2: Efficiencies for Various Sizes Power Supplies

Output in Watts Approximate Efficiency in Percent

20

30

4080100

200

70

75

808586

90

Note 1: The Interpolation

Step 1:

Find the range where the given value falls from the table.

Pout = 44.838 WattsRange available: 4080

Step 2:

Use the formula:


15/32

x1- d1

x1- x=

y1- d

y1- y

Where:

x1is the smallest from the range of independent valuesd1is the given value

x is the highest in the range of independent values

y1is the dependent value ofx1

d is the dependent value of the given value

y

is the dependent value ofx

Computation:

-

-1 = - d

-

(0.969545)(-1) = (85d)

-0.6047585 = -d

2 = 85.969545

VA =

VA =

VA = 128.4568 Volt-Amperes

Where:

Pout => Required Output Power computed previously

n => efficiency of the transformer

0.9 => power factor

VA => Power input in Volt-Amperes


16/32

Step 5:

Assume a square core area for convenience and compute

for the required area and the size of the EI lamination.

A =

A =1

A = 2.0312 in2

Where:

A => cross section area of the core in square

inches (n2).

Compute for the tongue width ( tw ) and the stacking height ( g ).

g = tw = g = tw = g = tw = 1.4252 in

Then use Table 3 to choose the size

of the EI lamination required.


17/32

Table 3. EI Core Data on , Tongue Lamination, and D Core Weight

At 60 Hertz EI Type

Lamination

Tongue Width of

Lamination (inches)

Core

Weight(lb)EI B (gauss)

3.9

5.813.0

17.024.0

37.0

54.082.0

110

145

195525

14,000

14,00014,000

14,00013,500

13,000

13,00012,500

12,000

12,000

11,00010,500

EI 21

EI 625EI 75

EI 75EI 11

EI 12

EI 12EI 125

EI 125

EI 13

EI 13EI 19

0.5

0.6250.75

0.750.875

1.00

1.001.25

1.25

1.5

1.51.75

0.199

0.3610.609

0.8120.966

1.43

2.142.83

3.97

4.92

6.569.75

Since 1.5 is the closest value to 1.4252, choose EI-13.

g = tw = 1.5 inches

The size of the EI lamination also gives the winding length ( wl ), which should

not exceed the tongue width ( tw ) and the allowable winding build-up which

should not be greater than the EI laminations window

W =1

tw

W =1

(1.5)

W = 0.75 inch

tl =

tw

tl =

(1.5)

tl = 2.25 inches


18/32

Step 6:

Compute for the number of the primary and secondary turns Np

and Ns.

Np = Ep 1

f

Np =1

1(1)Np = 473.9506 turns 474 turns

Ns = 1.05 Np

()

Ns = 1.05 (474)

Ns = 101.0105 turns 102 turns

Where:

Np => number of turns in the primary

Ns => number of turns in the secondary

Ep => primary voltage in RMS (22CV)

Es => secondary voltage in RMSf => frequency of operation in Hz (60Hz)

A => tongue area in square inches

B => flux density in Gauss (Table 3)

Step 7:

Compute for the size of the magnetic wires to be used for the coils.

Ip =

Ip =1

Ip = 0.5839 Ampere


19/32

Where:

Ip => current in the primary

Compute for the diameter of the wires for the primary and the

secondary coils.

d = 1.13 * I- login+

1

dp = 1.13 * -log1+

1

dp = 0.0246 inch diameter

ds = 1.13 * -log1+1

ds = 0.0479 inch diameter

Where:

d => diameter of the wire in inches

I => current in the wire in Amperes

Pin => required power in Volt-Amperes

The commercially available wire gauges based on the diameter of

the wire are shown in Table 4.


20/32

Table 4: Magnet wire data on diameter, resistance, and Core Weight.

AWG

B & S

Gauge

Diameter in inchesOhms per1000 Ft.

Pounds per1000 Ft.

Margin min inchesBare Enamelled

101112

13

14

15

16

17

1819

2021

22

2324

25

2627

28

29

30

31

3233

34

35

36

37

3839

40

0.10190.09270.808

0.0719

0.0641

0.0571

0.0508

0.0453

0.04030.0359

0.03200.0285

0.0253

0.04030.0359

0.0179

0.01590.0142

0.0126

0.0113

0.0100

0.0089

0.00800.0071

0.0063

0.0056

0.0050

0.0045

0.00400.0035

0.0031

0.10390.09270.0827

0.0738

0.0659

0.0588

0.0524

0.0469

0.04180.0374

0.03340.0299

0.0266

0.02390.213

0.0190

0.01690.0152

0.0135

0.0122

0.0109

0.0097

0.00880.0079

0.0070

0.0062

0.0056

0.0050

0.00450.0040

0.0036

0.99891.2601.588

2.003

2.525

3.184

4.016

5.064

6.3858.051

10.1512.80

16.14

20.3625.67

32.37

40.8151.47

64.90

81.83

103.2

130.1

164.1206.9

260.9

329.0

414.8

523.1

659.6831.8

1049

31.4324.9219.77

15.68

12.43

9.858

7.818

6.200

4.9173.899

3.0922.452

1.945

1.5421.223

0.9699

0.76920.6100

0.4837

0.3836

0.3042

0.2413

0.19130.1517

0.1203

0.0954

0.0757

0.0600

0.04760.0377

0.0299

0.250.250.25

0.25

0.25

0.25

0.1875

0.1875

0.18750.1562

0.15620.1562

0.125

0.1250.125

0.125

0.1250.125

0.125

0.125

0.125

0.125

0.09370.0937

0.0937

0.0937

0.0937

0.0937

0.06250.0625

0.0625


21/32

Primary wire size:

# 22 AWG, dp = 0.0253 inches

Secondary wire size:

# 16 AWG, ds = 0.0508 inches

Step 8:

Compute for the number of turns per layer. Initially compute for

the allowable winding length.

wl = tl2(margin)2(bobbin allowance)

wl = 2.252 (0.125)2 (0.032)

wl = 1.936 inches

Where:

wl => winding length

tl => tongue length

margin => 0.125 inches

bobbin allowance => 0.032 inches


22/32

The number of turns per layer:

turns

layer=

wl

diameter of wire with insulation

No. of Layers =turns

layer

Primary:

turns

layer=

turns

r

= 72.7820 turns/r 73 turns/r

No. of Layers =

No. of Lrs = 6.4932 7 rs

Secondary:

turns

layer=

turnsr

= 36.9466 turns/r 37 turns/r

No. of Layers =1

No. of Lrs = 2.7568 3 rs

Because some figures were rounded off, check if the values still

satisfy the number of turns in the primary winding.

Checking:

turns

layer=

p

o of ayers\

No. of Turns = turnslayer

(No. of Layers)


23/32

For Primary:

turns

layer=

turns

r= 67.714 turns/r 68 turns/r

No. of Turns = (68)(7)

No. of Turns = 476 turns

For Secondary:

turns

layer=

1

turnsr

= 34 turns/layer

No. of turns = (34)(3)

No. of turns = 102 turns

Step 9:

Compute the winding build up of the primary and secondary

winding.

p = o of layers in primary (dp inter layer

insulation)

WBp = 7 (0.0266 + 0.002)

WBp = o.2002 inch

s = o of layers in secondary (dsn inter layer

insulation)

WBs = 3 (0.0524 + 0.002)WBs = 0.1632 inch


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WBtot = 1.1 (WBp + WBs + bobbin thickness + inter

winding insulation)

WBtot = 1.1 (0.095 + 0.2002 + 0.1632 + 2 (0.002))

WBtot = 0.5086 inch

Where:

WBtot => total winding build up

WBp => winding build up in the primary

WBs => winding build-up in the secondary

Bobbin thickness => 0.095

Inter winding insulation => 0.002 number of winding

The total winding build up should not exceed 90% of the EI

laminations window If the computed value does exceed that

amount, choose a larger EI lamination and repeat steps 6, 8 and 9.


25/32

Step 10:

Compute for the total length of wires needed for the transformer.

Initially compute for the mean length of turns ( MLT ) of each

winding.

MTp = (tw g b) WBp

MTp = (1 1 ()) ()

MLTp = 7.3889 inches

MLTs = 2 (tw + g + 4b) + (2WBp + WBs)

MTs = (1 1 ()) ( () 1)

MLTs = 8.5306 inches

Where:

MLTp => mean length of turns in the primary

MLTs => mean length of turns in the secondary

g => stack height

b => bobbin thickness => 0.095

WB => winding build up


26/32

The actual length of wires is computed from:

L = MLT N

Primary:

Lp = 7.3889 474

Lp = 3,502.3386 inches

Secondary:

Ls = 8.5306 102

Ls = 870.1212 inches

Where:L => length of the wire

Step 11:

Since magnet wires are sold by the weight rather by length, the

total weight of the wires are computed using

W(lbs) =

1

f

1

Primary:

Wp =

1

1

1

Wp = 0.5677 lbs

Secondary:

Ws =

11

1

1

1

Ws = 0.5669 lbs

Where:

W => weight of the wire in lbs


f => conversion factor (see Table 4)


27/32

a. Winding resistances (From Table 4)Rp =

p

1

11

1

Rp =

1

11

1

Rp = 4.7106

Rs =s

1

1

1

Rs =11

1

1

1

Rs = 0.2912

Where:

Rp => winding resistance in the primary

Rs => winding resistance in the secondary

b. Copper lossesCp = Ip2 Rp

Cp = (0.5839)2 (4.7106)Cp = 1.606 Watts

Cs = Is2

Rs

Cs = (2.1)2 (0.2912)

Cs = 1.2842 Watt

Ctot = Cp + Cs

Ctot = 1.606 + 1.2842

Ctot = 2.8902 Watts


28/32

Where:

Cp => copper loss in the primary

Cs => copper loss in the secondary

Ctot => total copper loss

c. Core loss (From Table 5 and Table 3)Core loss = (Approx. Core loss) (Core Weight)

Table 5: Core characteristics data at different operating frequencies.

Freq.

inHertz

Lamination

Thickness(inches)

Core

Material

Core Flux

DensityBm in Gauss

Approximate

Core Lossin Watts/Lb

25

6060

400800

0.025

0.0140.014

0.0040.004

2.5% silicon

4% siliconGrain-oriented silicon

Grain-oriented siliconGrain-oriented silicon

14,000

12,00015,000

10,0006,000

0.65

1.01.0

4.54.5

Core loss =1 att

1 lb(4.92 lb)

Core loss = 4.92 Watts

Where:

Core loss is inatt

lb

Core weight is in lb

d. Voltage DropVDp = Ip Rp

VDp = (0.5839) (4.7106)

VDp = 2.7505 Volts

VDs = Is Rs

VDs = (2.1) (0.2912)

VDs = 0.6115 Volts


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Where:

VDp => voltage drop in the primary

VDs => voltage drop in the secondary

Step 12:

Determine the percent efficiency ( ) and voltage regulation ( r )

from:

=out 1

outcore loss(copper loss)

= 1

() = 92.71%

Vr = Is[ *+

]

Vr = (2.1) [ *+ ]Vr = 0.024 or 2.4%


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Summary of Design Computations.

Winding

DC

Voltage at

ratedcurrent

AmpereNumber of

TurnsAWG#

Weight

Needed

Primary

Secondary

220 V

19 V dc

0.58 A

2.1 A

474

102

22

16

EI lamination to be used is EI-13

The Output Transformer


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Step 6:

Immerse the core and the windings in a can of varnish for 10 to 15 minutes. This

is an insulation also. Remove and let the transformer dry

Step 7:

To complete everything, bake the transformer in the cover for about four hours or

let it dry for a couple of days to let the varnish dry and remove its sticky nature.

Step 8:

Now your transformer is ready for testing.

machine design 3

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