- Transformer design
"This page deals primarily with the design of electrical transformers. For the definition, function, and history, see "

Transformer .In

electrical engineering , practical**transformer design**requires knowledge of electrical principles, materials, and economics. Small transformers may be designed using handbook data and pencil-and-paper calculations,but larger or mass-produced units are often designed with extensive computer modelling. While the fundamental principles of transformer operation are deduced from Maxwell's equations and electromagneticfield theory , the ability to solve the fundamental equations for practical problems has greatly increased with the use of digital computers. Numerical techniques allow designers to study magnetic field distribution within the unit, and so allow them to make trade-offs between initial cost, material inputs, and the economic value of losses.**Power transformer design**The designer first needs several known factors to design a transformer. For a transformer using a sine or square wave, one needs to know the incoming line voltage, the operating frequency, the secondary voltage(s), the secondary current(s), the permissible temperature rise, the target efficiency, the physical size one can use, and the cost limitations. Once these factors are known, design can begin.

**Initial calculations**The designer first starts with the primary voltage and frequency. Since they are a known factor, they are the first numbers to be plugged into the equations. One then will find the power in watts (or volt-amperes) of each secondary winding by multiplying the voltage by the current of each coil. These are added together to get the total power the transformer must provide to the load(s).

s. These losses are dissipated as heat. Here, the permissible temperature rise must be kept in mind. Each type of core material will have a loss chart whereby one can find the loss in watts per pound by looking up the operating flux density and frequency. Next, one selects the type of iron by what efficiency is stated, and the value of losses to the user. Once the iron is selected, the flux density is selected for that material.

**Type of iron (steel)**(B). When using the equations, the two most important are the number of turns (N), and the core area (a). One needs to find the core area in square centimeters or inches, and match it to the total power in watts or volt-amperes. The larger the core, the more power it will handle. Once this core size is calculated, one then finds the number of turns for the primary. One then is looking at a transformer whose primary voltage will cause a flux density of a specified amount due to the number of turns in a certain type/size of core.

For sine wave operation, the designer then uses either the two short formulas, or begin using the long formulas which are more exact, and whereby all the factors can be changed. For

square wave operation, refer to the notes at the end of the equations section. Either way, it's time to use a transformer design sheet. The design sheet has places to write the details such as the flux density, the number of turns, calculate the turns per layer, and thickness of the coil.**Secondary turns calculation**Once the number of turns of the primary are calculated, the secondary windings numbers can be calculated with the same turns per volt figure. If the primary has 120 turns for 120 volts input, we would have 1 turn per volt. If we needed a 12 volt secondary, then we would require 12 turns on it. This is for a perfect transformer without losses though.

In reality, there are losses that have to be added, as the 12 turn coil will not produce 12 volts under load, but a lower voltage. A rule of thumb is to allow for 5% in losses. (Transformers below 300w often have higher regulation losses). In this case, we would multiply the 12 turns by 1.05 to get a new number of turns equaling 12.6 turns. Since fractional turns are not possible for line frequency transformers, 13 turns would be used. It is best to have a slightly higher voltage than one too low. Beware, smaller transformers which have a higher turns per volt, have higher losses, and the efficiency drops as the size goes down.

The turns per volt figure typically varies from 1 to about 4, with around 4 turns per volt common for small appliance transformers, and around 1 turn per volt used for intermittent duty fan cooled microwave oven transformers. Volts per turn is commonly used for larger transformers, distribution transformers are often limited by excessive insulation required between each turn.

Here's where cut-and-try still comes into transformer design. Since the primary coil has to be wound with a wire that is large enough to handle the total power the transformer will handle at a certain flux density, it must fit within the cores window(s) once the overall size is calculated after adding the bobbin and paper thickness of each layer. Most of the time, the design has to be played with over this because the coil is too big for the windows. If the coil doesn't fit, there are a few options. A larger core with larger window openings having the same core area can be used, or the flux density can be raised by reducing the turns on the primary. Once these turns are reduced, the turns in the secondary will be reduced. This since the number of volts per turn in the primary equal the number of volts per turn in the secondary minus losses. However, this is at the expense of raising the flux density, the magnetizing current, the temperature, and lowering the efficiency. It's much better to select a larger core which has larger windows to accept the coil. The depth or thickness of the new core can be adjusted to equal the old core area in square centimeters or square inches. This measurement is the cores tongue width multiplied by its depth or thickness. As the core size goes up, so does the tongue width.

**Thickness of windings**When calculating the coil thickness, several things need to be considered. The voltage that each layer of coil sees will determine the wires insulation thickness. Once this voltage is known, the diameter of the insulated wire can be used. By knowing the wire diameter, the number of turns per layer can be calculated, and the number of layers by using the window height. Next, one adjusts the thickness of the insulation paper for the layers of each winding due to the voltage between the coils. This thickness is added to the total coil thickness by multiplying the paper thickness by the number of layers. The paper that separates two different windings is always thicker than the layer paper to match the voltage difference between the windings and must support the wire. Last, the bobbin thickness is added. All is then added to the design sheet and the total calculated. This total thickness is compared to the window dimensions for a fit.

**Design should not exceed 80-85% of available opening to allow for manufacturing tolerances.**A smaller coil with few layers is always recommended. A coil with a large number of layers will run hotter than one with a few. Each winding has a "hot spot" which is always located mid-way at its center. If the winding has a number of layers, the heat will increase at this hot spot. The hot spot is almost always where the winding will fail due to heat. The heat from each winding has to travel through each layer and is dissipated from the outside of the coil. This means that the winding closer to the core will be hotter than the outer ones. Since this is the case, and most of the time the winding closest to the core is the primary, the largest wire that will fit should be used. The exception to having the primary here is using a winding with very small diameter wire. Since the coil will expand due to heat, a small wire coil on the outside could break because of the expansion. Being at the core, it would expand less and not break the wire. Most small transformers fail when the wire coming up from the bottom of the coil breaks near the terminal post. A good rule of thumb is to use 1000 circular mills per ampere on the primary winding. Most small bias windings, rated at a few miliamperes, used in vacuum tube circuits are wound in this manner.For using a two section bobbin (for a two winding transformer), the above is not necessary. These are used by jumble-winding the wire on each section of the bobbin. Jumble-winding by definition means that the wire is wound on the bobbin in a random way without layers separated by paper. However, the amount of wire used for each winding has to fit within the bobbin so it too will fit inside the cores windows. Most small transformers are manufactured this way to save cost, as it would be very difficult to neatly stack extremely fine wire.

**Wire selection**The wire generally used in transformer coils is (

magnet wire ). Magnet wire is generally a solid copper wire with an insulating jacket such as varnish. Other wires such aslitz wire are common use for r.f. transformers.The wire is selected by its ability to carry the proper amount of current without getting hot enough to melt the insulation or wire itself. It is sized by its cross sectional area measured in circular mils per ampere, or more commonly in engineering design circles, amperes per square meter. In transformer use, the circular mils per ampere runs anywhere from 500 cir. mils for intermittent operation, to 1500 cir. mils for heavy duty continuous operation. For most applications, 800 to 1000 cir. mils is a good starting point. The real value chosen is iterative, because heat is often not the limiting factor, as the desired regulation often puts the temperature rise well below insulation limits. Most wire charts have the wires cross sectional area in circular mils to make selection easier.

**The core stack**The core stack is the total amount of the steel laminations needed to produce the correct core area for the power in watts, or volt amperes that the transformer is required to handle.

The core can be stacked in two different ways when using steel laminations. The most common is the interleaved fashion where each lamination is staggered opposite to the other. This provides for the least air gap in the core and the highest efficiency. The other way is butt stacked. In this way, all the E type lams are stacked on one side, and all the I type lams are stacked on the other. This way though creates an air gap where the butt joint is created thus increasing the losses. However, when a DC current is superimposed on an AC current as in an audio transformer, the air gap can stop the core from saturating. A combination of the two stacking types can be used with good results obtaining the best properties of both.

When using the formulas on this page to calculate the core area, a stacking factor should be included. The stacking factor is given by the lamination manufacturer on the individual specification sheet for each size lamination. It varies by the thickness of the material, and whether it is either butt stacked, or interleaved. This factor ranges from around 0.90 to 0.98. The formulas here can be used as is, but one will have a slightly smaller core.

**Watts versus volt-amperes**A transformer's power handling ability is determined in two different ways. If the transformer's secondary is supplying a totally resistive load, one can simply use wattage, or the voltage multiplied by the current. However, if the transformer's secondary is supplying a reactive load containing capacitance or inductance, such as in most DC power supplies, one must use volt-amperes in place of watts. Volt-amperes is simply the wattage divided by a power factor. The power factor for most DC power supply circuits is around 0.90 to 0.95.

When calculating the power required for the secondary windings in DC power supplies, one must take into account the equivalent series resistance of the capacitors, and or the way they, or any inductors act in the filtering circuits. This is due to the current either leading or lagging the voltage in reactive circuits. The type of rectifier circuit also comes into play, and is described in the following section.

**Rectifier transformers**Rectifier transformers are transformers used to feed a rectifier circuit which converts an AC current into a DC current. Due to the small conduction angle when feeding a rectifier & reservoir, the rms AC current in the transformers secondary is somewhat higher than the DC load current. Each rectifier circuit has different conduction angle, and thus different rms current needs. The proper rms current the transformer needs to supply, and the volt amperes of each secondary, are calculated using the following formulae.

**Half Wave Rectifier (HWR):**"Without capacitor";

IAC = 1.6 x IDC

VA = 3.5 x (watts + IDC)

"With capacitor";

IAC = 2.6 x IDC

VA = 2.3 x (watts + IDC)

**Full Wave Center Tap (FWCT):**"Without capacitor";

IAC = 0.8 x IDC

VA = 1.4 x (watts + IDC)

"With capacitor";

IAC = 1.2 x IDC

VA = 1.7 x (watts + IDC)

**Full Wave Bridge (FWB):**"Without capacitor";

IAC = 1.1 x IDC

VA = 1.2 x (watts + (2 x IDC))

"With capacitor";

IAC = 1.8 x IDC

VA = 1.4 x (watts + (2 x IDC))

**Dual Complementary Rectifies (DCR):**"Without capacitor";

IAC = 1.1 x IDC

VA = 1.2 x (watts + (2 x IDC))

"With capacitor";

IAC = 1.8 x IDC

VA = 1.4 x (watts + (2 x IDC))

**Equations**There are two approaches used in designing transformers. One uses the long formulas, and the other uses the Wa product. The Wa product is simply the cores window area multiplied by the cores area. Some say it simplifies the design, especially in C-core (cut core) construction. Most manufacturers of C-cores have the Wa product added into the tables used in their selection. The designer takes the area used by a coil and finds a C-core with a similar window area. The Wa product is then divided by the window area to find the area of the core. Either way will bring the same result.

For a transformer designed for use with a sine wave, the universal voltage formula is:

:$E=\{frac\; \{2\; pi\; f\; N\; a\; B\}\; \{sqrt\{2\; \{10^\{-8\}\; !=4.44\; f\; N\; a\; B\; \{10^\{-8$

thus,

:$E=\{4.44\; f\; N\; a\; B\; \{10^\{-8\}\; !$

where,

* "E" is the sinusoidal rms or root mean square voltage of the winding,

* "f" is the frequency in hertz,

* "N" is the number of turns of wire on the winding,

* "a" is the cross-sectional area of the core in square centimeters or inches,

* "B" is the peak magnetic flux density in Teslas or Webers per square meter, gausses per square centimeter, or lines (maxwells) per square inch.

* "P" is the power in volt amperes or watts,

* "W" is the window area in square centimeters or inches and,

* "J" is the current density.

* Note: 10 kilogauss = 1 Tesla.This gives way to the following other transformer equations for cores in square centimeters:

:$N=\{frac\; \{E\; 10^8\}$4.44 f B a} !

:$B=\{frac\; \{E\; 10^8\}$4.44 f N a} !

:$a=\{frac\; \{E\; 10^8\}$4.44 f B N} !

:$P=\{0.707\; J\; f\; W\; a\; B\}\; !$

The derivation of the above formula is actually quite simple. The maximum induced voltage, "e", is the result of N times the time-varying flux:

"e" = N "dφ/dt"

If using RMS voltage values and E equal the rms value of voltage then:

e = E$\{sqrt\{2$

and

E = "dφ/dt"$\{frac\; \{N\}\; \{sqrt\{2\}$

Since the flux is created by a sinusoidal voltage, it too varies sinusoidally:

φ(t) = Φ$\_\{max\}\; sin\; wt$ = A$B\_\{max\}\; sin\; wt$, where A = area of the core

Taking the derivative we have:

dφ(t)/dt = wA$B\_\{max\}\; cos\; wt$

Substituting into the above equation and using

w = 2 $\{pi\}$f and the fact that we are only concerned with the maximum value yields

E = $frac$2pi}fNAB} {sqrt{2

**Imperial measurement system**The formulas for the imperial (inch) system are still being used in the United States by many transformer manufacturers. Most steel EI laminations used in the US are measured in inches. The flux is still measured in gauss or Teslas, but the core area is measured in square inches. 28.638 is the conversion factor from 6.45 x 4.44 (see note 1). The formulas for sine wave operation are below. For square wave operation, see Note (3):

:$E=\{28.638\; f\; a\; N\; B\}\; !$

:$N=\{frac\; \{E\; 10^8\}$28.638 f B a} !

:$T=\{frac\; \{10^8\}$28.638 f a B} !

:$B=\{frac\; \{E\; 10^8\}$28.638 f N a} !

:$a=\{frac\; \{E\; 10^8\}$28.638 f B N} !

To determine the power ("P") capability of the core, the core stack in inches ("D"), and the window-area ("Wa") product, the formulas are:

:$P=\{frac\; \{fBWa\}$17.26S} !

:$Wa=\{frac\; \{17.26SP\}$fB} !

:$D=\{frac\; \{17.26PS\}$WCfB} !

where,

* "P" is the power in volt amperes or watts,

* "T" is the volts per turn,

* "E" is the RMS voltage,

* "S" is the current density in circular mils per ampere (Generally 750 to 1500 cir mils),

* "W" is the window area in square inches,

* "C" is the core width in square inches,

* "D" is the depth of the stack in inches and,

* "Wa" is the product of the window area in square inches multiplied by the core area in square inches. This is especially useful for determining C-cores but can also be used with EI types.**impler formulae**A shorter formula for the core area (a) and the turns per volt (T) can be derived from the long voltage formula by multiplying, rearranging, and dividing out. This is used if one wants to design a transformer using a sine wave, at a fixed flux density, and frequency. Below is the short formulas for core areas in square inches having a flux density of 12 kilogauss at 60 Hz (see note 2):

:$a=\{0.1725\; \{sqrt\{P\}\; !$

:$T=\{frac\; \{4.85\}$a} !

And for 12 kilogauss at 50 Hz::$a=\{0.206\; \{sqrt\{P\}\; !$

:$T=\{frac\; \{5.82\}$a} !**Equation notes*** Note 1: The factor of 4.44 is derived from the first part of the voltage formula. It is from 4 multiplied by the form factor ("F") which is 1.11, thus 4 multiplied by 1.11 = 4.44. The number 1.11 is derived from dividing the rms value of a sine wave by the its average value, where "F" = rms / average = 1.11.

* Note 2: A value of 12 kilogauss per square inch (77,400 lines per sq. in.) is used for the short formulas above as it will work with most steel types used (M-2 to M-27), including unknown steel from scrap transformer laminations in TV sets, radios, and power supplies. The very lowest classes of steel (M-50) would probably not work as it should be ran at or around 10 kilogauss or under.

* Note 3: All formulas shown are for sine wave operation only. Square wave operation does not use the form factor (F) of 1.11. For using square waves, substitute 4 for 4.44, and 25.8 for 28.638.

* Note 4: None of the above equations show the stacking factor (Sf). Each core or lamination will have its own stacking factor. It is selected by the size of the core or lamination, and the material it is made from. At design time, this is simply added to the string to be multiplied. Example; "E" = 4.44 "f" "N" "a" "B" "Sf"

**Electrical steel types****ilicon steel**Ref: [

*Lowdon,Eric (1981). "Practical Transformer Design Handbook." ISBN 978-0-672-21657-2*] [*Fink,Donald (1969). "Standard Handbook For Electrical Engineers." ISBN 978-0-070-22005-8*] [*McPherson,W (1981). "Reference Data For Radio Engineers." ISBN 978-0-672-21218-5*] [*Eng. Staff of Massachusetts Institute of Technology (1949). "Magnetic Circuits and Transformers." John Wiley & Sons. ISBN 978-0-262-63063-4*]* Note 1: CRGO = Cold rolled, grain oriented, and CRNO = cold rolled, non oriented.

* Note 2: In the "M" numberining system set by the ASTM, the smaller number yields the highest efficiency, and lowest core losses. M-43 has a core loss at 12 kilogauss of approx. 2 watts per pound. M-15 at 12 kilogauss is approx. 0.75 watts per pound, and M-6 material has a loss of 0.64 watts per pound at 15 kilogauss. [

*Fink, Donald (1969). "Standard Handbook For Electrical Engineers." ISBN 978-0-070-22005-8*] [*Eng. Staff of Massachusetts Institute of Technology (1949). "Magnetic Circuits and Transformers." John Wiley & Sons. ISBN 978-0-262-63063-4*]**Other alloys**There are various iron alloys other than silicon-steel or low-carbon steel. These include alloys which contain nickel-iron (

Permalloy ), cobalt-nickel-iron (Perminvar ), cobalt-iron (Permendur ), and vanadium-cobalt-iron. Others includeSupermalloy , amorphous Metglas,Mu-metal ,Sendust , iron powder, andferrite types.Some of the

Permalloy types are processed to accentuate the squareness of the B-H loop and carry proprietary names like SuperSquare 80 (Magnetic Metals Corp.), and Square Permalloy Hy-Ra 80 (Carpenter Steel Co.). The squareness of the B-H loop helps in switching transformers as in inverter type power supplies with a square wave input. The nickel-iron content may range from about 45% to over 85%.Perminvar exhibits a substantially constant permeability and low hysteresis loss at low flux densities. This is mainly due to the addition of cobalt to the nickel and iron. In some cases it may have the odd property of low coercive force and remanence although the hysteresis loop area is still greater than zero. One type of Perminvar isFernico .Permendur is created by mixing cobalt with iron. It has a high permeability at high flux densities with a very high saturation point. It also has a high incremental permeability and is very good to use with a combination of AC and DC voltages combined such as in filter chokes.Vanadium-cobalt-iron has the very highest saturation point with low losses, but it is very expensive.

For descriptions of the other materials and shapes, refer to the Wikipedia section titled

Magnetic core .**References***"Practical Transformer Design Handbook," Lowdon, Eric (1981), ISBN 978-0-672-21657-2

*"Standard Handbook For Electrical Engineers," Fink, Donald (1969), ISBN 978-0-070-22005-8

*"Reference Data For Radio Engineers," McPherson, W (1981), ISBN 978-0-672-21218-5

*"Magnetic Circuits and Transformers," Engineering Staff, MIT (1949), John Wiley & Sons. ISBN 978-0-262-63063-4

* [*http://www.atc-frost.com/products/design/va.htm ATC-Frost Magnetics*]**Footnotes**

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