Determination of power transformer power

How to find out the power of a transformer?

For the manufacture of transformer power supplies, a single-phase power transformer is required, which reduces the alternating voltage of the 220 volt mains to the required 12-30 volts, which is then rectified by a diode bridge and filtered by an electrolytic capacitor. These transformations of electric current are necessary since any electronic equipment is assembled on transistors and microcircuits, which usually require a voltage of no more than 5-12 volts.

To assemble the power supply yourself. a novice radio amateur needs to find or purchase a suitable transformer for the future power supply. In exceptional cases, you can make a power transformer yourself. Such recommendations can be found on the pages of old books on radio electronics.

But nowadays it’s easier to find or buy a ready-made transformer and use it to make your own power supply.

Full calculation and independent production of a transformer for a beginning radio amateur is quite a difficult task. But there is another way. You can use a used but serviceable transformer. To power most home-made designs, a low-power power supply with a power of 7-15 watts is enough.

If the transformer is purchased in a store, then, as a rule, there are no special problems with selecting the right transformer. The new product has all its main parameters indicated, such as power. input voltage. output voltage. as well as the number of secondary windings, if there is more than one.

But what if you come across a transformer that has already worked in some device and you want to reuse it to design your own power supply? How to determine the power of a transformer, at least approximately? The power of the transformer is a very important parameter, since the reliability of the power supply or other device you assemble will directly depend on it. As you know, the power consumed by an electronic device depends on the current it consumes and the voltage required for its normal operation. Approximately this power can be determined by multiplying the current consumed by the device ( I n to the device supply voltage ( U n). I think many are familiar with this formula from school.

Let's look at determining the power of a transformer using a real example. We will train on the TP114-163M transformer. This is an armor-type transformer, which is assembled from stamped W-shaped and straight plates. It is worth noting that transformers of this type are not the best in terms of efficiency (Efficiency). But the good news is that such transformers are widespread, often used in electronics and can be easily found on the shelves of radio stores or in old and faulty radio equipment. In addition, they are cheaper than toroidal (or, in other words, ring) transformers, which have high efficiency and are used in fairly powerful radio equipment.

So, before us is the transformer TP114-163M. Let's try to roughly determine its power. As a basis for calculations, we will take recommendations from the popular book by V.G. Borisov "Young Radio Amateur".

To determine the power of a transformer, it is necessary to calculate the cross-section of its magnetic core. In relation to the TP114-163M transformer, the magnetic core is a set of stamped W-shaped and straight plates made of electrical steel. So, to determine the cross-section, it is necessary to multiply the thickness of the set of plates (see photo) by the width of the central lobe of the W-shaped plate.

When calculating, you must respect the dimensions. It is better to measure the thickness of the set and the width of the central petal in centimeters. Calculations must also be made in centimeters. So, the thickness of the set of the transformer under study was about 2 centimeters.

Next, measure the width of the central petal with a ruler. This is a more difficult task. The fact is that the TP114-163M transformer has a dense set and a plastic frame. Therefore, the central petal of the W-shaped plate is practically invisible; it is covered by the plate, and it is quite difficult to determine its width.

The width of the central petal can be measured at the side, the very first W-shaped plate in the gap between the plastic frame. The first plate is not complemented by a straight plate and therefore the edge of the central lobe of the W-shaped plate is visible. Its width was about 1.7 centimeters. Although the calculation given is indicative. but it is still desirable to carry out measurements as accurately as possible.

We multiply the thickness of the magnetic core set ( 2 cm.) and the width of the central lobe of the plate ( 1.7 cm.). We get the cross-section of the magnetic circuit - 3.4 cm 2. Next we need the following formula.

Where S— cross-sectional area of ​​the magnetic circuit; P tr— transformer power; 1,3 — average coefficient.

After some simple transformations, we obtain a simplified formula for calculating the power of a transformer based on the cross-section of its magnetic core. Here she is.

Let's substitute the value of the section into the formula S = 3.4 cm 2. which we received earlier.

As a result of calculations, we obtain the approximate value of the transformer power

7 Watt. Such a transformer is quite enough to assemble a power supply for a 3-5 watt monophonic audio amplifier, for example, based on the TDA2003 amplifier chip.

Here is another one of the transformers. Labeled as PDPC24-35. This is one of the representatives of transformers - “babies”. The transformer is very miniature and, naturally, low-power. The width of the central petal of the W-shaped plate is only 6 millimeters (0.6 cm).

The thickness of the set of plates of the entire magnetic circuit is 2 centimeters. According to the formula, the power of this mini-transformer is equal to about 1 W.

This transformer has two secondary windings, the maximum permissible current of which is quite small, amounting to tens of milliamps. Such a transformer can only be used to power circuits with low current consumption.

9zip.ru Tube sound hi-end and retro electronics Online calculator for calculating the overall power of a transformer based on the size of the magnetic circuit

It's no secret that radio amateurs often independently wind transformers to suit their needs. After all, it is not always possible to find, for example, a ready-made network transformer. This question becomes more relevant when you need an anode-filament or output transformer for a tube amplifier. All that remains is to stock up on wire and select good cores.

Sometimes it is not easy to get the necessary magnetic core and you have to choose from what is available. To quickly calculate the overall power, the online calculator given here was written. Based on the dimensions of the core, you can quickly carry out all the necessary calculations, which are performed using the formula below, for two types: PL and SHL.

Enter the dimensions of the transformer core magnetic circuit. Adjust other values ​​if necessary. Below you will see the calculated overall power of the transformer, which can be made on such a core, according to the formula:

And a small FAQ:

Is it possible to use iron from UPS transformers to make output transformers?

In these transformers, the plates have a thickness of 0.5 mm, which is not welcome in audio. But if you want, you can. When calculating outputs, one should proceed from the parameters of 0.5 T at a frequency of 30 Hz. When calculating security forces on this hardware, you should set no more than 1.2 T.

Is it possible to use plates from different transformers?

If they are the same size, then yes. To do this you should mix them.

How to properly assemble a magnetic circuit?

For a single-cycle output, you can place the two outer Sh-plates on the opposite side, as is often done in factory TVZs. Place I-plates in the gap through the paper, 2 pieces less. Taking the transformer so that the I-plates are at the bottom, place it on a thick, flat metal plate with a light blow. This can be done several times, monitoring the process with an inductance meter, to obtain the same pair of transformers.

How to determine the power of a transformer using a magnetic circuit?

For push-pull amplifiers, you need to divide the overall power of the iron by 6-7. For single-ended ones - 10-12 for a triode and 20 for a tetrode-pentode.

How to tighten a power transformer, is it necessary to glue the magnetic core?

If you want to glue, then use liquid glue. We apply a constant 5-15 volts to the primary winding to obtain a current of about 0.2A. In this case, the horseshoes will tighten without deformation. After this, you can put on the bandage, carefully tighten it and leave it until the glue dries.

How to remove the varnish that covers UPS transformers?

Soak for a couple of days in acetone or boil for a couple of hours in water. After this, the varnish should be removed. Mechanical removal of varnish is unacceptable, because burrs will appear and the plates will short-circuit with each other.

Are these transformers suitable anywhere without disassembling and rewinding?

If they have an additional winding (about 30 volts), then by connecting it in series with the primary, you can get a powerful incandescent transformer. But you need to look at the no-load current, because... these transformers are not designed to last long and are often not wound as we would like them to be.

Types of magnetic cores of power transformers.

The magnetic core of the low-frequency transformer consists of steel plates. Using laminations instead of a solid core reduces eddy currents, which increases efficiency and reduces heat.

Magnetic cores of type 1, 2 or 3 are produced by stamping.
Magnetic cores of types 4, 5 or 6 are produced by winding a steel tape onto a template, and magnetic cores of types 4 and 5 are then cut in half.

1, 4 – armored,
2, 5 – rod,
6, 7 – ring.

To determine the cross-section of the magnetic circuit, you need to multiply the dimensions “A” and “B”. For calculations in this article, the section size in centimeters is used.

Transformers with twisted rod position 1 and armored magnetic cores position 2.

Transformers with stamped armored magnetic cores, position 1, and core magnetic cores, position 2.

Transformers with twisted ring magnetic cores.

How to determine the overall power of a transformer.

The overall power of a transformer can be approximately determined by the cross-section of the magnetic core. True, the error can be up to 50%, and this is due to a number of factors. The overall power directly depends on the design features of the magnetic core, the quality and thickness of the steel used, the size of the window, the amount of induction, the cross-section of the winding wire and even the quality of the insulation between the individual plates.

The cheaper the transformer, the lower its relative overall power.
Of course, it is possible through experiments and calculations to determine the maximum power of a transformer with high accuracy, but there is not much point in this, since during the manufacture of the transformer, all this is already taken into account and reflected in the number of turns of the primary winding.
So, when determining the power, you can be guided by the cross-sectional area of ​​the set of plates passing through the frame or frames, if there are two of them.

P = B * S² / 1.69

Where:
P– power in Watts,
B– induction in Tesla,
S– cross-section in cm²,
1,69 – constant coefficient.

First, we determine the cross-section, for which we multiply the dimensions A and B.

S = 2.5 * 2.5 = 6.25 cm²

Then we substitute the cross-sectional size into the formula and get the power. I chose 1.5Tc induction, since I have an armored twisted magnetic circuit.

P = 1.5 * 6.25² / 1.69 = 35 Watt

If you need to determine the required cross-sectional area of ​​the manipulator based on the known power, you can use the following formula:

S = ²√ (P * 1.69 / B)

It is necessary to calculate the cross-section of an armored stamped magnetic circuit for the manufacture of a 50-watt transformer.

S = ²√ (50 * 1.69 / 1.3) = 8cm²

The magnitude of induction can be found in the table. You should not use maximum induction values, as they can vary greatly for magnetic cores of different quality.

Maximum indicative values ​​of induction.

In a household, it may be necessary to equip lighting in damp areas: basement or cellar, etc. These rooms have an increased risk of electric shock.

In these cases, you should use electrical equipment designed for a reduced supply voltage, no more than 42 volts.
You can use a battery-powered electric flashlight or use a step-down transformer from 220 volts to 36 volts.

As an example, let's calculate and manufacture a single-phase 220/36 volt power transformer.
To illuminate such rooms, a 36-volt electric light bulb with a power of 25-60 watts is suitable. Such light bulbs with a base for a standard socket are sold in electrical goods stores.

If you find a light bulb of a different power, for example 40 watts. There's nothing to worry about - she'll do too. It’s just that our transformer will be made with a power reserve.

LET'S MAKE A SIMPLER CALCULATION OF A 220/36 VOLT TRANSFORMER.

Secondary power: P2 = U2 I2 = 60 watts

Where:
P2– power at the output of the transformer, we set it to 60 watts;
U2- voltage at the transformer output, we set 36 volts;
I2- current in the secondary circuit, in the load.

Transformer efficiency up to 100 watts usually equal to no more &51; = 0,8 .
Efficiency determines how much of the power consumed from the network goes to the load. The remainder goes to heating the wires and core. This power is irretrievably lost.

Let's determine the power consumed by the transformer from the network, taking into account losses:

P1 = P2 / = 60 / 0.8 = 75 watts.

Power is transferred from the primary winding to the secondary winding through the magnetic flux in the magnetic core. Therefore, from the value of P1. power consumed from a 220 volt network. The cross-sectional area of ​​the magnetic circuit S depends.

The magnetic core is a W-shaped or O-shaped core made from sheets of transformer steel. The core will contain a frame with primary and secondary windings.

The cross-sectional area of ​​the magnetic circuit is calculated by the formula:

Where:
S- area in square centimeters,
P1- power of the primary network in watts.

S = 1.2 √75 = 1.2 8.66 = 10.4 cm².

By value S The number of turns w per volt is determined by the formula:

In our case, the cross-sectional area of ​​the core is S = 10.4 cm2.

w = 50 / 10.4 = 4.8 turns per 1 volt.

Let's calculate the number of turns in the primary and secondary windings.

Number of turns in the primary winding at 220 volts:

W1 = U1 w = 220 4.8 = 1056 turns.

Number of turns in the secondary winding at 36 volts:

W2 = U2 w = 36 4.8 = 172.8 turns, rounded to 173 turns.

In load mode, there may be a noticeable loss of part of the voltage across the active resistance of the secondary winding wire. Therefore, for them it is recommended to take the number of turns 5-10% more than calculated. Let's take W2 = 180 turns.

The magnitude of the current in the primary winding of the transformer:

I1 = P1 / U1 = 75 / 220 = 0.34 amperes.

Current in the secondary winding of the transformer:

I2 = P2 / U2 = 60 / 36 = 1.67 amperes.

The diameters of the wires of the primary and secondary windings are determined by the values ​​of the currents in them based on the permissible current density, the number of amperes per 1 square millimeter of conductor area. For transformers, the current density for copper wire is assumed to be 2 A/mm².

At this current density, the diameter of the wire without insulation in millimeters is determined by the formula:

For the primary winding, the wire diameter will be:

d1 = 0.8 √I 1 = 0.8 √0.34 = 0.8 * 0.58 = 0.46 mm. Let's take 0.5 mm.

Wire diameter for secondary winding:

d2 = 0.8 √I 2 = 0.8 √1.67 = 0.8 * 1.3 = 1.04 mm. Let's take 1.1 mm.

IF THERE IS NO WIRE OF THE REQUIRED DIAMETER. then you can take several thinner wires connected in parallel. Their total cross-sectional area must be no less than that corresponding to the calculated one wire.

The cross-sectional area of ​​the wire is determined by the formula:

Where: d - wire diameter.

For example: We could not find a wire for the secondary winding with a diameter of 1.1 mm.

The cross-sectional area of ​​a wire with a diameter of 1.1 mm is equal to:

s = 0.8 d² = 0.8 1.1² = 0.8 1.21 = 0.97 mm²

Let's round up to 1.0 mm².

From the table we select the diameters of two wires, the sum of the cross-sectional areas of which is equal to 1.0 mm².

For example, these are two wires with a diameter of 0.8 mm. and an area of ​​0.5 mm².

Or two wires:

The first one has a diameter of 1.0 mm. and cross-sectional area 0.79 mm²,
- the second with a diameter of 0.5 mm. and a cross-sectional area of ​​0.196 mm².
which adds up to: 0.79 + 0.196 = 0.986 mm².

The coil is wound with two wires simultaneously; an equal number of turns of both wires is strictly maintained. The beginnings of these wires are connected to each other. The ends of these wires are also connected.
It turns out like one wire with the total cross-section of two wires.

Program for calculating power transformers Trans50Hz v.3.7.0.0.

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The simplest calculation of a power transformer

The simplest calculation of a power transformer allows you to find the cross-section of the core, the number of turns in the windings and the diameter of the wire. The alternating voltage in the network is 220 V, less often 127 V and very rarely 110 V. For transistor circuits, a constant voltage of 10 - 15 V is needed, in some cases, for example, for powerful output stages of low-frequency amplifiers - 25 ÷ 50 V. To power the anode and screen circuits of electronic lamps most often use a constant voltage of 150 - 300 V, to power the incandescent circuits of lamps an alternating voltage of 6.3 V. All voltages necessary for any device are obtained from one transformer, which is called a power transformer.

The power transformer is made on a collapsible steel core from thin W-shaped, less often U-shaped plates isolated from each other, as well as hollowed-out strip cores of the ShL and PL types (Fig. 1).

Its dimensions, or more precisely, the cross-sectional area of ​​the middle part of the core, are selected taking into account the total power that the transformer must transmit from the network to all its consumers.

A simplified calculation establishes the following relationship: the cross-section of the core S in cm², squared, gives the total power of the transformer in W.

For example, a transformer with a core having sides of 3 cm and 2 cm (Sh-20 type plates, set thickness 30 mm), that is, with a core cross-sectional area of ​​6 cm², can consume 36 W of power from the network and “process” it. This simplified calculation gives quite acceptable results. And vice versa, if a power of 36 W is needed to power an electrical device, then taking the square root of 36, we find out that the cross-section of the core should be 6 cm².

For example, it should be assembled from Sh-20 plates with a set thickness of 30 mm, or from Sh-30 plates with a set thickness of 20 mm, or from Sh-24 plates with a set thickness of 25 mm, and so on.

The core cross-section must be matched to the power so that the core steel does not fall into the magnetic saturation region. And hence the conclusion: the cross-section can always be taken in excess, say, instead of 6 cm², take a core with a cross-section of 8 cm² or 10 cm². It won't get any worse. But it is no longer possible to take a core with a cross-section smaller than the calculated one, since the core will fall into the saturation region, and the inductance of its windings will decrease, their inductive resistance will drop, the currents will increase, the transformer will overheat and fail.

A power transformer has several windings. Firstly, network, connected to a network with a voltage of 220 V, it is also primary.

In addition to the network windings, a network transformer can have several secondary windings, each with its own voltage. A transformer for powering tube circuits usually has two windings - a 6.3 V filament winding and a step-up winding for the anode rectifier. In a transformer for powering transistor circuits, there is most often one winding that powers one rectifier. If a reduced voltage needs to be supplied to any stage or circuit node, it is obtained from the same rectifier using a quenching resistor or voltage divider.

The number of turns in the windings is determined by an important characteristic of the transformer, which is called the “number of turns per volt,” and depends on the cross-section of the core, its material, and the grade of steel. For common types of steel, you can find the "number of turns per volt" by dividing 50-70 by the core cross-section in cm:

So, if you take a core with a cross-section of 6 cm², then the “number of turns per volt” will be approximately 10.

The number of turns of the primary winding of the transformer is determined by the formula:

This means that the primary winding at a voltage of 220 V will have 2200 turns.

The number of turns of the secondary winding is determined by the formula:

If a 20 V secondary winding is needed, it will have 240 turns.

Now we select the winding wire. For transformers, copper wire with thin enamel insulation (PEL or PEV) is used. The wire diameter is calculated based on low energy losses in the transformer itself and good heat dissipation using the formula:

If you take a wire that is too thin, then, firstly, it will have high resistance and generate significant thermal power.

So, if we take the primary winding current to be 0.15 A, then the wire needs to be 0.29 mm.

More posts on the topic

The simplest calculation of power transformers and autotransformers

Sometimes you have to make your own power transformer for the rectifier. In this case, the simplest calculation of power transformers with a power of up to 100-200 W is carried out as follows.

Knowing the voltage and the maximum current that the secondary winding (U2 and I2) should give, we find the power of the secondary circuit: If there are several secondary windings, the power is calculated by adding the powers of the individual windings.

Power is transferred from the primary winding to the secondary winding through the magnetic flux in the core. Therefore, the cross-sectional area of ​​the core S depends on the power value P1, which increases with increasing power. For a core made of normal transformer steel, S can be calculated using the formula:

where s is in square centimeters, and P1 is in watts.

The value of S determines the number of turns w" per volt. When using transformer steel

If you have to make a core from steel of poorer quality, for example from tin, roofing iron, steel or iron wire (they must first be annealed so that they become soft), then S and w" should be increased by 20-30%.

In load mode there may be a noticeable loss of part of the voltage across the resistance of the secondary windings. Therefore, for them it is recommended to take the number of turns 5-10% more than calculated.

Primary current

The diameters of the winding wires are determined by the current values ​​and based on the permissible current density, which for transformers is taken on average 2 A/mm2. At this current density, the diameter of the wire without insulation of any winding in millimeters is determined from the table. 1 or calculated by the formula:

When there is no wire of the required diameter, you can take several thinner wires connected in parallel. Their total cross-sectional area must be no less than that corresponding to the calculated one wire. The cross-sectional area of ​​the wire is determined according to the table. 1 or calculated by the formula:

For low voltage windings, which have a small number of turns of thick wire and are located on top of other windings, the current density can be increased to 2.5 and even 3 A/mm2, since these windings have better cooling. Then in the formula for the wire diameter, the constant coefficient instead of 0.8 should be 0.7 or 0.65, respectively.

Finally, you should check the placement of the windings in the core window. The total cross-sectional area of ​​the turns of each winding is found (by multiplying the number of turns w by the cross-sectional area of ​​the wire equal to 0.8d2iz, where diz is the diameter of the wire in insulation. It can be determined from Table 1, which also indicates the mass of the wire. The cross-sectional areas of all windings are added To take into account the approximate looseness of the winding, the influence of the frame of insulating spacers between the windings and their layers, it is necessary to increase the found area by 2-3 times.The area of ​​the core window should not be less than the value obtained from the calculation.

As an example, let's calculate a power transformer for a rectifier that powers a device with vacuum tubes. Let the transformer have a high-voltage winding designed for a voltage of 600 V and a current of 50 mA, as well as a winding for incandescent lamps with U = 6.3 V and I = 3 A. The mains voltage is 220 V.

We determine the total power of the secondary windings:

Primary circuit power

Find the cross-sectional area of ​​the transformer steel core:

Number of turns per volt

Primary current

The number of turns and diameter of the wires of the windings are equal:

For primary winding

For boost winding

For winding filament lamps

Let us assume that the core window has a cross-sectional area of ​​5×3 = 15 cm2 or 1500 mm2, and the selected wires have the following insulated diameters: d1iz = 0.44 mm; d2iz = 0.2 mm; d3iz = 1.2 mm.

Let's check the placement of the windings in the core window. Find the cross-sectional area of ​​the windings:

For primary winding

For boost winding

For winding filament lamps

The total cross-sectional area of ​​the windings is approximately 430 mm2.

As you can see, it is more than three times smaller than the window area and, therefore, the windings will fit.

The calculation of an autotransformer has some features. Its core should not be calculated for the full secondary power P2, but only for that part of it that is transmitted by the magnetic flux and can be called the transformed power Pm.

This power is determined by the formulas:

For step-up autotransformer

For a step-down autotransformer, and

If the autotransformer has taps and will operate at different values ​​of n, then in the calculation it is necessary to take the value of n that is most different from unity, since in this case the value of Pm will be the largest and it is necessary that the core can transmit such power.

Then the design power P is determined, which can be taken equal to 1.15 Rt. The 1.15 multiplier here takes into account the efficiency of the autotransformer, which is usually slightly higher than that of the transformer. D

Next, the formulas for calculating the cross-sectional area of ​​the core (based on power P), the number of turns per volt, and wire diameters indicated above for the transformer are applied. It should be borne in mind that in the part of the winding that is common to the primary and secondary circuits, the current is equal to I1 - I2 if the autotransformer is step-up, and I2 - I1 if it is step-down.

Power transformer calculation

A transformer is a passive energy converter. Its coefficient of performance (efficiency) is always less than one. This means that the power consumed by the load, which is connected to the secondary winding of the transformer, is less than the power consumed by the loaded transformer from the network. It is known that power is equal to the product of current and voltage, therefore, in the step-up windings the current is less, and in the step-down windings the current is greater than the current consumed by the transformer from the network.

Transformer parameters and characteristics.

Two different transformers with the same mains voltage can be designed to produce the same secondary winding voltages. But if the load of the first transformer consumes more current, and the load of the second one is small, it means that the first transformer is characterized by greater power in comparison with the second. The greater the current in the windings of the transformer, the greater the magnetic flux in its core, so the core must be thicker. In addition, the greater the current in the winding, the thicker the wire it must be wound, and this requires an increase in the core window. Therefore, the dimensions of the transformer depend on its power. Conversely, a certain size core is suitable for making a transformer only up to a certain power, which is called the overall power of the transformer. The number of turns of the secondary winding of the transformer determines the voltage at its terminals. But this voltage also depends on the number of turns of the primary winding. At a certain value of the supply voltage of the primary winding, the voltage of the secondary winding depends on the ratio of the number of turns of the secondary winding to the number of turns of the primary. This ratio is called the transformation ratio. If the voltage on the secondary winding depends on the transformation ratio, you cannot arbitrarily select the number of turns of one of the windings. The smaller the core dimensions, the greater the number of turns of each winding should be. Therefore, the size of the transformer core corresponds to a very certain number of turns of its windings per one volt of voltage, less than which cannot be taken. This characteristic is called the number of turns per volt.

Like any energy converter, a transformer has an efficiency factor - the ratio of the power consumed by the transformer load to the power that the loaded transformer consumes from the network. The efficiency of low-power transformers, which are usually used to power consumer electronic equipment, ranges from 0.8 to 0.95. Higher power transformers have higher values.

Electrical calculation of the transformer

Before calculating a transformer, it is necessary to formulate the requirements that it must satisfy. They will be the initial data for the calculation. The technical requirements for the transformer are also determined by calculation, as a result of which the voltages and currents that must be provided by the secondary windings are determined. Therefore, before calculating the transformer, the rectifier is calculated to determine the voltages of each of the secondary windings and the currents consumed from these windings. If the voltages and currents of each of the windings of the transformer are already known, then they are the technical requirements for the transformer. To determine the overall power of the transformer, it is necessary to determine the power consumed from each of the secondary windings and add them up, also taking into account the efficiency of the transformer. The power consumed from any winding is determined by multiplying the voltage between the terminals of this winding by the current consumed from it:

P – power consumed from the winding, W;

U is the effective value of the voltage removed from this winding, V;

I is the effective value of the current flowing in the same winding, A.

The total power consumed, for example, by three secondary windings is calculated by the formula:

P S = U 1 I 1 +U 2 I 2 +U 3 I 3

To determine the overall power of the transformer, the resulting value of the total power P S must be divided by the efficiency of the transformer: P g = , where

P g – overall power of the transformer; η – transformer efficiency.

It is impossible to calculate the efficiency of a transformer in advance, since for this you need to know the amount of energy losses in the windings and in the core, which depend on the parameters of the windings themselves (diameters of the wires and their length) and the parameters of the core (length of the magnetic power line and steel grade). Both parameters become known only after calculating the transformer. Therefore, with sufficient accuracy for practical calculation, the efficiency of the transformer can be determined from Table 6.1.

Table 6.1

The most common are two core shapes: O - shaped and W - shaped. There are usually two coils located on an O-shaped core, and one on a W-shaped core. Knowing the overall power of the transformer, find the cross-section of the working core of its core on which the coil is located:

The cross-section of the working core of the core is the product of the width of the working core a and the thickness of the package c. Dimensions a and c are expressed in centimeters, and the cross-section is expressed in square centimeters.

After this, the type of transformer steel plates is selected and the thickness of the core package is determined. First, find the approximate width of the working core core using the formula: a = 0.8

Then, based on the obtained value a, the type of transformer steel plates is selected from among those available and the actual width of the working core a is found. then determine the thickness of the core package with:

The number of turns per 1 volt of voltage is determined by the cross-section of the working core of the transformer core according to the formula: n = k/S, where N is the number of turns per 1 V; k – coefficient determined by the properties of the core; S - cross-section of the working core core, cm 2.

From the above formula it is clear that the lower the coefficient k, the fewer turns all windings of the transformer will have. However, the coefficient k cannot be chosen arbitrarily. Its value usually ranges from 35 to 60. First of all, it depends on the properties of the transformer steel plates from which the core is assembled. For C-shaped cores twisted from a thin tape, you can take k = 35. If you use an O-shaped core assembled from U- or L-shaped plates without holes in the corners, take k = 40. The same value of k and for plates of type Ш, in which the width of the side cores is more than half the width of the middle core.. If plates of type Ш are used without holes in the corners, in which the width of the middle core is exactly twice the width of the outer cores, it is advisable to take k = 45, and if Ш - shaped the plates have holes, then k = 50. Thus, the choice of k is largely arbitrary and can be varied within certain limits, taking into account that decreasing k makes winding easier, but toughens the transformer mode. When using plates made of high-quality transformer steel, this coefficient can be slightly reduced, but with low-quality steel it is necessary to increase it.

Knowing the required voltage of each winding and the number of turns per 1 V, it is easy to determine the number of turns of the winding by multiplying these values: W = Un

This relationship is valid only for the primary winding, and when determining the number of turns of the secondary windings, it is necessary to additionally introduce an approximate correction to take into account the voltage drop on the winding itself from the load current flowing through its wire: W = mUn

The coefficient m depends on the current flowing through a given winding (see table 6.2). If the current strength is less than 0.2 A, you can take m = 1. The thickness of the wire with which the transformer winding is wound is determined by the current strength flowing through this winding. The higher the current, the thicker the wire must be, just as increasing the flow of water requires using a thicker pipe. The winding resistance depends on the thickness of the wire. The thinner the wire, the greater the resistance of the winding, therefore, the power released in it increases and it heats up more. For each type of winding wire there is a limit of permissible heating, which depends on the properties of the enamel insulation. Therefore, the diameter of the wire can be determined by the formula: d = p, where d is the copper wire diameter, m; I - current strength in the winding, A; p - coefficient (Table 6.3) which takes into account the permissible heating of a particular brand of wire.

Table 6.2: Determination of coefficient m

Table 6.3: Selecting wire diameter.

By choosing the coefficient p, you can determine the diameter of the wire of each winding. The found diameter value is rounded to a larger standard value.

The current strength in the primary winding is determined taking into account the overall power of the transformer and the network voltage:

Practical work:

U 1 = 6.3 V, I 1 = 1.5 A; U 2 = 12 V, I 2 = 0.3 A; U 3 = 120 V, I 3 = 59 mA

There was a need for a powerful power supply. In my case, there are two magnetic cores: armored tape and toroidal. Armor type: ШЛ32х50(72х18). Toroidal type: OL70/110-60.

INITIAL DATA for calculating a transformer with a toroidal magnetic core:

• primary winding voltage, U1 = 220 V;
• secondary winding voltage, U2 = 36 V;
• secondary winding current, l2 = 4 A;
• core outer diameter, D = 110 mm;
• core inner diameter, d = 68 mm;
• core height, h = 60 mm.

Calculation of a transformer with a magnetic core type ШЛ32х50 (72х18) showed that the core itself is capable of producing a voltage of 36 volts with a current strength of 4 amperes, but it may not be possible to wind the secondary winding due to insufficient window area. Let's start calculating a transformer with a magnetic core of type OL70/110-60.

Software (on-line) calculation will allow you to experiment with parameters on the fly and reduce development time. You can also calculate using the formulas, they are given below. Description of the input and calculated fields of the program: a light blue field - the initial data for calculation, a yellow field - data selected automatically from the tables, if you check the box to adjust these values, the field changes color to light blue and allows you to enter your own values, green field - calculated value.

Formulas and tables for manual calculation of a transformer:

1. Secondary winding power;

2. Overall power of the transformer;

3. The actual cross-section of the steel of the magnetic core at the location of the transformer coil;

4. The calculated cross-section of the magnetic core steel at the location of the transformer coil;

5. Actual cross-sectional area of ​​the core window;

6. The value of the rated current of the primary winding;

7. Calculation of the wire cross-section for each of the windings (for I1 and I2);

8. Calculation of the diameter of the wires in each winding without taking into account the thickness of the insulation;

9. Calculation of the number of turns in the transformer windings;

n - winding number,
U’ is the voltage drop in the windings, expressed as a percentage of the nominal value, see table.

In toroidal transformers, the relative value of the total voltage drop in the windings is significantly less compared to armored transformers.

10. Calculation of the number of turns per volt;

11. Formula for calculating the maximum power that the magnetic circuit can deliver;

Sst f - the actual cross-section of the steel of the existing magnetic circuit at the location of the coil;

Sok f - actual window area in the existing magnetic circuit;

Vmax - magnetic induction, see table No. 5;

J - current density, see table No. 3;

Kok - window fill factor, see table No. 6;

Kst is the coefficient of filling of the magnetic circuit with steel, see table No. 7;

The magnitudes of electromagnetic loads Vmax and J depend on the power removed from the secondary winding of the transformer circuit and are taken for calculations from tables.

Having determined the value of Sst*Sok, you can select the required linear size of the magnetic circuit, having an area ratio no less than that obtained as a result of the calculation.

INITIAL DATA

2. Supply voltage frequency f= 400 Hz.

4. Currents of secondary windings I 2 =0.16 A

5. Electrical steel grade E340.

6. Type of magnetic circuit – core.

SHAPE AND GEOMETRIC DIMENSIONS OF MAGNETIC CIRCUIT

The design data of the transformer are determined from the following dependencies known from theory for the effective values ​​of the primary voltage U 1 and primary current I 1:

E 1 = 4,44 f w 1 F m; I 1 = δ 1 S pr 1

Where δ 1 - current density in the primary winding, A/mm 2;

S pr1 - copper cross-section of the primary winding wire, mm 2.

Substituting the expressions into these formulas

F m = B m k st F st and S pr1 = F O k m/[ w 1 (1 +η n)]

and using the rationalized SI system of units, we obtain:

U 1 E 1 = 4,44 f w 1 k st F st 10 - 4, B (1)

I 1 = (δ 1 F O k m 10 2) /[ w 1 (1 +η n)] (2)

Between values U 1 and E 1 (supply voltage and emf of the primary winding) in expression (1) an approximate equality sign is substituted because in transformers of normal design U 1 only slightly exceeds E 1, because The voltage drops in the primary winding are small compared to E 1 .

In expressions (1) and (2):

f- voltage frequency U 1 , Hz;

w 1 - number of turns of the primary winding;

B m is the amplitude value of magnetic induction in the magnetic core of the transformer, T;

F st - cross-sectional area of ​​the magnetic circuit, cm 2;

k st - steel coefficient, taking into account the presence of plate insulation and loose assembly of the magnetic circuit package.

k st = F st act / F st is the ratio of the cross-sectional area of ​​all sheets of the magnetic core rod without insulation to the product of the width of the rod and the thickness of the magnetic core package;

F st act - active cross-section of magnetic core steel, cm 2;.

I 1 - primary current;

F o - area of ​​the magnetic circuit window, cm 2;

k m is the coefficient of filling the magnetic circuit window with copper (the ratio of the total cross-sectional area of ​​all the wires of the transformer windings piercing its window to the window area);

η n - efficiency transformer in nominal mode;

1/ (1 + η n) - coefficient taking into account the copper area of ​​the window per primary winding (approximately equal to two);

Accepting that U 1 I 1 = P n /(η n cos 1n), where P n is the active power supplied by the transformer to the consumer and solving together (I) and (2), we have:

F O F st =  P n (1 + η n) 10 2  / 4.44 f B m η n cos 1n δ 1 k m k st  (3)

Where P n = , W, and:

i- number of the secondary winding;

n - number of secondary windings;

cos i- taken equal to unity (active load);

cos 1н - power factor of the transformer.

P n = U 2 I 2 cos 2 =460.2*0.16= 73.63 W.

P n =73, 63 Tue

Let us make a preliminary selection of the quantities included in the basic calculation formula (3) of the transformer:

a) Magnitude of induction B m

Table 1

B m = 1. 4 Tl ,

b) Current density δ i

table 2

δ i =5, 9 A/mm 2

c) Window filling coefficient with copper k m and the filling factor of the magnetic core section with steel k st

Table 3

k m = 0,23 ,

Table 4

k st = 0,9,

d) Efficiency values η n and cos 1 n

Table 5

η n = 0,84 ,

cos  1n = 0,84 .

Using formula (3), the calculated value of the product is found F O F st:

F O F st =  P n (1 + η n ) 10 2  /  4.44fB m η n cos 1 n δ 1 k m k st  =

= (138,06*1,84*10 2 )/(4,44 * 400*1,4*0,84*0,84*5 , 9 * 0,23 * 0,9) = 11,86 cm 4

Using the table, we will find the nearest larger value F O F st and the required standard size of the magnetic circuit:

Table 7

DETERMINATION OF NO-LOAD CURRENT

Let's find the values ​​of total losses in steel R st , magnetizing power Q st, absolute and relative values ​​of the no-load current.

The relative value is the no-load current 1 0 , expressed as % from the primary rated current.

Total losses in steel can be determined by the formula:

R st =  R st G st (4)

where  R st - specific losses, W/kg;

G st - weight of the magnetic circuit, kg (It was previously determined from the table that G st = 203 kg).

Size  R st We determine from experimental curves the dependence of specific losses in transformer steels on induction:

We get that when B m = 1.4 T R st = 20 W/kg. ,

Thus R st =  R st G st = 20*230*10 -3 =4.6 W.

The absolute and relative values ​​of the active component of the no-load current are determined by the formulas:

I 0 A = P st / U 1 ; I 0 A % = (I 0 A / I 1 n ) . 100 = (P st / S 1n ) . 100 (5)

Where I 1 n = S 1n / U 1 = P n /(U 1 η n cos 1n )  A

Then I 0 A = R st / U 1 = 4,6/36 = 0.128 A,

I 1 n = P n /(U 1 η n cos 1n ) = 73,63/(36 * 084*084) =2.04 A,

I 0 A % = (I 0 A / I 1 n ) . 100 = (0,128/2,04)*100 ≈ 6,27%.

The total magnetizing power is determined by the formula:

Q st =  Q st . G st  INAR . (6)

where  Q st - total specific magnetizing power, VAR/kg.

Value  Q st is determined by the curve shown in the figure:

We get at B m = 1.4 T Q st = 150 VAR/kg.

Q st =Q Art. G st = 150*230*10 -3 = 34.5 VAR ,

The absolute and relative values ​​of the reactive component of the no-load current will be found using the formulas:

I 0r = Q st /U 1 =34,5/36 ≈0,96 A ,

I 0r % = (I 0r /I 1 n )*100 = (Q st /S 1 n ) *100 = (0,96/2,04) * 100 =47,06% (7)

The value of the relative no-load current based on I 0 a % And I 0 r % is equal to:

I 0 % ≈ 47,48%

CALCULATION OF WINDINGS

The calculation of the transformer windings consists of determining the number of turns and the wire diameter of each of them.

1. Based on formula (1) we have:

w 1 = (E 1 10 4 )/(4,44 f B m F Art. act );

w 2 = (E 2 10 4 )/(4,44 f B m F Art. act ); (9)

w 3 = (E 3 10 4 )/(4,44 f B m F Art. act ) etc.

All quantities included in the right-hand sides of the above expressions are known, with the exception of E.M.F.

If we designate the magnitude of the voltage drops in the windings, expressed in % from nominal, through  U 1 % , U 2 % etc., then e.m.f. windings can be found from the expressions

;

; (10)

etc. Approximate magnitude values U 1 % i U 2 % we find from table 8:

Table 8

U 1 % =2 %;

U 2 %= 2,5 %.

Since a rod magnetic circuit is used, we reduce the indicated values ​​by 25%, i.e.:

U 1 % = 1,5 %;

U 2 %= 1,875 %.

Thus we have:

w 1 = (E 1 · 10 4 )/(4,44 f B m F Art. act ) = (35,46*10 4 )/(4,44 * 400 * 1,4*1,7) = 83,9 ,

w 2 = (E 2 · 10 4 )/(4,44 f B m F Art. act ) = (468,83*10 4 )/(4,44 * 400*1,4 * 1,7) = 1109,2 .

2. The cross-sections of the winding wires are determined by the formula

, [mm 2 ] (11)

The current of the primary winding, necessary to determine the cross-section of the wire of this winding, is found by the formula

(12)

I 1n =73,63 / (36*0,84*0,84) = 2.04 A ,

I 2n = 0.23A.

Then the cross-sections of the wires, taking into account the fact that the current density value in the primary winding was reduced by 20%, and in the secondary winding increased by 15%:

S pr1 =2,04/4,72 = 0,43 mm 2 ,

S pr2 =0,23/6,79 = 0 ,034 mm 2 .

The wire diameter is found using the formula:

, [mm] (13)

d pr1 = 1,13*0,43 = 0.49 mm,

d pr2 = 1,13*0,034 = 0.04 mm .

3. Select the brands of winding wires from Table 9a:

Table 9a

Round wires

Let's choose the wire insulation according to table 9b:

Double-sided thickness of wire insulation (rounded)

Table 9b

d from 1 = d pr1 + 0,08=0,49+0,05= 0.54 mm,

d from 2 = d pr2 + 0,02=0,04+0,02= 0.06 mm.

WINDING CONSTRUCTION

a) the number of turns in the layer w With according to dependency:

Where h" = h –1 - frame height (less by I mm than the height of the magnetic circuit window), mm;

δ" - thickness of brushes and frame walls (usually 1.5 - 4.3 mm depending on the diameter of the wire);

k at - laying coefficient determined according to table 10 (takes into account the looseness of the winding),

k c is a coefficient that takes into account the buckling of windings during winding;

d from - diameter of wire with insulation, mm;

h" = h –1 =40 – 1 = 3 9 mm ,

Table 10

k y1 = 1,15;

k y2 = 1,05;

k in 1 = 1,05;

k at 2 = 1,12;

δ" = 1,5 , then the number of turns in the layer

w c1 = (39 – 3 )/(1,15*0,54) = 57,97,

w s2 = (39 – 3 )/(1,05*0,06) =571,43.

b) thickness of each winding:

Where: w - number of turns of each winding;

 from - thickness of gaskets (insulation) between layers, mm;

N = w/w With - rounds to the nearest larger integer and determines the number of rows in the layer.

N 1 = w 1 / w c1 =83,9/57,97 = 2 ,

N 2 = w 2 / w s2 =1109,2/571,43= 2.

As spacers between layers, we will choose telephone paper with a thickness of 0.05 mm.

δ 1 =2*0,54 +1* 0,05 = 1.13 mm,

δ 2 =2*0,06 + 1*0,05 = 0.17 mm .

c) the total thickness of the winding of one coil for core magnetic cores, in which the windings are located on both rods (two coils):

 = δ 1 /2 + δ 2 /2 + ……. + δ n /2 + (δ" +1) +n . d mo , mm

Where d mo - thickness of inter-winding insulation, mm;

n - number of windings.

 =1,13/2 +0,17/2 + (1,5+1)+2*0,2 = 3,55 mm.

d) gap between the coil and the magnetic core:

X = s -2k V  = 16 –2* 1,1*3,55= 8.19 mm

DETERMINATION OF WINDING OVERHEATING TEMPERATURE

Once the geometric dimensions of the transformer windings have been found, you can proceed to determining their operating temperature. First of all, it is necessary to find the value of the total power loss in the windings of each coil,

(18)

where, in addition to what is indicated above:

r - winding wire resistance, Ohm;

ρ m - resistivity of copper wire at operating temperature, Ohm. cm.

In formula (18) δ in A/mm 2, S pr in cm 2, l pr - total length of the winding wire in cm.

Replacing in (18) the product S etc l etc its meaning from

G m = γ m S etc l etc (19)

Where G m - winding wire weight, g;

γ m - specific gravity of copper ( γ m = 8.9 g/cm 3), we get:

(20)

The temperature of the wire in a heated state reaches 100 - 110˚C. Substituting into (20) the value ρ m for this temperature ρ m = O.0214. 10 -4 Ohm. cm, we get:

P m = 2,4 δ 2 G m , W (21)

δ - current density in A/mm 2

G m - wire weight, kg.

The copper weight of each winding can be found from the expression:

G m = l Wed at . w . g m . 10 -3 , kg.(22)

Where lср в - average length of a winding turn;

w- the total number of turns of the winding.

g m - weight I m ​​of wire, g.

For a two-coil rod transformer, we take half the number of winding turns w, calculated earlier, since the windings are distributed equally across two coils.

Without introducing a significant error into the calculations, instead of calculating the average lengths of turns for each winding ( l Wed v1, l cf b2, ....... etc.) take the same average length for the windings l cf in, calculating it from the dependence:

l Wed at  2(a + b +2  ) (23)

where  - full thickness of the coil winding.

l Wed at  2 *(12.5 +16 + 2*3.55) =71.2 mm=71,2*10 -3 m .

Since the transformer is a two-coil rod transformer, we take half the number of turns w, since the windings are distributed equally across two coils:

G m1 = l Wed at . w 1 /2 . g m . 10 -3 = 71,2 *83,9*1,68*10 -6 /2 = 0.005 kg,

G m2 = l Wed at . w 2 /2 . g m . 10 -3 = 71,2*1109,2*0,0144* 10 -6 /2 = 0.0006 kg,

P m1 = 2,4 δ 2 G m1 = 2,4*34,81 * 0,005 = 0.42 W,

P m2 = 2,4 δ 2 G m2 = 2,4*34,81 * 0,0006 = 0.05 W.

Total losses in the coil:

P m cat = P m 1 + P m 2 +............+ P m n (24)

P m cat =0,42+0,05= 0.47 W.

The overheating temperature can be determined by the formula:

(25)

Where P m cat losses in copper of one coil, W;

F m cat - cooling surface of a given coil, cm 2;

 m - heat transfer coefficient, W/cm 2 С.

Due to part of the end surfaces of the coil and part of its side surfaces, closed by the magnetic circuit, practically do not participate in the process of heat transfer to the environment; we can assume that the cooling surface in formula (25) includes only the open side surfaces of this coil:

F m cat = 2h(a + b +4 ), cm 2 (26)

F m cat = 2*40*(12.5 +16 + 4*3.55) =3416 mm 2 = 34.16 cm 2

 m = 1,2 . 10  3 W/cm 2  WITH,

t m = 0,47 /(34,16 * 1,2* 10  3 ) = 11,47  WITH,

DETERMINATION OF TRANSFORMER WEIGHT

Previously, the weight of the magnetic core (steel) of the calculated transformer was written out from the table G st, g . Using formula (22), the copper weights of each winding were calculated G m1, G m2 etc. Therefore, the weight of copper windings of one coil is equal to:

G m cat = G m 1 + G m 2 +............+ G m n , G

Since when determining this weight the weights of wire insulation, interlayer and interwinding insulation, as well as the weight of the frame were not taken into account, it is necessary G m cat increase by 5%, getting the weight of the coil with windings G cat .

G tr = G st + G cat , G(27)

G tr =2*1,05 * (0,005 + 0,0006) *10 3 +203 =214.76 g.

DEFINITION OF EFFICIENCY TRANSFORMER

Efficiency value transformer is determined by the formula:

% = P n/( P n + P st + P m cat)

% = 73,63/(73,63 + 4,6 +0,47) * 100 = 93,6 %.

Dear colleagues!!

I have already told you how to build a pulse transformer on a ferrite ring in my lessons. Now I’ll tell you how I make a transformer using a W-shaped ferrite core. For this, I use ferrites of suitable size from old “Soviet” equipment, old computers, televisions and other electrical equipment that I have lying around in the corner “on demand”.

For a UPS using a push-pull half-bridge generator circuit, the voltage on the primary winding of the transformer, according to the circuit, is 150 volts, under load we will take 145 volts. The secondary winding is made according to a full-wave rectification circuit with a midpoint.
See diagram.

I will give examples of calculation and manufacture of transformers for a small power UPS of 20 - 50 watts for this circuit. I use transformers of this power in switching power supplies for my LED lamps. The transformer diagram is below. It is necessary to pay attention that the W-core, folded from two halves, does not have a gap. A magnetic core with a gap is used only in single-cycle UPSs.

Here are two examples of calculating a typical transformer for different needs. In principle, all transformers for different powers have the same calculation method, almost the same wire diameters and the same winding methods. If you need a transformer for a UPS with a power of up to 30 watts, then this is the first calculation example. If you need a UPS with a power of up to 60 watts, then the second example.

First example.

We will choose from ferrite cores No. 17, Ш - a shaped core Ш7.5 × 7.5. Cross-sectional area of ​​the middle rod Sк = 56 mm.sq. = 0.56 cm2
Window Sо = 150 mm.sq. Rated power 200 watts.
The number of turns per 1 volt of this core will be: n = 0.7/Sk = 0.7 / 0.56 = 1.25 turns.
The number of turns in the primary winding of the transformer will be: w1 = n x 145 = 1.25 x 145 = 181.25. Let's take 182 turns.
When choosing the wire thickness for the windings, I proceeded from the “” table.
In my transformer, I used a wire with a diameter of 0.43 mm in the primary winding. (a wire with a large diameter does not fit in the window). It has a cross-sectional area S = 0.145 mm2. Allowable current (see table) I = 0.29 A.
The power of the primary winding will be: P = V x I = 145 x 0.29 = 42 watts.
A communication winding must be placed on top of the primary winding. It should produce a voltage v3 = 6 volts. The number of turns will be: w3 = n x v3 = 1.25 x 6 = 7.5 turns. Let's take 7 turns. Wire diameter 0.3 - 0.4 mm.
Then the secondary winding w2 is wound. The number of turns of the secondary winding depends on the voltage we need. The secondary winding, for example at 30 volts, consists of two equal half-windings, w3-1 and w3-2).
Current in the secondary winding, taking into account the efficiency (k=0.95) of the transformer: I = k xP/V = 0.95 x 42 watts / 30 volts = 1.33 A;
Let's select a wire for this current. I used a wire I had in stock with a diameter of 0.6 mm. Its S = 0.28 mm.sq.
The permissible current of each of the two half-windings is I = 0.56 A. Since these two secondary half-windings work together, the total current is 1.12 A, which is slightly different from the calculated current of 1.33 A.
The number of turns in each half-winding for a voltage of 30 volts: w2.1 = w2.2 = n x 30 = 1.25 x 30 = 37.5 vit.
Let's take 38 turns in each half-winding.
Transformer output power: Pout = V x I = 30 V x 1.12 A = 33.6 Watt, which, taking into account losses in the wire and core, is quite normal.

All windings: primary, secondary and communication winding fit perfectly into the window Sо = 150 mm2.

The secondary winding can thus be designed for any voltage and current, within a given power.

Second example.
Now let's experiment. Let's add two identical cores No. 17, W 7.5 x 7.5.

In this case, the cross-sectional area of ​​the magnetic core “Sk” will double. Sk = 56 x 2 = 112 mm2 or 1.12 cm2
The window area will remain the same “So” = 150 mm2. The indicator n (the number of turns per 1 volt) will decrease. n = 0.7 / Sk = 0.7 /1.12 = 0.63 vit./volt.
Hence, the number of turns in the primary winding of the transformer will be:
w1 = n x 145 = 0.63 x 145 = 91.35. Let's take 92 turns.

In the feedback winding w3, for 6 volts, there will be: w3 = n x v3 = 0.63 x 6 = 3.78 turns. Let's take 4 turns.
Let us take the voltage of the secondary winding as in the first example, equal to 30 volts.
Number of turns of secondary half-windings, each 30 volts: w2.1 = w2.2 = n x 30 = 0.63 x 30 = 18.9. Let's take 19 turns.
I used a wire for the primary winding with a diameter of 0.6 mm. : wire cross-section 0.28 mm2, current 0.56 A.
With this wire, the power of the primary winding will be: P1 = V1 x I = 145 V x 0.56 A = 81 Watts.
I wound the secondary winding with a wire with a diameter of 0.9 mm. 0.636 mm.sq. for a current of 1.36 amperes. For two half-windings, the current in the secondary winding is 2.72 amperes.
Secondary winding power P2 = V2 x I = 30 x 2.72 = 81.6 watts.
Wire with a diameter of 0.9 mm. a little big, fits with a large margin, that's not bad.

I use the wire for the windings at the rate of 2 A per square millimeter (this way it heats up less and the voltage drop across it will be less), although all “factory” transformers are wound at the rate of 3 - 3.5 A per mm2. and install a fan to cool the windings.
The general conclusion from these calculations is:
- when adding two identical Sh-shaped cores, the area “Sk” doubles with the same window area “So”.
- the number of turns in the windings (in comparison with the first option) changes.
- the primary winding w1 from 182 turns is reduced to 92 turns;
- the secondary winding w2 from 38 turns is reduced to 19 turns.

This means that in the same “So” window, with a decrease in the number of turns in the windings, it is possible to place a thicker wire of the windings, that is, to double the real power of the transformer.

I wound such a transformer, with folded cores No. 17, and made a frame for them.

It must be borne in mind that transformers, according to first and second For example, you can use it under a smaller load, down to 0 watts. The UPS maintains voltage quite well and stably.

Compare the appearance of transformers: example-1, with one core and example-2, with two folded cores. The actual sizes of transformers vary slightly.

The analysis of ferrite cores #18 and #19 is similar to the previous examples.
All our calculations are theoretical estimates. In fact, it is quite difficult to obtain such power from a UPS on transformers of these sizes. The design features of the switching power supply circuits themselves come into force. Scheme.
The output voltage (and therefore the output power) depends on many factors:
- capacity of the network electrolytic capacitor C1,
- containers C4 and C5,
- power drops in the winding wires and in the ferrite core itself;
- power drops on the key transistors in the generator and on the output rectifier diodes.
The overall efficiency “k” of such switching power supplies is about 85%.
This figure is still better than that of a rectifier with a steel core transformer, where k = 60%. Despite the fact that the size and weight of the UPS on ferrite is significantly less.

The procedure for assembling a ferrite Ш - transformer.

Whether ready-made or assembled, a new frame is made to fit the dimensions of the core.
See how to make "" here. Although this article talks about a frame for a transformer with a steel core, the description is quite suitable for our case.
The frame must be placed on a wooden frame. Winding of the transformer is done manually.
The primary winding is first wound onto the frame. The first row is filled turn by turn, then a layer of thin paper, varnished cloth, then the second row of wire, etc. A thin PVC tube is placed at the beginning and end of the wire (insulation from the installation wire can be used) to stiffen the wire so that it does not break off.
Two layers of paper are applied on top of the primary winding (inter-winding insulation), then you need to wind the turns of the communication winding w3. Winding w3 has few turns, and therefore it is placed at the edge of the frame. Then the turns of the secondary winding are applied. Here it is advisable to proceed in such a way that the turns of the secondary winding w2 are not located on top of the turns w3. Otherwise, malfunctions of the switching power supply may occur.
Winding is carried out with two wires at once (two half-windings), turn to turn in a row, then a layer of paper or tape and a second row of two wires. There is no need to put a PVC tube on the ends of the wire, because The wire is thick and will not break. The finished frame is removed from the mandrel and placed on a ferrite core. First check the core for any play.
If the frame is tight on the core, be very careful, the ferrite breaks very easily. A broken core can be glued together. I glue with PVA glue, followed by drying.
The assembled ferrite transformer is secured at the end with tape for strength. It is necessary to ensure that the ends of the core halves coincide without gap or shift.