Calculation of the network transformer of the power supply. Encyclopedia of radio electronics and electrical engineering

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In linear power supplies, which have already become "classic", the main element is a mains transformer, usually a step-down one, which reduces the mains voltage to the required level. How to calculate it correctly (choose a magnetic circuit, calculate the diameter of the winding wire, the number of turns in the windings, etc.) will be discussed in the proposed article.

How to choose a magnetic core

By design, magnetic circuits for network transformers are divided into armored, rod and toroidal, and according to manufacturing technology - into lamellar (Fig. 1) and tape (Fig. 2). On fig. 1 and 2 magnetic circuits are indicated: a) - armored, b) - rod, c) - toroidal.

In transformers of small (up to 00 W) and medium power (up to 1000 W), tape magnetic circuits are more often used [1]. And among tape cores, core magnetic circuits are most applicable. They have a number of advantages compared, for example, with armor [2]:

1. Approximately 25% less weight for the same transformer power.
2. Approximately 30% less leakage inductance.
3. Higher efficiency.
4. Less sensitivity to external electromagnetic fields, since the interference EMF induced in the windings, which are located on different rods, have opposite signs and are mutually compensated.
5. Large winding cooling surface.

However, rod magnetic circuits also have disadvantages:

1. Still significant leakage inductance.
2. The need to manufacture two coils.
3. Less protection of the coils from mechanical impact.

In toroidal transformers, almost the entire magnetic flux passes through the magnetic circuit, so their leakage inductance is minimal, but the complexity of manufacturing the windings is very high.

Based on the foregoing, we choose a rod tape magnetic circuit [3]. Similar magnetic circuits are made of the following types: PL-rod tape; PLV - rod tape of the smallest mass; PLM - rod tape with reduced copper consumption; PLR - rod tape of the lowest cost.

On fig. 3 shows the designations of the overall dimensions of the magnetic circuit: A - width; H - height; a is the thickness of the rod; b - tape width; c - window width; h - window height; h1 - yoke height.

Rod magnetic circuits are given an abbreviated designation, for example, PL8x 12,5x16, where PL is a U-shaped tape, 8 is the thickness of the rod, 12,5 is the width of the tape, 16 is the height of the window. The dimensions of the magnetic cores of the PL and PLR are given in Table. 1 and 2.

Options for placing coils on a magnetic circuit

We compare various options for the arrangement of coils on the cores of the magnetic circuit in terms of one of the main parameters of transformers - leakage inductance, which we calculate using the formula from [2]

where μ0 = 4π 10-7 H/m is the magnetic constant; w, - number of turns of the primary winding; vsr.ob - the average length of a coil of windings, cm; b - winding thickness, cm; h is the height of the winding, cm. This formula is obtained under the condition that the windings are cylindrical, not sectioned and arranged concentrically. Winding connection diagrams for all options are shown in fig. 4.

We will carry out comparative calculations for a transformer on a magnetic circuit PLx10x12,5x40, which has one primary and one secondary winding. In order for all calculation options to be in the same conditions, we take the thickness of the windings b = c/4 and the number of turns of the primary winding w1 = 1000.

Consider the first option, when the primary and secondary windings are located on the same rod (Fig. 4, a). The drawing of the coil is shown in fig. 5. First, calculate the average length of the winding turn

and then the leakage inductance of the coil of the first option

In the second version, the primary and secondary windings are divided into two equal parts, which are placed on two rods (Fig. 4, b). Each coil consists of half winding W1 and half winding w2. The drawing of the coils is shown in fig. 6. Calculate the leakage inductance of one coil (W1 = 500), and then double the result, since the coils are the same:

The two primary windings in the third version are located in two coils on different rods, each of which contains 1000 turns. Both primary windings are connected in parallel. The secondary winding is also placed in two coils on different rods, and two cases are possible: two half-windings with a full number of turns, connected in parallel (Fig. 4, c), or the secondary winding is divided into two half-windings with half the number of turns, connected in series (Fig. 4, d). The drawing of the coils is shown in fig. 6. In this option, the leakage inductance is the same as in the second option: LS3 = LS2 = 2,13 mH.

It should be remembered that in the second and third options, the primary and secondary windings and half-windings must be connected in accordance with the magnetic fluxes they create in the magnetic circuit have the same direction. In other words, magnetic fluxes must be added, not subtracted. On fig. 7, a shows an incorrect connection, and in fig. 7b is correct.

The need to comply with the rules for connecting windings and half-windings is a disadvantage of the second and third options. In addition, in the third variant, the total magnetic flux from the primary winding is twice as high as in the others, which can lead to saturation of the magnetic circuit and, as a result, to distortion of the sinusoidal voltage waveform. Therefore, the use of the third option for turning on the windings in practice should be done with caution.

In the fourth version, the primary winding is completely located on one core of the magnetic circuit, and the secondary winding is located on the other (Fig. 4, e). The drawing of the coils is shown in fig. 8. Since the windings are not located concentrically, to calculate the leakage inductance, we use the formula from [2]:

where b \u4d c / 2 - thickness of the windings, cm; Rvn \u2d wob / (2π) - outer radius of the winding, cm; vob \u2d 6,5a + 1,04b + 4πb - the outer length of the winding turn, cm. Calculate the outer length of the turn and the outer radius of the winding: \u88,2d XNUMX cm; Rin = XNUMX cm. Substituting the calculated values ​​into the formula for calculating the leakage inductance, we obtain LSXNUMX = XNUMX mH.

In addition to the four considered, there are many other options for the location of the windings on the rods of the magnetic circuit, however, in all other cases, the leakage inductance is greater than in the second and third options.

Analyzing the obtained results, we can draw the following conclusions:

1. The leakage inductance is minimal in the second and third winding arrangements and is in the following ratio: LS4>>LS1>>LS2 = LS3.
2. Transformers of the third option have two identical primary windings, so they are heavier, more laborious and expensive than in the second option.

Therefore, in the manufacture of low power transformers, one should choose the connection scheme and winding arrangement considered in the second option. The secondary half-windings can be connected in series if a higher output voltage is required, or in parallel if a higher output current is required.

Brief information about the materials of magnetic circuits

Until now, we have not taken into account losses in a real transformer, which are made up of losses in the magnetic circuit - for eddy current and magnetization reversal (hysteresis): in calculations they are taken into account as power losses in steel Rst, and losses in windings - as power losses in copper Rm. So, the total power loss in the transformer is:

P∑ = Рst + Рm = Рv.t + Рg + Рm,

where Рв.т - eddy current loss power; Рг - power loss for hysteresis.

To reduce them, steel is subjected to heat treatment - carbon is removed, and also alloyed - silicon, aluminum, copper and other elements are added. All this increases the magnetic permeability, reduces the coercive force and, accordingly, the hysteresis loss. In addition, steel is subjected to cold or hot rolling to obtain the desired structure (rolled texture).

Depending on the content of alloying elements, structural state, magnetic properties, steels are marked with four-digit numbers, for example, 3412.

The first digit means the class of electrical steel in terms of structural state and rolling class: 1 - hot-rolled isotropic; 2 - cold-rolled isotropic; 3 - cold-rolled anisotropic with rib texture.

The second digit is the percentage of silicon content: 0 - unalloyed steel with a total mass of alloying elements not more than 0,5%; 1 - alloyed with a total mass of more than 0,5, but not more than 0,8%; 2 - 0,8...1,8%; 3 - 1,8 ... 2,8%; 4 - 2,8...3,8%; 5 - 3,8...4,8%.

The third digit is the group according to the main normalized characteristic (specific losses and magnetic induction): 0 - specific losses at a magnetic induction of 1,7 T at a frequency of 50 Hz (Pij/so); 1 - losses at a magnetic induction of 1,5 T at a frequency of 50 Hz (P1,5 / 50); 2 - at an induction of 1 T at a frequency of 400 Hz (P1/400); 6 - induction in weak magnetic fields at a strength of 0,4 A/m (B0,4); 7 - induction in medium magnetic fields at a strength of 10 A/m (B10) or 5 A/m (B5).

The first three digits indicate the type of electrical steel.

The fourth digit is the serial number of the steel type.

The magnetic circuits of transformers for household appliances are made of cold-rolled textured steel grades 3411-3415 [3] with normalized specific losses at a magnetic induction of 1,5 T at a frequency of 50 Hz and a resistivity of 60 10-8 Ohm m. The parameters of some grades of electrical steel are given in table. 3.

Cold rolled electrical steel has higher magnetic characteristics. In addition, a smoother surface makes it possible to increase the fill factor of the magnetic core volume (kT) up to 98% [4].

Initial data for calculating the transformer

Let us calculate a transformer having a primary and two identical secondary windings, with the following parameters: effective (effective) voltage of the primary winding U1 = 220 V; effective (effective) voltage of the secondary windings U2 = U3 = 24 V;

effective (effective) current of the secondary windings l2 = I3 = 2A. Line voltage frequency f = 50 Hz.

The transformation ratio is equal to the ratio of the voltage on the primary to the voltage on the open (EMF) secondary winding. In this case, the error arising due to the difference between the EMF and the voltage on the primary winding is neglected:

where w1 and w2 are the number of turns, respectively, of the primary and secondary windings; E1 and E2 - EMF of the primary and secondary windings.

The current in the primary winding is:

The overall power of the transformer is:

In the process of calculation, it is necessary to determine the dimensions of the magnetic circuit, the number of turns of all windings, the diameter and approximate length of the winding wire, power losses, the total power of the transformer, efficiency, maximum dimensions and weight.

Calculation of the transformer magnetic circuit

The methodology for calculating sizes and other parameters is taken mainly from [1].

First, we calculate the product of the cross-sectional area of ​​the rod and the area of ​​the magnetic circuit window. The rod is called the section of the magnetic circuit (axbxh), on which the coil is placed:

where B - magnetic induction, T; j - current density in the windings, A/mm2; η is the efficiency of the transformer, n is the number of cores of the magnetic core; ks is the filling factor of the magnetic circuit section with steel; km is the filling factor of the magnetic circuit window with copper.

Recommended values ​​of magnetic induction and average values ​​of current density, efficiency and window filling factor for frequency f - 50 Hz are given in Table. 4.

The filling factor of the magnetic circuit section for steels 3411-3415 is 0,95 ... 0,97, and for steels 1511-1514 - 0,89 ... 0,93.

For calculation, we take B \u1,35d 2,5 T; j = 2 A/mm0,95; η = 0,96; Kc = 0,31; km = 2; n=XNUMX:

The thickness of the core of the magnetic circuit is calculated by the formula

A suitable magnetic circuit is selected according to the table. 1 and 2. When choosing, one should strive to ensure that the cross section of the magnetic circuit is close to a square, since in this case the consumption of the winding wire is minimal.

The width of the magnetic circuit tape is calculated by the formula

We select the PLR18x25 magnetic circuit, in which a is 1,8 cm; b = 2,5 cm; h = 7,1 cm;

Calculation of transformer windings

Calculate the EMF of one turn by the formula

Calculate the approximate voltage drop across the windings:

Then we calculate the number of turns of the primary winding:

secondary windings:

Calculate the diameter of the winding wire without insulation using the formula

Substituting the numerical values, we get the diameter of the primary wire:

and secondary windings:

According to the table 5 select the brand and diameter of the winding wire in insulation [5]: for the primary winding - PEL or PEV-1 di = 0,52 mm; for secondary - PEL or PEV-1 d2 = d3 = 1,07 mm.

We specify the number of turns of the windings. To do this, we first specify the voltage drop across the windings:

Calculate the average length of the coil, using Fig. 5 or 6:

and then the length of the wire in the windings:

The specified values ​​of the voltage drop across the windings are:

Taking into account the obtained values, we calculate the number of turns of the primary:

and secondary windings:

Calculate the mass of the winding wire:

where m1 and m2 are the linear mass of wires, respectively, of the primary and secondary windings from Table. 5.

The mass of the magnetic circuit is determined from the table. 2: Mm = 713 g.

The mass of the transformer without taking into account the mass of the fastening parts is M = = 288+2-165+713 = 1331 g. Maximum dimensions: (b+c)x(A+c)xH = 43x72x107 mm. Transformation ratio k \u1d W2 / W1640 \u192d 8,54/XNUMX \uXNUMXd XNUMX.

Power loss calculation

Losses in the magnetic circuit are equal to:

where ore - specific losses in the magnetic circuit from the table. 3. Suppose that the magnetic circuit is made of steel tape 3413 with a thickness of 0,35 mm, then according to Table. 3 we find that the specific losses in such a magnetic circuit are 1,3 W/kg. Accordingly, the losses in the magnetic circuit Рst = 0,713-1,3 = 0,93 W.

Losses in the winding - on the active resistance of the wires - we calculate by the formula

where r1, r2 - active resistance, respectively, of the primary and secondary windings, I'1 - current of the primary winding, taking into account losses:

where r1m, r2m - linear resistance of the wires, respectively, of the primary and secondary windings from the table. 5.

We recalculate the current of the secondary windings into the current of the primary winding:

The current of the primary winding, taking into account losses, is equal to:

where η = 0,95 - transformer efficiency from table. 4 for 100W power. Winding losses are:

The total power of the transformer, taking into account losses, is equal to:

The efficiency of the transformer is calculated by the formula

Making a transformer

We will manufacture the transformer according to the second option discussed above. The location of the coils is shown in fig. 6. To do this, it is necessary to make two coils, each of which contains half of the turns of the primary and each of the secondary windings: w'1 = 820 turns of PEL wire (or PEV-1) with a diameter of 0,52 mm; w'2=w'3= 96 turns of PEL wire (or PEV-1) with a diameter of 1,07 mm.

Since the transformer has low power and dimensions, the coils can be made frameless. Coil thickness b ≤ c/2 = 9 mm, its height hK ≤ 71 mm.

Number of turns in the primary layer

number of layers

Number of turns in the secondary layer

number of layers

The windings are wound on a wooden mandrel, made in exact accordance with the dimensions of the section of the magnetic circuit on which the coils will be located (18x25x71 mm). Cheeks are attached to the ends of the mandrel.

Despite the fact that the winding wires are covered with enamel insulation and therefore have a high electrical strength, usually additional, for example, paper insulation is laid between the winding layers. Most often, transformer paper 0,1 mm thick is used to isolate the windings from the magnetic circuit and among themselves. Calculate the maximum voltage between two adjacent layers of the primary winding

Since the stress between the layers is small, additional insulation can be laid through the layer or made thinner, such as using capacitor paper. Between the primary and secondary, a shielding winding should be placed - one open turn of thin copper foil or one layer of winding wire, which prevents the penetration of interference from the network into the secondary windings and vice versa.

First, the mandrel is wrapped with three layers of paper tape (Fig. 9), the petals of the tape are glued to the cheeks. Then the primary winding is wound, laying each layer with insulation. Two layers of insulation are laid between the primary, shielding and secondary windings. The total thickness of the manufactured coils does not exceed 8 mm.

Transformer check

The assembled transformer is first checked in idle mode - without load. At a mains voltage of 220 V, the current in the primary winding

secondary winding voltage

The voltage across the secondary windings can only be accurately measured with a high impedance voltmeter. Finally, the voltage on the secondary windings of the transformer is measured at rated load.

Literature

1. Linde D. P. et al. Handbook of radio electronic devices. Ed. A. A. Kulikovsky. T. 2. - M.: Energy, 1978.
2. Gorsky A. N. et al. Calculation of electromagnetic elements of secondary power sources. - M.: Radio and communication, 1988.
3. Sidorov IN et al. Small-sized magnetic circuits and cores. Directory. - M.: Radio and communications. 1989.
4. Gerasimov V. G. and others. Electrotechnical reference book. T. 1. - M.: Energy, 1980.
5. Malinin R. M. Handbook of a radio amateur designer. - M.: Energy, 1978

Author: V. Pershin, Ilyichevsk, Odessa region, Ukraine

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