Calculation of the overall power of a pulse transformer by core. Calculation of transformers of switching power supplies. Why is it so good and is it really better than ferrite

Various types of transformer equipment are used in electronic and electrical circuits, which are in demand in many areas of economic activity. For example, pulse transformers (hereinafter referred to as IT) are an important element installed in almost all modern power supplies.

Design (types) of pulse transformers

Depending on the shape of the core and the placement of coils on it, IT are produced in the following designs:

• rod;
  
• armored;
  
• toroidal (does not have coils, the wire is wound on an insulated core);
  
• armored rod;
  

The figures show:

• A - a magnetic circuit made of transformer steel grades made using the technology of cold or hot rolled metal (with the exception of a toroidal core, it is made of ferrite);
• B - coil of insulating material
• C - wires that create an inductive connection.

Note that electrical steel contains few additives of silicon, since it causes power loss from the effect of eddy currents on the circuit of the magnetic circuit. In IT of toroidal design, the core can be made from rolled or ferrimagnetic steel.

Plates for a set of an electromagnetic core are selected in thickness depending on the frequency. With an increase in this parameter, it is necessary to install plates of smaller thickness.

Principle of operation

The main feature of pulse type transformers (hereinafter referred to as IT) is that they are supplied with unipolar pulses with a constant current component, and therefore the magnetic circuit is in a state of constant bias. Shown below circuit diagram connecting such a device.

As you can see, the connection diagram is almost identical with conventional transformers, which cannot be said about the timing diagram.

The primary winding receives pulse signals having a rectangular shape e (t), the time interval between which is quite short. This causes an increase in the inductance during the interval t u , after which its decline is observed in the interval (T-t u).

The induction drops occur at a rate that can be expressed in terms of the time constant by the formula: τ p =L 0 /R n

The coefficient describing the difference of the inductive difference is determined as follows: ∆V=V max - V r

• In max - level maximum value induction;
• In r - residual.

More clearly, the difference in inductions is shown in the figure showing the shift of the operating point in the IT magnetic circuit.

As can be seen in the timing diagram, the secondary coil has a voltage level U 2 in which there are reverse surges. This is how the energy accumulated in the magnetic circuit manifests itself, which depends on the magnetization (parameter i u).

The current pulses passing through the primary coil are trapezoidal in shape, since the load and linear currents (caused by core magnetization) are combined.

The voltage level in the range from 0 to t u remains unchanged, its value e t =U m . As for the voltage on the secondary coil, it can be calculated using the formula:

wherein:

• Ψ is the flux linkage parameter;
• S is a value that displays the cross section of the magnetic core.

Considering that the derivative characterizing the changes in the current passing through the primary coil is a constant value, the increase in the level of induction in the magnetic circuit occurs linearly. Based on this, it is permissible, instead of the derivative, to introduce the difference of indicators made after a certain time interval, which allows you to make changes to the formula:

in this case, ∆t will be identified with the parameter t u , which characterizes the duration with which the input voltage pulse flows.

To calculate the area of ​​the pulse with which the voltage is formed in the secondary winding of IT, it is necessary to multiply both parts of the previous formula by t u. As a result, we will come to an expression that allows us to obtain the main IT parameter:

U m x t u =S x W 1 x ∆V

Note that the value of the pulse area directly depends on the parameter ∆В.

The second most important value that characterizes the operation of IT is the induction drop, it is influenced by such parameters as the cross section and magnetic permeability of the core of the magnetic circuit, as well as the number of turns on the coil:

Here:

• L 0 - induction difference;
• µ a is the magnetic permeability of the core;
• W 1 - the number of turns of the primary winding;
• S is the cross-sectional area of ​​the core;
• l cp - length (perimeter) of the core (magnetic circuit)
• B r is the value of the residual induction;
• In max - the level of the maximum value of the induction.
• H m - Magnetic field strength (maximum).

Considering that the IT inductance parameter completely depends on the magnetic permeability of the core, the calculation must be based on the maximum value of µ a, which is shown by the magnetization curve. Accordingly, for the material from which the core is made, the level of the parameter B r , which reflects the residual induction, should be minimal.

Video: detailed description principle of operation of a pulse transformer

Based on this, a tape made of transformer steel is ideal for the role of the IT core material. You can also use permalloy, in which such a parameter as the coefficient of squareness is minimal.

Ferrite alloy cores are ideal for high-frequency IT because this material has low dynamic losses. But because of its low inductance, it is necessary to make IT of large sizes.

Calculation of a pulse transformer

Consider how it is necessary to calculate IT. Note that the efficiency of the device is directly related to the accuracy of the calculations. As an example, let's take a conventional converter circuit that uses a toroidal type IT.

First of all, we need to calculate the IT power level, for this we use the formula: P \u003d 1.3 x P n.

The value of R n displays how much power the load will consume. After that, we calculate the overall power (P gb), it should be no less than the load power:

Parameters required for calculation:

• S c - displays the cross-sectional area of ​​the toroidal core;
• S 0 - the area of ​​​​its window (as a hint, this and the previous value are shown in the figure);

• B max is the maximum peak induction, it depends on which brand of ferromagnetic material is used (the reference value is taken from sources describing the characteristics of ferrite grades);
• f is a parameter characterizing the frequency with which the voltage is converted.

The next step is to determine the number of turns in the primary winding Tr2:

(results are rounded up)

The value of U I is determined by the expression:

U I \u003d U / 2-U e (U is the voltage supply to the converter; U e is the voltage level supplied to the emitters of transistor elements V1 and V2).

We proceed to the calculation of the maximum current passing through the primary winding of IT:

The parameter η is equal to 0.8, this is the efficiency with which our converter must work.

The diameter of the wire used in the winding is calculated by the formula:

If you are having trouble defining the basic parameters of IT, you can find thematic sites on the Internet that allow you to online mode calculate any pulse transformers.

In the calculation method described in, to determine the minimum number of turns of the primary winding W 1 and the overall P gab (maximum allowable) power of the push-pull converter transformer, the formulas are used:

where U1 is the voltage on the primary winding of the transformer, V; f - conversion frequency, Hz; B max - maximum magnetic induction in the magnetic circuit, T; S c and S w, - cross-sectional area and window area, cm 2.

These formulas allow you to perform an approximate calculation of the transformer. But formally following the calculation given in the example and ignoring the resulting errors can give an erroneous result, which may result in the failure of the transformer and switching transistors.

Consider, for example, a K40x25x11 ring magnetic circuit made of 2000NM1 ferrite. The recommended maximum value of the magnetic induction should be equal to the saturation induction: B max \u003d B us \u003d 0.38 T. Probably concluded. that under load, the rectified mains voltage of 310 V will drop to 285 V. Therefore, for a half-bridge converter, the voltage on the primary winding of the transformer (minus the saturation voltage on the switching transistor, which is assumed to be 1.6 V): U 1 \u003d 285 / 2-1.6≈141 V. From the calculation according to formula (1) we obtain W 1 =11.24≈12 turns of the primary winding.

Let's say you need to get in the load D.C. l n \u003d 4 A at a voltage U n \u003d 50 V, which corresponds to the useful power P n \u003d 200 W. With efficiency η≈0.8, the power used is P app =P n /η=200/0.8=250 W. The overall power of the selected transformer, calculated by formula (2), is more than four times the required one, so it should function without problems. In accordance with the maximum current in the primary winding is l 1max \u003d P test / U 1 \u003d 1.77 A. We select switching transistors with a current margin of 50%, then the maximum allowable collector (drain) current I to add \u003d 1.77 * 1.5 \u003d 2.7 A. For the primary winding of the transformer, a wire with a diameter of 0.8 mm is required. The secondary winding should contain five turns of wire with a diameter of 1.2 mm. This completes the calculation of the transformer according to the method. But will the converter work normally with this Transformer?

Consider the process of transferring energy to the load using a pulse transformer, the switching circuit of which is shown in Fig. 1a. The directions of currents in the primary i 1 and secondary i 2 windings of the transformer and the polarity of the voltage and the considered half-cycle of the input pulse voltage u 1 are shown, the rectangular shape of which is shown in Fig. 1b.

Note that the shape of the current in the primary winding is not rectangular. This current is the sum of the useful rectangular component with the amplitude l 1max =1.77 A and the triangular component of the magnetizing current. The last component can be estimated by the formula

The magnitude of the magnetizing current is determined by the duration of the half-cycle ∆t:

Figure 1c shows how, during one half-cycle, the magnetization current i μ increases from the value -l max to +l max, and the other - decreases in the same interval. Even in the absence of saturation of the magnetic circuit, only due to an increase in the magnetization current, the total current l ∑max shown in Fig. 1b can increase to dangerous values ​​for transistors.

Consider the effect of hysteresis. The magnetization and magnetization reversal of the magnetic circuit occurs in accordance with the curves shown in Fig.2. The abscissa shows the strength of the magnetic field H created by the primary winding of the transformer, the ordinate shows the magnetic induction B in the magnetic circuit. On fig. 2 shows the limiting hysteresis loop and the partial (internal) hysteresis loop corresponding to fig. 1b and 1c.

Fig.2

The curve in Fig. 2, emanating from the point of intersection of the coordinate axes, corresponds to the initial section of the magnetization curve and characterizes the operation of the transformer in weak magnetic fields. Since, as indicated, the strength of the magnetic field H created by the primary winding of the transformer is proportional to the magnetizing current i μ, it is quite legitimate to combine its diagram in one figure with a change in the magnetic induction B in the magnetic circuit.

If a tangent is drawn at any point of the hysteresis loop (in the figure, this is the tangent AC at point A), then its slope will determine the change in the magnetic induction of the LV with respect to the change in the magnetic field strength ∆Н at the selected point, i.e. ∆В/ ∆N. This is dynamic magnetic permeability. At the point of intersection of the coordinate axes, it is equal to the initial magnetic permeability. For ferrite 2000NM1, it is nominally 2000, but its real value can be in a very wide range: 1700 ... 2500.

For the example shown in the figure, in which the magnetization reversal of the magnetic circuit occurs along a partial hysteresis loop with a peak at point D, the change in the magnetization current i μ1 is determined by formula (3). will occur almost linearly. If the conversion frequency f does not exceed 50 kHz, the energy loss for heating the magnetic circuit due to its magnetization reversal is negligible. As for the regime with the entry of the value of the magnetic induction into the saturation region of the material of the magnetic circuit (B max =B us). selected in , the picture will be completely different. In this case, the main magnetization curve corresponds to the shape of the current i μ2, which is very far from linear. The tangent at point E with coordinates (H us, V us) is almost horizontal, which is equivalent to a significant decrease in the inductance of the primary winding, and therefore, in accordance with formula (3), the magnetizing current increases sharply, which is illustrated by the graph i μ2. If the switching transistor is selected without sufficient current margin, it will inevitably be damaged. To exclude saturation of the magnetic circuit, it is necessary to fulfill the condition: at the maximum possible supply voltage, the maximum magnetic induction must correspond to the inequality B max ≤ (0.5 ... 0.75) * V us. Often, when designing a push-pull converter, they also use another criterion - relative value magnetizing current. The parameters of the primary winding are chosen as follows. so that the amplitude of the magnetizing current ∆l corresponds to no more than 5 ... 10% of the amplitude of the rectangular component of the current in the primary winding l 1max, then the total current can be approximately considered rectangular.

The inductance of the primary winding of the transformer, containing 12 turns in our example, is 0.3 mH. Magnetizing current amplitude calculated by formula (4). - 1.18 A. If now for a payload of 200 W we compare the obtained maximum value of the total switching current l ∑max \u003d l 1max + l max \u003d 1.77 + 1.18 \u003d 2.95≈3 A (Fig. 1, b) with the maximum allowable current of the switching transistor 2.7 And, the fact of the wrong choice of the transistor and the discrepancy between the calculated diameter of the conductor of the primary winding and the required value becomes quite obvious. This discrepancy will be further aggravated in the case of a quite possible increase in the input voltage by 20%. Since at the rated supply voltage the mode is selected with the value of the magnetic induction entering the saturation region of the magnetic circuit material (B max \u003d B us), in the event of an increase in the mains voltage, the maximum value of the current in the primary winding of the transformer l ∑ max will significantly exceed even its refined value of 3 A.

The conversion frequency of 100 kHz, arbitrarily chosen in the calculation example, as the experiment shows, is the maximum possible for ferrite 2000NM1, while it is necessary to take into account the energy losses for heating the transformer. Even if they are not taken into account, the number of turns of the primary winding should be significantly larger. If the mains voltage is increased by 20%, the voltage amplitude on the primary winding will reach 180 V. If we assume that at this voltage the maximum magnetic induction in the magnetic circuit does not exceed V max \u003d 0.75 * V us \u003d 0.285 T, then the number of turns of the primary winding, calculated by formula (1) should be equal to 20, but not 12.

Thus, an insufficiently substantiated choice of initial values ​​in formula (1) can lead to inaccurate or even erroneous calculation of the pulse transformer. In order to avoid doubts about the legitimacy of using formula (1), we justify it analytically.

The maximum magnetic induction B m ax (Tl) in a closed magnetic circuit can be calculated using the well-known formula

where μ 0 = 4π·10 7 Gn/m is the absolute magnetic permeability of the vacuum; μ EFF - effective magnetic permeability of the material of the magnetic circuit; l max - amplitude of the magnetizing current, A; W 1 - the number of turns of the primary winding; lEFF- effective length of the magnetic field line in the magnetic circuit, m. Substitute in (5) l max from (4), using the well-known formula for the inductance of the toroidal winding

and moving from meters to centimeters, we get a formula for calculating the number of turns

As you can see, formula (6) differs from (1) only in that it includes the effective cross-sectional area of ​​the magnetic circuit, and not the geometric one. Detailed methodology for calculating effective parameters various types of magnetic circuits is given in [3]. At practical use of this formula, the value of W should be rounded up to the nearest integer N 1 .

Let us pay attention to the features of the application of the ratios used in the design of transformers for various push-pull converters.

Self-oscillating converters with one transformer, similar to that described in (4), operate with entry into the saturation region of the magnetic circuit material (points E and E "in Fig. 2). Formulas (1) and (2) are used at B max \u003d V us. Several otherwise, these formulas are used in the case of designing self-oscillating converters with two transformers, such as that described in. In it, the coupling winding on a powerful transformer is connected to a low-power transformer in the control circuit of the bases of switching transistors. The impulse voltage induced in the coupling winding creates saturation in a low-power transformer, which sets the conversion frequency in accordance with formula (1).This frequency is selected so as to avoid saturation in a powerful transformer, the size of which is determined according to formula (2).In such power supplies, control signals generated by a saturable low-power transformer are minimized through current in switching transistors.

Along with oscillators, push-pull converters with external excitation are very popular with radio amateurs. To exclude the through switching current, external excitation signal generators form a protective time interval between turning off the open and turning on the closed switching transistors. After selecting the conversion frequency and the maximum value of the magnetic induction in the magnetic circuit, usually, first, on the basis of (2), the required magnetic circuit of the transformer is determined, and then, using formula (1), the number of turns of the primary winding of the transformer is calculated.

For approximate calculations and preliminary selection of the required size of the magnetic circuit made of ferrite 2000NM1, there is a table in which, for several values ​​of the conversion frequency f, the results of calculations of the minimum number of turns N 1 of the primary winding according to formula (6), the amplitude value of the magnetization current I max according to formula (4) and the maximum possible useful power P max . When calculating the latter, the overall power was first calculated by formula (2) using the effective cross-sectional area of ​​the magnetic circuit instead of the geometric one, then it was multiplied by the efficiency value equal to 0.8. Sum

I ∑max = l 1 max + l max

gives the basis for choosing a switching transistor according to the maximum allowable collector (drain) current. The same current value can also be used to determine the diameter of the wire of the primary winding of the transformer in accordance with the formula given in

The calculations are made under the condition that the maximum magnetic induction Vmax does not exceed 0.25 T, even if the mains voltage is 20% higher than the nominal voltage, as a result of which the voltage on the primary winding of the transformer of a push-pull half-bridge inverter can reach 180 V (taking into account the voltage drop across the current-limiting resistor and rectifier diodes). The magnetic circuit should be selected with a margin of 20 ... 40% according to the maximum output power indicated in the table. Although the table was compiled for a half-bridge converter, its data can be easily modified for a bridge converter. In this case, the voltage on the primary winding of the transformer will be twice as high, and the amplitude of the rectangular current component of the primary winding will be half as much. The number of turns should be twice as much. The winding inductance will increase four times, and the current > I max will be halved. It is possible to use a magnetic circuit from two ferrite rings of the same size stacked together, which will lead to a twofold increase in the cross-sectional area of ​​​​the magnetic circuit S c and the inductance coefficient A L . According to formula (2), the overall and useful output power will also double. The minimum number of turns of the primary winding, calculated by formula (6), will remain unchanged. Its inductance will double, and the magnetizing current I max determined by formula (4) will remain the same.

In power supplies with an output from the midpoint of the primary winding of the transformer, the full voltage of the network is applied to half of this winding, so the number of turns of the winding must be twice as large as compared to the bridge converter, all other things being equal.

We emphasize that due to the significant spread of the real values ​​of the parameters of ferromagnetic materials compared to their reference data, the table can only be used for preliminary selection of the magnetic circuit, and then, after experimental measurement of its characteristics, it is required to carry out an updated calculation of the transformer. For example, for the K40x25x11 magnetic circuit, the table shows the value of the inductance coefficient A L = 2.08 μH per turn. Let us experimentally refine the magnetic properties of a particular instance of the magnetic circuit: for a test winding of N samples = 42 turns, the measured inductance is ≈3.41 mH, and the inductance coefficient

But the differences can be more significant, so the value of the inductance coefficient given in the table should still be considered as a guideline. In our case, it is necessary either to increase the number of turns so that the winding inductance is not less than that calculated from the tabular data, or when choosing transistors, take into account that the current l max will be 2.08 / 1.93≈1.1 times more than the tabular one.

At the manufacturing stage, it will most likely turn out that the recommended minimum number of turns of the primary winding will only partially fill the first layer of the transformer. In order for the magnetic field created by such a winding in the magnetic circuit to be uniform, its turns are either "discharged" or fill the entire layer with them, and then, taking into account the new number of turns, the final calculation of the transformer is carried out.

Let's complete the calculation of the transformer chosen as an example. It follows from the table that at a frequency of 50 kHz, the maximum useful power will be 265 W, the minimum number of turns of the primary winding N 1 is 45. Approximately the maximum value of the switched current: 1.77 + 0.21 = 1.98 A. Determine the diameter of the wire of the primary winding of the transformer. As indicated, we will choose the nearest in diameter from the nomenclature produced by the industry d 1 \u003d 0.83 mm, and taking into account the insulation d 1 \u003d 0.89 mm. If we take into account the electrical insulation of the magnetic circuit with several layers of varnished fabric with a total thickness of 0.25 mm, the inner diameter of the magnetic circuit will decrease to 25-0.5=24.5 mm. In this case, the length of the inner circumference will be π·24.5≈80 mm. Taking into account the fill factor of 0.8, 64 mm is available for winding the first winding layer, which corresponds to 64 / 0.89 = 71 turns. Thus, there is enough space for 45 turns. We wind them "in a row".

When determining the number of turns of the secondary winding, it is necessary to know the voltage drop across the primary winding. If we take into account that the length of one turn is 40.5-24.5+2-11.5=39 mm, then the total length of the wire in the primary winding is 45*39=1.755 m. and the voltage drop on the primary winding will reach U 1nad \u003d 1.77 * 0.06 \u003d 0.1 V.

Obviously, such a small value can be neglected. If we assume that the losses on the rectifier diode are approximately equal to 1 V, then we get the calculated number of turns of the secondary winding N 2 \u003d 45 * (51/150) \u003d 15.3 ≈ 16 turns. Secondary wire diameter

Filling the transformer window with copper

which corresponds to the fill factor

Taking into account the need for interlayer and interwinding insulation, the average value of the fill factor can reach K m =0.35, and the maximum - K m = 0.5. Thus, the condition for placing the windings is fulfilled.

Let us specify the maximum value of the magnetization current, taking into account the fact that the measured value of the inductance coefficient turned out to be 1.1 times less than the table value. Therefore, the magnetization current I max will be 1.1 times greater and will be 0.23 A, which in our example does not differ much from the table value, 0.21 A. The total switching current in the primary winding at the maximum mains voltage is l Σmax =1.77+0.23=2 A. Based on this, it is necessary to select switching transistors with a maximum allowable collector (drain) current of at least l add \u003d 1.5 * 2 \u003d 3 A. The maximum voltage on the switching transistors (in the closed state) is equal to the full rectified voltage of the network, therefore the maximum allowable collector voltage ( drain) must be at least U dop = 1.2 * 360 = 432 V. This completes the calculation of the pulse transformer.

LITERATURE

1. Zhuchkov V. Calculation of the transformer of the switching power supply. - Radio, 1987, No. 11. p. 43.

2. reference Information. Reference book on ferrites. ferromagnetic materials. - http://www.qrz.ru/reference/ferro/ferro.shtml

3. Mikhailova M. M., Filippov V. V., Muslekov V.P. Soft magnetic ferrites for radio-electronic equipment. Directory. - M.: Radio and communication, 1983.

four . Knyazev Yu., Sytnik G., Sorkin I. ZG block and power supply of the IK-2 set. - Radio, 1974, No. 4, p. 17.

5. Bereboshkin d. Improved economical power supply. - Radio, 1985. No. 6, p. 51.52.

6. Pershin V. Calculation of the network transformer of the power source. - Radio, 2004, No. 5, p. 55-57.

S. KOSENKO, Radio, 2005, No. 4, pp. 35-37,44.

Various types of transformer equipment are used in electronic and electrical circuits, which are in demand in many areas of economic activity. For example, pulse transformers (hereinafter referred to as IT) are an important element installed in almost all modern power supplies.

Design (types) of pulse transformers

Depending on the shape of the core and the placement of coils on it, IT are produced in the following designs:

• rod;
  
• armored;
  
• toroidal (does not have coils, the wire is wound on an insulated core);
  
• armored rod;
  

The figures show:

• A - a magnetic circuit made of transformer steel grades made using the technology of cold or hot rolled metal (with the exception of a toroidal core, it is made of ferrite);
• B - coil of insulating material
• C - wires that create an inductive connection.

Note that electrical steel contains few additives of silicon, since it causes power loss from the effect of eddy currents on the circuit of the magnetic circuit. In IT of toroidal design, the core can be made from rolled or ferrimagnetic steel.

Plates for a set of an electromagnetic core are selected in thickness depending on the frequency. With an increase in this parameter, it is necessary to install plates of smaller thickness.

Principle of operation

The main feature of pulse type transformers (hereinafter referred to as IT) is that they are supplied with unipolar pulses with a constant current component, and therefore the magnetic circuit is in a state of constant bias. The schematic diagram for connecting such a device is shown below.

As you can see, the connection diagram is almost identical with conventional transformers, which cannot be said about the timing diagram.

The primary winding receives pulse signals having a rectangular shape e (t), the time interval between which is quite short. This causes an increase in the inductance during the interval t u , after which its decline is observed in the interval (T-t u).

The induction drops occur at a rate that can be expressed in terms of the time constant by the formula: τ p =L 0 /R n

The coefficient describing the difference of the inductive difference is determined as follows: ∆V=V max - V r

• B max - the level of the maximum value of induction;
• In r - residual.

More clearly, the difference in inductions is shown in the figure showing the shift of the operating point in the IT magnetic circuit.

As can be seen in the timing diagram, the secondary coil has a voltage level U 2 in which there are reverse surges. This is how the energy accumulated in the magnetic circuit manifests itself, which depends on the magnetization (parameter i u).

The current pulses passing through the primary coil are trapezoidal in shape, since the load and linear currents (caused by core magnetization) are combined.

The voltage level in the range from 0 to t u remains unchanged, its value e t =U m . As for the voltage on the secondary coil, it can be calculated using the formula:

wherein:

• Ψ is the flux linkage parameter;
• S is a value that displays the cross section of the magnetic core.

Considering that the derivative characterizing the changes in the current passing through the primary coil is a constant value, the increase in the level of induction in the magnetic circuit occurs linearly. Based on this, it is permissible, instead of the derivative, to introduce the difference of indicators made after a certain time interval, which allows you to make changes to the formula:

in this case, ∆t will be identified with the parameter t u , which characterizes the duration with which the input voltage pulse flows.

To calculate the area of ​​the pulse with which the voltage is formed in the secondary winding of IT, it is necessary to multiply both parts of the previous formula by t u. As a result, we will come to an expression that allows us to obtain the main IT parameter:

U m x t u =S x W 1 x ∆V

Note that the value of the pulse area directly depends on the parameter ∆В.

The second most important value that characterizes the operation of IT is the induction drop, it is influenced by such parameters as the cross section and magnetic permeability of the core of the magnetic circuit, as well as the number of turns on the coil:

Here:

• L 0 - induction difference;
• µ a is the magnetic permeability of the core;
• W 1 - the number of turns of the primary winding;
• S is the cross-sectional area of ​​the core;
• l cp - length (perimeter) of the core (magnetic circuit)
• B r is the value of the residual induction;
• In max - the level of the maximum value of the induction.
• H m - Magnetic field strength (maximum).

Considering that the IT inductance parameter completely depends on the magnetic permeability of the core, the calculation must be based on the maximum value of µ a, which is shown by the magnetization curve. Accordingly, for the material from which the core is made, the level of the parameter B r , which reflects the residual induction, should be minimal.

Video: a detailed description of the principle of operation of a pulse transformer

Based on this, a tape made of transformer steel is ideal for the role of the IT core material. You can also use permalloy, in which such a parameter as the coefficient of squareness is minimal.

Ferrite alloy cores are ideal for high-frequency IT because this material has low dynamic losses. But because of its low inductance, it is necessary to make IT of large sizes.

Calculation of a pulse transformer

Consider how it is necessary to calculate IT. Note that the efficiency of the device is directly related to the accuracy of the calculations. As an example, let's take a conventional converter circuit that uses a toroidal type IT.

First of all, we need to calculate the IT power level, for this we use the formula: P \u003d 1.3 x P n.

The value of R n displays how much power the load will consume. After that, we calculate the overall power (P gb), it should be no less than the load power:

Parameters required for calculation:

• S c - displays the cross-sectional area of ​​the toroidal core;
• S 0 - the area of ​​​​its window (as a hint, this and the previous value are shown in the figure);

• B max is the maximum peak induction, it depends on which brand of ferromagnetic material is used (the reference value is taken from sources describing the characteristics of ferrite grades);
• f is a parameter characterizing the frequency with which the voltage is converted.

The next step is to determine the number of turns in the primary winding Tr2:

(results are rounded up)

The value of U I is determined by the expression:

U I \u003d U / 2-U e (U is the voltage supply to the converter; U e is the voltage level supplied to the emitters of transistor elements V1 and V2).

We proceed to the calculation of the maximum current passing through the primary winding of IT:

The parameter η is equal to 0.8, this is the efficiency with which our converter must work.

The diameter of the wire used in the winding is calculated by the formula:

If you have problems with determining the main IT parameters, you can find thematic sites on the Internet that allow you to calculate any pulse transformers online.

It's about "lazy winding". This is when too lazy to count the turns. https://website/

The most interesting videos on Youtube

The choice of the type of magnetic circuit.

The most versatile magnetic cores are W-shaped and cup-shaped armored cores. They can be used in any switching power supply, thanks to the ability to set the gap between the parts of the core. But, we are going to wind a pulse transformer for a push-pull half-bridge converter, the core of which does not need a gap and therefore a ring magnetic circuit will fit perfectly. https://website/

For the ring core, it is not necessary to make a frame and make a winding device. The only thing you have to do is to make a simple shuttle.

The picture shows the M2000NM ferrite magnetic core.

It is possible to identify the standard size of the ring magnetic circuit by the following parameters.

D is the outer diameter of the ring.

d is the inner diameter of the ring.

Obtaining initial data for a simple calculation of a pulse transformer.

Supply voltage.

I remember when our power grids had not yet been privatized by foreigners, I was building a switching power supply. The work dragged on into the night. During the last tests, it suddenly turned out that the key transistors began to get very hot. It turned out that the mains voltage jumped up to 256 volts at night!

Of course, 256 Volts is too much, but you should not focus on GOST 220 + 5% -10% either. If you choose for the maximum voltage of the network 220 Volts + 10%, then:

242*1.41=341.22V(we consider the amplitude value).

341.22 - 0.8 * 2 ≈ 340V(subtract the drop across the rectifier).

Induction.

We determine the approximate value of induction according to the table.

Example: M2000NM - 0.39T.

Frequency.

The frequency of generation of a converter with self-excitation depends on many factors, including the magnitude of the load. If you choose 20-30 kHz, then you are unlikely to make a big mistake.

Boundary frequencies and induction values ​​of widespread ferrites.

Manganese-zinc ferrites.

Nickel-zinc ferrites.

How to choose a ferrite ring core?

You can choose the approximate size of the ferrite ring using a calculator for calculating pulse transformers and a guide to ferrite magnetic cores. Both can be found in .

We enter the data of the proposed magnetic circuit and the data obtained in the previous paragraph into the calculator form in order to determine the overall power of the core.

You should not choose the dimensions of the ring close to the maximum load power. It is not so convenient to wind small rings, and you will have to wind a lot more turns.

If a free space in the case of the future design is enough, then you can choose a ring with a obviously higher overall power.

At my disposal was the M2000NM ring, size K28x16x9mm. I entered the input data into the calculator form and got an overall power of 87 watts. This is more than enough for my 50 watt power supply.

Run the program. Select "Calculation of the transformer half-bridge converter with master oscillator".

To prevent the calculator from “swearing”, fill in the windows that are not used to calculate the secondary windings with zeros.

How to calculate the number of turns of the primary winding?

We enter the initial data obtained in the previous paragraphs into the form of a calculator and obtain the number of turns of the primary winding. By changing the size of the ring, the brand of ferrite and the generation frequency of the converter, you can change the number of turns of the primary winding.

It should be noted that this is a very, very simplified calculation of a pulse transformer.

But, the properties of our wonderful self-excited power supply are such that the converter itself adapts to the parameters of the transformer and the load by changing the generation frequency. So, with an increase in the load and an attempt by the transformer to enter saturation, the generation frequency increases and the work normalizes. In the same way, small errors in our calculations are compensated. I tried to change the number of turns of the same transformer by more than one and a half times, which is reflected in the examples below, but I could not find any significant changes in the operation of the PSU, except for a change in the generation frequency.

How to calculate the wire diameter for primary and secondary windings?

The wire diameter of the primary and secondary windings depends on the PSU parameters entered into the form. The greater the winding current, the larger the required wire diameter. The primary current is proportional to the "Usable Power of the Transformer".

Features of winding pulse transformers.

The winding of pulse transformers, and especially transformers on ring and toroidal magnetic cores, has some features.

The fact is that if any winding of the transformer is not evenly distributed around the perimeter of the magnetic circuit, then separate sections magnetic circuits can enter saturation, which can lead to a significant decrease in the power of the PSU and even lead to its failure.

We are trying to wind the “lazy winding”. And in this case, the easiest way is to wind a single-layer winding "coil to coil".

What is needed for this?

It is necessary to select a wire of such a diameter that it fits “turn to turn”, in one layer, into the window of the existing ring core, and even so that the number of turns of the primary winding does not differ much from the calculated one.

If the number of turns obtained in the calculator does not differ by more than 10-20% from the number obtained in the formula for calculating the laying, then you can safely wind the winding, not counting the turns.

True, for such winding, most likely, it will be necessary to choose a magnetic core with a slightly overestimated overall power, which I already advised above.

1 - ring core.

2 - gasket.

3 - winding turns.

The picture shows that when winding "coil to coil", the calculated perimeter will be much smaller than the inner diameter of the ferrite ring. This is due to the diameter of the wire itself and the thickness of the gasket.

In fact, the actual perimeter that will be filled with wire will be even smaller. This is due to the fact that the winding wire does not adhere to the inner surface of the ring, forming a gap. Moreover, there is a direct relationship between the diameter of the wire and the size of this gap.

It is not necessary to increase the tension of the wire when winding in order to reduce this gap, as this can damage the insulation, and the wire itself.

Using the empirical formula below, you can calculate the number of turns, based on the diameter of the existing wire and the diameter of the core window.

The maximum calculation error is approximately -5% + 10% and depends on the density of the wire laying.

w = π(D - 10S - 4d) / d, where:

w- the number of turns of the primary winding,

π – 3,1416,

D is the inner diameter of the annular magnetic circuit,

S- the thickness of the insulating gasket,

d- diameter of the wire with insulation,

/ - fractional line.

How to measure the diameter of the wire and determine the thickness of the insulation - told.

To make it easier, check out this link:

Several examples of calculation of real transformers.

● Power - 50 watts.

Magnetic circuit - K28 x 16 x 9.

Wire - Ø0.35mm.

w \u003d π (16 - 10 * 0.1 - 4 * 0.39) / 0.39 ≈ 108 (turns).

Really fit - 114 turns.

● Power - 20 watts.

Magnetic circuit - K28 x 16 x 9.

Wire - Ø0.23mm.

w \u003d π (16 - 10 * 0.1 - 4 * 0.25) / 0.25 ≈ 176 (turns).

Really fit - 176 turns.

● Power - 200 Watts.

Magnetic circuit - two rings K38 x 24 x 7.

Wire - Ø1.0mm.

w \u003d π (24 - 10 * 0.1 - 4 * 1.07) / 1.07 ≈ 55 (turns).

Really fit 58 turns.

In the practice of a radio amateur, it is rarely possible to choose the diameter of the winding wire with the required accuracy.

If the wire turned out to be too thin for winding "turn to turn", and this often happens when winding secondary windings, then you can always slightly stretch the winding by pushing the turns apart. And if there is not enough wire cross-section, then the winding can be wound in several wires at once.

How to wind a pulse transformer?

First you need to prepare the ferrite ring.

In order for the wire not to cut through the insulating gasket, and not to damage itself, it is advisable to blunt the sharp edges of the ferrite core. But, this is not necessary, especially if the wire is thin or a reliable gasket is used. For some reason, I always do this.

Using sandpaper, round off the outer sharp edges.

We do the same with the inner faces of the ring.

To prevent breakdown between the primary winding and the core, an insulating gasket should be wound around the ring.

As an insulating material, you can choose varnished cloth, fiberglass, keeper tape, lavsan film or even paper.

When winding large rings using a wire thicker than 1-2 mm, it is convenient to use a keeper tape.

Sometimes, in the manufacture of home-made pulse transformers, radio amateurs use fluoroplastic tape - FUM, which is used in plumbing.

It is convenient to work with this tape, but fluoroplasts have cold fluidity, and the pressure of the wire in the area of ​​the sharp edges of the ring can be significant.

In any case, if you are going to use the FUM tape, then lay a strip of electric cardboard or plain paper along the edge of the ring.

When winding the gasket on rings of small sizes, it is very convenient to use a mounting hook.

The mounting hook can be made from a piece of steel wire or a bicycle spoke.

We carefully wind the insulating tape around the ring so that each next turn overlaps the previous one from the outside of the ring. Thus, the insulation outside the ring becomes two-layer, and inside - four or five layers.

To wind the primary winding, we need a shuttle. It can be easily made from two pieces of thick copper wire.

It is quite easy to determine the required length of the winding wire. It is enough to measure the length of one turn and multiply this value by the required number of turns. A small allowance for conclusions and a calculation error also does not hurt.

34 (mm) * 120 (turns) * 1,1 (times) = 4488 (mm)

If a wire thinner than 0.1 mm is used for winding, then stripping the insulation with a scalpel can reduce the reliability of the transformer. It is better to remove the insulation of such a wire with a soldering iron and an aspirin (acetylsalicylic acid) tablet.

Be careful! When melting acetylsalicylic acid, toxic fumes are released!

If a wire with a diameter of less than 0.5 mm is used for any winding, then it is better to make the leads from a stranded wire. We solder a piece of stranded insulated wire to the beginning of the primary winding.

We isolate the place of soldering with a small piece of electric cardboard or plain paper with a thickness of 0.05 ... 0.1 mm.

We wind the beginning of the winding so as to securely fix the junction.

We perform the same operations with the output of the end of the winding, only this time we fix the junction with cotton threads. So that the tension of the thread does not weaken while tying the knot, we fasten the ends of the thread with a drop of molten rosin.

If a wire thicker than 0.5 mm is used for winding, then the conclusions can be made with the same wire. At the ends you need to put on pieces of PVC or other tube (cambric).

Then the conclusions together with the tube must be fixed with cotton thread.

Over the primary winding we wind two layers of varnished cloth or other insulating tape. This winding gasket is necessary for reliable isolation of the secondary circuits of the power supply from the lighting network. If a wire with a diameter of more than 1 millimeter is used, then it is a good idea to use keeper tape as a gasket.

If you intend to use, then you can wind the secondary winding in two wires. This will ensure complete symmetry of the windings. The turns of the secondary windings must also be evenly distributed around the perimeter of the core. This is especially true for the most powerful windings in terms of power take-off. Secondary windings, which take a small, compared to the total, power, can be wound at random.

If there was no wire of sufficient cross section at hand, then you can wind the winding with several wires connected in parallel.

On the picture secondary winding wound with four wires.

Where: t and, T is the duration and period of the pulse.

For single-cycle circuits: γ = 0..0.5; γ max = 0.5;

where: , - minimum and maximum voltage on the primary winding.

Where

FET open channel resistance;

Bipolar transistor saturation voltage;

Voltage drop across the rectifier forward-biased diode in the output circuit (for Schottky diodes= 0.5..0.6V ). For the bridge, the value is 2

j \u003d 5 10 6 A / m 2 maximum current density in the wire;

B - magnetic induction in the magnetic circuit;

B = 0.2 T (selected with a margin);

η = 0.93..0.95.

(mm).

(mm); S c \u003d [cm 2]; μ 0 = 4π∙10 -8 = 1.256637∙10 -7 .

S 0 \u003d 0.2827 cm 2;

S c \u003d 0.05 cm 2.

S 0 \u003d 1.1308 cm 2;

S c \u003d 0.24 cm 2.

S 0 \u003d 2.0102 cm 2;

S c \u003d 0.54 cm 2.

S 0 \u003d 3.1416 cm 2;

S c \u003d 0.6 cm 2.

S 0 \u003d 0.825 cm 2;

S c \u003d 0.36 cm 2;

Bm = 0.38 T;

L cf = 2.9 cm.

S 0 S c \u003d 0.297 cm 4.