What is a pulse transformer and how to calculate it? Calculation and modeling of a high-frequency transformer as part of a single-cycle flyback converter Calculation of a pulse transformer using a bridge circuit

It largely depends on how correctly it is executed. A slight deviation of its parameters from the optimal ones for a particular power source leads to a decrease in efficiency and deterioration in performance.

The procedure for calculating a pulse transformer

Risp = 1.3 Рн (Рн is the power consumed by the load). Next, having specified the overall power Prab, which must satisfy the condition Prab ≥ Risp, it is necessary to select a suitable toroidal ferrite magnetic core. The parameters of the magnetic circuit are related to Pgab by the relation Pgab = ScS0fBmax/150, W.

Here f is the voltage conversion frequency, Hz; Sc = (D-d)h/2 is the cross-sectional area of ​​the magnetic circuit, cm2 (D and d are the outer and inner diameters, respectively, h is the height of the ring, cm); S0 = p d2/4—magnetic core window area, cm2; Bmax is the maximum value of induction (in Tesla), which depends on the grade of ferrite and can be determined from a reference book containing information about ferromagnetic materials.

After this, knowing the voltage on the primary winding of the transformer U1, find the number of turns w1=0.25x104U1/fBmaxSc.

For the converter (see figure) U1 = Upit/2- Uke us, where Upit is the supply voltage of the converter, and Uke us is the saturation voltage of the collector - emitter of transistors VT1, VT2.

The calculated value of w1 must be rounded up (to avoid saturation of the magnetic circuit).

Then find the number of turns of the secondary winding of the transformer: w2 = w1U2/U1 and the wire diameter: d2 = 0.6√I2 (U2 and I2 are the voltage and current of the secondary winding, respectively).

Now, to consolidate the material covered, let’s consider the calculation of a transformer for a specific example.

• Let's calculate the high-frequency transformer of the power supply of a stereo amplifier [3] having the following output voltages and currents:
• U2 = (25+25) V
• I2 = 3 A
• U3 = 20 V
• I3 = 1 A
• U4 = 10 V
• I4 = 3 A

Load power Pn = 200 W. The used power of this transformer Risp = 1.3 · 200 = 260 W.

Let us choose the conversion frequency f equal to 10 5 Hz. As a magnetic core we use a ring of standard size K38x24x7 made of ferrite grade 2000NN (Bmax = 0.25 T).

Let's determine the cross-sectional area Sc = (3.8 - 2.4) · 0.7/2 = 0.49 cm 2 and the window area of ​​the selected magnetic circuit So = p · 2.4 2 ÷ 4 = 4.5 cm 2, calculate the overall transformer power Rgab = 0.49 4.5 10 5 0.25/150 = = 367 W. The condition Prab ≥ Risp is satisfied.

Now we determine the voltage on the primary winding of the transformer and the number of turns:

U1 = (285/2) - 1.6 = 141 V; w1 = (0.25 ∙ 10 4 ∙ 141) ÷ (10 5 ∙ 0.25 ∙ 0.49) ≈ 29.

To avoid saturation of the magnetic circuit, select w1 = 30.

Imax = 200/0.8 141 = 1.75 A; d1 = 0.6√1.75 = 0.8 mm..

And in conclusion, we determine the number of turns and the diameter of the wire of the output windings:

w2 = 30 25/141 = 5; d2 = 0.6√3 = 1 mm;

w3 = 30 20/141 = 4; d3 = 0.6√1 = 0.6 mm;

w4 = 30 10/141 = 2; d4 = 0.6√3 = 1 mm.

Push-pull converter is a voltage converter that uses a pulse transformer. The transformation ratio of the transformer can be arbitrary. Although fixed, in many cases the pulse width can be varied, expanding the available voltage regulation range. The advantage of push-pull converters is their simplicity and the ability to increase power.

In a properly designed push-pull converter, there is no direct current through the winding and no core bias. This allows you to use a full magnetization reversal cycle and get maximum power.

The following simplified technique allows you to calculate the main parameters of a pulse transformer made on a ring magnetic core.

1. Calculation of overall transformer power

where Sc is the cross-sectional area of ​​the magnetic circuit, cm2; Sw—core window area, cm2; f - f - oscillation frequency, Hz; Bmax is the permissible induction value for domestic nickel-manganese and nickel-zinc ferrites at frequencies up to 100 kHz.

Limit frequencies and induction values ​​of widespread ferrites

Manganese-zinc ferrites.

Nickel-zinc ferrites.

To calculate the cross-sectional area of ​​the magnetic core and the window area of ​​the magnetic core, the following formulas are used:

Sc = (D - d) ⋅ h / 2

Sw=(d / 2)2 π

where D is the outer diameter of the ferrite ring, cm; d—inner diameter; h is the height of the ring;

2. Calculation of the maximum power of the transformer

We select the maximum power of the transformer as 80% of the overall power:

Pmax = 0.8 Pgab

3. Calculation of the minimum number of turns of the primary winding W1

The minimum number of turns of the primary winding W1 is determined by the maximum voltage on the winding U1 and the permissible core induction Bmax:

4. Calculation of the effective value of the current in the primary winding:

The effective value of the primary winding current is calculated by the formula:

I1 = Pmax / Ueff

It should be taken into account that Ueff = U1 / 1.41 = 0.707U1, since Ueff is the effective voltage value, and U1 is the maximum voltage value.

5. Calculation of the wire diameter in the primary winding:

where I1 is the effective value of the current in the primary winding, A; j—current density, A/mm2;

The current density depends on the power of the transformer, the dissipated amount of heat is proportional to the area of ​​the winding and the temperature difference between it and the environment. As the size of the transformer increases, the volume grows faster than the area, and for the same overheating, the specific losses and current density must be reduced. For transformers with a power of 4..5 kVA, the current density does not exceed 1..2 A/mm².

For reference, the table shows current density data depending on the power of the transformer

6. The effective value of the secondary winding current (I2), the number of turns in the secondary winding (W2) and the diameter of the wire in the secondary winding (d2) are calculated using the following formulas:

I2 = Pmax / U2eff

where Uout is the output voltage of the secondary winding, Pmax is the maximum output power of the transformer, it should also be taken into account that the Pmax value can be replaced by the load power, provided that the load power is less than the maximum output power of the transformer.

W2 = (U2eff*W1) / Ueff

Based on all the above formulas (taking into account the current density, which depends on the power of the transformer), you can approximately calculate the main parameters of the pulse transformer; for the convenience of calculations, you can use an online calculator.

This article is a simplified method for calculating a pulse transformer for a push-pull converter; all formulas and an online calculator allow you to calculate approximate Pulse transformer winding data, since the transformer has many interdependent parameters.

If you find errors in the formulas, methods of their application and other comments, please leave them in the comments.

After determining the diameter of the wire, it should be taken into account that the diameter of the wire is calculated without insulation, use the winding wire data table to determine the diameter of the wire with insulation.

Winding wire data table.

In the calculation method described in, to determine the minimum number of turns of the primary winding W 1 and the overall P gain (maximum permissible) power of the transformer of a push-pull converter, 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 may give an erroneous result, which may result in failure of the transformer and switching transistors.

Consider, for example, a ring magnetic core K40x25x11 made of 2000NM1 ferrite. The recommended maximum value of magnetic induction should be equal to the saturation induction: B max =B us =0.38 Tesla. Probably the conclusion has been drawn. 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 = 285/2-1.6≈141 V. From the calculation using formula (1) we obtain W 1 =11.24≈12 turns of the primary winding.

Let's say it is necessary to obtain a direct current l n = 4 A in the load at a voltage U n = 50 V, which corresponds to useful power P n = 200 W. With efficiency η≈0.8, the power used is P used =P n /η=200/0.8=250 W. The overall power of the selected transformer, calculated using formula (2), is more than four times higher than required, so it should function without problems. In accordance with the maximum current in the primary winding is equal to l 1max =P use /U 1 =1.77 A. Let's choose switching transistors with a current reserve of 50%, then the maximum permissible collector (drain) current I to additional = 1.77*1.5=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?

Let's consider the process of transferring energy to the load using a pulse transformer, the connection diagram of which is shown in Fig. 1, a. 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, the rectangular shape of which is shown in Fig. 1, b, are shown.

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 amplitude l 1max = 1.77 A and the triangular component of the magnetizing current. The last component can be estimated using the formula

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

Figure 1,c shows how during one half-cycle the magnetizing 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 magnetizing current, the total current l ∑max shown in Fig. 1b, can increase to values ​​dangerous for transistors.

Let's consider the influence of hysteresis. Magnetization and magnetization reversal of the magnetic circuit occurs in accordance with the curves shown in Fig. 2. The abscissa axis is the magnetic field strength H created by the primary winding of the transformer; the ordinate axis is the magnetic induction B in the magnetic core. In Fig. Figure 2 shows the limiting hysteresis loop and the private (internal) hysteresis loop corresponding to Fig. 1,b and 1,c.

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 magnetic field strength 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 the change in magnetic induction B in the magnetic circuit.

If you draw a tangent 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 LP in relation to the change in the magnetic field strength ∆H at the selected point, i.e. ∆B/ ∆H. 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 within very wide limits: 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 the apex 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, energy losses for heating the magnetic core due to its magnetization reversal are negligible. As for the mode with the value of magnetic induction entering the saturation region of the magnetic circuit material (B max = B us). chosen in , the picture will be completely different. In this case, the main magnetization curve corresponds to a current shape i μ2 that is very far from linear. The tangent at point E with coordinates (H us, B 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, as illustrated by the graph i μ2. If the switching transistor is selected without sufficient current reserve, it will inevitably be damaged. To prevent saturation of the magnetic circuit, it is necessary to fulfill the following 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, another criterion is used - the relative value of the magnetizing current. The parameters of the primary winding are chosen as follows. so that the magnitude 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, which in our example contains 12 turns, is 0.3 mH. Magnetizing current amplitude calculated using 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 =l 1max +l max =1.77+1.18=2.95≈3 A (Fig. 1,b) with the maximum permissible current of the switching transistor 2.7 And, it becomes completely obvious that the transistor was chosen incorrectly and the calculated diameter of the primary winding conductor does not correspond to the required value. This discrepancy will be further exacerbated by a possible 20% increase in input voltage. Since at the rated supply voltage the mode with the value of magnetic induction entering the saturation region of the magnetic core material (B max = B us) is selected, in the case 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 specified value of 3 A.

The conversion frequency of 100 kHz, arbitrarily chosen in the calculation example, as experiment shows, is the maximum possible for 2000NM1 ferrite, and it is necessary to take into account 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 greater. If the network voltage increases 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 will not exceed V max = 0.75 * V us = 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 justified choice of initial values ​​in formula (1) can lead to inaccurate or even erroneous calculation of a pulse transformer. To avoid any doubts about the legality of applying formula (1), we will justify it analytically.

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

where μ 0 = 4π·10 7 H/m - absolute magnetic permeability of vacuum; μ EFF - effective magnetic permeability of the magnetic core material; l max - magnetizing current amplitude, A; W 1 - number of turns of the primary winding; lEFF- effective length of the magnetic field line in the magnetic core, m. Let us substitute l max from (4) into (5), 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 we see, formula (6) differs from (1) only in that it includes the effective cross-sectional area of ​​the magnetic core, and not the geometric one. A detailed methodology for calculating the effective parameters of various types of magnetic circuits is given in [3]. When using this formula in practice, 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 relationships used in the design of transformers for various push-pull converters.

Self-oscillating converters with one transformer, similar to those described in (4), operate by entering the saturation region of the magnetic circuit material (points E and E" in Fig. 2). Formulas (1) and (2) are used at B max = V us. Several Otherwise, the indicated formulas are used in the case of designing self-oscillating converters with two transformers, such as those described in. In it, the coupling winding on a powerful transformer is connected to a low-power transformer in the control circuit of the switching transistor bases. The pulse voltage induced in the coupling winding creates saturation in the low-power transformer. which sets the conversion frequency in accordance with formula (1). This frequency is selected to avoid saturation in a powerful transformer, the size of which is determined according to formula (2). current in switching transistors.

Along with self-generators, push-pull converters with external excitation are very popular among radio amateurs. To eliminate 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 magnetic induction in the magnetic circuit, usually first, based on (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 standard size of a magnetic core 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 are presented according to formula (6), the amplitude value of the magnetizing 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 using formula (2) using the effective cross-sectional area of ​​the magnetic core 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

provides a basis for selecting a switching transistor based on the maximum permissible 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

Calculations were performed under the condition that the maximum magnetic induction Vmax will not exceed 0.25 Tesla, even if the network 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 core should be selected with a margin of 20...40% of the maximum output power indicated in the table. Although the table is 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 large, and the amplitude of the rectangular component of the primary winding current will be half as large. The number of turns should be twice as large. The winding inductance will increase fourfold, and the current >I max will decrease by half. You can use a magnetic core made of two ferrite rings of the same size folded together, which will lead to a twofold increase in the cross-sectional area of ​​the magnetic core 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 output from the middle point of the primary winding of the transformer, the full mains voltage is applied to half of this winding, so the number of winding turns must be twice as large as in a bridge converter, all other things being equal.

We emphasize that due to the significant scatter in the actual values ​​of the parameters of ferromagnetic materials in comparison with their reference data, the table can only be used for preliminary selection of the magnetic core, and then, after experimental measurement of its characteristics, it is necessary to carry out a refined calculation of the transformer. For example, for the magnetic circuit K40x25x11 the table shows the value of the inductance coefficient A L = 2.08 µH per turn. Let us experimentally clarify the magnetic properties of a specific 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 approximate. In our case, we need to either increase the number of turns so that the winding inductance is no 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 greater than the tabulated 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 core to be uniform, its turns are placed either “discharged” or they fill the layer entirely, 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. From the table it follows 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. Let us determine the diameter of the wire of the primary winding of the transformer. As indicated, we will choose the closest diameter from the industrially produced nomenclature d 1 = 0.83 mm, and taking into account the insulation d 1 = 0.89 mm. If we take into account the electrical insulation of the magnetic circuit with several layers of varnished cloth with a total thickness of 0.25 mm, the internal diameter of the magnetic circuit will decrease to 25-0.5 = 24.5 mm. In this case, the length of the inner circle will be π·24.5≈80 mm. Taking into account the fill factor of 0.8, 64 mm are available for winding the first layer of winding, which corresponds to 64/0.89 = 71 turns. Thus, there is enough space for 45 turns. We wind them “discharged”.

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. Taking into account the linear resistance of the wire, we obtain R exchange1 = 0.0324 * 1.755 = 0.06 Ohm, and the voltage drop on the primary winding will reach U 1nad = 1.77 * 0.06 = 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 obtain the calculated number of turns of the secondary winding N 2 = 45 * (51/150) = 15.3 ≈ 16 turns. Secondary wire diameter

Filling a 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 met.

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

LITERATURE

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

2. Background information. Ferrite Handbook. 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.

4. Knyazev Yu., Sytnik G., Sorkin I. ZG block and power supply of the IK-2 kit. - 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 a network transformer of a power source. - Radio, 2004, No. 5, p. 55-57.

S. KOSENKO, Radio, 2005, No. 4, pp. 35-37.44.

And yet I was invited! Now things will go more quickly with the articles. Initially, I wanted to focus on the circuit design of some block for the next part, but what are you waiting for? But then I remembered my school youth and the great problem that I faced - how to make a beast device unknown to me at that time - pulse transformer . Ten years have passed and I understand that many (and not just beginners) radio amateurs, electronics engineers and students have such difficulties - they are simply afraid of them, and as a result they try to avoid powerful switching power supplies (further IIP).
After these thoughts, I came to the conclusion that the first topic should be about the transformer and nothing else! I would also like to make a reservation: what I mean by the concept of “powerful SMPS” is power from 1 kW and above, or in the case of amateurs, at least 500 W.

Figure 1 - This is the kind of 2 kW transformer we will eventually get for the H-bridge

The Great Battle or Which Material to Choose?

Once upon a time, having introduced pulse technology into my arsenal, I thought that transformers could only be made using ferrite, which was accessible to everyone. Having assembled the first designs, the first thing I decided to do was present them to the judgment of more experienced comrades and very often heard the following phrase: “Your shitty ferrite is not the best material for an impulse generator.”. I immediately decided to find out from them what alternative could be opposed to it and they told me - alsifer or whatever they call it sindust.

Why is it so good and is it really better than ferrite?

First you need to decide what an almost ideal material for a transformer should be able to do:
1) must be soft magnetic, that is, it is easy to be magnetized and demagnetized


Figure 2 - Hysteresis cycles of ferromagnets: 1) hard cycle, 2) soft cycle

2) the material must have the highest possible saturation induction, which will either reduce the dimensions of the core, or, while maintaining them, increase the power

Saturation

The phenomenon of transformer saturation is that, despite the increase in current in the winding, the magnetic flux in the core, having reached a certain maximum value, then practically does not change.
In a transformer, the saturation mode leads to the fact that the transfer of energy from the primary winding to the secondary winding partially stops. Normal operation of a transformer is possible only when the magnetic flux in its core changes in proportion to the change in current in the primary winding. To fulfill this condition, it is necessary that the core is not in a state of saturation, and this is only possible when its volume and cross-section are not less than a certain value. Therefore, the greater the power of the transformer, the larger its core must be.

3) the material must have as little loss as possible due to magnetization reversal and Foucault currents

4) the properties of the material should not change significantly under external influences: mechanical forces (compression or tension), changes in temperature and humidity.

Now let's look at the properties of ferrite and how well it meets the requirements presented above.

Ferrite is a semiconductor, which means it has its own high electrical resistance. This means that at high frequencies, eddy current losses (currents Foucault) will be quite low. It turns out that at least one condition from the list above has already been met. Let's move on...
Ferrites can be thermally stable or unstable, but this parameter is not decisive for the SMPS. The important thing is that ferrites work stably in the temperature range from -60 to +100 o C, and this is for the simplest and cheapest brands.


Figure 3 - Magnetization curve at a frequency of 20 kHz at different temperatures

And finally, the most important point - in the graph above we saw a parameter that will determine almost everything - saturation induction. For ferrite it is usually taken as 0.39 Tesla. It is worth remembering that under different conditions this parameter will change. It depends on both frequency and operating temperature and other parameters, but special emphasis should be placed on the first two.

Conclusion: ferrite is good! perfect for our purposes.

A few words about alsifer and how it differs

1) alsifer works in a slightly wider range of temperatures: from -60 to +120 o C - is it suitable? Even better than ferrite!
2) the loss coefficient for hysteresis in alsifers is constant only in weak fields (at low power), in a powerful field they increase very strongly - this is a very serious disadvantage, especially at powers of more than 2 kW, so it loses here.
3) saturation induction up to 1.2 Tesla!, 4 times more than ferrite! - the main parameter is already ahead, but not everything is so simple... Of course, this advantage will not go anywhere, but point 2 weakens it very much - definitely a plus.

Conclusion: Alsifer is better than ferrite, this guy didn’t lie to me.

Result of the battle: Anyone who reads the description above will say give us Alsifer! And rightly so... but try to find an alsifer core with an overall power of 10 kW? Here usually a person comes to a dead end, it turns out that they are not really on sale, and if they are, then they are ordered directly from the manufacturer and the price will scare you.
It turns out that we use ferrite, especially if we evaluate it as a whole, it loses very little... ferrite is estimated relative to alsifer at "8 out of 10 parrots."

I wanted to turn to my favorite matan, but decided not to do so, because... I consider +10,000 characters to the article excessive. I can only recommend a book with very good calculations by B. Semenov, “Power Electronics: From Simple to Complex.” I don’t see the point in retelling his calculations with some additions.

And so we proceed to the calculation and manufacture of the transformer

First of all, I would like to immediately recall a very serious point - the gap in the core. It can “kill” all the power or add another 30-40%. I want to remind you what we do transformer for H-bridge, and it refers to forward converters (forward in bourgeois). This means that the gap should ideally be 0 mm.
Once, while studying for a 2-3 course, I decided to assemble a welding inverter and turned to the topology of Kemppi inverters. There I saw a gap of 0.15 mm in the transformers. I wondered what it was for. I didn’t approach the teachers, but instead called the Russian representative office of Kemppi! What to lose? To my surprise, I was connected to a circuit engineer and he told me several theoretical points that allowed me to “crawl” beyond the 1 kW ceiling.
In short - a gap of 0.1-0.2 mm is simply necessary! This increases the rate of demagnetization of the core, which allows more power to be pumped through the transformer. The maximum effect of such a feint with the ears of the gap was achieved in the topology "oblique bridge", there the introduction of a gap of 0.15 mm gives an increase of 100%! In our H-bridge this increase is more modest, but I think 40-60% is not bad either.

To make a transformer we need the following kit:

A)
Figure 4 - Ferrite core E70/33/32 made of 3C90 material (slightly better analogue of N87)

b)
Figure 5 - Frame for core E70/33/32 (the larger one) and choke D46 made of atomized iron

The overall power of such a transformer is 7.2 kW. We need such a reserve to provide starting currents 6-7 times higher than the rated ones (600% according to technical specifications). It’s true that such starting currents only occur in asynchronous motors, but everything needs to be taken into account!
Suddenly, a certain choke “surfaced”; it will be needed in our further scheme (as many as 5 pieces) and therefore I decided to show how to wind it.

Next, you need to calculate the winding parameters. I use a program from a well-known friend in certain circles Starichok51 . A man with enormous knowledge and always ready to teach and help, for which I thank him - at one time he helped me take the right path. The program is called - Excellent IT 8.1 .

Here is an example of a calculation for 2 kW:


Figure 6 - Calculation of a pulse transformer using a bridge circuit for 2 kW step-up

How to calculate:

1) Highlighted in red. These are the input parameters that are usually set by default:
a) maximum induction. Remember for ferrite it is 0.39 T, but our transformer operates at a fairly high frequency, so the program sets 0.186 itself. This is saturation induction in the very worst conditions, including heating up to 125 degrees
b) conversion frequency, it is set by us and how it is determined in the diagram will be in the following articles. This frequency should be from 20 to 120 kHz. If less, we will hear the trance and whistle, if higher, then our switches (transistors) will have large dynamic losses. And even expensive IGBT switches operate up to 150 kHz
c) coefficient window filling is an important parameter, because the space on the frame and core is limited, you should not make it more than 0.35, otherwise the windings will not fit
d) current density - this parameter can be up to 10 A/mm 2. This is the maximum current that can flow through a conductor. The optimal value is 5-6 A/mm 2 - under severe operating conditions: poor cooling, constant operation at maximum load, etc. 8-10 A/mm 2 - can be set if your device is perfectly ventilated and several coolers cost over 9000.
e) food at the entrance. Because we calculate the transformer for DC->DC 48V to 400V, then we set the input voltage as in the calculation. Where did the figure come from? In a discharged state, the battery produces 10.5V, further discharging will reduce the service life, multiply by the number of batteries (4 pcs) and get 42V. Let's take 40V with a margin. 48V is taken from the product 12V * 4 pcs. 58V is taken from the consideration that in a charged state the battery has a voltage of 14.2-14.4V and, by analogy, multiply by 4.

2) Highlighted in blue.
a) set 400V, because this is a reserve for voltage feedback and for cutting a sine wave a minimum of 342V is required
b) rated current. We choose from consideration 2400 W / 220 (230) V = 12A. As you can see, everywhere I take a reserve of at least 20%. This is what any self-respecting manufacturer of quality equipment does. In the USSR, such a reserve was the standard 25%, even for the most difficult conditions. Why is 220 (230) V the voltage at the output of a pure sine wave?
c) minimum current. Selected from real conditions, this parameter affects the size of the output choke, so the higher the minimum current, the smaller the choke, and therefore the cheaper the device. Again, I chose the worst option 1A, this is the current for 2-3 light bulbs or 3-4 routers.
d) drop on diodes. Because We will have ultra-fast diodes at the output, then the drop across them will be 0.6V in the worst conditions (the temperature is exceeded).
d) wire diameter. I once bought a 20 kg copper coil for such a case and just with a diameter of 1 mm. Here we put the one you have. I just don’t recommend setting it to more than 1.18 mm, because... the skin effect will begin to affect

Skin effect

Skin effect is the effect of reducing the amplitude of electromagnetic waves as they penetrate deep into a conducting medium. As a result of this effect, for example, high-frequency alternating current when flowing through a conductor is not distributed evenly over the cross-section, but mainly in the surface layer.
If we speak not like Google, but in my collective farm language, then if you take a conductor with a large cross-section, it will not be fully used, because currents at a higher frequency flow along the surface, and the center of the conductor will be “empty”

3) Highlighted in green. Everything is simple here - we plan a “full bridge” topology and select it.

4) Highlighted in orange. The core selection process takes place, everything is intuitive. A large number of standard cores are already in the library, like ours, but if anything can be added by entering the dimensions.

5) Highlighted in purple. Output parameters with calculations. The coefficient was highlighted in a separate window. filling the window, remember - no more than 0.35, and preferably no more than 0.3. All the necessary values ​​are also given: the number of turns for the primary and secondary windings, the number of wires of a previously specified diameter in the “braid” for winding.
Parameters for further calculation of the output choke are also given: inductance and voltage ripple.

Now you need to calculate the output choke. It is needed to smooth out ripples, as well as to create a “uniform” current. The calculation is carried out in the program of the same author and it is called ThrottleRing 5.0. Here is the calculation for our transformer:


Figure 7 - Calculation of the output choke for a boost DC-DC converter

In this calculation, everything is simpler and clearer, it works on the same principle, the output data is: the number of turns and the number of wires in the braid.

Manufacturing stages

Now we have all the data for manufacturing the transformer and inductor.
The main rule for winding a pulse transformer is that all windings, without exception, must be wound in one direction!

Stage 1:

Figure 8 - Winding process of the secondary (high-voltage) winding

We wind the required number of turns of 2 wires with a diameter of 1 mm onto the frame. We remember the direction of winding, or better yet, mark it with a marker on the frame.

Stage 2:

Figure 9 - Isolate the secondary winding

We insulate the secondary winding with fluoroplastic tape 1 mm thick, this insulation can withstand at least 1000 V. We also additionally impregnate it with varnish, this is another +600V to the insulation. If there is no fluoroplastic tape, then we insulate it with ordinary plumbing foam in 4-6 layers. This is the same fluoroplastic, only 150-200 microns thick.

Stage 3:

Figure 10 - We begin to wind the primary winding, solder the wires to the frame
We wind in one direction with the secondary winding!

Stage 4:

Figure 11 - Drawing out the tail of the primary winding

He wraps the winding and insulates it with fluoroplastic tape. It is also advisable to impregnate it with varnish.

Stage 5:


Figure 12 - Impregnate with varnish and solder the “tail”. Winding winding is completed
Stage 6:

Figure 13 - We complete the winding and insulation of the transformer with keeper tape with final impregnation in varnish

Keeper tape

Kiper tape - cotton (less often silk or semi-silk) braid made of kiper fabric with a width of 8 to 50 mm, twill or diagonal weave; harsh, bleached or plain-dyed. The tape material has a high density due to the weave, it is thicker than its closest analogue - plain tape - due to the use of thicker threads.
Thanks to Wikipedia.

Stage 7:


Figure 14 - This is what the finished version of the transformer looks like

A gap of 0.15 mm is established during the gluing process by inserting a suitable film between the core halves. The best option is printing film. The core is glued together with instant glue (good) or epoxy resin. The 1st option is forever, the 2nd allows you to disassemble the transformer without damage if something happens, for example, if you need to wind another winding or add more turns.

Choke winding

Now, by analogy, you need to wind the inductor; of course, winding it on a toroidal core is more difficult, but this option will be more compact. All the data we have is from the program, the core material is atomized iron or permalloy. The saturation induction of this material is 0.55 Tesla.

Stage 1:


Figure 15 - Wrap the ring with fluoroplastic tape

This operation allows you to avoid the case of breakdown of the winding on the core, this happens rarely, but we do it for ourselves for quality!

Stage 2:

Figure 16 - Wind the required number of turns and insulate

In this case, the number of turns will not fit into one winding layer, so after winding the first layer, it is necessary to insulate and wind the second layer, followed by insulation.

Stage 3:

Figure 17 - Insulate after the second layer and impregnate with varnish

Epilogue

I hope my article will teach you the process of calculating and manufacturing a pulse transformer, as well as give you some theoretical concepts about its operation and the materials from which it is made. I tried not to load this part with unnecessary theory, keep everything to a minimum and focus exclusively on practical aspects. And most importantly, on the key features that affect performance, such as clearance, winding directions, etc.
To be continued...

Samples of conversion and rectification circuits are given. Some program input fields and some calculation results that require comments are provided with tooltips.

More about the program

1. The main work in the program takes place in the “Optimization” group.
Automatic calculation is used when selecting a different core or when changing any input data (outside the Optimization group) to obtain a starting point when optimizing the winding data of the transformer.

2. In the “Optimization” group, when changing values ​​using the arrows, optimization starts automatically.
But if a new value is entered “manually”, then optimization should be started with this button.

3. For PWM controllers, the frequency is set equal to half the frequency of the master oscillator of the microcircuit. The master oscillator pulses are supplied to the outputs in turn, so the frequency at each output (and at the transformer) is 2 times lower than the frequency of the master oscillator.
IR2153 microcircuits, and similar ones of this family of microcircuits, are not PWM controllers, and the frequency at their outputs is equal to the frequency of the master oscillator.
You shouldn't chase a high frequency. At high frequencies, switching losses in transistors and diodes increase. Also, at a high frequency, due to the small number of turns, the magnetizing current is too high, which leads to a large no-load current and, accordingly, low efficiency.

4. The window fill factor characterizes how much of the core window will be occupied by the copper of all windings.

5. The current density depends on the cooling conditions and the size of the core.
With natural cooling, you should choose 4 - 6 A/mm2.
For ventilation, the current density can be selected higher, up to 8 - 10 A/mm2.
Large current densities correspond to small cores.
With forced cooling, the permissible current density depends on the cooling intensity.

6. If output voltage stabilization is selected, then the first output is the leading one. And the output with the highest consumption must be assigned to it.
The remaining outputs are counted according to the first one.
To truly stabilize all outputs, a group stabilization choke should be used.

7. With unipolar rectification, despite the higher copper consumption, the rectification circuit with a midpoint has an advantage, since the losses on two diodes will be 2 times less than on four diodes in a bridge circuit.

8. For proper operation of the inductor, there should be no capacitors in the rectifier after the diodes before the inductor! Even a small denomination.

9. On the number of winding turns in the calculation results, pop-up tips are placed with the number of layers occupied by the winding.

10. On the numbers of wires in the windings in the calculation results, pop-up tips with the current density in the winding are placed.

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