How to calculate a core choke. Scheme, description

ENCYCLOPEDIA OF RADIO ELECTRONICS AND ELECTRICAL ENGINEERING
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Electric welding. How to calculate a core choke. Encyclopedia of radio electronics and electrical engineering

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Encyclopedia of radio electronics and electrical engineering / welding equipment

A necessary element of the DC-DC converter is throttle.

The purpose of this section, without going beyond the school physics course, is to give a method for calculating the most common choke - a choke that works with bias. To begin with, we consider that a direct current with a slight ripple flows in the inductor winding.

The inductor winding usually completely occupies the core window. Therefore, knowing the magnitude of the current I and the current density J (A / mm2) in the winding, as well as the area of the core window S_o (cm2) and its filling factor K_o, it is possible to determine maximum number of turns, which can be placed in the core window:

Flux linkage choke windings can be determined if the turns are known, the maximum induction B_m (T), Core cross section S_c (cm2) and its fill factor K_m:

Substituting (18.10) into (18.11), we get:

It is known that

From (18.12) and (18.13) we find choke inductance:

From the inductance formula, it is easy to obtain the overall dimensions of the core, which will allow you to obtain the required choke inductance:

To select B, J, K_c, K_o Table 18.5 can be used. XNUMX. At the same time, the overall power R_gab can be equated to 1,25 • S_cS_c.

For aluminum wire, the current density should be reduced by a factor of 1,6.

Attention! In order to avoid saturation, the inductor core must have a non-magnetic gap.

We believe that, compared with a non-magnetic gap, the inductor core is an ideal magnetic conductor and all winding ampere-turns are applied to the non-magnetic gap. Due to the long non-magnetic gap, the induction in the core varies from almost zero to V_m.

The length of the non-magnetic gap with known ampere-turns can be determined by the formula:

or:

From (18.10), (18.13) and (18.17) we derive a formula for finding choke inductance:

Often we see that steel core chokes are used in inverter sources at a higher frequency than it would seem to be acceptable for them. There is a reasonable explanation for this.

Losses in the steel core of the transformer are determined by the formula:

where P_c - loss in the core; R_ud - specific losses for a given material at given values of maximum induction B_у and frequency f_у sinusoidal magnetic induction; G_с - mass of the core; IN_m - maximum induction in the core; α and β - frequency indicators.

In the transformer, the induction range reaches twice the value of the maximum induction B_m (induction changes from -B_m to +B_m). And in the inductor, even in the mode of discontinuous currents, the range does not exceed the value of V_m (induction changes from 0 to V_m). So, for the throttle, the formula can be rewritten in the following form:

ΔB is the range of induction in the core of the inductor.

It follows from the formula that the losses in the core increase along with an increase in the induction range ΔB and with increasing operating frequency f. However, if, by increasing the frequency, we reduce the range of induction, then the losses will not increase.

From here it is possible to determine maximum range of induction for higher operating frequency:

Consider practical examples of the calculation of the throttle.

Choke Calculation Example #1

Let's say we are building an adjustable welding source. The source is powered by a single-phase network 220 V, 50 Hz. Adjustment of welding current ranging from I_min = 50 A to I_Max = 150 A is carried out using a controlled thyristor rectifier.

Load frequency PN = 40%. So that the welding arc does not go out during voltage pauses, at minimum current and for the maximum control angle, it is necessary that the current does not fall below I_Art = 10 A.

From here you can determine the minimum inductance of the inductor:

We will wind the throttle on a W-shaped core made of steel 3411 (E310).

Let's first choose:

B = 1,42 T;
J = 5 A/mm2 (taking into account the specified duty cycle);
К_o - 0,35;
К_c = 0,95.

Find the overall size of the core:

For the choke, you can use two cores ШЛ40х80 (S_c = 32 cm2, S_o = 40 cm2).

Determine the number of turns of the winding:

The winding is carried out with a wire section:

Let us determine the length of the non-magnetic gap:

Let's define the resulting inductance:

The result can be considered satisfactory, despite the fact that the obtained inductance is somewhat lower than required.

Choke Calculation Example #2

As mentioned in the first example, the inductor is mainly needed to maintain the current in the pauses caused by the operation of the rectifier (controlled or uncontrolled). There is no great need for the absence of a pause in the throttle.

Consequently, it is possible to significantly reduce the dimensions of the inductor if it is made non-linear and saturable. That is, when the current in the inductor is below the saturation current 1nap, the inductor has a significant inductance sufficient to maintain the current in pauses, and when the current becomes greater than I_us the inductor is turned off, because its core enters saturation.

Let us calculate a nonlinear two-winding saturable choke for a welding source with a thyristor controller. The main primary winding of the inductor to saturation should have an inductance of 0,3 mH, and an additional secondary winding - 7,5 mH.

The maximum current of the primary winding is I₁ = 180 A, and secondary - I₂ = 13 A. The inductor core must enter saturation if the primary current exceeds I_us = 132 A.

We preliminarily believe that the primary winding of the inductor will be wound with aluminum, and the secondary with copper. Previously, we determined that at PV = 20% for copper, the current density J_Cu = 8 A/mm2.

Since aluminum has a higher resistivity than copper, it is necessary to choose a current density 1,6 times less for it, i.e. J_Al = 5 A/mm2.

Since the inductance of the inductor windings is known, the transformation ratio of the inductor can be found by the formula:

The formulas derived earlier are valid for a single-winding inductor with a minimum current ripple in the windings. To take into account the difference between the effective current and the saturation current, it is necessary to multiply the value of the current density J by the saturation factor:

To allocate space in the core window for additional winding, it is necessary to multiply the core size by a factor:

As a core for the inductor, we will choose a W-shaped tape core made of steel 3411 (E310). According to the modified formula (18.15) we find:

For the choke, you can use one core ШЛ32х50 (S_c =16 cm2, S_o = 26 cm2, S_cS_o = 416 cm4).

Let's determine the number of turns of the primary winding according to the modified formula (18.10):

Determine the number of turns of the secondary winding:

The primary winding is wound with a wire with a cross section:

The secondary winding is wound with a wire with a cross section:

Let us determine the length of the non-magnetic gap:

Let's determine the resulting inductance of the primary winding of the inductor:

The inductance turned out to be more than necessary. To obtain the required inductance, we reduce the number of primary windings to Wt \u18d 2. Accordingly, W90 \u5d 2 turns and XNUMX \uXNUMXd XNUMX mm.

Choke Calculation Example #3

Let's calculate the inductor L2 ERST. The maximum inductor current is 315 A, the minimum is -10 A.

The current ripple frequency in the inductor corresponds to the PWM frequency and is equal to F_PWM = 25000 Hz.

Let us determine the parameters of the inductor necessary to ensure the continuity of the welding current. On fig. 18.25 shows the shape of the current in the inductor L2, corresponding to the boundary of continuity.

How to Calculate a Core Choke
Rice. 18.25. Current shape corresponding to the continuity boundary

During the open state of the ERST key, the current in the inductor increases from zero to the amplitude value. Further, during the pause, the current decreases to zero. The danger of going beyond the boundaries of continuity exists at a minimum welding current I_{sv min} = 10 A and the maximum input voltage ERST. Determine the arc voltage for the minimum welding current:

Let us determine the relationship between the amplitude and average value of the triangular current. The mean value of a function is the integral of this function, or, simply put, the area bounded by this function and the zero level line.

The area of a triangle is defined as the product of the height of the triangle and half the length of the base:

From here we find the relationship between the average and amplitude values of the current:

If the key is open, then voltage is applied to the throttle:

The current in the inductor increases from 0 to I_a.

During a pause, voltage -U is applied to the throttle_{d min}, and the current in it decreases to 0.

Since the change in current () in both cases will have the same value, but a different sign, then

Let's say that as the material of the inductor core, we intend to use electrical steel with a sheet thickness of 0,08 mm, which at a frequency f_y = 1000 Hz, at induction B_y = 1 T and a rectangular voltage has losses P_y = 22 W/kg.

Frequency indicators of steel α = 1,4 and β = 1,8. Let's find the permissible range of induction for a frequency of 25000 Hz, which will provide the same level of losses as at a frequency of 1000 Hz:

Let us first determine that the induction in the core for direct current can reach B = 1,42 T, current density J = 3,5 A / mm2, K_o = 0,35 and K_c = 0,10. Find the overall size of the core:

The size fits the core ШЛ25х50 (S_c = 12,5 cm2, S_o = 16 cm2). Core size S_cS_o = 12,5 x 16 = 200 cm4.

Let's determine the number of turns:

The winding is carried out with a copper bus with a cross section:

Let's define the non-magnetic gap:

Let's define the resulting inductance:

Now you should make sure that the amplitude of the high-frequency induction ripple does not exceed ΔB = 0,16 T

The maximum range of induction in the core of the inductor takes place at the maximum input voltage U_{in max} = 80 V and pulse duty D = 0,5, and can be found by the formula:

which does not exceed the allowable value.

Author: Koryakin-Chernyak S.L.

See other articles Section welding equipment.

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