How to calculate the leakage inductance of a welding transformer. Scheme, description

ENCYCLOPEDIA OF RADIO ELECTRONICS AND ELECTRICAL ENGINEERING
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Electric welding. How to calculate the leakage inductance of a welding transformer. Encyclopedia of radio electronics and electrical engineering

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Transformer leakage inductancemainly depends on:

from the relative position of the windings;
from the configuration of the windings;
from extraneous factors, such as objects made of magnetic materials close to the transformer.

How to Calculate the Leakage Inductance of a Welding Transformer
Rice. 18.6 Pattern of the distribution of force magnetic fields in a rod transformer with disk windings

Unfortunately, an accurate calculation of the leakage inductance is practically impossible. Usually, in practice, the calculation is carried out by the method of successive approximations with the refinement of winding and design data on a practical sample. Let us develop a method for calculating the leakage inductance of a transformer with disk windings.

On fig. 18.6 schematically shows the pattern of the distribution of force magnetic fields in a rod transformer with disk windings. Here is a schematic representation of the total magnetic flux Ф and winding leakage fluxes - Ф_s1 and F_s2. These flows arise due to ampere-winds.

created by the windings of the transformer.

In the core window, the current of the primary winding of the transformer is directed away from the viewer, and the current of the secondary winding is vice versa. Due to this, for leakage fluxes, the primary and secondary windings are a kind of gap solenoid δcalled main scattering channel (hereinafter referred to as the channel).

In this channel, the main leakage fluxes of the primary and secondary windings pass. Due to the fact that the transformer windings are not concentrated at a point, but are distributed in space in a certain way, part of the leakage flux passes inside the windings. When shifted to the edge of the winding, the leakage flux weakens, since it is created by smaller ampere-turns of the winding (Fig. 18.6).

The stray fluxes in the channel between the coils, as well as inside the coils, are summed up and create a common stray flux. To determine these components, we take a number of assumptions.

Assumption 1. Since the core of the transformer has a very small magnetic resistance, we will assume that all ampere-turns of the windings are applied to the leakage channel.

Assumption 2. We will accept the same assumption for sections of the coils located on the outer side of the core, because outside the channel between the coils, the magnetic flux closes through a space that has an incomparably larger cross section and, consequently, much less resistance. This assumption will lead to a somewhat overestimated calculated value of the leakage flux, which can subsequently be taken into account by introducing a correction factor.

Let us determine the leakage flux created by the secondary winding in the leakage channel δ. To simplify our work, we will assume that the transformer windings have an equal number of turns and the transformation ratio n=1.

Assuming that the power lines of one winding occupy half of the channel, we find its magnetic resistance for one secondary winding:

where: F - winding ampere-turns, A; Ф - magnetic flux, Wb; H - magnetic field strength, A/m; c - channel length, m; S - channel area, m2; B - magnetic induction, Tl.

Tension and magnetic induction are interconnected through absolute magnetic permeability of a substance

which, in turn, is equal to the product

where - magnetic constant (vacuum permeability); μ - relative permeability of the medium.

Since for air that The area of the channel can be found by the formula:

where p is the channel perimeter, m.

Substituting the obtained values into the formula for magnetic resistance, we get

Magnetic flux in the channel for one winding can be found by the formula:

where w is the number of turns of the winding; I - current in the winding, A.

Flux linkage of one winding with flux in the channel can be found by the formula:

To calculate the flux linkage with the flow passing through the thickness of the secondary winding, we select a power tube with a width dx (Fig. 18.6) and a length equal to the average length of the winding turn. Its magnetic resistance can be found by the formula:

Highlighted coils:

In this way, tube flow equals:

А tube flux linkage is

General flux linkage of similar tubes according to the width of the secondary winding δ₂ will

General winding flux linkage can be found by summing the flux linkage in the channel and the flux linkage in the thickness of the winding:

Dividing the flux linkage by the current, we get secondary leakage inductance:

Leakage inductance of the primary winding, reduced to the secondary:

Total leakage inductance, reduced to the secondary winding:

For sections of the coils located on the outer side of the core, the leakage flux closes through the space, and therefore the real leakage inductance turns out to be less than the calculated one by about 30%:

Author: Koryakin-Chernyak S.L.

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