Inductors, Difficult Finishes of Transformers

A group of friends in the msPLC/msOS group mentioned that for inductors, especially magnetic materials, there is no way to start. Indeed, many people learn about power supplies. The first thing that comes across is the inductance problem. Compared to capacitors, there are many standard parts, and they are slightly larger. A little inductance, you need to do it yourself, especially transformers. During the university period, the book of knowledge about inductance was just a matter of fury. It couldn't be used at all. Here's a look at inductors, where the difficulties of transformers are.

The energy storage formula of the inductor W = 1/2*L*I^2.

This formula is essentially equivalent to the form of the energy storage capacitor: W = 1/2*C*U^2.

Think about it again, and everyone will find that the kinetic energy formula of the mechanics of moving with high school students is the same: E = 1/2*M*V^2.

Since there is kinetic energy, there is potential energy, the potential energy of the spring, E = 1/2*K*X^2. In the above four formulae, the first two capacitive inductances are the electromagnetic energy formulas, and the latter two are the formulas of the material mechanical energy.

Everyone knows that in mechanical energy, the conversion of kinetic energy and potential energy is vibration, and the conversion of kinetic energy and potential energy is a mechanical wave, such as water waves, vibration waves, sounds, and so on. Waves are a basic way of energy transfer. Unfortunately, the study of waves in college physics, many second- and third-tier colleges have not conducted in-depth development, so many people do not understand waves.

Similar to the mechanical energy of materials, the container of electric energy is capacitance, the container of magnetic energy is inductance (the exact name should be magnetic capacitance), the mutual conversion of electric energy and magnetic energy is LC oscillation, and the conversion of electric energy and magnetic energy is electromagnetic waves. In addition, everyone should note that all these formulas have a 1/2 in front of them. This is very interesting. For example, when charging a capacitor at a constant voltage, the capacitor can only get 1/2 of the power supply end, and half the loss. On the resistance, this rule is also mechanically. The typical case is that when the car starts, it consumes oil, and after uniform speed, the fuel consumption is the lowest. At the start, the speed of the car was 0, and the burning gas did work on the piston. These work were almost converted to heat energy wasted. So if you want to reduce the loss, you need to reduce the gear, start slowly, let the difference between the piston speed and the car speed drop (the middle of the reducer and clutch, start by relying on the clutch to achieve speed drop damping match, similar to the RC charging circuit resistance, Damping voltage difference). The common indicators of capacitance are capacity and withstand voltage, followed by loss, frequency characteristics, loss and frequency characteristics, and have a great relationship with dielectric materials.

Inductance also has similar indicators, inductance, current resistance, followed by loss, frequency characteristics, and the same reason also has a great relationship with magnetic materials. Inductance, previously said, should belong to the magnetic capacity, contain magnetic things, compared to the capacitor to accommodate the electric field, the inductor is to accommodate the magnetic field, according to the Ampere loop law, the current is magnetic, because there will be a magnetic field generated by the current, so with the capacitance, Inductors store magnetism, which is the current.

The formula for the charge of the capacitor is Q = C*U, and the formula for the flux of the inductor is Φ = L*I. (In multi-turn coils, divide by turns, Φ = L * I / N)

The amount of electric charge in the capacitor must not increase unrestrictedly. If too much, the voltage rises, causing dielectric breakdown E = U / d

The magnetic flux of the inductor can't increase unrestrictedly. If too much, the magnetic pressure will increase, which will lead to the magnetic saturation of the magnetic material. This is the difficulty of inductance learning. Many inductors are developed around the inductance and magnetic saturation. The amount of inductance and magnetic saturation is a pair of contradictory bodies. Next, the core talks about this pair of contradictory bodies. Let's take a magnetic ring inductor as an example.

The formula of inductance: L = kN^2, k is the coefficient, N is the number of turns, and the inductance is proportional to the square of the turns of the coil. Note that it is proportional to the square.

The formula for magnetic flux: Φ = L*I/N = k*N*I. The magnetic flux is proportional to the current and the number of turns.

Inductance is proportional to the square of the turns, and the magnetic flux is proportional to the number of turns. After determining the shape, different magnetic materials, the maximum magnetic flux has a limit. In order to unite to the material, regardless of the shape, the magnetic flux density is generally used, that is, magnetic induction. Intensity: B = Φ / S, which is the magnetic flux per unit area. Therefore, different materials, the maximum magnetic induction intensity is different, such as ferrite materials, manganese zinc, nickel zinc, the maximum magnetic saturation is 0.5T, which is 5000 Gauss, generally less than 0.3T, iron core magnetic saturation of 1T.

The inductors we make ourselves need to meet the following conditions:

1, the amount of inductance;

2, magnetic induction intensity can not reach saturation;

3, volume, heat.

All problems with the inductor are based on this development. To design the inductor, we always want to use the smallest magnetic ring, the shortest copper wire to achieve the goal, and there is no heat. If you want to achieve this, then the magnetic material loss must be low, in addition to the shorter copper lines, copper heat is also less, especially at high frequencies, copper heat because of the skin effect.

Based on the above conditions, high permeability and low loss MnZn or NiZn materials are generally used. MnZn is commonly used in PC40 materials, and the magnetic permeability is 2500. Nickel and zinc are suitable for higher frequencies, but the magnetic permeability is low. Is 800. With high magnetic permeability, it is easy to obtain high inductance, so it is very suitable for high-frequency transformers. Now the switching power supply basically works at 100KHz, and the transformers are made of manganese and zinc. However, in some cases, such as transformer rectification output, the need for inductive filtering, this time there are both AC and DC power, high permeability, it is easy to make the DC lead to magnetic saturation, similar to a horn, no sound, it should be The center of the vibration of the voice coil, and because of the existence of direct current, deviates from the center, so that the range of AC fluctuations is greatly reduced. So this time we need to adjust the strategy. First we must choose a magnetic core with high saturation. For example, the magnetic saturation of iron powder core can reach 1T, which is higher than that of MnZn, and secondly, the magnetic permeability of iron powder core is low. At 60 or even lower, the magnetic permeability is low, and the number of windings around must be higher in order to achieve the desired inductance. Perhaps we will ask, the inductance is low, the number of turns is more, and magnetic saturation is also required. Ah, I don't know if you noticed that there is none at all. The inductance is proportional to the square of the number of turns, and the magnetic induction is proportional to the number of turns, that is to say, as the number of turns increases, the inductance will soon reach what we want. The desired value, but the magnetic induction did not reach saturation. Using the square of the inductance and the number of turns, and the linear relationship between the magnetic induction and the number of turns, it is very important to achieve a good balanced inductance and magnetic saturation. Like the current high-power power supply has PFC in the front section, that is, power factor calibration, all need to use the inductor for boosting. The current flowing through the inductor is AC-DC composite, so iron powder core material is generally better. An iron powder core is named FeSiAl with a permeability of around 125 and a low calorific value. It can be said that most of the current switching power supply on the market, transformers generally use high permeability ferrite materials Mn-Zn, and step-up, buck filtering, with a DC component are all with a low permeability iron powder core material.

Electric Furnace Transformer

Resistance furnace transformer
Electric resistance furnace and salt bath furnace for heating, heat treatment of mechanical parts, powder metallurgy sintering, non-ferrous metal smelting, etc. Because the resistance of the heater is too small, or the resistance of the heater changes too much during the heating process, a resistance furnace transformer needs to be equipped between the furnace and the power grid to reduce and adjust the input voltage of the furnace.

The resistance furnace transformer and salt bath furnace transformer with small capacity and low voltage are mostly dry-type transformers, with a box shell and natural cooling; Resistance furnace transformers with medium capacity (hundreds to thousands of volt amperes) are mostly oil immersed self cooling transformers; Large capacity is forced oil circulating water-cooled transformer.


Electric arc furnace transformer
A special transformer used to supply power to electric arc furnaces for iron and steel smelting. Large capacity, complex structure and high technical requirements. Its secondary side voltage is low, generally from tens of volts to hundreds of volts, and it is required to be adjustable in a large range; The secondary side current often reaches thousands to tens of thousands of amperes. In addition, in iron and steel smelting, large power is required during the melting period, and the transformer is required to have 20% overload capacity within 2 hours. In the process of steelmaking, electrode short circuit is easily caused by the collapse of furnace charge, so the primary side of the electric arc furnace transformer should be connected to the current limiting reactor in series or have a large impedance to limit the short circuit current.

Special transformer for intermediate frequency furnace
The special transformer for medium frequency furnace produced by the company is applicable to medium frequency electric furnace below 30 tons. The rated voltage at the primary side can be divided into 10KV or 35KV. The rated voltage of the secondary side can be divided into: 400V, 600V, 660V, 690V, 700V, 720V, 750V, 800V, 850V, 900V, 950V, 1000V, 1050V, 1100V, 1250V, etc. Rectification mode can be divided into 6 pulse, 12 pulse, 24 pulse or more. The current popular rectifier string, inverter string, rectifier parallel, etc

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