# Lowering transformer 220 to 36 volts

Transformer 220 to 36 volts idealfor lighting circuit supply in bathhouses, saunas, bathrooms, cellars. Safety requirements impose some restrictions on the use of AC voltages of 220 volts in rooms with high humidity. Therefore, the ideal way out of this situation is the use of alternating current of 12, 24 or 36 volts. If there is a person's contact with the electrical network, he will feel only a faint blow. It will not do any harm to the human body.

## Transformer construction

Like any other, the step-down transformer from 220 to 36 volts consists of three main components:

- Primary winding.
- Secondary winding.
- Magnetic wire.

Both the primary and secondary windings consist ofa certain number of turns of a copper wire. Usually it is the copper wire used in lacquered insulation, as it is much better in characteristics than aluminum. If you use aluminum, it will require several times more, which significantly increases the dimensions of the transformer. True, aluminum wires in lacquer insulation used to be used in voltage regulators. The magnetic line can be made both of transformer steel and of a ferromagnet. This material is much better than any metal.

## Power and transformation ratio

You can visually even assess the power of anytransformer - the larger the dimensions, the higher it is. But for exact calculation of power it is necessary to use special formulas. The simplest method of calculating the power of a transformer is to multiply the voltage of the secondary winding by the current in it. Get the real power value of the transformer under investigation. To work on creating and designing such a device, you need to know one more basic parameter that characterizes the transformer.

This is nothing more than a coefficient of transformation. It is the ratio of the number of turns of the secondary winding to the primary winding. The same value can be obtained by dividing I2 / I1, as well as the voltage U2 / U1. In any of these three cases, you will get the same value. You may need it when calculating the exact number of turns for the primary and secondary windings.

## Calculation of the transformer

If it is necessary to manufacture a transformer 220 for36 volts (1000 watts), it is desirable to use the formula to calculate the power in the secondary winding. It was mentioned above, the power is equal to the product of the current to the voltage. In this case, there are two parameters that are known to be known: this is directly the power P2 (1000 W) and the voltage in the secondary circuit U2 (36 V). From this formula it is now necessary to calculate the current that flows through the primary circuit.

One of the important parameters is the coefficienta useful action that transformers do not exceed 0.8. It shows how much power directly consumed from the network goes into the load connected to the secondary winding (in this case it is only 80%). The difference in power goes to heating the magnetic circuit and windings. It is lost, and irrevocably. The power that is consumed from the AC network is equal to the ratio of P2 to the efficiency.

## Transformer magnetic line

All power is transferred from the primary winding tosecondary by means of a magnetic flux, which is created in a magnetic core (core). It is from the power P1 that the cross-section area of the core S depends. Most often, a set of plates in the shape of the letter "SH" is used for the core. In this case, the cross-sectional area is equal to the product of the square root of P1 by a factor of 1.2. Knowing the area value, you can determine the number of turns W to 1 V. To do this, you need to divide 50 by area.

Voltage in the primary and secondary windingsknown - this 220 and 36 volts. The number of turns for each winding is determined by multiplying the voltage by W. In the event that the decimal values are obtained, it is necessary to round them up. It should also be taken into account that when the load of the secondary circuit is connected, a voltage drop occurs. For this reason, it is desirable to increase the number of turns by about 10% of the calculated one.

## Winding wires

And now you need to calculate the current in the primaryand secondary windings. The current is equal to the ratio of power to voltage. If a transformer 220 is made at 36 volts (500 watts), then a current equal to the ratio of 500/36 = 13.89 A will flow in the secondary circuit. The power in the primary circuit will be 625 W, and the current will be 17.36 A.

Next, the current density is calculated. This parameter specifies the current value for each square millimeter of the wire cross-sectional area. Typically, transformers assume a current density of 2 A / sq. mm. The diameter of the wire needed for winding can be determined by a simple formula: the efficiency multiplied by the square root of the current. Therefore, in the secondary circuit it is necessary to use a wire whose diameter will be equal to 0.8 to 3.73, which is 2.9 mm (rounded to 3 mm). In the primary winding, a wire with a diameter of 3.33 mm should be used. In the event that you do not have wires with the correct diameter, you can use a simple trick. Wind the wire simultaneously with several wires connected in parallel. In this case, the sum of the cross sections should not be less than the one that was calculated by you. The cross-section of the wire is equal to the ratio of the efficiency to the square of the diameter.

## Conclusion

Knowing all these simple formulas, you canindependently produce a reliable transformer that will work in perfect mode. But you still need to know how to connect a 220 volt transformer to 36 volts. In this there is nothing complicated, it is enough to connect the primary winding with the 220 V alternating current network, and the secondary one with the load, the lighting system, for example. When starting for the first time, try to connect the transformer with the maximum power load to determine if the core and windings are overheated.

## Comments (0)

## Add a comment