Analysis of Transformer Loss Measurement

1.Introduction

With the development of China's modernization, transformer is increasingly applied in each scientific research field. Large-scale power grid as well as electronic products adopt various transformers. In addition, production equipment in all industrial sectors is increasingly utilizing transformer due to electrification improvement. No mater what circuit or power grid is, they demand the assistance of high-quality and stable transformer. The transformer quality and working efficiency directly determine work quality of other electric equipment[AC Dielectric Test System]. Hence, transformer has been attracting more and more attention.

2.Working Principle of Transformer

The transformer is a device which can exchange AC voltage, current and impedance. When AC current passes through primary coil, AC magnetic flux generates in iron core , which the voltage or current appears in the secondary coil. The transformer is composed of iron core (or magnetic core) and coil. The coil has two or above two winding. The winding for connecting power source is called primary coil while the rest is called secondary coil.

Basic principle of transformer is electromagnetic induction. Take single-phase double-winding transformer as an example to illustrate basic principle. When primary winding is applied voltage U and current I, alternating flux ?will generate in the iron core, which is called main flux. Under its force, induced potential of both sides winding is Eand E The formula of induced potential:

E = 4.44fNΦ

In the formula:

E - effective value of induced potential

f - frequency

N - number of turns

Φ- maximum value of main flux

Because number of primary side turn is different from secondary side, Eand Eare not the same. If voltage drop of inner impedance is omitted, voltage Uand Uare not the same.

When secondary side of transformer is no load, current Iof main flux passes through primary side. This current is named as magnetizing current. When secondary side and load pass through load current I, flux also generates in the iron core to alter main flux. However, if one-time voltage does not change, main flux remains the same. Two currents will pass through primary side. One part is magnetizing current Iwhile the other is used to balance I. Hence this current varies with I. Current multiplying number of turns is magnetomotive force.

As a matter of fact, the above balance effect is magnetomotive force. The transformer realizes energy transmission of primary and secondary sides through magnetomotive force balance.

When one sine-wave AC voltageUis applied on both sides of primary coil, alternating current Iand alternating flux Φ generate in the wire. The alternating flux passes through primary coil and secondary coil along iron core, forming enclosed magnetic circuit. In the secondary coil, mutual inductance potential Uis induced out. Meanwhile, Φwill also induce out one self-induction potential E in the primary coil. Eand applied voltage U are in the opposite direction while their amplitude is close, thus restricting the value of I. In order to remain flux Φ, there needs a certain electric energy consumption. In addition, the transformer has some dissipation of its own. Although secondary stage is not connected with load at this time, there is still certain current in the primary coil, which is called “no-load current”.

If secondary stage is connected with load, secondary coil will generate current Iand flux Φ. Φand Φare in the opposite direction. They are offset with each other, causing a  fall in total quantity of flux in the iron core. Therefore, primary self-induction potential E falls but Iincreases on the contrary. We can conclude that primary current is closely related to secondary load. When secondary load current increases, Iand Φincrease as well. In addition, increasing part of Φfills the gap of flux, which is offset by Φ, to keep the total quantity of flux the same in the iron core. If the loss of transformer were neglected, we can assume that consumed power of secondary load of one ideal transformer is the electric power that primary side takes from the power source. The transformer can change secondary stage voltage via altering the number of secondary coils but cannot change the power of allowable load consumption.

Set the resistance of primary winding is R; electromotive force of magnetic flux leakage is E. We can obtain voltage balance equation of primary and secondary winding:

 

 

(Z= R+ jX leakage impedance of primary winding)

When the transformer is from no-load to load, applied voltage U does not change while the current changes from I to I. Hence, voltage balance equation of primary and secondary winding is as follows when the transformer operates in load:

 

Set Z=R+jX

Zis called inner impedance of secondary winding; Iis secondary current; we can get the voltage balance equation of load secondary winding:

 

 

3.Transformer Loss

After primary winding of transformer is on, flux generated by coils is flowing in the iron core. Because iron core itself is conductor, the flatness vertical to magnetic line will induce potential, which forms enclosed circuit on iron core section and generate the current. It is like a vortex so this current is called “vortex current”. The vortex current increases transformer loss, makes iron core of transformer heat and improves the transformer temperature. In the iron core, “magnetic hysteresis” will cause power loss. The loss resulted from “vortex current” and “magnetic hysteresis” is named as “iron loss”. In addition, producing the transformer needs large quantities of copper wires. These copper wires have the resistance. When the current passes, these resistance will consume certain power. This part of loss always becomes the heat and is consumed. We call this loss as “copper loss”. Therefore, so transformer loss is mainly composed of iron loss and copper loss.

According to the law of conservation of energy, electric power Pshould be equal to loss power (iron loss Pand P) plus output power P of secondary winding.

P= P+ P+ P

4.Transformer Loss Measurement Test

4.1 Test Items

4.1.1 Measuring iron loss of transformer - P

4.1.2 Measuring copper loss of transformer - P

4.1.3 By virtue of changing the number of secondary load lights of transformer, we can obtain data, draw the efficiency curve and the curve of relation between voltage and current of transformer secondary side.   

4.2 Test principle

• The no-load test can be used to measure the iron loss of transformer - P, which is shown in Fig.1. Make secondary winding of transformer in open circuit and apply rated voltage Uon primary winding. When it is no-load, secondary output power of transformer is zero and no-load current is also low. The copper loss Pcan be neglected. Hence, the power is iron loss P.
  

               Fig.1

• When the transformer is in rated load, the loss is called full copper loss, which is expressed by P. The short circuit test can be used to measure the copper loss of transformer - P, which is shown in Fig.2. Make secondary winding of transformer short circuit; connect primary winding to regulating transformer; arise the voltage of power source to make the current reach rated value I. At this time, displayed value on the voltmeter is called short circuit voltage. Based on Fig.2, when secondary side is short circuit and rated current passes through primary side, the primary voltage is voltage drop of short circuit impedance. Therefore, the iron loss is low and can be neglected and displayed value on the power meter is full copper loss P.
  

Fig.2

• We connect the circuit based on Fig.3 and ensure that primary applied voltage is always rated voltage. At this point, number of secondary side load lights is altered to measure many groups of data. The data mainly includes primary power of the transformer -P, current I, secondary power of the transformer - P, current Iand secondary voltage U. The transformer efficiency ? can be calculated. ? =. In addition, two curves are drawn based on obtained data.
  

Fig.3

5.Test Data

The data of JMB transformer is as follows via above tests:

Iron loss P= 32W;

No-load current of primary side I = 0.78A;

Full copper loss P = 30W;

Short circuit voltage U = 10V

Many groups of data are obtained according to test principle 3 , which is shown in Tab.1. 

We draw transformer efficiency curve, which is shown in Fig.4 and relation curve between transformer secondary side voltage and current, which is illustrated in Fig.5

Fig.4

Fig.5 Curve of relation between secondary-side voltage and current

Tab.1

P: primary power

I: current

P: secondary power

I: current

U: secondary voltage

?: efficiency  

? =P/P

6.Conclusions

The method presented in this paper is the most common method to measure the transformer in the lab and daily life. When using the transformer, please keep the following in mind: suitable measuring device should be selected according to rated voltage, rated current and turn ratio of measured transformer; when measuring transformer power curve, select suitable secondary side load device and estimate the quantity of primary and secondary electricity to choose measuring instrument with reasonable range and avoid damaging the instrument. In addition, input voltage should be no higher than rated voltage of the transformer; otherwise, its inner winding will be broken. Please carefully examine the circuit connection before the test in order to protect the instrument. Be sure that the circuit is correctly connected. If measured transformer belongs to high-power one, radiator must be installed to avoid overheating and damaging the transformer. With the continuous development of scientific technologies, higher requirement is put forward for designing the transformer, which needs us to continuously make conclusions in the practices.

 

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