Harmonic control plan for small current loads

The AHF produced by Xi’an CoEpower Electric is mainly aimed at 400V , and the minimum AHF capacity is 30A.The load it targets is mainly a load current with a harmonic content of 10% to 30%, which means that the load current is 50A or above. It will have better effect.

If the load current is relatively small, for example, when the load current is smaller than the AHF capacity, the filtering effect will not be obvious, and the waveform will even become worse.

This is because when the AHF is working, its output current will always have a certain small proportion of ripple while controlling harmonics. This ripple proportion is relatively small. When the load current is relatively large, the ripple can be ignored, but when When the load current is very small, the ripple ratio will be very objective, resulting in little improvement in the waveform or even deterioration.

In this regard, Xi’an CoEpower Electric Co., Ltd. has adopted a new scheme to compensate for harmonics of small load currents.

An Introduction:

1. Main principles

The compensation principle of AHF is as shown in the figure above. The AHF outputs and load harmonics and other large reverse harmonic currents to offset the load’s harmonic current.

Among them, IAHF=Ih+Iw, Iload=IF+Ih, Ih represents the load harmonic current, Iw represents the ripple current caused by the AHF output, and IF represents the fundamental wave current of the load. Is=Iload-IAHF=IF-Iw.

Therefore, when the load current is large enough, that is, IF is large enough, the proportion of Iw in Is can be ignored, but when the load current is too small, the negative effect of Iw is more obvious.

Usually the content of Iw is about 3% of the total capacity of AHF.

For example, if the load is 10A and the AHF installation capacity is 30A, then the ripple 3% is 0.9A. At this time, the ripple relative to the load ratio is 0.9/10A=9%. In the same way, for example, if the load is 10A and the harmonic content of the device is 20%, then the harmonic current is 2A and the ripple is 0.9A. Even if we completely filter out the 2A harmonic, the ripple is caused by the APF itself. And it cannot be completely eliminated, and the final harmonic content will still be no less than 9%. In the same way, for example, if the load harmonic content is 5%, the harmonic content will actually increase after AHF operation.

2. Plan ideas

After the above analysis, for small current load currents, AHF cannot effectively reduce the harmonic content index, mainly due to the ripple current and sampling error of the AHF, and the harmonic demand of small current loads is very small, for example, within 5A. In this regard, we can use a transformer to reduce the current of a conventional APF in equal proportions. For example, if we reduce the current of a 30A AHF by 10 times, we will get a 3A AHF with a ripple content of 0.09A, and the impact on the load can be ignored.

3. Concrete implementation

The specific implementation is as shown in the figure below. The AHF capacity is configured as 30A. A 10:1 transformer is connected in series to the AHF output. The primary side of the transformer is connected to the system 230V and the secondary side voltage is 23V. When the apf outputs a rated current of 30A, there will be a voltage on the primary 230V bus. A current of 3A is generated. Through this 3A current, we can adapt to the harmonic current of the load by compensating it.

4. Several issues that need attention

A The transformation ratio of this transformer needs to be determined, generally 10 times is selected, and the rated capacity of the transformer is 1.5 times the operating capacity. For example, if 3A management capacity is required on the bus side, then the operating capacity is 3A*230V*3 phases = 2070w, then a 3kva transformer needs to be actually used.

B In order to ensure that there is no phase difference between the output current of the AHF and the output current of the bus side, the wire system of the transformer is Yyn0, and the midpoints of the primary and secondary of the transformer need to be short-circuited.

C Since the working voltage of AHF has been reduced by 10 times, the parameters of AHF need to be modified, including operating parameters, protection parameters, etc., which need to be set correctly according to the voltage.

D Since the AHF output current is reduced by the transformer, the actual transformer ratio needs to be enlarged by 10 times during setting. For example, if we use a 20:5 transformer, we need to set it to 200:5, so that the AHF will output according to 10 times the sampling value, and after being reduced by 10 times, it will just adapt to the harmonics of the load.

E In this solution, the AHF transformer must be hung on the load side.

F Since the AHF working port voltage is reduced by 10 times to 23V phase voltage, this voltage level cannot allow the AHF control power supply to operate normally. Therefore, an external switching power supply is required. The switching power supply takes power from 230V and outputs 24V to the power supply interface inside the AHF. The power of this power supply requires 150W.

5.The following is a case

The load current is 5A and the harmonic content is 31%. Since the current is too small, the current loop range of the power analyzer is 500A. The error of testing the load current of 5A is too large to see the waveform shape clearly, so use an oscilloscope to observe the current waveform.

When using standard APF treatment, due to the influence of ripple noise, the waveform is very messy and cannot be observed at all. The effect of the treatment cannot be seen, and it has a counterproductive effect on the current waveform.

After the final transformation, a 10:1 transformer was used to reduce the compensation current by 10 times. It can be seen that the compensated current waveform has been greatly improved.

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