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How to design an output sine wave filter based on a frequency converter? [Copy link]

How to design a sine wave filter based on the power and carrier frequency of the inverter? Taking the inverter with rated power of 100kW, rated voltage of 400V, rated current of 200A and carrier frequency of 4K as an example, how to choose the capacity of the reactor and capacitor?

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Obviously, RC filtering cannot be used because the loss on the resistor is too large. Active filtering cannot be used because there is no amplifier with such a high power. As a result, only LC filtering is available. When using LC filtering, it is necessary to pay attention to the frequency response curve of the filter not having too large a peak, that is, the quality factor of LC must be controlled.   Details Published on 2024-9-16 11:39
 
 

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[Take the inverter rated power of 100kW, rated voltage of 400V, rated current of 200A, and carrier frequency of 4K as an example, how to choose the capacity of the reactor and capacitor? 】

The rated voltage is 400V, the rated current is 200A, how did you get the rated power of 100kW?

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The rated current is twice the rated power.  Details Published on 2024-9-15 14:12
 
 
 

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maychang published on 2024-9-15 14:08 [Take the inverter rated power of 100kW, rated voltage of 400V, rated current of 200A, and carrier frequency of 4K as an example, how to choose the capacity of the reactor and capacitor? 】 ...

The rated current is twice the rated power.

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[It is derived from the fact that the rated current is twice the rated power] Current and power are not the same physical quantity. The unit of current is ampere, and the unit of power is watt. There is absolutely no multiple relationship between the two. You buy rice by the pound, and you buy cloth by the foot. There is no multiple relationship between the two either.  Details Published on 2024-9-15 14:23
 
 
 

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[The rated current is twice the rated power]

Current and power are not the same physical quantities. The unit of current is ampere, and the unit of power is watt. There is absolutely no multiple relationship between the two.

Rice is sold by the pound, and cloth is sold by the meter. There is no multiple relationship between the two.

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The rated output current is twice the rated power and is designed as  Details Published on 2024-9-15 14:47
 
 
 

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maychang posted on 2024-9-15 14:23 [It is derived from the fact that the rated current is twice the rated power] Current and power are not the same physical quantity. The unit of current is ampere, and the unit of power is...

The rated output current is twice the rated power and is designed as

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[In terms of numerical value, the rated output current is twice the rated power and is designed accordingly] Electricians often say "two currents for one kilowatt", right?  Details Published on 2024-9-15 15:04
 
 
 

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Luan Shi Zhu Jiu Lun Tian Xia published on 2024-9-15 14:47 The numerical value is 2 times, and the rated output current is twice the rated power and designed

[The rated output current is designed to be twice the rated power based on the numerical value of 2]

Electricians often say "two currents for one kilowatt", right?

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Yes   Details Published on 2024-9-15 15:22
 
 
 

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maychang posted on 2024-9-15 15:04 [In terms of numerical value, the rated output current is twice the rated power and designed according to 2 times] Electricians often say "one kilowatt has two currents" ...

Yes

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Let's not talk about "one kilowatt, two flows" nonsense, let's talk about filter design first. The 8th post said it very accurately: maintain the transmission efficiency of the fundamental wave (sine wave) and reduce the high-frequency harmonics in the inverter output. What you need is a low-pass filter. The carrier frequency must be outside the band of the low-pass filter, and the attenuation must be as large as possible. You need to convert the frequency  Details Published on 2024-9-16 11:39
Let's not talk about "one kilowatt, two flows" nonsense, let's talk about filter design first. The 8th post said it very accurately: maintain the transmission efficiency of the fundamental wave (sine wave) and reduce the high-frequency harmonics in the inverter output. What you need is a low-pass filter. The carrier frequency must be outside the band of the low-pass filter, and the attenuation must be as large as possible. You need to convert the frequency  Details Published on 2024-9-16 11:36
 
 
 

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It is important to consider that the goal of the filter is to reduce the high-frequency harmonics in the inverter output while maintaining the transmission efficiency of the fundamental wave (sine wave).

 
 
 

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Without further ado about the [one kilowatt, two flows] nonsense, let’s first talk about filter design.

The 8th floor said it very accurately: maintain the transmission efficiency of the fundamental wave (sine wave) and reduce the high-frequency harmonics in the inverter output.

What you need is a low-pass filter. The carrier frequency must be outside the band of the low-pass filter, and the attenuation must be as large as possible. The power frequency you want the inverter to output must be within the band, and the attenuation must be as small as possible.

 
 
 

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Obviously, RC filtering cannot be used because the loss on the resistor is too large. Active filtering cannot be used because there is no amplifier with such a high power. As a result, only LC filtering is available. When using LC filtering, it is necessary to pay attention to the frequency response curve of the filter not having too large a peak, that is, the quality factor of LC must be controlled.

 
 
 

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