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Suppose a induction motor is connected through a inverter. Inverter is fed from a DC-DC boost converter. My question is: to boost converter how load current will look like? Will it look like DC or AC?

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The Current will most likely flow in one direction as pulsating DC throughout the whole system. I'll assume a basic 3-Phase inverter and simple DC-DC boost converter for thinking this through. I also am assuming the converter is connected directly to the inverter's DC bus.

The current through the induction motor will be roughly sinusoidal cycling at what ever frequency is appropriate for the speed the motor is supposed to be turning. The current flowing out of the inverter's DC bus will be pulsating DC, cycling at a rate 6 times higher then the frequency of the inverters output (The increased frequency is due to the three switching branches feeding the motor's phases).

The current flowing out of the inverter's DC bus will lower the DC bus voltage. The DC-DC converter will act to prevent that dip in voltage. The converter does this by supplying the needed current to keep the voltage steady into the inverter. The only way to keep the inverter's DC bus voltage stable is too feed into it whatever current is flowing out.

Assuming the DC-DC converter is able too it will supply pulsating DC current into the inverter.

EDIT: The inverter output voltage will switch at whatever switching frequency the inverter's PWM frequency is set to, typically somewhere between 1k Hz - 40k Hz. The motor will act as a low pass filter due to its high inductance, which results in a lower frequency sinusoidal current waveform.

In short the inverter's output voltage will switch at the PWM's high frequency, and the current will more smoothly oscillate at 50Hz. The 50Hz 3 phase motor current will look like pulsating DC at 300 Hz to the DC-DC converter.

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I hope this helps.

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  • \$\begingroup\$ Thank you Clipboard, I appreciate your answer.. So 50 hz wave will produce 300 hz dc ripple as you say. I am doing space vector modulation. Here inverter is considered a single unit which will produce sample in a voltage space. Let me tell you, I will produce 48 sample during a full vector rotation of 360 degree. Inverter has 8 standard vector according to switching pattern. The vectors are (100),(110),(010),(011),(001),(101),(111),(000).1 means upper switch on,where 0 means off.So,111 and 000 will not produce any active vector.Other 6 are active vector. \$\endgroup\$ – Zarzisur Jan 19 at 20:01
  • \$\begingroup\$ In a sample making time, two of active vector and one inactive vetor is used. For symmetric pattern a mirror image of combination is used. So, to make a sample total 6 vector will be used. So,time for single vector is 1/6 th of sampling time. So, will the DC ripple follow the frequency of changing a vector? Again thank you for your nice comment. I have upvoted it \$\endgroup\$ – Zarzisur Jan 19 at 20:01

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