Calculate Inductor for 3-phase Rectifier Ripple <20%

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Discussion Overview

The discussion revolves around calculating the appropriate inductor size for smoothing the ripple in a three-phase rectifier circuit, specifically aiming for a ripple of less than 20% of the output frequency. Participants explore the relationship between supply frequency and output ripple frequency, as well as the formulas used for these calculations.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant states that in three-phase rectification at 50 Hz, the DC output ripple will be six times the supply frequency and seeks a formula for selecting an inductor to achieve a ripple of less than 20% of the output frequency.
  • Another participant questions whether "20%" refers to the output average or the input level, indicating a need for clarification on the initial query.
  • A participant clarifies that "20% of the output frequency" means calculating 20% of 300 Hz, resulting in a target ripple of 60 Hz, and provides a formula for maximum ripple based on inductor and resistance values.
  • Concerns are raised about the assumptions made regarding resistance in the calculations, specifically questioning the use of a value of 10Ω without justification.
  • Further clarification is provided regarding the output voltage and current, with a participant stating the output as 150 volts and 15 amps, while suggesting to ignore inductor losses.

Areas of Agreement / Disagreement

Participants express differing views on the assumptions and calculations related to the inductor sizing, particularly regarding the resistance value used and the interpretation of the ripple percentage. The discussion remains unresolved with multiple competing views on the methodology and assumptions involved.

Contextual Notes

Participants have not reached a consensus on the appropriate resistance value to use in calculations, and there are unresolved questions about the assumptions made in the initial problem statement.

commelion
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in three phase rectification at say 50 hz the dc output ripple will be in 6 times the supply frequency, when choosing an inductor to smooth this ripple to <20% of the output frequency,what formula should be used.

also if say the supply frequency was 100 hz is it correct to persume that the out put would be 600 hz ect

cheers
 
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commelion said:
in three phase rectification at say 50 hz the dc output ripple will be in 6 times the supply frequency, when choosing an inductor to smooth this ripple to<20% of the output frequency[/color],what formula should be used.

What do you mean here? 20% of the output average? 20% of the input level?

I doubt that I can help much, but someone else may be able to once you fix this description. Inductors are not so common for smoothing nowadays. Have you covered this topic in class?
also if say the supply frequency was 100 hz is it correct to persume that the out put would be 600 hz ect
Yes, I think so.
 
hi

thanks for your time

it means 20 % of the output frequency ie 300 x 0.2 = 60 hz
I got an answer as


max ripple=1/((2pi (L /R)) 60= 10/2pi.L = 26 milli h

looks ok

regards
 
hi again

i should have mentioned the input as being 3 phase 50 hz

regards
 
commelion said:
it means 20 % of the output frequency ie 300 x 0.2 = 60 hz
I got an answer as

max ripple=1/((2pi (L /R)) 60= 10/2pi.L = 26 milli h

looks ok
Hmm, there are some things not clear in your rule of thumb there. I surmize that the method you are following is to design a low-pass LR filter having a bandwidth = 20% of the output frequency. That's fair enough.

Your formula 1/((2pi (L /R)) looks appropriate, too. But from where did you pluck your value for R? You seem to be assuming R equals 10Ω, though I can see nothing in the question that could lead to this or any other value. :confused:
 
NascentOxygen said:
Hmm, there are some things not clear in your rule of thumb there. I surmize that the method you are following is to design a low-pass LR filter having a bandwidth = 20% of the output frequency. That's fair enough.

Your formula 1/((2pi (L /R)) looks appropriate, too. But from where did you pluck your value for R? You seem to be assuming R equals 10Ω, though I can see nothing in the question that could lead to this or any other value. :confused:

appoligies the question states the output voltage as being 150 volts/ current 15 amps, and ignore inductor losses

cheers
 
Fair enough.
 

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