Calculate Inductor for 3-phase Rectifier Ripple <20%

[commelion]

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

 

[NascentOxygen]

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.

 

[commelion]

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

 

[commelion]

hi again

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

regards

 

[NascentOxygen]

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.

 

[commelion]

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.

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

cheers

 

[NascentOxygen]

Fair enough.