Harmonic filter design question
Harmonic filter design question
(OP)
Hello,
I just bought a harmonic filter to mitigate the distortion produced by a 400 HP VFD. The filter consists of some capacitors and 2 line reactors. Here's how it is wired.
One line reactor is connected immediately downstream of the main breaker. From there it is connected to the VFD. Between the VFD and the line reactor, a second filter is connected phase to phase. The second reactor is wired in series with some capacitors, and that series combination is connected line to line. So, between each two phases, there is an LC circuit (not including the resistance of the reactor).
So, I have a couple of questions. The first is very basic.
1. How does a line reactor absorb high frequency current? I understand that the impedance varies with frequency. But in order to absorb the high frequency current, something has to absorb real (not reactive) power. There is a tiny DC resistance in the line reactor. Is this the load which absorbs the high frequency current? Where does the high frequency energy go?
Question 2. Why would the circuit be wired such as it is? If I had designed this, I would have put an LC shunt off of each phase, tying one end to ground. The filter I bough makes the connection line to line. And why the two line reactors?
thanks in advance
I just bought a harmonic filter to mitigate the distortion produced by a 400 HP VFD. The filter consists of some capacitors and 2 line reactors. Here's how it is wired.
One line reactor is connected immediately downstream of the main breaker. From there it is connected to the VFD. Between the VFD and the line reactor, a second filter is connected phase to phase. The second reactor is wired in series with some capacitors, and that series combination is connected line to line. So, between each two phases, there is an LC circuit (not including the resistance of the reactor).
So, I have a couple of questions. The first is very basic.
1. How does a line reactor absorb high frequency current? I understand that the impedance varies with frequency. But in order to absorb the high frequency current, something has to absorb real (not reactive) power. There is a tiny DC resistance in the line reactor. Is this the load which absorbs the high frequency current? Where does the high frequency energy go?
Question 2. Why would the circuit be wired such as it is? If I had designed this, I would have put an LC shunt off of each phase, tying one end to ground. The filter I bough makes the connection line to line. And why the two line reactors?
thanks in advance





RE: Harmonic filter design question
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If we learn from our mistakes I'm getting a great education!
RE: Harmonic filter design question
Are you saying that the 5th order harmonic is of interest assuming that the drive is a 6-pulse drive? What about the 7th harmonic in this case?
Can you explain how the series reactor acts as a low pass filter. I understand that the reactor impedence represented by XL = jwL will increase with increased frequencies as you mentioned, however what will this increased impedance do to this high frequency current?
RE: Harmonic filter design question
RE: Harmonic filter design question
Yes, low order odd harmonics are of interest if it's a 6-pulse front end. If it's tuned to the 6th and the Q-factor isn't too high then it will trap 5th and 7th without being resonant at either. It's reasonable to expect that even order harmonics will be negligible in a properly behaving rectifier so the resonance at 6th isn't a problem.
Low pass filter was perhaps not the best description. It will serve to slow down commutation edges and reduce the production of high order harmonics inherent in fast switching edges.
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If we learn from our mistakes I'm getting a great education!
RE: Harmonic filter design question
2. The reason the circuit is being wired this way is to have these resonant frequency currents circulate inside the filter only, if you ground the filter you will provide a path back to the system canceling its very purpose. .
"Throughout space there is energy. Is this energy static or kinetic! If static our hopes are in vain; if kinetic — and this we know it is, for certain — then it is a mere question of time when men will succeed in attaching their machinery to the very wheelwork of nature". – Nikola Tesla
RE: Harmonic filter design question
----------------------------------
If we learn from our mistakes I'm getting a great education!
RE: Harmonic filter design question
"Throughout space there is energy. Is this energy static or kinetic! If static our hopes are in vain; if kinetic — and this we know it is, for certain — then it is a mere question of time when men will succeed in attaching their machinery to the very wheelwork of nature". – Nikola Tesla
RE: Harmonic filter design question
Came across this while looking for something else on CDA's website: CDA filter link . I haven't read it properly but might be of interest.
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If we learn from our mistakes I'm getting a great education!
RE: Harmonic filter design question
If I understand you correctly the LC circuit is wired inbetween phases so that the resonant harmonic currents will be trapped and fluculate between the reactor and the cap. It sounds like you are saying that this LC circuit is designed with values such that at the 5th harmonic there would be a resonant frequency between the cap and the reactor and thus set up a resonant current trapped inbetween these two devices? Do I understnad this correctly?
If you ground this LC circuit would the harmonic resonant current not still flow back and forth between the cap and reactor or would it flow to ground and back to the source?
RE: Harmonic filter design question
----------------------------------
If we learn from our mistakes I'm getting a great education!
RE: Harmonic filter design question
"It sounds like you are saying that this LC circuit is designed with values such that at the 5th harmonic there would be a resonant frequency"
Not exactly to resonate at 5th for the reasons mentioned above, but close to it, and yes that is my understanding of it.
And as far as grounding the filter,
Yes the current will flow back and forth through your system, but the questions is do you really want this current in your grounding system? Note that you cannot completely eliminate these harmonic currents rather just reduce the THD.
Refer the attachment for a diagram that shows a phase to phase connection.
"Throughout space there is energy. Is this energy static or kinetic! If static our hopes are in vain; if kinetic — and this we know it is, for certain — then it is a mere question of time when men will succeed in attaching their machinery to the very wheelwork of nature". – Nikola Tesla
RE: Harmonic filter design question
If you look at the components, they are probably not tuned to a frequency close to 250 or 350 Hz (in 50 Hz grids) or 300 or 420 Hz (in 60 Hz grids), but a frequency that is a lot higher.
The filter simply shorts out frequencies starting at around 50 - 100 kHz (150 kHz is where the European standards atart).
Also, to answer the OPs first question; they do not absorb much of the energy. They just short the frequency components so they cannot propagate out into he grid.
Another point to consider is that real good filters also have capacitors going to ground in order to keep common mode interference low - not just normal mode interference.
The resistors seen in some filters are mostly there to discharge the capacitors so the filters don't carry lethal voltages after disconnection from grid.
It is only in some motor reactor applications (other side of the inverter) that you find damping resistors parallel to the coils. They are there to avoid ringing between reactor and cable capacitance.
There is a lot of confusion about these filters. Manufacturers, salespersons, users are among the confused. Not saying that all are confused, but a large percentatge.
Gunnar Englund
www.gke.org
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100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...
RE: Harmonic filter design question
Thanks for the circuit diagram. I'm trying to understand how the harmonic current propogates in this circuit. I'm assuming Ih in this circuit is the total harmonic current and If is the current that is passed through the filter, while It is the current that goes back to the transformer.
Is the Ih current the current consisting of the 5th and 7th harmonic?
Why does only the harmonic currents flow as If and not back through It. In other words how does the LC combination divert and trap the harmonic current. When we talked above about the resonant current going back and forth between the cap and reactor, does it travel back and forth on the short path on the branch that they are on, or does it travel back and forth back through the transformer.
RE: Harmonic filter design question
Ih Current is the total current drawn by the vfd which includes the fundamental and harmonic currents .Now by getting the different values for Xt, Xl and Xc at different frequencies (5th, 7th, 11th, 13th...), you can calculate approximately how much current will be "diverted" by the filter using the equation provided in previous attachment. For example, lets assume 1MVA xfmr 5%Z, 60Hz, 4160V pri/480V sec. feeding a VFD driving a 350hp motor. 6 Pulse drive.
Ifla=1200A
Im=414A
Calculate Zeff = %Z(Im/Ifla) = 1.75%
From Table (Prediction of Harmonic Spectrum based upon Effective Source Impedance) I get
(at Zeff=1.75, Individual Harmonic magnitudes (% fundamental) gives 1st at 100%, 5th at 48%, 7th at 27%, 11th at 10%, 13th at 6%, and THD at 59%.)
Now these are your expected harmonics before the filter is applied.
Next step then would be to design the filter.
First thing to do is to convert the hp to approx. Kvars.
Kvar=hp*0.30 = 105
Then calculate C
Cf=Var/(2pi60V^2) = 1209 uF
Xc then would be approx 2.195ohms in our setup.
Now having calculated the C you can calculate L and design it to resonate at 282Hz (4.7*fund.)
Lh=1/[Cf(2pif)^2] = 263uH
Xl=.099
Those are your values for the filter.
Now to check the harmonic distortion after the filter has been applied. Refer to diagram previously attached.
Now next thing to do is to find Xt.(from previous diagram)
Xt=(kV^2/MVA)*Zpu
Xt=0.01152ohms at fundamental
Now all of the values are in respect to the fundamental.
Calculate Xt, Xl and Xc for 5th harmonic by multiplying fundamental values by 5.
You get Xt= 0.0576, Xl=0.495, and Xc=.438
Do the same for 7th, 11th, 13th....
Now again refer to the diagram and equation attached. Using the equation you can calculate It current, and from this example I get
5th at 23%, 7th at 20%, 11th at 8%, 13th at 5% and total with 33%.
So as you can see you reduced but not completely eliminated the current distortion.
FYI:I also get that the total voltage distortion with at approx. 3.98% which meets IEEE-519 voltage distortion standard, I believe.
I hope this helps.
"Throughout space there is energy. Is this energy static or kinetic! If static our hopes are in vain; if kinetic — and this we know it is, for certain — then it is a mere question of time when men will succeed in attaching their machinery to the very wheelwork of nature". – Nikola Tesla