I am working on a design of harmonic rejection filter to reject 3rd harmonics present in neutral. I plan to use a capacitor of 120 uF and inductor of around 9 mH in parallel connection. The analysis shows that this combination has high impedance at 3rd harmonic (150 Hz)but the programme also shows a very high voltage dropping around few KVs across this combination. Is it true that there will be very high circulating currents between the capacitor and inductor and because of this there will be very high voltage. Do I need to design capacitor and inductor to carry such high currents and voltages. Then the design will become unviable. As the impedance is very high will it not block the current fully ? Please advise.
Eng-Tips is the largest engineering community on the Internet
Intelligent Work Forums for Engineering Professionals
harmonic rejection filtering 2
- Thread starter deeveeyes
- Start date
electricpete
Electrical
The loop circuit through L and C has total impedance of zero. So if your program is telling you that you have a high voltage it sounds very reasonable this could be caused by a high circulating current. Perhaps if you detune the resonance just a little bit you might greatly reduce the circulating current without losing effectiveness?
How about instead of using a resonannt circuit, try to create a low pass filter which will pass that 50hz but not much ofthe 150hz?
I'm no expert. I'm hoping some other folks will chime in.
How about instead of using a resonannt circuit, try to create a low pass filter which will pass that 50hz but not much ofthe 150hz?
I'm no expert. I'm hoping some other folks will chime in.
-
1
- #3
I'm scratching my head about the assumptions behind your design. I am no expert on this, but the following is my understanding of what the problem is and what you should try to accomplish with your filter:
1. Harmonic loads (PC's, VFD's, etc.) act like current sources at harmonic frequencies. Usually the problem will be worst from about 3rd harmonic (180 Hz) or 7th harmonic (420 Hz) up to about 12th harmonic (720 Hz), depending on type of power supply.
2. Power systems typically have a pretty high impedance at high frequencies. Therefore, when they are hit with the current requirements of your harmonic loads, there is a high frequency voltage imposed on them equal to V=IxZ where V=harmonic voltage, I=harmonic load current requirement, and Z=system impedance.
3. This leads to two problems:
3a. Voltage THD may be unacceptable.
3b. Harmonic currents may lead to unacceptable conductor or transformer neutral temperatures.
4. The typical solution is therefore to provide a LOW neutral impedance at high frequency (low-pass filter) to effectively "short out" high frequency currents. Provide oversized neutral conductors in order to supply harmonic currents as required and avoid overheating. This may be accomplished in various ways, including LC circuits or phase-shift transformers.
Hope this helps.
1. Harmonic loads (PC's, VFD's, etc.) act like current sources at harmonic frequencies. Usually the problem will be worst from about 3rd harmonic (180 Hz) or 7th harmonic (420 Hz) up to about 12th harmonic (720 Hz), depending on type of power supply.
2. Power systems typically have a pretty high impedance at high frequencies. Therefore, when they are hit with the current requirements of your harmonic loads, there is a high frequency voltage imposed on them equal to V=IxZ where V=harmonic voltage, I=harmonic load current requirement, and Z=system impedance.
3. This leads to two problems:
3a. Voltage THD may be unacceptable.
3b. Harmonic currents may lead to unacceptable conductor or transformer neutral temperatures.
4. The typical solution is therefore to provide a LOW neutral impedance at high frequency (low-pass filter) to effectively "short out" high frequency currents. Provide oversized neutral conductors in order to supply harmonic currents as required and avoid overheating. This may be accomplished in various ways, including LC circuits or phase-shift transformers.
Hope this helps.
-
1
- Thread starter
- #4
Thanks PEEBEE and ELECTRICPETE. Can someone tell me the configuration for a low pass filter ?? I thought tank circuit as envisaged by me acts effectively to reject 3rd harmonic, but the problem is to design the components to carry high currents. Is there any way to overcome this problem ??
It's been about 10 years since I last played with filters, but here's what I remember:
+ Capacitors (C) look like short circuits to high frequencies, and like open circuits to low freq's.
+ Inductors (L) are the opposite, short circuits to low freqs, open circuits to high freqs.
+ Your tank circuit (LC), a cap and ind in parallel, is a band-pass filter, similar to a radio tuner. It will pass voltage and block current over a narrow range. When you try to block current drawn by a current source, you get super-high voltages, which is just what you experienced.
+ Put the cap and ind in series (LC) and you get a band-block filter, it will short current and give you zero voltage over a small range of frequencies. Voltage will approach zero over the bandwidth of the filter, current will be about equal to that drawn by your current source.
+ Caps and resistors (RC) or Inds and res (RL) will give you high pass or low pass depending on the configuration. Resistors burn real power, though, so you probably do not want to add your own resistance to the circuit, you'd want to use only caps and inds in your design. In a low-pass configuration, voltage will approach zero above the break frequency ("knee"
of the filter , current will be about equal to that drawn by your current source.
This said, and given the assumption that you are trying to short (band-block) frequencies from 180-720 Hz, or from 420-720Hz, or trying to low-pass (high-block) all freqs below 180Hz or 420Hz, I'd recommend trying the following configurations:
1. Capacitor in parallel with harmonic load. Shorts high frequencies (the cap supplies the high freq current required by the load). Actually an RC circuit, where you provide the cap as part of your filter, and the harmonic load is the resistance.
2. Capacitor in parallel with the harmonic load, with an inductor in series between the hot source and the cap/load. The inductor would help prevent any harmonic currents from making their way back to the panelboard. This is an LC or an LRC circuit.
You will have high currents with either configuration. You cannot avoid this, as the load looks like a current source. The RMS value of the harmonic current will be roughly of the same magnitude as the rated current of your load. The instantaneous harmonic current could be much higher, though; I don't know for sure, but it would not surprise me if the instantaneous harmonic current was 10x or 100x the RMS current. FUSE YOUR CAPACITOR TO KEEP IT FROM BLOWING UP.
Other problems with capacitors on AC systems in general are that they can raise voltages due to transient switching, and they can resonate with the inductance in the rest of the power system causing steady-state overvoltages. Carful with this, you don't want your 120-volt PC to get hit with 200 or 400 volts. You may need to detune your filter to avoid such problems and make it appear resistive or slightly inductive at 60Hz, careful selection of the component sizes used in design #2 above should help with that.
The whole problem with switchmode power supplies (by far the largest source of harmonics) is that they draw all their current over a very short period, maybe only 10% duty cycle. "Nice" loads like motors and incandescent lights use 100% of the cycle, and have a nice sinusoidal current waveform. PC's and other harmonic loads have "spikey" current waveforms. In a quick burst or pulse, they draw all of their energy, and the rest of the cycle they are essentially shut off. When they are turned on, the high current draw causes a voltage drop, which looks like a "bite" out of the voltage waveform. Your filter should provide that pulse current, rather than forcing an upstream transformer to do it.
Hope this helps.
+ Capacitors (C) look like short circuits to high frequencies, and like open circuits to low freq's.
+ Inductors (L) are the opposite, short circuits to low freqs, open circuits to high freqs.
+ Your tank circuit (LC), a cap and ind in parallel, is a band-pass filter, similar to a radio tuner. It will pass voltage and block current over a narrow range. When you try to block current drawn by a current source, you get super-high voltages, which is just what you experienced.
+ Put the cap and ind in series (LC) and you get a band-block filter, it will short current and give you zero voltage over a small range of frequencies. Voltage will approach zero over the bandwidth of the filter, current will be about equal to that drawn by your current source.
+ Caps and resistors (RC) or Inds and res (RL) will give you high pass or low pass depending on the configuration. Resistors burn real power, though, so you probably do not want to add your own resistance to the circuit, you'd want to use only caps and inds in your design. In a low-pass configuration, voltage will approach zero above the break frequency ("knee"
of the filter , current will be about equal to that drawn by your current source.This said, and given the assumption that you are trying to short (band-block) frequencies from 180-720 Hz, or from 420-720Hz, or trying to low-pass (high-block) all freqs below 180Hz or 420Hz, I'd recommend trying the following configurations:
1. Capacitor in parallel with harmonic load. Shorts high frequencies (the cap supplies the high freq current required by the load). Actually an RC circuit, where you provide the cap as part of your filter, and the harmonic load is the resistance.
2. Capacitor in parallel with the harmonic load, with an inductor in series between the hot source and the cap/load. The inductor would help prevent any harmonic currents from making their way back to the panelboard. This is an LC or an LRC circuit.
You will have high currents with either configuration. You cannot avoid this, as the load looks like a current source. The RMS value of the harmonic current will be roughly of the same magnitude as the rated current of your load. The instantaneous harmonic current could be much higher, though; I don't know for sure, but it would not surprise me if the instantaneous harmonic current was 10x or 100x the RMS current. FUSE YOUR CAPACITOR TO KEEP IT FROM BLOWING UP.
Other problems with capacitors on AC systems in general are that they can raise voltages due to transient switching, and they can resonate with the inductance in the rest of the power system causing steady-state overvoltages. Carful with this, you don't want your 120-volt PC to get hit with 200 or 400 volts. You may need to detune your filter to avoid such problems and make it appear resistive or slightly inductive at 60Hz, careful selection of the component sizes used in design #2 above should help with that.
The whole problem with switchmode power supplies (by far the largest source of harmonics) is that they draw all their current over a very short period, maybe only 10% duty cycle. "Nice" loads like motors and incandescent lights use 100% of the cycle, and have a nice sinusoidal current waveform. PC's and other harmonic loads have "spikey" current waveforms. In a quick burst or pulse, they draw all of their energy, and the rest of the cycle they are essentially shut off. When they are turned on, the high current draw causes a voltage drop, which looks like a "bite" out of the voltage waveform. Your filter should provide that pulse current, rather than forcing an upstream transformer to do it.
Hope this helps.
- Status
Similar threads
- Locked
- Question
- Locked
- Question
- Locked
- Question
- Locked
- Question