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Inverter Filter Desgin

thannara123

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Inverter Filter Desgin

I have a full bridge sinewave inverter .(SPWM signal is used to drive an H bridge to get a 50 Hz)
The full wave bridge is fed to the primary of the low frequency transformer .

How do I calculate the filter to get a smooth sinewave ,I mean the basic calculation and consideration .

Most of the ( low frequency transformer) inverter uses a capacitor accorss the secondary of the transformer (ie ,a capacior is connected accross the outut)

most of the threads or internet forum says better filter is LC filter .
which coil may i take as filter inductor is it secondary one ? (or the primary one or the both one )

is there any text book for the filter desgin ?
 
Hi,

Just recently I answered about the same question here in this forum.
There are many discussions here and even more in the internet.

Also many semiconductor manufacturers provide very informative and reliable application notes / design notes. Just do a search on your own.

Klaus
 
The big advantage of SPWM is that it operates switching in all-On-all-Off fashion. It reduces parasitic resistance and allows maximum Amperes in the system.

This is important for the low voltage circuit of your system. Another series inductor robs electron flow. A parallel capacitor robs electron flow. I think it's wiser to omit the C or LC filter in the low voltage side. The transformer being an inductive element has some effect in smoothing SPWM.

The LC low-pass filter conventionally goes closest to the load. LC values need to be tailored to suit power and frequency.

A larger inductor L value reduces amplitude. Adjust its value so that your load receives the desired amplitude and resembles a sine wave .

The capacitor C further reduces jaggies in the waveform, and carries the benefit of improving power factor. It makes things easy on upstream components if you correct power factor to get the Ampere waveform to align with the Voltage waveform. Inductors have the tendency to create unwanted voltage spikes when their voltage is shut off abruptly while current is flowing.
 
What power ? 230Vac out ? what is the sw freq ? 2 level, quasi 2 level, 3 level switching ?

Yes you can use the leakage of the LV pri to HV sec as part of the filter - have you measured this leakage ?

If you don't know what it is - it is very hard to use it in a calculation.

Adding choke(s) to the HV sec side is generally easier as they are lower current and easier to construct.

One can also fairly easily model the filtering effect in LTspice.

How are you going to arrange your volt feed back loop to take account of the amplitude change caused by the filter ?

A single LC filter is generally the best choice, for example, for 20kHz sw freq and 1kW, the peak current is 6.15 A plus switching ripple.

If we assume a 2 level inverter, +/- 400V say, then the ripple is worst at zero volts, and given by V/L = di/dt, here dt = 25uS as 20kHz

So for 1mH in total for the L, the ripple ( worst case ) = 10A ( +/- 5A )

Since our load is only 6.15A pk, we might like the ripple to be +/- 2A worst case

thus the total L in our LC filter will be 1mH x 5/2 = 2.5mH

We now need a cap to filter the 20kHz and leave a nice clean sine wave at 50 Hz.

Again looking at zero volts out, the current ripple in the cap is +/- 2Apk or 1.414 amp rms

If we want to reduce the 20kHz ripple to 500mV rms say, then the cap must be 22.5uF ( 2 x 10uF will be close enough )

[ from Vc/Ic = Xc = 1. ( 2. pi. freq. C ) ]

This gives the filter a low pass res freq of 1/ 2.pi.SQRT(LC) = 711 Hz - so we must never modulate at this freq or we will get very high voltages across the filter components.

The filter choke(s) must handle 8.5Apk without saturating, and 6.5 A rms, and the caps ( ideally 630VDC rated ) must handle the 50Hz that will be in them ( Ic = Vc / Xc @ 50Hz = 1.45A plus the vector sum of the ripple current @ 20kHz ( = 1.414 A ) which vector sum is = 2.1A rms.

There is a slight drop in the fundamental due to the 50Hz impedance of the L and C, 230Vac becomes 228.87 at light load - you can easily calc this your self. There will be more droop at full power as we drop approx 3.37 V across the combined L due to the 4.3A of fundamental ( 50 Hz ) flowing in this inductance.

So there you have it

- you're welcome.
 
Last edited:
What power ? 230Vac out ? what is the sw freq ? 2 level, quasi 2 level, 3 level switching ?

Yes you can use the leakage of the LV pri to HV sec as part of the filter - have you measured this leakage ?

If you don't know what it is - it is very hard to use it in a calculation.

Adding choke(s) to the HV sec side is generally easier as they are lower current and easier to construct.

One can also fairly easily model the filtering effect in LTspice.

How are you going to arrange your volt feed back loop to take account of the amplitude change caused by the filter ?

A single LC filter is generally the best choice, for example, for 20kHz sw freq and 1kW, the peak current is 6.15 A plus switching ripple.

If we assume a 2 level inverter, +/- 400V say, then the ripple is worst at zero volts, and given by V/L = di/dt, here dt = 25uS as 20kHz

So for 1mH in total for the L, the ripple ( worst case ) = 10A ( +/- 5A )

Since our load is only 6.15A pk, we might like the ripple to be +/- 2A worst case

thus the total L in our LC filter will be 1mH x 5/2 = 2.5mH

We now need a cap to filter the 20kHz and leave a nice clean sine wave at 50 Hz.

Again looking at zero volts out, the current ripple in the cap is +/- 2Apk or 1.414 amp rms

If we want to reduce the 20kHz ripple to 500mV rms say, then the cap must be 22.5uF ( 2 x 10uF will be close enough )

[ from Vc/Ic = Xc = 1. ( 2. pi. freq. C ) ]

This gives the filter a low pass res freq of 1/ 2.pi.SQRT(LC) = 711 Hz - so we must never modulate at this freq or we will get very high voltages across the filter components.

The filter choke(s) must handle 8.5Apk without saturating, and 6.5 A rms, and the caps ( ideally 630VDC rated ) must handle the 50Hz that will be in them ( Ic = Vc / Xc @ 50Hz = 1.45A plus the vector sum of the ripple current @ 20kHz ( = 1.414 A ) which vector sum is = 2.1A rms.

There is a slight drop in the fundamental due to the 50Hz impedance of the L and C, 230Vac becomes 228.87 at light load - you can easily calc this your self. There will be more droop at full power as we drop approx 3.37 V across the combined L due to the 4.3A of fundamental ( 50 Hz ) flowing in this inductance.

So there you have it

- you're welcome.
Hai @Easy peasy .Thanks for the reply,
The transformer output is 230 Vac 50Hz ,SW Freq - 16Khz ,2 Level H -Bridge Topology
Not measured yet , May I measure it by using short test ?
I dont have much more in mathamatical calculation ,So now i dont want the calculation Only require the practicale measurement and its calculation.
Adding choke is not intrested .
Want to be a filter with a capacitor + leakage Inductor of the HV coil.
sir ,I am not familer with LTspice
.Feebdback taken by using a opamp + ADC

I didnt get well the example due to my low knowldege .
I wi will try to understand it .
 
Simulation of 12V power inverter. The load gets sine waves 240VAC about 1 Ampere. I used a 1400 Hz carrier in order to clarify all waveforms.

A capacitor a few uF is in the range of a suitable filter in the secondary side. It can be assembled from other capacitors, in order to make it non-polarized and able to tolerate a few Amperes.

The primary should be sufficiently low milli-Henries so it can admit enough Amperes on the low-voltage side. Yet If it is too low then Ampere level rises alarmingly, wasting power.

transformer steps up 12V SPWM fm 2 op amps to 230VAC sinewave to load.png
 
Hai @Easy peasy .Thanks for the reply,
The transformer output is 230 Vac 50Hz ,SW Freq - 16Khz ,2 Level H -Bridge Topology
Not measured yet , May I measure it by using short test ?
I dont have much more in mathamatical calculation ,So now i dont want the calculation Only require the practicale measurement and its calculation.
Adding choke is not intrested .
Want to be a filter with a capacitor + leakage Inductor of the HV coil.
sir ,I am not familer with LTspice
.Feebdback taken by using a opamp + ADC

I didnt get well the example due to my low knowldege .
I wi will try to understand it .
I wish you luck - but without real engineering and mathematical knowledge - it is very hard to make these inverters work well
--- Updated ---

Simulation of 12V power inverter. The load gets sine waves 240VAC about 1 Ampere. I used a 1400 Hz carrier in order to clarify all waveforms.

A capacitor a few uF is in the range of a suitable filter in the secondary side. It can be assembled from other capacitors, in order to make it non-polarized and able to tolerate a few Amperes.

The primary should be sufficiently low milli-Henries so it can admit enough Amperes on the low-voltage side. Yet If it is too low then Ampere level rises alarmingly, wasting power.

View attachment 184390
Can you program extra inductance into the output side of the tx ?
 
I wish you luck - but without real engineering and mathematical knowledge - it is very hard to make these inverters work well
--- Updated ---


Can you program extra inductance into the output side of the tx ?
The transformer Edit window has fields to change primary inductance and windings ratio. I believe that's an indirect method to change secondary inductance because secondary inductance is one of the variables in the formula that determines step-up ratio (I think).

To run my simulation (post #6), click on this (shortened) link:


1) Opens the website falstad.com/circuit,
2) Loads my schematic into the simulator,
3) Runs it on your computer.

The simulator is animated and interactive. Choose Toggle full screen under File menu. Right-click on components when you wish to change values.

The recent version contains also a center-tap xfmr and a custom xfmr (with 3 windings). All have only a few parameters for the user to adjust.

Screenshot showing Edit window:

Falstad xfmr with Edit window.png
 

LaTeX Commands Quick-Menu:

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