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LC Filter Design for H Bridge Inverter

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sudhinpk

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I need to design a LC Filter for my H bridge inverter 230 Vac, 100 W, 50 Hz switching frequency 5KHz.
How I decide cut off Frequency and L,C Values? Plz help
 

mrinalmani

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What makes you limit your speed only to 5kHz. Are you using BJTs?
Filter design is more effected by the minimum load, and not maximum. Is 100W the minimum or maximum? Also, how much ripple are you willing to tolerate?
 

sudhinpk

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What makes you limit your speed only to 5kHz. Are you using BJTs?
Filter design is more effected by the minimum load, and not maximum. Is 100W the minimum or maximum? Also, how much ripple are you willing to tolerate?

I am using MOSFET with SPWM switching. Minimum load will be 10 W. It's for mtech project work
 

mrinalmani

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Lower frequency will require larger filter. With MOSFETs you can use frequencies as high as 200kHz. 25kHz is a good starting point.
Leave apart cutoff frequency, etc etc. Here's a logical and simple way, consider the following....

1. calculate the current at 10W load.
2. Calculate the resistance of the load at 10W.
3. Let the inductor be L, then the time constant it L/R
4. The drop in inductor current in a time interval DT is, DI = Io*(L/R)*DT, where Io is the initial current through inductor. (you may verify this by differentiation, assuming that the time constant L/R >> DT)
5. If the inductor was not there, the current across the load would fall to zero as soon as the input becomes zero. With the inductor, the current will not fall suddenly to zero, but will be governed by the above equation.
6. DI is nothing but the ripple current. So assuming you want your ripple not to exceed 10% of the load current, the above equation becomes:
0.1*Io = Io*(L/R)*DT
From the above, find the value of L. (DT = 1/F)
7. For your design, I think the value should be approx 1mH.

If you have doubts till here, ask. The capacitor is optional, will discuss about it when it is clear to u till here.
 

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Tahmid

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Lower frequency will require larger filter. With MOSFETs you can use frequencies as high as 200kHz. 25kHz is a good starting point.
Leave apart cutoff frequency, etc etc. Here's a logical and simple way, consider the following....

1. calculate the current at 10W load.
2. Calculate the resistance of the load at 10W.
3. Let the inductor be L, then the time constant it L/R
4. The drop in inductor current in a time interval DT is, DI = Io*(L/R)*DT, where Io is the initial current through inductor. (you may verify this by differentiation, assuming that the time constant L/R >> DT)
5. If the inductor was not there, the current across the load would fall to zero as soon as the input becomes zero. With the inductor, the current will not fall suddenly to zero, but will be governed by the above equation.
6. DI is nothing but the ripple current. So assuming you want your ripple not to exceed 10% of the load current, the above equation becomes:
0.1*Io = Io*(L/R)*DT
From the above, find the value of L. (DT = 1/F)
7. For your design, I think the value should be approx 1mH.

If you have doubts till here, ask. The capacitor is optional, will discuss about it when it is clear to u till here.

How does this handle the case where the load is not of resistive nature but of, say, inductive nature? Then, the external L adds to the filter L and the R is only the series R of the two inductors. What happens then?

I need to design a LC Filter for my H bridge inverter 230 Vac, 100 W, 50 Hz switching frequency 5KHz.
How I decide cut off Frequency and L,C Values? Plz help

A rule of thumb is that the cutoff frequency be in the region between 10x the output frequency to 0.1x the switching frequency. This brings your design requirement to 500Hz. That would be a good starting point. From there on, it's a matter of trial and error. Do remember the effect of using too large an inductance (increased output impedance) and too large a capacitance (increased inrush).

Hope this helps.
Tahmid.
 

mrinalmani

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Then things become even better!
Now we have extra inductance and hence even lower ripple!
 

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