Inverter LC Filter Calculator

Let’s say we have a push-pull or full bridge inverter and its output goes to the primary of a transformer, and the secondary gives you high voltage AC like 220V square wave.

Now to make this AC waveform smooth and more like a sine wave we connect an LC low-pass filter just after this secondary, like this:

Connections Details of the LC Filter:

• Take the two output wires from the transformer secondary — this is your 220V AC (square wave).
• Now we connect a series inductor (L) in one of the wires — just one wire, like in series, not both wires.
• Then after the inductor, we put a high voltage capacitor (C) across the output, like between the two wires, that is parallel to the load.

Now to caalculate the values of the L and C, you can use the following calculator:

How to Decide this fc Value?

We do not want to block 50Hz because that is our main signal. We want to block the switching frequency harmonics like:

• 300Hz
• 600Hz
• 1kHz
• And higher junk frequencies

So we choose fc just above 50Hz but much below the unwanted high frequencies.

Rule of Thumb:

You can select cut-off (fc) between:

3 × main frequency......... upto 10 × main frequency

Example:

If your inverter PWM has big amount of noise like around 1kHz to 3kHz, then you pick fc = 300Hz, so that:

• 50Hz passes cleanly.
• Above 300Hz (like 1kHz, 2kHz) get blocked fast.
• Gives smooth sine output.

Final Tip:

So just take 5 or 6 times the 50Hz, then:

fc ≈ 50 × 6 = 300Hz

It is a safe and effective choice for LC filter to clean inverter waveform.

Example Calculations:

Now let me give you a clear manual example calculation using our default values so visitors can verify the tool step-by-step.

Let us assume:

Vrms = 220V
P = 500W
f = 50Hz
fc = 300Hz

STEP 1: Calculate Load Current

I = P / Vrms

I = 500 / 220
I = 2.27 A

So the load current is:

I ≈ 2.27 A

STEP 2: Calculate Inductive Reactance Term

XL = 2 × pi × fc

XL = 2 × 3.1416 × 300
XL = 1884.96

STEP 3: Calculate Inductor Value

Your formula:

L = Vrms / (XL × I)

Substituting:

L = 220 / (1884.96 × 2.27)

First multiply denominator:

1884.96 × 2.27 = 4278.86

Now divide:

L = 220 / 4278.86
L = 0.0514 H

Convert to mH:

L = 0.0514 × 1000
L = 51.4 mH

So,

Suggested Inductor ≈ 51 mH

STEP 4: Calculate Capacitor Using Resonance Formula

Formula used in your code:

C = 1 / ((2 × pi × fc)^2 × L)

First calculate:

2 × pi × 300 = 1884.96

Now square it:

1884.96^2 = 3,553,057

Now multiply with L:

3,553,057 × 0.0514 = 182,694

Now invert:

C = 1 / 182,694
C = 0.00000547 F

Convert to microfarads:

C = 0.00000547 × 1,000,000
C = 5.47 uF

So,

Suggested Capacitor ≈ 5.5 uF

FINAL VERIFIED RESULTS

Load Current ≈ 2.27 A
Inductor ≈ 51 mH
Capacitor ≈ 5.5 uF

Important Practical Note

This is a simplified theoretical LC design. In real inverter systems:

• Inductor current rating must be higher than 2.27 A (preferably 1.5× to 2× margin).
• Capacitor must be AC rated (polypropylene recommended).
• ESR and inductor DCR will slightly shift cutoff.
• Final tuning is usually done experimentally.