# Shunt Filter Design: Short Guide

## Description, working principle and design with illustrative example

Contents :

## What are harmonics?

According to the Fourier Theorem, a periodical waveform could be reduced to its Fourier components as sinusoidal waveforms with different frequencies and amplitudes.

In the case of a** pure sinusoidal waveform, there will be no harmonics other than the first harmonic** of the waveform itself. Abrupt changes in a voltage or current waveform indicate a high harmonic content in the waveform.

Especially in parts of the waveform, where these **abrupt changes take place, there will be a constructive addition of the amplitudes of the harmonics**. Harmonics will be described as integer multiples of the frequency of the first harmonic.

**Example**:

A waveform with a fundamental frequency of 50Hz could have a second harmonic with 100Hz and a** third harmonic with 150Hz**. With different amplitudes of every harmonic, the total voltage could assume changing waveforms.

To make it more concrete the practical consequences of harmonics will be explained in the following subsection!

## How are harmonics generated?

Injection of Harmonics into the transmission system is caused by different factors.

The **most prevalent factor could be defined as static power converters which represent the largest nonlinear loads in the transmission system**.

The **fast-switching action of these converters results in a distorted voltage waveform **of the transmission system by drawing rapidly changing currents.

The measurement of the distortion in a waveform caused by harmonics will be expressed by the **total harmonic distortion THD**, which in general will cover harmonics until the 50th harmonic in terms of their percentage contribution.

In the following subsection problems related to the harmonics are mentioned.

## Negative effects of harmonics on grid components

The **most prominent degrading effect of harmonics could be summarized as losses resulting from heat and vibration**. Especially components like transformers, generators, and motors of the grid suffer from iron losses, eddy current losses, and hysteresis losses.

Other than additional losses, **harmonics also cause the malfunction of circuit breakers and fuses**. A non-linear load might require a higher RMS current for the same amount of power, which results in more heating of the trip mechanism in fuses and circuit breakers. This might result in a false breaking of the circuit breaker, even though the amount of power delivered is the same.

There are additional harmful effects of harmonics, which will not be covered in this article (read this for more details!).

In this article, we presented different solutions to reduce harmonics in electrical systems. In the sequel, **shunt filters as a solution for harmonics reduction** will be explained.

## Shunt Filters description

Attenuation of harmonics will be required by the majority of non-linear equipment in the grid. There are different approaches to solve this issue. **Power equipment might come as an all-in-one package, including own harmonic mitigation, or it might be necessary to include a discrete mitigation mechanism**.

The above figure illustrates a **shunt filter structure** that operates as a notch filter tuned for certain harmonic filterings such as 3rd, 5th, or 11th harmonics.

## Working Principle of Shunt Filters

The working principle is simple!

At a designed filter resonance frequency f_R, the amplitude of the impedance of the filter goes to zero, providing a short circuit path for harmonic currents.

**Shunt filters could be arranged as single tuned or multiple tuned filters.** Multiple tuned filters provide attenuation for multiple frequencies by compromising the amount of attenuation in comparison to the single tuned type.

The resonance frequency could be calculated as follows.

\displaystyle f_R=\frac{1}{2\pi \sqrt{L_sC_s}}

It should be noted that there is also the ESR of the capacitor and the resistance of the inductance which together **influence the quality factor of the filter**.

The impedance of inductance at a given frequency *f* is given as follows:

X_L=j\omega L

and the impedance of a capacitor is:

\displaystyle X_C=-\frac{j}{\omega C_s}

It is trivial to see that,** at a certain frequency namely f_R the sum of the impedances goes to zero**. This means that the only limiting factor for the current will be the resistance resulting from the line, capacitor, and inductor.

If the total contribution of resistances is neglected, the whole harmonic current will be flowing through the path of the “trapping” filter, hindering any pass of the current of the harmonic frequency to any other consumer on the line.

## Design Steps of a Shunt Filter

According to the IEEE Std 141-1993 there are** 4 steps to a design of a shunt filter**. Those consider the:

- maximum voltage rating of the filter capacitor,
- total RMS current of the filter inductance, and
- series resistance value required to have a desired quality factor.

The steps are listed below:

**Choice of capacitance to improve the power factor.****Selection of a reactor to tune the filter.****Calculation of the peak voltage across the capacitor and the total RMS current through the reactor.****Choice of off-the-shelf components.**

## Application example

A shunt filter will be designed for a system with the following characteristics:

- a line to neutral voltage of 230 Vrms
- a load of 1187 kVA per phase having a lagging power factor of 0.539.
- It is assumed, that the power factor should be upgraded to a value of 0.95.

### 1. Calculate filter capacitance

Firstly, the required capacitance will be calculated:

\displaystyle S=(640+1000i)\text{VA}=1187.2\angle{57.38^{\circ}}\text{VA}

To improve the power factor, the capacitor should provide a reactive power of 326 VAR.

\displaystyle 326\text{VAR}=\frac{230V^2}{X_C} \implies C=19.675\mu F

### 2. Calculate filter inductance

By choosing the reactance, the filter can be tuned to the desired frequency. In this example the desired frequency is 150Hz.

\displaystyle 150\text{Hz}=\frac{1}{2\pi \sqrt{LC}} \implies L=57.34 \text{mH}

### 3. Capacitor voltage constraint

The peak voltage can be calculated separately for 1st and 3rd harmonics. To accommodate the possibility of a worst-case condition, it will be assumed that the 3rd harmonic will be constitutive to the 1st causing a voltage increase. The calculation is as follows.

\displaystyle I_1=\frac{V_{L-N}}{\sqrt{(X_X-X_L)^2+R^2}}=\frac{230V}{144 \Omega} =1.59A

\displaystyle I_3=\frac{1}{3}\frac{640 W}{230V}=0.92A

V_C=I_1X_C+I_3X_{C_3}=1.59*162+0.92*54=307.26V

A standard capacitor with 240 V 1kVAR would suffice in this example. The per-unit voltage would be determined as:

\displaystyle V_{pu}=\frac{307.26}{240}=1.28pu

The reactor should be chosen according to its total RMS current, which could be calculated as follows:

\displaystyle I_{rms}=\sqrt{I_1^2+I_3^2}=1.83A

### 4. Simulation & Validation

A further assessment of the special condition is required to implement safety factors that are responsible to accommodate other load conditions the filter will undergo.

The above-calculated example was simulated using LT Spice. The l**oad condition was simulated using phase-shifted current sources in series with resistive loads**, fig. 2. To emphasize the effect of the filter, a switch is added to the circuit, which will be triggered after the first second.

The result is represented in fig. 3; where:

- Before the switching action, the phase shift and current magnitude could be read as 57.35° and 1.2 Arms.
- After the filter is connected, the RMS current decreases to 1 Arms. There is a remaining 26.46° phase shift due to imperfect power factor correction.

That’s all for today’s tutorial!

## Conclusion

In this article, we considered the problem of harmonics reduction using passive filters. We considered the shunt filter topology, we describe it and highlight it in different design steps. To complete the tutorial, we added an application example with simulations!