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Harmonic current filter

**Library:**Simscape / Electrical / Passive

The Passive Harmonic Filter (Three-Phase) block suppresses system harmonic currents and decreases voltage distortion by providing low-impedance paths for the harmonics. At the rated frequency, the passive shunt filters are capacitive and provide reactive power, which can improve electrical power factor.

Each of the four models for the block corresponds to an option for the **Filter
type** parameter:

At the tuned frequency, LC resonance occurs and the impedance of the filter reaches its minimum, which equals the value of the resistance.

The filter tuned frequency is defined by this equation:

$${f}_{1}=n{f}_{0}=\frac{1}{2\pi \sqrt{LC}},$$

where *f _{1}* and

The quality factor is defined as the ratio between the inductive or capacitive reactance at the tuned frequency and the resistance, as described by this equation:

$${Q}_{f}=\frac{{X}_{LN}}{R}=\frac{{X}_{CN}}{R},$$

where:

R is the resistance.

`X`

is the impedance of the inductor at the tuned frequency $${X}_{LN}=2\pi {f}_{1}L.$$_{LN}`X`

is the impedance of the capacitor at the tuned frequency $${X}_{CN}=\frac{1}{2\pi {f}_{1}C}.$$_{CN}

Higher quality factor values result in sharper frequencies. However, this produces high-power dissipation at the base frequency due to a relative low resistance.

The rated reactive power is given by:

$${Q}_{r}=\frac{{V}^{2}}{{X}_{L0}-{X}_{C0}}=\frac{{V}^{2}}{\frac{{X}_{C0}}{{n}^{2}}-{X}_{C0}}=\frac{{V}^{2}}{{X}_{C0}}\cdot \frac{{n}^{2}}{{n}^{2}-1},$$

where *Q _{r}* is the rated
reactive power for one phase and

A double-tuned filter has two tuned frequencies,
*f _{1}* and

The double-tuned filter comprises a series LC and a parallel RCL circuit, each
tuned at frequencies *f _{s}* and

$${f}_{m}=\sqrt{{f}_{1}{f}_{2}}=\sqrt{{f}_{s}{f}_{p}},$$

where

$${f}_{s}=\frac{1}{2\pi \sqrt{LC}}$$

$${f}_{p}=\frac{1}{2\pi \sqrt{{L}_{2}{C}_{2}}}.$$

The quality factor of this filter is defined as the quality factor of the parallel
`R`

and `L`

elements at the mean geometric
frequency, *f _{m}*:

$${Q}_{f}=\frac{R}{2\pi {f}_{m}{L}_{2}}.$$

The second order high-pass filter shunts a large percentage of the harmonics at and above the tuned frequency. The filter is designed to have a flat impedance for high-order harmonics.

The tuned frequency is described by this equation:

$${f}_{1}=n{f}_{0}=\frac{1}{2\pi \sqrt{LC}}.$$

The quality factor is the reciprocal of the band-pass, single-tuned filter:

$${Q}_{f}=\frac{R}{{X}_{LN}}=\frac{R}{{X}_{CN}}.$$

The rated reactive power is the same of the band-pass, single-tuned filter:

$${Q}_{r}=\frac{{V}^{2}}{{X}_{L0}-{X}_{C0}}=\frac{{V}^{2}}{\frac{{X}_{C0}}{{n}^{2}}-{X}_{C0}}=\frac{{V}^{2}}{{X}_{C0}}\cdot \frac{{n}^{2}}{{n}^{2}-1}.$$

Compared to the single-tuned version, the C-type, high-pass filter has lower losses at the fundamental frequency, because the capacitor and inductor are parallel with the resistor.

To prevent fundamental currents from passing through the resistor, the resonance
frequency of *L _{2}* and

$${f}_{0}=\frac{1}{2\pi \sqrt{{L}_{2}{C}_{2}}}.$$

The quality factor is calculated using this equation:

$${Q}_{f}=\frac{R}{2\pi {f}_{0}n{L}_{2}}.$$

Single-Tuned | Double-Tuned | Second-Order, High-Pass | C-type, High-Pass | |
---|---|---|---|---|

R |
$$\frac{1}{2\pi {f}_{0}C{Q}_{f}}$$ |
$$2\pi {f}_{m}{L}_{2}{Q}_{f}$$ |
$$2\pi n{f}_{0}L{Q}_{f}$$ |
$${Q}_{f}\left(2\pi \text{\hspace{0.17em}}n\text{\hspace{0.17em}}{f}_{0}\text{\hspace{0.17em}}{L}_{2}\right)$$ |

L |
$$\frac{1}{C{(2\pi n{f}_{0})}^{2}}$$ |
$$\frac{{V}^{2}}{{Q}_{r}\pi {f}_{0}\left({n}_{1}^{2}+{n}_{2}^{2}-2\right)}$$ |
$$\frac{1}{C{(2\pi n{f}_{0})}^{2}}$$ | None |

C |
$$\frac{{Q}_{r}}{2\pi {f}_{0}{V}^{2}}\cdot \frac{{n}^{2}-1}{{n}^{2}}$$ |
$$\frac{{Q}_{r}}{4\pi {f}_{0}{V}^{2}}\left(\frac{{n}_{1}^{2}-1}{{n}_{1}^{2}}+\frac{{n}_{2}^{2}-1}{{n}_{2}^{2}}\right)$$ |
$$\frac{{Q}_{r}}{2\pi {f}_{0}{V}^{2}}\cdot \frac{{n}^{2}-1}{{n}^{2}}$$ |
$$\frac{{Q}_{r}}{2\pi {f}_{0}{V}^{2}}$$ |

L_{2} | None |
$$\begin{array}{c}\frac{\left(1-\frac{{f}_{1}^{2}}{{f}_{s}^{2}}\right)\left(1-\frac{{f}_{1}^{2}}{{f}_{p}^{2}}\right)}{C(2\pi {f}_{1})}\\ where\text{\hspace{0.17em}}\text{\hspace{0.17em}}{f}_{s}=\frac{1}{2\pi \sqrt{LC}},\text{\hspace{0.17em}}{f}_{p}=\frac{{f}_{1}{f}_{2}}{{f}_{s}}\end{array}$$ | None |
$$\frac{1}{{C}_{2}{(2\pi n{f}_{0})}^{2}}$$ |

C_{2} | None |
$$\frac{1}{{L}_{2}{\left(2\pi {f}_{p}\right)}^{2}}$$ | None |
$$C({n}^{2}-1)$$ |

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