Hybrid filters

ELL 758 Power Quality

(July-Dec 2024)
Anandarup Das
Department of Electrical Engineering, IIT Delhi.
anandarup@ee.iitd.ac.in
Content
• Concept of hybrid filters
• Many combinations
• Operating principle
• Controller

Harmonic Sources and Filtering approaches

F. Z. Peng, article in IEEE Industrial Electronics Magazine,
1999 IEEE/IAS Annual Meeting.

ANANDARUP DAS, INDIAN INSTITUTE OF TECHNOLOGY, DELHI, INDIA. FOR INTERNAL USE ONLY, NOT TO BE COPIED. 2
Passive and active filters
• Passive filters
• Pros:
• Cheap, easy to design, no controller design
• Reactive power support
• Cons:
• Parameter variation, aging issue
• System parallel resonance can happen
• Dependent on system impedance Ls (which
may change over time)
• Harmonics at the voltage source

ANANDARUP DAS, INDIAN INSTITUTE OF TECHNOLOGY, DELHI, INDIA. FOR INTERNAL USE ONLY, NOT TO BE COPIED. 3
Passive and active filters
• Active filters
• Pros:
• Much more accurate
• Dynamic adjustment is possible
• Cons:
• Costly
• Controller design and development
• Reactive power is usually not supplied

ANANDARUP DAS, INDIAN INSTITUTE OF TECHNOLOGY, DELHI, INDIA. FOR INTERNAL USE ONLY, NOT TO BE COPIED. 4
Purpose
• A hybrid filter tries to combine a passive filter and an active filter to get the
benefit of both.
• Many combinations are possible
• Series active, parallel passive
• Parallel active, series passive
• Series active, series passive
• Parallel active, parallel passive
• Etc.

ANANDARUP DAS, INDIAN INSTITUTE OF TECHNOLOGY, DELHI, INDIA. FOR INTERNAL USE ONLY, NOT TO BE COPIED. 5
Some circuits

Parallel active, parallel passive

Current source type load

Series active, Series passive

Voltage source type load

ANANDARUP DAS, INDIAN INSTITUTE OF TECHNOLOGY, DELHI, INDIA. FOR INTERNAL USE ONLY, NOT TO BE COPIED. 6
Some circuits

Series active, parallel passive

Current source type load
(discussed later)

Parallel active, Series passive

Voltage source type load

ANANDARUP DAS, INDIAN INSTITUTE OF TECHNOLOGY, DELHI, INDIA. FOR INTERNAL USE ONLY, NOT TO BE COPIED. 7
Some circuits

Series combination of parallel

passive, parallel active
Current source type load

Parallel combination of
series active, Series passive
Voltage source type load

ANANDARUP DAS, INDIAN INSTITUTE OF TECHNOLOGY, DELHI, INDIA. FOR INTERNAL USE ONLY, NOT TO BE COPIED. 8
Few considerations
• Active filters are expensive, and it is better to use them for removing lower
order harmonics (5th and 7th). This will keep the switching frequency low
and increase the efficiency.
• Passive filters can be used to remove higher order harmonics and provide
reactive power at fundamental frequency.
• It is also desirable to reduce the power rating of active filter.

ANANDARUP DAS, INDIAN INSTITUTE OF TECHNOLOGY, DELHI, INDIA. FOR INTERNAL USE ONLY, NOT TO BE COPIED. 9
Series active parallel passive filter
• The circuit is shown.
• It has the following advantages.
• Greatly increases the system impedance seen from the load terminal.
• Acts as a harmonic isolator for source and load harmonic sources.
• Parallel resonance is highly minimized.

ANANDARUP DAS, INDIAN INSTITUTE OF TECHNOLOGY, DELHI, INDIA. FOR INTERNAL USE ONLY, NOT TO BE COPIED. 10
Series active parallel passive filter
• The equivalent circuit is shown.
𝑉𝑐
• Ideally, the impedance 𝐺 = is 0 at At fundamental
𝐼𝑠 frequency
fundamental frequency and very high
at harmonic frequency.

At harmonic
frequency

ANANDARUP DAS, INDIAN INSTITUTE OF TECHNOLOGY, DELHI, INDIA. FOR INTERNAL USE ONLY, NOT TO BE COPIED. 11
Operating principle
• At harmonic frequency, source current: 𝐼𝑠ℎ =
𝑍𝑓 1
𝐼𝐿ℎ + 𝑉𝑠ℎ
𝑍𝑓 +𝑍𝑠 +𝐺 𝑍𝑓 +𝑍𝑠 +𝐺
𝑉𝑐
• Here, 𝐺 =
𝐼𝑠
• If 𝐺 ≫ 𝑍𝑓 + 𝑍𝑠 and 𝐺 ≫ 1, at the harmonic
frequency, then 𝐼𝑠ℎ ≅ 0.

ANANDARUP DAS, INDIAN INSTITUTE OF TECHNOLOGY, DELHI, INDIA. FOR INTERNAL USE ONLY, NOT TO BE COPIED. 12
Operating principle
𝑍𝑓 1
• Source current: 𝐼𝑠ℎ = 𝐼𝐿ℎ + 𝑉𝑠ℎ
𝑍𝑓 +𝑍𝑠 +𝐺 𝑍𝑓 +𝑍𝑠 +𝐺
• For example, the active filter is used for 5th
harmonic compensation.
• Consider 𝑍𝑠 = 𝑗0.05 𝑝𝑢 at fundamental
frequency, so 𝑍𝑠 = 𝑗0.25 𝑝𝑢 at 5th harmonic.
• 𝑍𝑓 = 0.01 pu and 𝐺 = 10 𝑝𝑢 at 5th harmonic.
0.01 1
• 𝐼𝑠ℎ = 𝐼𝐿ℎ + 𝑉𝑠ℎ
0.01+𝑗0.25+10 0.01+𝑗0.25+10
• Very small impact of 𝐼𝐿ℎ , less impact of 𝑉𝑠ℎ on the
source current harmonic.

ANANDARUP DAS, INDIAN INSTITUTE OF TECHNOLOGY, DELHI, INDIA. FOR INTERNAL USE ONLY, NOT TO BE COPIED. 13
Enhancement of passive filter performance
• Series active filter is now adding to the source
impedance; thus, the passive filter will absorb
greater portion of load harmonics.
• Series active filter is also acting as a high impedance
to the voltage harmonics present upstream.
Otherwise, these harmonics will appear across the
passive filter and will be absorbed by it, in addition
to load generated harmonics.
• Parallel resonance problem is minimized.
• Series active filter therefore enhances the
performance of the passive filter, rather than as a
compensator for voltage harmonics.

ANANDARUP DAS, INDIAN INSTITUTE OF TECHNOLOGY, DELHI, INDIA. FOR INTERNAL USE ONLY, NOT TO BE COPIED. 14
Output voltage of AF
• Output voltage of series active filter: 𝑉𝑐 = 𝐺𝐼𝑠ℎ
𝐼𝐿ℎ 𝑍𝑓 +𝑉𝑠ℎ
• 𝑉𝑐 = 𝐺
𝑍𝑓 +𝑍𝑠 +𝐺
• If, 𝐺 ≫ 𝑍𝑓 + 𝑍𝑠 then, 𝑉𝑐 = 𝐼𝐿ℎ 𝑍𝑓 + 𝑉𝑠ℎ
• This indicates that the voltage rating of the AF
will reduce if the quality factor of the passive
filter improves (𝑍𝑓 reduces).

ANANDARUP DAS, INDIAN INSTITUTE OF TECHNOLOGY, DELHI, INDIA. FOR INTERNAL USE ONLY, NOT TO BE COPIED. 15
Voltage across PF
• Harmonic voltage across passive filter:
𝑍𝑓 𝑍𝑠 +𝐺
• 𝑉𝑃𝐹ℎ = 𝑉𝑠ℎ − 𝐼𝐿ℎ 𝑍
𝑍𝑓 +𝑍𝑠 +𝐺 𝑍𝑓 +𝑍𝑠 +𝐺 𝑓
• If, 𝐺 ≫ 𝑍𝑓 + 𝑍𝑠 then, 𝑉𝑃𝐹ℎ ≅ −𝐼𝐿ℎ 𝑍𝑓
• This indicates that negligible harmonic voltage
across the passive filter appears due to source
voltage harmonics.
• However, load generated harmonics will appear.

ANANDARUP DAS, INDIAN INSTITUTE OF TECHNOLOGY, DELHI, INDIA. FOR INTERNAL USE ONLY, NOT TO BE COPIED. 16
Filtering characteristics
• The figure is taken from a paper. Shows
the relationship between ISh /ILh for
different values of K (or G).
• Two passive filters are tuned at 7th and
11th harmonics. At two resonance
frequencies of the passive filters, i.e., 350
and 550 Hz, the gain is very low.
• K (or G) is ~zero at fundamental frequency.
• When no active filter is connected (K = 0),
harmonic amplification appears between
100 and 400 Hz, due to parallel resonance.
• When the active filter is connected (K >0),
no harmonic amplification occurs and the
harmonic damping increases with greater
values of K.
ANANDARUP DAS, INDIAN INSTITUTE OF TECHNOLOGY, DELHI, INDIA. FOR INTERNAL USE ONLY, NOT TO BE COPIED. 17
Waveforms
• Before the AF started, there was
reasonable harmonics in source
current, and the PF alone was not
sufficient.
• The compensating voltage 𝑉𝑐ℎ from
the AF is only 2.5V rms, so the VA
rating of the AF is very small.
• Harmonic spectrum of source
current is shown before and after the
introduction of the AF.

ANANDARUP DAS, INDIAN INSTITUTE OF TECHNOLOGY, DELHI, INDIA. FOR INTERNAL USE ONLY, NOT TO BE COPIED. 18
 Parallel Active, parallel passive filter

• Another type of hybrid filter is shown: parallel active, parallel passive.

• This has been used in practice but has some challenges.

ANANDARUP DAS, INDIAN INSTITUTE OF TECHNOLOGY, DELHI, INDIA. FOR INTERNAL USE ONLY, NOT TO BE COPIED. 19
 Parallel Active, parallel passive filter

𝑍𝐹 𝑉𝑠ℎ
• Recall that, 𝐼𝑆ℎ = 𝑍𝐹 𝐼𝐿𝑂ℎ + 𝑍
𝐹
𝑍𝑠 +1−𝐺 𝑍𝑠 +1−𝐺
• Consider 𝑍𝑠 = 𝑗0.05 𝑝𝑢 at fundamental frequency, and 𝑍𝐹 = 0.01 pu and 𝐺 =
0.8 𝑝𝑢 at 5th harmonic frequency.
0.01 𝑉𝑠ℎ 0.04 4
• 𝐼𝑆ℎ = 𝐼
0.01 𝐿𝑂ℎ + 0.01 = 𝐼𝐿𝑂ℎ + 𝑉𝑠ℎ
𝑗0.25+ 0.2 𝑗0.25+ 0.2 0.2+𝑗 0.2+𝑗

• This HF has performance issues: if the source voltage has high harmonic content,
considerable source harmonic current can flow.
ANANDARUP DAS, INDIAN INSTITUTE OF TECHNOLOGY, DELHI, INDIA. FOR INTERNAL USE ONLY, NOT TO BE COPIED. 20
Example
• A hybrid filter is used for harmonic
compensation of the 415V, 50Hz system
supplying a non-linear load, as shown in
the single line diagram. The load current
waveform is indicated. The passive filter is
tuned near 5th harmonic frequency and has
R=0.3mohm, L= 0.5mH and C= 0.8mF. The
active filter is tuned to compensate 7th and
higher harmonics in the system.

ANANDARUP DAS, INDIAN INSTITUTE OF TECHNOLOGY, DELHI, INDIA. FOR INTERNAL USE ONLY, NOT TO BE COPIED. 21
Example

• (a) Find the peak magnitude of the 5th harmonic current flowing into the
source if none of the filters are used.
50 5𝜋
• If no filters are used, then current flowing into source is 4 × cos =
5𝜋 6
11A peak.

ANANDARUP DAS, INDIAN INSTITUTE OF TECHNOLOGY, DELHI, INDIA. FOR INTERNAL USE ONLY, NOT TO BE COPIED. 22
 Example
• A hybrid filter is used for harmonic compensation
of the 415V, 50Hz system supplying a non-linear
load, as shown in the single line diagram. The load
current waveform is indicated. The passive filter is
tuned near 5th harmonic frequency and has
R=0.3mohm, L= 0.5mH and C= 0.8mF. The active
filter is tuned to compensate 7th and higher
harmonics in the system.
• (b) Find the peak magnitude of the 5th harmonic current flowing into the source if both
the filters are used.
1
• 𝑍𝑓5 = 0.0003 +j 5 × 314.15 × 0.5 × 10−3 − = 0.0003 − j0.0107
5×314.15×0.8×10−3
• 𝑍𝑠5 = j 5 × 314.15 × 0.5 × 10−3 = j0.785
𝑍𝑓5
• 𝐼𝑠5 = × 11 = 0.152A
𝑍𝑓5 +𝑍𝑠5

ANANDARUP DAS, INDIAN INSTITUTE OF TECHNOLOGY, DELHI, INDIA. FOR INTERNAL USE ONLY, NOT TO BE COPIED. 23
 Example
• A hybrid filter is used for harmonic compensation
of the 415V, 50Hz system supplying a non-linear
load, as shown in the single line diagram. The load
current waveform is indicated. The passive filter is
tuned near 5th harmonic frequency and has
R=0.3mohm, L= 0.5mH and C= 0.8mF. The active
filter is tuned to compensate 7th and higher
harmonics in the system.
• (c) What is the peak magnitude of 5th harmonic voltage appearing across the passive filter
if both the filters are used?
𝑍𝑠5
• Voltage appearing across PF is 𝑍𝑓5 𝐼𝑓5 = 𝑍𝑓5 × × 11 = 0.1188 V
𝑍𝑓5 +𝑍𝑠5

ANANDARUP DAS, INDIAN INSTITUTE OF TECHNOLOGY, DELHI, INDIA. FOR INTERNAL USE ONLY, NOT TO BE COPIED. 24
 Example
• A hybrid filter is used for harmonic compensation
of the 415V, 50Hz system supplying a non-linear
load, as shown in the single line diagram. The load
current waveform is indicated. The passive filter is
tuned near 5th harmonic frequency and has
R=0.3mohm, L= 0.5mH and C= 0.8mF. The active
filter is tuned to compensate 7th and higher
harmonics in the system.
• (d) What is the peak magnitude of the 7th harmonic voltage produced by the active filter,
if negligible 7th harmonic current flows into the source? Consider both filters are
operational.
𝑍𝑓7 1
• 7th harmonic source current is: 𝐼𝑠7 = 𝐼7 × + 𝑉𝑐7 ×
𝑍𝑓7 +𝑍𝑠7 𝑍𝑓7 +𝑍𝑠7
50 7𝜋
• Since, 𝐼𝑠7 = 0, so 𝑉𝑐7 = −𝐼7 𝑍𝑓7 . Here𝐼7 = 4 × cos = 7.87A peak.
7𝜋 6
• Thus, 𝑉𝑐7 = 4.19 V. The entire 7th harmonic load current will flow through the PF.

ANANDARUP DAS, INDIAN INSTITUTE OF TECHNOLOGY, DELHI, INDIA. FOR INTERNAL USE ONLY, NOT TO BE COPIED. 25