Document 191-1

CommonModeFilterDesignGuide
Introduction
The selection of component values for common mode
filters need not be a difficult and confusing process. The
use of standard filter alignments can be utilized to achievea relatively simple and straightforward design process,though such alignments may readily be modified to utilize
pre-defined component values.

GeneralLine filters prevent excessive noise from being conducted
between electronic equipment and the AC line; generally,
the emphasis is on protecting the AC line. Figure 1 shows
the use of a common mode filter between the AC line (via
impedance matching circuitry) and a (noisy) power converter. The direction of common mode noise (noise on bothlines occurring simultaneously referred to earth ground) isfrom the load and into the filter, where the noise common
to both lines becomes sufficiently attenuated. The resulting common mode output of the filter onto the AC line (viaimpedance matching circuitry) is then negligible.

ACLineInZ-Matching

Filtering

Converter

Any inductance encountered by the differential signal isthen the result of imperfect coupling of the two chokes;
they perform as independent components with their leakage inductances responding to the differential signal: theleakage inductances attenuate the differential signal.
When the inductors, L1 and L2, encounter an identicalsignal of the same polarity referred to ground (commonmode signal), they each contribute a net, non-zero flux inthe shared core; the inductors thus perform as independent components with their mutual inductance responding to the common signal: the mutual inductance thenattenuates this common signal.

The First Order Filter
The simplest and least expensive filter to design is a firstorder filter; this type of filter uses a single reactive
component to store certain bands of a spectral energy
without passing this energy to the load. In the case of alow pass common mode filter, a common mode choke isthe reactive element employed.
The value of inductance required of the choke is simply theload in Ohms divided by the radian frequency at and above
which the signal is to be attenuated. For example, attenuation at and above 4000 Hz into a 50⏲ load would requirea 1.99 mH (50/(2π x 4000)) inductor. The resulting common
mode filter configuration would be as follows:

Figure 1. Generalized line filtering

1.99 mH

The design of a common mode filter is essentially thedesign of two identical differential filters, one for each of
the two polarity lines with the inductors of each side
coupled by a single core:
L1+
VCMin
–

50 Ω+
VCMout
–
50 Ω–
VCMin
+

50 Ω–
VCMout+1.99 mH(A)(B)

L2

Figure 2. The common mode inductor

For a differential input current ( (A) to (B) through L1 and(B) to (A) through L2), the net magnetic flux which is
coupled between the two inductors is zero.

Figure 3. A first order (single pole) common mode filter

The attenuation at 4000 Hz would be 3 dB, increasing at
6 dB per octave. Because of the predominant inductor
dependence of a first order filter, the variations of actual
choke inductance must be considered. For example, a±20% variation of rated inductance means that thenominal 3 dB frequency of 4000 Hz could actually beanywhere in the range from 3332 Hz to 4999 Hz. It is
typical for the inductance value of a common mode choke

Document 191-1© Coilcraft, Inc. 2007

Revised 11/08/07

Document 191-2

to be specified as a minimum requirement, thus insuringthat the crossover frequency not be shifted too high.
However, some care should be observed in choosing achoke for a first order low pass filter because a muchhigher than typical or minimum value of inductance may
limit the choke’s useful band of attenuation.

Second Order Filters
A second order filter uses two reactive components andhas two advantages over the first order filter: 1) ideally, asecond order filter provides 12 dB per octave attenuation(four times that of a first order filter) after the cutoff point,and 2) it provides greater attenuation at frequencies
above inductor self-resonance (See Figure 4).
LRL

RL+
VCMout–+
VCMout–One of the critical factors involved in the operation ofhigher order filters is the attenuating character at thecorner frequency. Assuming tight coupling of the filter
components and reasonable coupling of the choke itself(conditions we would expect to achieve), the gain near thecutoff point may be very large (several dB); moreover, thetime response would be slow and oscillatory. On the otherhand, the gain at the crossover point may also be lessthan the presumed -3 dB (3 dB attenuation), providing agood transient response, but frequency response nearand below the corner frequency could be less thanoptimally flat.
In the design of a second order filter, the damping factor(usually signified by the Greek letter zeta (ζ )) describes
both the gain at the corner frequency and the time
response of the filter. Figure (5) shows normalized plotsof the gain versus frequency for various values of zeta.

CDifferential
Load

20+VCMin–AC

10L==1Ls + LCs 21+RL1⎛ L⎞
1− LCω 2 + jω ⎜ ⎟⎝ RL ⎠CGain (dB)

VCMout ( s )=VCMin ( s )B

0D-10E-201ω ⎛ω⎞
1 + j 2ζ−⎜ ⎟ωn ⎝ ωn ⎠2ω ≡ radian frequency
R L ≡ the noise load resistance1ωn ≡LCLζ≡
2RL LC-30

1

Gain =
1+ j-40
0.1Wn2ζω ⎛ ω ⎞−⎜⎟ωn ⎝ ω n ⎠2dB

1Wn

10Wn

Radian Frequency, WA ≡ ζ = 0.1; B ≡ ζ = 0.5; C ≡ ζ = 0.707; D ≡ ζ = 1.0; E ≡ ζ = 4.0
Figure 5. Second order frequency response for various
damping factors (ζ)

Figure 4. Analysis of a second order (two pole) common mode
low pass filter

The design of a second order filter requires more care andanalysis than a first order filter to obtain a suitable
response near the cutoff point, but there is less concern
needed at higher frequencies as previously mentioned.

As the damping factor becomes smaller, the gain at thecorner frequency becomes larger; the ideal limit for zero
damping would be infinite gain. The inherent parasitics ofreal components reduce the gain expected from ideal
components, but tailoring the frequency response within
the few octaves of critical cutoff point is still effectively afunction of ideal filter parameters (i.e., frequency, capacitance, inductance, resistance).

Document 191-2© Coilcraft, Inc. 2007

Revised 11/08/07

Document 191-3

For some types of filters, the design and damping characteristics may need to be maintained to meet specificperformance requirements. For many actual line filters,however, a damping factor of approximately 1 or greaterand a cutoff frequency within about an octave of thecalculated ideal should provide suitable filtering.
The following is an example of a second order low passfilter design:
1) Identify the required cutoff frequency:
For this example, suppose we have a switching power
supply (for use in equipment covered by UL478) thatis actually 24 dB noisier at 60 KHz than permissible forthe intended application. For a second order filter (12dB/octave roll off) the desired corner frequency would
be 15 KHz.
2) Identify the load resistance at the cutoff frequency:
Assume RL = 50 Ω3) Choose the desired damping factor:
Choose a minimum of 0.707 which will provide 3 dB
attenuation at the corner frequency while providing
favorable control over filter ringing.
4) Calculate required component values:

ωn = 2πfn = 94248 rad / sec
C=1
Lωn 2Lω nζ = 0.707 =2RL
L = 750μH
5) Choose available components:
C = 0.05 µF (Largest standard capacitor value thatwill meet leakage current requirements for UL478/
CSA C22.2 No. 1: a 300% decrease from design)
L = 2.1 mH (Approx. 300% larger than design tocompensate for reduction or capacitance: Coilcraft
standard part #E3493-A)
6) Calculate actual frequency, damping factor, and attenuation for components chosen:
1= 15532 Hz (very nearly 15 KHz)
2π LC
ζ = 2.05 (a damping factor of about 1 or more is
acceptible)
Attenuation = (12 dB/octave) x 2 octaves = 24 dB

Note: Damping factors much greater than 1 may cause
unacceptably high attenuation of lower frequencies whereas a damping factor much less than
0.707 may cause undesired ringing and the filter
may itself produce noise.

Third Order Filters
A third order filter ideally yields an attenuation of 18 dB per
octave above the cutoff point (or cutoff points if the threecorner frequencies are not simultaneous); this is theprominently positive aspect of this higher order filter. The
primary disadvantage is cost since three reactive components are now required. Higher than third order filters aregenerally cost-prohibitive.
L1L2+VCMin–CRLCRL

L1+VCMout–DifferentialLoad–VCMout+L2(RL + L2 s )⎛⎞⎟VCMout (s) ⎛ RL ⎞ ⎜sC⎟=⎜⎟⎜(RLs)
Ls+VCMin(s)⎝ R L + L2 s ⎠ ⎜ L2⎜+ RL L1s + L2 L1s 2 + 1 ⎟⎟⎝sCsC ⎠=1
L1 + L2
LL C
1+s + L1Cs 2 + 1 2 s 3
RLRL

12s 2s 2 s 31+++ωn1 ωn 2 ωn3

Butterworth →(L1 + L2 ) = 2 ; ωRL

L1C =

ω n12
ω n2 2n1 =; ω n2 =2RL
L
( 1 + L2 )1414.

(L1C )

L 1L 2 C
12 RL= ωn2 2ω n 4=3 ; ω n3 = ωn2RL
2L 2
ωn3
ωn4 =RL2L 2Figure 6. Analysis of a third order (three pole) low pass filter
where ω1, ω2 and ω4 occur at the same -3dB frequency of ω07) The resulting filter is that of figure (4) with:
L = 2.1 mH; C = 0.05 µF; RL = 50 ΩDocument 191-3© Coilcraft, Inc. 2007

Revised 11/08/07

Document 191-4First Order

Second Order

L2

Filter Schematic+ei–Third Order

L1RL+eO–+ei–

Filter Transfer
Functioneo1=
Lsei+1RL

eo=ei

Butterworth
Alignmenteo
1=sei+1ωneo=eiC

L1RL+eO–

1LCs 2 +Ls+1RL

1LCs 2 +Ls+1RL

L2+ei–

CRL+eO–eo1=
L1L2 3(L + L )seis + L1Cs 2 + 1 2 + 1RLRLeo= 3eisω n31+22s2s++1ω n2 ω n1Figure 7. The first three order low pass filters and their Butterworth alignments

The design of a generic filter is readily accomplished byusing standard alignments such as the Butterworth (“maximally flat”) alignments. Figure (6) shows the general
analysis and component relationships to the Butterworthalignments for a third order low pass filter. Butterworthalignments provide an inherent ζ of 0.707 and a -3 dB pointat the crossover frequency. The Butterworth alignments
for the first three orders of low pass filters are shown in
Figure (7).
The design of a line filter need not obey the Butterworthalignments precisely (although such alignments do provide a good basis for design); moreover, because ofleakage current limits placed upon electronic equipment(thus limiting the amount of filter capacitance to ground),adjustments to the alignments are usually required, butthey can be executed very simply as follows:
1) First design a second order low pass with ζ ≥ 0.5
2) Add a third pole (which has the desired corner frequency) by cascading a second inductor between thesecond order filter and the noise load:
L = R/ (2 π fc )Where fc is the desired corner frequency.

Design Procedure
The following example determines the required component values for a third order filter (for the same requirements as the previous second order design example).
1) List the desired crossover frequency, load resistance:
Choose fc = 15000 Hz
Choose RL = 50 Ω2) Design a second order filter with ζ = 0.5 (see secondorder example above):3) Design the third pole:
RL/(2πfc ) = L250/(2π15000) = 0.531 mH4) Choose available components and check the resulting
cutoff frequency and attenuation:
L2 = 0.508 mH (Coilcraft #E3506-A)
fn = R/(2πL1 )= 15665 Hz
Attenuation at 60 KHZ: 24 dB (second order filter) +2.9 octave × 6 = 41.4 dB5) The resulting filter configuration is that of figure (6)with:L1 = 2.1 mH
L2 = 0.508 mH
RL = 50 ΩConclusions
Specific filter alignments may be calculated by manipulating the transfer function coefficients (component values) of a filter to achieve a specific damping factor.
A step-by-step design procedure may utilize standard
filter alignments, eliminating the need to calculate thedamping factor directly for critical filtering. Line filters,with their unique requirements, yet non-critical characteristics, are easily designed using a minimum allowable
damping factor.
Standard filter alignments assume ideal filter components; this does not necessarily hold true, especially athigher frequencies. For a discussion of the non-ideal
character of common mode filter inductors refer to theapplication note “Common Mode Filter Inductor Analysis,”
available from Coilcraft.

Document 191-4© Coilcraft, Inc. 2007

Revised 11/08/07