Differential Signal and Common Mode Signal

in Time Domain

Most of multi-Gbps IO technologies use differential signaling, and their typical signalpath impedance is 100ohm differential. Two 50ohm cables, however, are usually used inparallel to connect a DUT to/from test and measurement instruments such as oscilloscopeand BERT. One may wonder why two 50ohm cables in parallel make 100ohm differentialpath. Or, one may wonder why it is often said that the two cables’ characteristic such ascable type and length must be matched well. Starting with the definition of differential sig-nal, we will discuss 100ohm differential signal path vs. dual 50ohm signal path in detail inthis article. In the end, we will study why the two 50ohm cables’ length must be matchedwell when they are used to send differential signal.

1.0 Mode of Signal

1.1 Differential Mode and Common Mode

Most of practical high speed IOs use two signal lines plus ground to send differentialsignal. Such a transmission medium is considered as a four-port network. The volt-ages and the currents at the four ports are denoted here as shown in Fig.1.

Assuming that v1 and v2 are input voltages and i1 and i2 are input currents, the differ-ential voltage, the common mode voltage, the differential current and the commonmode current are defined by (Eq.1) through (Eq.4) respectively.

While (Eq.1) and (Eq.2) are intuitively understandable as the two voltages’ differ-ence and the average respectively, one may wonder why the differential current isthe two currents’ difference divided by two, and the common mode current is thesum of the two currents. The reason is because the total power calculated with (v1, i1,v2, and i2) must be equal to the total power calculated with (vd, id, vc, and ic) asexpressed by (Eq.5) and (Eq.6).

v3

i3

v4

i4v1

i1

v2

i24-PortNetwork

Fig.1 A Network with Two Input and Two Output Ports

vd v1 v2–≡

vcv1 v2+

2----------------≡

idi1 i2–

2--------------≡

ic i1 i+≡ 2

(Eq.1)

(Eq.2)

(Eq.3)

(Eq.4)

1.2 Odd Mode and Even Mode

Defining even and odd voltage vectors by (Eq.7) and even and odd current vectorsby (Eq.8), v1 and v2 are expressed by (Eq.9), and i1 and i2 are expressed by (Eq.10).From (Eq.9), vc and vd can be considered as the coefficients of even voltage vectorand odd voltage vector respectively. From (Eq.10), ic and id can be considered as thecoefficients of even current vector and odd current vector respectively.

The electric fields of the even and odd modes of a coupled microstrip lines are illus-trated in Fig.2.

Although even and odd modes may seem to be introduced only for convenience inthe discussion above, these two modes are the eigenvectors of the circuit (i.e., thereis a significant physical meaning) when the signal path consisting of two signal linesis symmetric. Refer to the subsection 3.1.

P1 v1 i1.= P2 v2 i2

.=

Pd vd id= Pc vc ic.=

(Eq.5)

Pd Pc+ v1 v2–(i1 i2–

2--------------

v1 v2+2

---------------- i1 i2+(+=

v1 i1 v2 i2+= P1 P2+= (Eq.6)

veven11

vodd11–

ieven11

iodd11–

(Eq.7)

(Eq.8)

v1

v2vc

11

vd2----- 1

1–+=

i1

i2

ic2--- 1

1id

11–

+=

(Eq.9)

(Eq.10)

(a) Even Mode (b) Odd Mode

Fig.2 E-field Cross Sections of Even and Odd Modes of a Coupled Microstrip Lines

.

) . . )

. .

≡ ≡

≡≡

2.0 Impedance and Signal Path

Having defined differential and common mode voltages and currents, let’s study theimpedance of each mode.

2.1 Geometrical Characteristic of Signal Path

2.1.1 Asymmetric Signal PathThe voltage and current pairs at the input of an asymmetric signal path is related viaan impedance matrix as expressed by (Eq.11). To obtain the common mode imped-ance, force even mode current, then v1 and v2 are measured as expressed by (Eq.12).From the definition of common mode voltage and current, (Eq.13) and (Eq.14) areobtained, from which the common mode impedance is obtained as expressed by(Eq.15). Forcing odd mode current, the differential mode impedance is obtained like-wise as expressed by (Eq.19).

2.1.2 Symmetric Signal Path

Symmetric signal path is used for differential signal in practice, where the voltage andcurrent relation is expressed by (Eq.20). Since Za is the single-ended impedance whenonly one signal line is excited, and Zb is related to the coupling of the two lines, let’srename them as expressed by (Eq.21). Then the voltage and current relation is ex-pressed by (Eq.22). Forcing even mode current, the common mode impedance is ob-tained as expressed by (Eq.23). Forcing odd mode current, the differential modeimpedance is obtained as expressed by (Eq.24).

v1

v2

Z11 Z12

Z21 Z22

i1

i2=

i1 i2 i0= =v1 Z11 Z12+( i0=

ic i1 i2+ 2i0=

vd v1 v2– Z11 Z22+ Z12 Z21+– i0=

v2 Z21 Z22+ i0=

vcv1 v2+

2----------------

Z11 Z12 Z21 Z22+ + +2

---------------------------------------------------- i0=

Zcvcic----

Z11 Z12 Z21 Z22+ + +4

----------------------------------------------------=

idi1 i2–

2-------------- i0=

v1 Z11 Z12– i0=

v2 Z21 Z22– i0=

Zdvdid----- Z11 Z22+ Z12 Z21+–=

For even mode current

i1 i2– i0= =For odd mode current

(Eq.11)

(Eq.12)

(Eq.13)

(Eq.14)

(Eq.15)

(Eq.16)

(Eq.17)

(Eq.18)

(Eq.19)

)

( )

≡ ×

×

≡

∴ ≡

( )

( )

≡ [ ( ) )(

]×

≡

∴ ≡ ( ) ( )

When even mode current is forced, the even mode voltage is obtained as expressedby (Eq.25), and the even mode impedance is obtained as expressed by (Eq.26). Whenodd mode current is forced, the odd mode voltage is obtained as expressed by (Eq.27),and the odd mode impedance is obtained as expressed by (Eq.28). From (Eq.23) and(Eq.26), the relation between the common mode impedance and the even mode im-pedance is obtained as expressed by (Eq.29). From (Eq.24) and (Eq.28), the relationbetween the differential mode impedance and the odd mode impedance is obtained asexpressed by (Eq.30).

When two 50ohm cables are used in parallel, there is no coupling between the twocables. Therefore, the odd mode impedance of this dual 50ohm cables path becomes50ohm by (Eq.28). Then the differential impedance becomes 100ohm by (Eq.30).

v1

v2

Za Zb

Zb Za

i1

i2= (Eq.20)

Za Z0

Zb Z0(Eq.21)

Zcvcic---- 1

2---Z0 1 +=

Zdvdid----- 2Z0 1 –=

(Eq.22)

(Eq.23)

v1

v2

Z0 Z0Z0 Z0

i1

i2=

(Eq.24)

0

vevev v1 v2 Z0 1 + ieven= = =

Zevenvevevieven------------ Z0 1 + Z0=

i1 i2 ieven=

i1 i– 2 iodd=

Zc12--- Zeven=

vodd v1 v– 2 Z0 1 – iodd= = =

Zoddvoddiodd---------- Z0 1 – Z0=

Zd 2 Zodd=

(Eq.25)

(Eq.26)

(Eq.27)

(Eq.28)

(Eq.29)

(Eq.30)

≡

≡

α × (α ≥ )

αα

≡ ( α)

≡ ( α)

≡

( α)

∴ ≡ ( α) ≥

≡

( α)

∴ ≡ ( α) ≤

×

×

2.2 Impedance Matching at Receiver

With differential signaling, our intention is usually such that differential mode isused to send useful information and common mode is used to provide DC bias. Inthis case, signal path termination is considered only for differential signal, that is,only differential impedance matching is considered. AC common mode, however,usually exists in real life. In this case, if common mode impedance is not matched,AC common mode is reflected, and it could cause problems such as generating dif-ferential noise through mode conversion and EMI. Let’s analyze differential andcommon mode impedance matching with the following three examples. Note thatdifferential impedance matching is equivalent to odd mode impedance matching,and common mode impedance matching is equivalent to even mode impedancematching. We use even/odd modes impedance matching here for simplicity.

2.2.1 LVDS Type

LVDS type termination scheme is shown in Fig.3. The even mode impedance at thedevice input is infinite. If AC even mode exits, it is 100% reflected at the device input.The odd mode impedance matching is achieved by matching Ra/2 to the odd modeimpedance of the signal path.

In order to achieve impedance matching for both odd mode and even mode, two moreresistors (Rb x 2) are required as shown in Fig.4. Since the even mode signal does notsee Ra, and Rb directly terminates the even mode signal, Rb should be equal to theeven mode impedance of the signal path. Since the odd mode signal sees Ra/2 and Rbin parallel, the odd mode impedance matching condition is expressed by (Eq.31),from which Ra is obtained as expressed by (Eq.32).

Ra

Fig.3 LVDS Type Termination

Zeven =

ZoddRa2

------=

Rb Zeven=

RbRa2

------ Zodd=

Zeven Ra2Zeven Ra+----------------------------- Zodd=

Ra2Zeven ZoddZeven Zodd–--------------------------------=

(1) even mode impedance matching

(2) odd mode impedance matching

Ra

Rb

Rb

Fig.4 Modified LVDS Type Termination

(Eq.31)

(Eq.32)

∞

| | .

∴.

2.2.2 CML Type

CML type termination is shown in Fig.5. Both the even and the odd mode signals di-rectly see Ra at the device input. When there is no coupling between the two symmet-ric signal lines, the even mode impedance and the odd mode impedance become thesame as expressed by (Eq.26) and (Eq.28). In this case, impedance matching for bothodd mode and even mode is achieved by matching Ra to the even/odd mode signalpath impedance.

When the even mode impedance and the odd mode impedance are different, one moreresistor (Rb) is required to achieve the impedance matching for both modes as shownin Fig.6. Since the odd mode signal directly sees Ra at the device input, Ra needs tobe equal to the odd mode signal path impedance. The even mode impedance at thedevice input is expressed by (Eq.33), from which Rb is obtained as expressed by(Eq.34).

2.2.3 Test and Measurement Instruments

Typical outputs or inputs terminals of test and measurement instruments are two50ohm connectors, one for true signal and another for complement signal. UtilizingCML type termination, impedance matching for both differential mode and commonmode is achieved.

Ra

Ra

Zeven Ra=

Zodd Ra=

Fig.5 CML Type Termination

VT

Ra Zodd=

Zevenveie---- Zodd 2Rb+= =

RbZeven Zodd–

2------------------------------=

Rb

Ra

Ra

Fig.6 Modified CML Type Termination

(1) odd mode impedance matching

(2) even mode impedance matching

(Eq.33)

(Eq.34)

ve Ra ie Rb 2ie+ Zodd 2Rb+ ie= =. . ( )

∴

3.0 Mode Conversion

Assume that useful information is sent by differential mode signal (i.e. odd modesignal) at a transmitter. Ideal situation is that the differential signal propagates to areceiver without distortion. Part of the original differential signal, however, could beconverted to common mode signal during propagation, and the receiver might notfully receive the original signal.

3.1 Condition for No Mode ConversionLet’s study the condition under which mode conversion does not occur. Then we willknow when mode conversion would occur too. Applying even mode current to(Eq.11), the resulting differential voltage is obtained by (Eq.35). Applying odd modecurrent to (Eq.11), the resulting common mode voltage is obtained by (Eq.36).

In order for mode conversion not to occur, the differential mode voltage expressed by(Eq.35) and the common mode voltage expressed by (Eq.36) must be zero. Then theno-mode-conversion condition is obtained as expressed by (Eq.39).

Therefore the impedance matrix needs to be symmetric for mode conversion not tooccur as expressed by (Eq.40). Then one can recognize from (Eq.41) and (Eq.42) thateven mode current and odd mode current are the eigenvectors of the symmetric im-pedance matrix. Likewise even mode voltage and odd mode voltage are the eigenvec-tors of the associated admittance matrix.

i1 i2 i0= =For even mode current

i1 i2– i0= =For odd mode current

vd v1 v2– Z11 Z12+ Z21 Z22+– i0= (Eq.35)

vcv1 v2+

2----------------

Z11 Z12– Z21 Z22–+2

------------------------------------------------------------- i0= (Eq.36)

For no mode conversion to occur,Z11 Z12+ Z21 Z22+– 0=

Z11 Z12– Z21 Z22–+ 0=

(Eq.37)

(Eq.38)

(Eq.39)Equivalent to Z12 Z21= Z11 Z22=and

Z11 Z12

Z21 Z22

Za Zb

Zb Za=

Za Zb

Zb Za

11

Za Zb+ 11

=

Za Zb

Zb Za

11–

Za Zb– 11–

=

(Eq.40)

(Eq.41)

(Eq.42)

≡

≡

( ) ] ×[

(

(

(

)

) )×

(

(

(

(

) )

) )

(

(

)

)

3.2 Mode Conversion due to Signal Path SkewFrom the discussion above, when two signal lines are not symmetric, mode conver-sion occurs. As an asymmetric signal path example, let’s examine non coupled dualtransmission lines whose characteristic impedance are the same (50ohm for exam-ple), but their line length is different. When odd mode sinusoidal signal is sent fromone end of the signal path, the outputs at the two lines at another end are expressed by(Eq.43) and (Eq.44). Then the differential mode signal and the common mode signalare expressed by (Eq.45) and (Eq.46).

With the sinusoidal signals’ period normalized to 1 and the skew of 10% of the signalperiod, the two single-ended signals are shown in Fig.7. The corresponding differen-tial and common mode signals are shown in Fig.8. Note that the common mode signaldoes not exist with zero skew. It is the skew that converts part of the original differ-ential signal energy into common mode energy.

v1 t 2 tT

--------cos=

v2 t tskew,2 t tskew+

T-------------------------------cos–=

vd t tskew, v1 t v2 t tskew,–

vc skew 2---------------------------------------------

(Eq.43)

(Eq.44)

(Eq.45)

(Eq.46)

0 0.5 1 1.5 2 2.5 32

1

0

1

22

2?

v d t 0.1,( )

v c t 0.1,( )

30 t

0 0.5 1 1.5 2 2.5 31

0.5

0

0.5

11

1?

v 1 t( )

v 2 t 0.1,( )

30 t

Fig.7 Skewed Signals at Two Output Ports

Fig.8 Differential and Common Modes Signals at Output Ports

( )

π

( )π ( )

( ) ≡ ( ) ( )

t t,( ) ≡v1 t v2 t tskew,( ) ( )+

Normalizing each line impedance to 1 as expressed by (Eq.47), the differential signalpower as the function of the skew is expressed by (Eq.48) and the common modepower is expressed by (Eq.49).

The calculated differential and common modes power is shown in Fig.9 with the sig-nal path skew varying from 0% to 100% of the sinusoidal signal period. Since v1 andv2 without signal path skew are the already 50% skewed sinusoidal signals, additional50% skew due to the signal path completely eliminates the originally intended differ-ential mode.

We have discussed how the differential and the common modes energy varies de-pending on the skew between the two sinusoidal signals of a given frequency. Thisalso means that the amount of the differential and the common modes energy variesdepending on the frequency of the two sinusoidal signals of a given absolute skew.Normalizing the line impedance to 1 as done for (Eq.48) and (Eq.49), the differentialsignal power as the function of the frequency is expressed by (Eq.50), and the com-mon mode power is expressed by (Eq.51). Let’s assume the same transmission linesas discussed in Fig.7 and Fig.8, where their characteristic impedance is the same, buttheir line length is different by 0.1 second. The calculated differential and commonmodes power is shown in Fig.10 with the signal frequency varying from 0.1Hz to10Hz.

Pd tskewvd t tskew

2

Zd-----------------------------

Z0Zd2----- 2Zc 1= = =

12--- 1

T--- vd t tskew

2 td0

T=

Pc tskewvc t tskew

2

Zc-----------------------------

2 1T--- vc t tskew

2 td0

T= (Eq.49)

(Eq.47)

(Eq.48)

0 0.25 0.5 0.75 10

0.25

0.5

0.75

11

0

P d t skew

P c t skew

10 t skewFig.9 Differential and Common Modes Power as the Function of Skew

( ) ≡ ⟨ ⟩ ( , )

× ∫ ( , )

⟨ ( ) ≡ ⟩( , )

× ∫ ( , )

(

(

)

)

3.3 Bandwidth Reduction due to Signal Path Skew

The frequency domain information such as the one shown in Fig.10 is critical for ourapplications because high speed digital signal has broad frequency spectrum. Fig.10indicates that skewed signal path intended for differential signal transmissionbehaves as a low pass filter even if each signal line does not cause signal loss such asdielectric loss and skin effect loss. Using two skewed single-ended signals expressedby (Eq.43) and (Eq.44), the corresponding differential signal can be obtained asexpressed by (Eq.52). Thus, the cut of frequency of this signal path for differentialsignal can be obtained as a function of the skew by solving (Eq.52). The -3dB band-width and -1dB bandwidth vs. the dual lossless transmission lines skew is plotted inFig.11. For example, 25ps skew makes the -3dB bandwidth of a differential signalpath only 10GHz although each transmission line has no loss.

0.1 1 1060

40

20

00

60?

Pd dB f( )

Pc dB f( )

50.1 f

f 1T---=

PddB f 20 12--- f vd t tskew

2 td0

1f---

log

PcdB f 20 2 f vc t tskew2 td

0

1f---

log

Fig.10 Differential and Common Modes Power as the Function of Frequency

(Eq.50)

(Eq.51)

vd t tskew v1 t v2 t tskew–

2 2 f ttskew

2------------+cos ftskewcos=

vd t f tskew= (Eq.52)

v1 t 2 tT

--------cos=

v2 t tskew2 t tskew+

T-------------------------------cos–=

(Eq.43)

(Eq.44)

( ) ≡ × × ∫ ( , )

( ) ≡ × × ∫ ( , )

( ) π

( , )π ( )

( , )≡ ( ) ( , )

π

(π )

)( , ,

. .

0 5 10 15 20 25 30 35 40 45 500

5

10

15

20

25

30

35

40

45

5050

0

f 3dB dt( )

f 1dB dt( )

502 dt

Fig.11 Skew of Dual Transmission Lines vs. Differential Signal Path Bandwidth

[ps]

[GHz]