Smith Charts

8/9/2019 Smith Charts

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*Note: The information below can be referenced to: Bowick, C., RF Circuit Design,Howard W. ams ! Com"an#, $ndiana"olis, $%, &'(). $ B%: *+ -)+)&( (+).

The Smith Chart

The mith Chart is "robabl# one of the most useful gra"hical tools a ailable to RF circuitdesigners toda#. This chart was de elo"ed back in the Thirties b# a Bell /aboratoriesengineer b# the name of 0hilli" mith. mith wanted an easier method of sol ing thetedious re"etiti e e1uations that often a""ear in RF theor#. Toda# the de elo"ment of the

mith Chart is still widel# in use. ee Figure 2+)' for the basic mith Chart.

Plotting Impedance Values

The mith Chart at an# "oint re"resents a series combination of resistance and reactanceof the form:

3 4 R 5 67

To locate the im"edance 3 4 & 5 6&• First find the R 4 & constant resistance circle.• Then follow it until it crosses the 7 4& constant reactance circle.• 8t the 6unction of these two circles will re"resent the needed im"edance alue.• Refer to Figure 2+9* and notice that this "articular "oint is located in the u""er

half of the chart. Wh# Because 7 is a "ositi e reactance or an inductor.• %ow locate the "oint & ; 6&. 8s #ou can see this "oint is in the lower half of the

chart. Wh# Because 7 is a negati e 1uantit# and thus re"resents a ca"acitor.

Therefore, the 6unction of R 4 & is a constant resistance circle and the 7 4 +& is a constantreactance.

o, to find an# series im"edance of the form R < 67 on the mith Chart, sim"l# find the 6unction of the R 4 constant and the 7 4 constant circles.$n Figure 2+9& the im"edance alues "lotted are er# small numbers. $f #ou ha e to "lotan im"edance of 3 4 &** 5 6&=* ohms, #ou will not be able to do so accuratel#. o, inthis case #ou must use a "rocess called normalization . This in ol es di iding theim"edance to be "lotted b# a con enient number that will "lace the new normali>edim"edance near the center of the chart where increased accurac# in "lotting is achie ed.

Example:

0lot the im"edance 3 4 &** 5 6&=*.

/et?s di ide 3 b# &** resulting in an im"edance alue of 3 4 & 5 6&.=.This im"edance alue is much easier to "lot. @%ote: 8ll im"edances must be di ided b#the same number to correctl# meet the normalization "rocess.

%otes b#: Debbie 0restridge &


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Impedance Manipulation on the Smith Chart

What ha""ens when a series ca"aciti e reactance of ;6&.* ohm is added to an im"edanceof 3 4 *.= 5 6*.- ohm Aathematicall#, #ou will end of with a result of:

3 4 *.= 5 6*.- ; 6&.* 4 *.= ;6*.9 ohm

This re"resents a series RC 1uantit#. Refer to Figure 2+9) for a gra"hical illustration.What has ha""ened is that we ha e mo ed downward along the R 4 *.=+ohm constantresistance circle for a distance of 7 4 +6&.* ohm. The new "lotted im"edance "oint is 3 4 *.= ; 6*.9 ohm, as shown in Fig. 2+9). $n Figure 2+99 we ha e added a seriesinductance to a "lotted im"edance alue which causes a mo e upward along a constant

resistance circle to the new im"edance alue. Therefore, the addition of a series ca"acitor to a "lotted im"edance mo es that im"edance downward counterclockwise along aconstant resistance circle for a distance that is e1ual to the reactance of the ca"acitor. Theaddition of an# series inductor to a "lotted im"edance mo es that im"edance u"ward

clockwise along a constant resistance circle for a distance that is e1ual to the reactanceof the inductor.

Conversion o Impedance to !dmittance

The mith Chart can also be used to con ert an# im"edance 3 to an admittance , andice+ ersa. $n mathematical terms, an admittance is sim"l# the in erse of an im"edance,

or

Υ Ζ

=&

2+'

The admittance contains both a real and an imaginar# "art, similar to the im"edance3 . Therefore,

4 E < 6B 2+&*

E 4 the conductance in mhos.B 4 the susce"tance in mhos.

Figure 2+92 illustrates the circuit re"resentation.@%ote: usce"tance + is "ositi e for a ca"acitor and negati e for an inductor

Reactance ; is negati e for a ca"acitor and "ositi e for an inductor

%otes b#: Debbie 0restridge )


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+ jB

Y = G + j B

F i g u r e 4 - 3 4 . C i r c u i t r e p r e s e n t a t i o n f o r a d m i t t a n c e .

o r - jBGG

Y = G - j B

Take e1uation 2+' to find the in erse of a series im"edance of the form 3 4 R 5 67. Howcan #ou use the mith Chart to "erform the calculation for #ou without the use of acalculator

• Fist work the "roblem out mathematicall#• 0lot the results on the mith Chart• Com"are to see how the two functions are related

Example:

Take the series im"edance 3 4 & 56&.The in erse of 3 is:

Υ =+

&

& & j

4&

& 2&2 2=. ∠

4 * -*-& 2=. ∠ −

4 * = * =. .− j mho

0lot the "oints & 5 6& and *.= ; 6*.= on the mith Chart. %otice the gra"hical relationshi" between the two. This is illustrated in Figure 2.9=. %otice that the two "oints are locatedat e actl# the same distance d from the center of the mith Chart but in o""ositedirections &(*G from each other. The same relationshi" will hold true for an#im"edance and its in erse. Therefore, #ou can find the reci"rocal of an im"edance or an

admittance b# sim"l# "lotting the "oint on the mith Chart, measuring the distance dfrom the center of the mith Chart to that "oint, and then "lotting the measured result thesame distance from the center but in the o""osite direction &(*G from the original.There is another a""roach that can be used to achie e the same result which is to rotatethe mith Chart itself &(*G while fi ing the starting "oint in s"ace. This "rocedure isillustrated in Figure 2+9 mith Chart Form 3 ; *& ; % G where the rotated chart isdone in black. ou can see that the im"edance "lotted solid lines on the red coordinatesis located at 3 4 & 5 6& ohms, and the dotted lines on the black coordinates as 4 *.= ;

%otes b#: Debbie 0restridge 9


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6*.=. $n order to obtain the admittance coordinates the chart has been rotated &(*G. Theu""er half of the admittance chart re"resents negative susceptance +6B which isinductive and the lower half is of the admittance chart is positive susceptance 56B whichis ca"aciti e. 8s #ou can see nothing has been lost in the rotation "rocess.

!dmittance Manipulation on the Smith Chart

Figure 2+9- illustrates the addition of a shunt ca"acitor. %ow we will begin with anadmittance of 4 *.) ; 6*.= mho and add a shunt ca"acitor with a susce"tance reci"rocalof reactance of 56*.( mho. 0arallel susce"tances are added together in order to find thee1ui alent susce"tance. The result of this calculation is:

4 *.) ; 6*.= 56*.( 4 *.) 5 6*.9 mho

0lot this "oint on the admittance chart and notice what has been done. ou ha e mo ed

along a constant conductance circle E downward clockwise a distance of 6B 4 *.(mho. $n other words, the real "art of the admittance has not changed, onl# the imaginar# "art has. $n Figure 2+9( a shunt inductor is added to an admittance which mo es the "oint along a constant conductance circle u"ward counterclockwise a distance +6Be1ual to the alue of its susce"tance.

%ow su"erim"ose the im"edance and admittance coordinates and combine Figs. 2+9), 2+99, 2+9-, and 2+9( #ou can obtain the useful chart that is illustrated in Figure 2.9'. Thischart is a gra"hical illustration of the direction of tra el along the im"edance andadmittance coordinates, which results when the "articular t#"e of com"onent that isindicated is added to an e isting im"edance or admittance. ee am"le 2+ which willillustrate this "oint.

Example "#$:

What is the im"edance looking into the network shown in Figure 2+2* %oticethat the "roblem has been sim"lified, due to the fact that shunt susce"tances are beingillustrated rather that shunt reactances.

Solution:This "roblem is er# easil# handled on a mith Chart. o, there will be no

calculations to be "erformed. The solution is illustrated in Figure 2+2).te"s to Follow:

&. Break the circuit down into indi idual branches down as shown in Figure 2+2&.). 0lot the im"edance of the series R/ branch where 3 4 & 5 6& ohm. This is "oint 8

in Figure 2+2).9. %e t, follow the rules diagrammed in Figure 2+9'I begin b# adding each

com"onent back into the circuit+one at a time.

@%ote: The following constructions Figure 2+2) should be "erformed:

%otes b#: Debbie 0restridge 2


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• 8rc 8B 4 shunt / 4 +6B 4 *.9 mho• 8rc BC 4 series C 4 +67 4 &.2 ohms• 8rc CD 4 shunt C 4 56B 4 &.& mhos• 8rc D 4 series / 4 567 4 *.' ohm

The im"edance at "oint Figure 2+2) can be read directl# off the chart 3 4 *.) 5 6*.= ohm.

Impedance Matching on the Smith Chart:

The mith Chart is also an e cellent candidate for an im"edance matching tool. $f a loadim"edance is gi en and gi en the im"edance that the source would like to see, sim"l# "lotthe load im"edance and begin adding series and shunt elements on the chart until the desiredim"edance is achie ed. This same techni1ue was a""lied in e am"le 2+ .

T%o#Element Matching:

/et?s consider e am"le 2+- to begin our stud# of a mith Chart im"edance+matching "rocedure. The following e1uations ma# be used to sim"lif# the "rocedure:

For a series+C com"onent:

C X

=&

ω Ν2+&&

%otes b#: Debbie 0restridge

jX = 0.9 jX = 1-jX = 1.4

R = 1- jB = 0.3

+ jB = 1.1

Figure 4-40. Circuit for Examp e 4-!.

3

Figure 4-4". Circuit is #ro$en do%n into indi&idua #ranc' e ements.

-jB = 0.3

jX = 0.9

R = 1

jX = 1

+jB = 1.1

-jX = 1.4

DB

C

8

=


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For a series+/ com"onent:

L X

=Ν

ω 2+&)

For a shunt+C com"onent:

C =Β

Νω 2+&9

For a shunt+/ com"onent:

L =Ν

Βω 2+&2

Where:ω π = ) f ,7 4 the reactance as read from the chart,B 4 the susce"tance as read from the chart,

% 4 the number used to normali>e the original im"edances that are matched.

Example "#&:

Design a two+element im"edance+matching network on a mith Chart to match a )= ; 6&=ohm source to a &** ; 6)= ohm load at * AH>. The matching network must also act as a

low+"ass filter between the source and the load.

Solution

This source is a com"le im"edanceI therefore it wants to JseeK a load im"edance that ise1ual to its com"le con6ugate. o the first ste" is to force the &** ; 6)= ohm load to a""earas an im"edance of )= 5 6&= ohms.8ctuall#, the source and the load im"edances are both too large to "lot on the mith Chart, sonormali>ation is necessar# in this case.

te"s for sol ing this "roblem:&. First choose a con enient number % 4 =* and di ide all im"edances b# this number.

our results should be *.= 5 6*.9 ohm for the source im"edance and ) ; 6*.= for theactual load im"edance.). %ow "lot these two alues on the mith Chart, as illustrated in Figure 2+22, where at

"oint 8, 3 / is the n ormalized com"le con6ugate of the load im"edance, and at "ointc, 3 @ is the normalized com"le con6ugate of the source im"edance.

9. %e t #our matching network must also be a low+"ass filter that will force #ou to usesome form of series+/, shunt+C arrangement. The onl# wa# to accom"lish this is totake the "ath as shown in Figure 2+22. B# following the rules of Figure 2+9', the arc



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8B of Figure 2+22 is a shunt ca"acitor with a alue of 56B 4 *.-9 mho. The arc BC isa series inductor with a alue of 567 4 &.) ohms.

2. %ow change the shunt ca"acitor from a susce"tance into an e1ui alent reactance b#taking the reci"rocal

X jc

=+

&

Β

=&

* -9 j mho.

= − j ohms& 9-..

=. %ow unnormali>e all im"edance alues b# multiplying them b# the number % 4 =*+the alue originall# used in the normali>ation "rocess. Therefore:

X ohms L = ,* X ohmsc = ,( = .

The com"onent alues are:

( )

L X

n

L=

=∗

=

ω

π

,*

) ,* &*

&='

,

Η

( ) ( )

C X

pF

c=

=∗

=

&

&

) ,* &* ,( =

9( -

,

ω

π .

.

ee Figure 2+29 for the final circuit.

%otes b#: Debbie 0restridge -


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F i g u r e 4 - 4 3 . F i n a c i r c u i t f o r E x a m p e 4 - ( .

) * - j " * o ' m

1 5 9 n H

" 0 0 - j ) * o ' m

3 8 . 7 p F

Three#Element Matching

For three+element matching #ou are able to choose the loaded L for the network. Howcan circuit L be re"resented on a mith ChartThe L of a series+im"edance circuit is sim"l# e1ual to the ratio of its reactance to itsresistance. Therefore, an# "oint on a mith Chart has a L associated with it. 8lternatel#,if #ou were to s"ecif# a certain L, #ou could find an infinite number of "oints on thechart that could satisf# that L re1uirement.

Example:

The following im"edances located on a mith Chart ha e a L of =:

R 5 67 4 & < 6=4 *.= < 6).=4 *.) < 6&4 *.& < 6*.=4 *.*= < 6*.)=

These alues are "lotted in Figure 2+2= and form the arcs shown. Therefore if an#im"edance located on these arcs must ha e a L of =. imilar arcs for other alues of Lcan be drawn with the arc of infinite L being located along the "erimeter of the chart andthe L 4 * arc actuall# a straight line l#ing along the "ure resistance line located at thecenter of the chart.

The "rocedure for designing a three+element im"edance matching network for a s"ecifiedL is summari>ed as follows:

&. 0lot the constant+L arcs for the s"ecified L.). 0lot the load im"edance and the com"le con6ugate of the source im"edance.9. Determine the end of the network that will be used to establish the loaded L of the

design. For networks, the end with the smaller terminating resistance determines theL. For 0i networks, the end with the larger terminating resister sets the L.

2. For T networks: ee Figure 2+&( for a three+element T network. R R s L>

%otes b#: Debbie 0restridge (


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Ao e from the load along a constant+R circle series element and intersect the Lcur e. The length of this mo e determines #our first element. Then, "roceedfrom this "oint to ( )Ζ Ζ Ζ s s s conjugate∗ ∗ = in two mo es. tart with the shunt

first and then do the series element. R R s L<

Find the intersection $ of the cur e and the source im"edance?s R 4 constant circle,

and "lot that "oint. Ao e from the load im"edance to "oint $ with twoelements. The series element is first and then the shunt element. Ao e from "oint$ toΖ s

∗ along the R 4 constant circle with another series element.=. For the 0i networks ee Figure 2+&-

R R s L>

Find the intersection $ of the L cur e and the source im"edance?s E 4 constantcircle, and "lot that "oint. Ao e from the load im"edance to "oint $ with twoelements. tart first with a shunt element and then with a series element. Ao e from

"oint $ to Ζ s∗ along the E 4 constant circle with another shunt element.

R R s L<

Ao e from the load along a constant E circle shunt element and intersect the Lcur e. The length of this mo e will be determined b# #our first element. Then

"roceed from this "oint to Ζ s∗ in two mo es. our first mo e will be with the series

element and then with the shunt element.


+"+" +3+"

+)

Figure 4-",. 'e t'ree-e ement net%or$.

R L R S

Figure 4-"(. 'e t'ree-e ement i net%or$.

X &

X )

X 9

'

R L

RS


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Example "#':

Design a T network to match a 3 4 &= 5 6&=+ohm source to a ))=+ohm load at9*AH> with a loaded L of =.

Solution Follow the ste"s below:

• Draw arcs for L 4 =.• $f necessar# normali>e the im"edances if the# are too large to be located on the

chart. Di ide b# a con enient alue choose % 4 -= for normali>ation.Ζ s

∗ 4 *.) ; 6*.) ohmΖ L 4 9 ohms

@ %ote: The construction details for the design are shown in Figure 2+2 .• 0lot the load im"edance and the com"le con6ugate of the source im"edance.• ince the design statement s"ecifies a T network, the source termination will

determine the network L. R R s L< + Follow ste" 2 abo e.o 0lot "oint $ . This is the intersection of the L 4 = cur e and the R 4

constant circuit that "asses through Ζ s∗ .

• Ao e from the load im"edance to "oint $ with two elementso lement & 4 arc 8B 4 series / 4 6).= ohmso lement ) 4 arc B$ 4 shunt C 4 6&.&= mhos

• Then, mo e from "oint $ to Ζ s∗ along the R 4 constant circle.

o lement 9 4 arc $C 4 series / 4 6*.( ohm• Mse e1uations 2+&& through 2+&2 to find the actual element alues.

o lement & 4 series /:( )

( ) L

nH

=∗

=

) = -=

) 9* &*

''=

,

.

π

o lement 4 shunt C:

( )C

pF

=∗

=

&&=

) 9* &*

(&

,

.

π

o lement 9 4 series /:

( )

( ) L

nH

=∗

=

*( -=

) 9* &*

9&(

,

.

π

• The final network is illustrated in Figure 2+2-.

%otes b#: Debbie 0restridge &*


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F i g u r e 4 - 4 ( . F i n a c i r c u i t f o r E x a m p e 4 - , .

9 9 5 n H

2 2 5 o h m

8 1 p F

" * + j " . *3 1 8 n H

Multi#Element Matching

There is no L constraint, in multi+element matching networksI therefore the mith Chart becomes an actual treasure tro e of infinite number of "ossible solutions. There is ano"timum solution for getting from "oint 8 to "oint B on the mith ChartI howe er the

o"timum solution will not be the onl# solution. 8 two element network will get #ou from "oint 8 to "oint b with the least number of com"onents and the three element networkwill "ro ide a s"ecified L b# sim"l# taking another route. Howe er, if #ou are notinterested in a L, there are 9+, 2+, =+, &*+, and )* element and more im"edance matchingnetworks that are easil# designed on a mith Chart b# following the constant+conductance and constant resistance circles until #ou arri e at "oint B, which, in our case,is usuall# then com"le con6ugate of the source im"edance. The illustration of this "ointcan be found in Figure 2+2(. 8s #ou can see:

• olution & starts with a series+/ configuration and taking ' elements in order toarri e at "oint B.

• olution ) starts with a shunt+/ "rocedure and takes ( elements to arri e at "oint

B.• olution 9 starts with a shunt+C arrangement and takes fi e elements to reach

"oint B.The element reactances and susce"tances can be read directl# from the mith Chart, ande1uations 2+&& through 2+&2 can be used to calculate the actual com"onent alues.

%otes b#: Debbie 0restridge &&

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Smith Charts