Chapter 3 Matching and Tuning By Professor Syed Idris Syed Hassan Sch of Elect. & Electron Eng Engineering Campus USM Nibong Tebal 14300 SPS Penang ©Prof Syed Idris Syed Hassan

Why matching or tuning is important? To maximize power delivery and minimize power loss. To improve signal to noise ratio as in sensitive receiver components such as LNA, antenna, etc. To reduce amplitude and phase error as in distributed network such as antenna array. Basic idea of impedance matching

Concept of maximum power transfer In lump circuit Power deliver at ZL is Power maximum whence ZL = Zo

continue In transmission line The important parameter is reflection coefficient No reflection whence ZL = Zo , hence The load ZL can be matched as long as ZL not equal to zero (short-circuit) or infinity (open-circuit)

Factors in selecting matching network Complexity: simpler, cheaper, more reliable and low loss circuit is preferred. Bandwidth: match over a desirable bandwidth. Implementation: depend on types of transmission line either cable, stripline, microstripline, waveguide, lump circuit etc. Adjustability:some network may need adjustment to match a variable load.

Matching with lumped elements The simplest matching network is an L-section using two reactive elements Configuration 1 Whence RL>Zo ZL=RL+jXL Configuration 2 Whence RL<Zo

continue If the load impedance (normalized) lies in unity circle, configuration 1 is used.Otherwise configuration 2 is used. The reactive elements are either inductors or capacitors. So there are 8 possibilities for matching circuit for various load impedances. Configuration 2 Configuration 1 Matching by lumped elements are possible for frequency below 1 GHz or for higher frequency in integrated circuit(MIC, MEM).

Impedances for serial lumped elements Serial circuit Reactance relationship values +ve X=2pfL L=X/(2pf) -ve X=1/(2pfC) C=1/(2pfX)

Impedances for parallel lumped elements Parallel circuit Susceptance relationship values +ve B=2pfC C=B/(2pf) -ve B=1/(2pfL) L=1/(2pfB)

Lumped elements for microwave integrated circuit Planar resistor Chip resistor Loop inductor Spiral inductor Interdigital gap capacitor Metal-insulator-metal capacitor Chip capacitor

Matching by calculation for configuration 1 For matching, the total impedance of L-section plus ZL should equal to Zo,thus Rearranging and separating into real and imaginary parts gives us * **

continue Solving for X from simultaneous equations (*) and (**) and substitute X in (**) for B, we obtain +ve capacitor -ve inductor Since RL>Zo, then argument of the second root is always positive, the series reactance can be found as +ve inductor -ve capacitor Note that two solution for B are possible either positive or negative

Matching by calculation for configuration 2 For matching, the total impedance of L-section plus ZL should equal to 1/Zo,thus Rearranging and separating into real and imaginary parts gives us * **

continue Solving for X and B from simultaneous equations (*) and (**) , we obtain +ve inductor -ve capacitor +ve capacitor -ve inductor Since RL<Zo,the argument of the square roots are always positive, again two solution for X and B are possible either positive or negative

Matching using Smith chart (+ve) (+ve) (-ve) (-ve)

Serial L Matching using lumped components Parallel C Serial C 10W 50W Parallel L

Example Design an L-section matching network to match a series RC load with an impedance ZL=200-j100 W, to a 100 W line, at a frequency of 500 MHz. Solution Normalized ZL we have : ZL= 2-j1 Parallel L (-j0.7) Serial C (-j1.2) ZL= 2-j1 Parallel C (+j0.3) Serial L (j1.2) Solution 1 Solution 2

continue Solution 2 seems to be better matched at higher frequency

Single stub-matching Parallel configuration Short-stub matching Open-stub matching

Example Solution Smith Chart Design two single–stub shunt tuning networks to match a load ZL=15+j10W to 50W at 2GHz. The load consists of a resistor and inductor in series. Plot the reflection magnitude from 1 GHz to 3 GHz for each solution. Solution Normalized the load zL=0.3+j0.2 Construct SWR circle and convert load to admittance Then move from load to generator to meet a circle (1+jb) to obtain two points i.e y1=1-j1.33 and y2=1+j1.33. The distance d from the load to stub is obtained either of these two points i.e. d1= 0.044l and d2=0.387l. To improve bandwidth, the stub is chosen as close as possible to the load. The tuning requires a stub of j1.33 for y1 and –j1.33 for y2. For open circuit-stub i1 =0.147l and i2=0.353l. Smith Chart

Continue Convert j0.2 to inductance value, thus Use this value to calculate reflection coefficient

Formulas for calculation Let ZL=RL+jXL, then the impedance at distance d from the load is Admittance at this point is Thus, by substituting Zd and separating real and imaginary, we have

continue Equating G = Yo = 1/Zo,and , thus we have Solving

continue The two principle solution are To find the stub length,

Single stub-matching Serial configuration Short-stub matching i Open-stub matching i

Example Solution Smith Chart Match a load impedance of ZL=100+j80 to a 50 W line using a single series open-stub.Assuming the load consists of resistor and inductor in series at 2GHz. Plot the reflection coefficient from 1 GHz to 3 GHz. Solution Normalized the load zL=2+j1.6 Construct SWR circle Then move from load to generator to obtain two points on unity circle(1+jx) z1=1-j1.33 and z2=1+j1.33. The distance d from the load to stub is obtained either of these two points i.e. d1= 0.120l and d2=0.463l. To improve bandwidth, the stub is chosen as close as possible to the load. The tuning requires a stub of j1.33 for z1 and –j1.33 for z2. For open circuit-stub i1 =0.397l and i2=0.103l. Smith Chart

Continue Convert j1.6 to inductance value, thus Use this value to calculate reflection coefficient

Formulas for calculation Let YL=GL+jBL, then the load admittance at distance d from the load is Where t = tan bd Impedance at this point is Thus, by substituting Yd and separating real and imaginary, we have

continue Equating R = Zo = 1/Yo , thus we have Solving

continue The two principle solution are To find the stub length,

Double-stub matching Open or short stubs The advantage of this technique is the position of stubs ( d and x) are fixed. The matching are done by changing the length of stubs. The disadvantage of this technique is not all impedances can be matched.

A A’ B B’ A= load admittance A’=admittance of A at stub 2 B= Adjust of stub S2 to bring to the S2 circle B’=Admittance of B at S1 Then by adjusting stub 1 will bring the admittance to the center of the chart

Example Design a double-stub shunt tuner to match a load ZL=60-j80 to a 50 W line. The stubs are to be short circuited stubs, and are spaced l/8 apart. The load consists of a series resistor and capacitor that match at 2 GHz. Plot the reflection coefficient magnitude versus frequency from 1 GHz to 3GHz. Solution Plot normalized load zL=1.2 –j 1.6 W and convert to admittance we have yL= 0.3 +j0.4. Construct a rotated 1+ j b circle which is a distance d=l/8 from a 1+jb circle. Get two possible points on the rotated 1+jb circle, y1 and y1’ by adding susceptance of the first stubs. In this case we take x=0(match section immediate after the load . I.e b1=1.314 and b1’=-0.114.The length of stub will be i1=0.396l and i1’=0.232l. Now transform both point onto 1+jb circle along SWR circles.This bring two solution y2 =1-j3.38 and y2’=1+j1.38. Then the second stub should be b2= 3.38 and b2’=-1.38. The length of stub will be i2=0.454l and i2’=0.100l.

continue yL y1 y1’ b1 b1’ y2’ y2 b2’ b2 Rotated 1+jb circle

continue The shorter the stubs the wider will be the bandwidth

Formulas for calculation Let YL=GL+jBL, then the load admittance at distance x from the load is x is distance between stub and load Admittance at first stub After transforming to stub 2 , we have Where t = tan bd

continue At this point the ral part of Y2 =Yo ,thus Solving Since GL’ is real, the quantity in square root must be nonnegative, thus Simplified to

continue So susceptances of stubs are and To find the stub length , B either B1 or B2

continue The two principle solution are To find the stub length,

Graphical method Suitable for load impedance laying inside the unity circle. Plot ZL normalised to 50 ohm Find bisector and point intersect the chart axis (I.e A) Draw circle at center A touching the center of the chart Obtain R1 and calculated new characteristic impedance of the line Renormalised the ZL to ZT, thus we have ZL2 . Then plot on the chart. Obtain x for the length of the matching section by moving towards generator.

Graphical method

Transmission line transformer Z1 Quarter-wave transformer Matching at one frequency .For other frequency the impedance at the input of matching section will be Where t = tan bd At matched frequency tan bd = tan (p/2)

continue Reflection coefficient Since Zo2=Z1Z2, this reduces to qm qm p-qm p

continue Fractional bandwidth is given by Reflection coefficient with different ratio of Z1:Z2

Example Design a single section quarter-wave matching transformer to match a 10 W load to a 50 W line at fo =3GHz. Determine the percentage bandwidth for which the SWR < 1.5 For a pure resistive load we can just calculate directly and the length of transformer is l/4 Then determine reflection coefficient for SWR=1.5 Hence fractional bandwidth is or 29%

Multisection transformer Binomial multisection Chebyshev multisection Two types of transformer Reflection coefficient for multisection can be written as Or its magnitude At center frequency (i.e =l/4)

Binomial transformer For binomial expansion the reflection coefficient can be written as where Note CnN=CN-nN, C0N=1 and C1N=N=CN-1N Let then Solving for to obtain value of A or Hence

continue For bandwidth Therefore The fractional bandwidth, thus

Example Design a three section binomial transformer to match 50W load to a 100W line Solution N=3 , ZL=50W , Z0=100W then But Z1=91.7 W Z2=70.7 W Z3=54.5 W

Chebyshev transformer Chebyshev polinomial in general Useful forms of Chebyshev polinomial are

continue For N section,Chebyshev expansion the reflection coefficient can be written as Solving for to obtain value of A or For maximum reflection coefficient magnitude Therefore or

Example Design a three section Chebyshev transformer to match a 100W load to a 50W line. Taking Gm=0.05. Solution * ** Equating * and ** and A=Gm=0.05 cos3q

continue cosq For symmetrical and But Then Start n=0 n=1 n=2

Simple form of multisection transformers 2-quarter-wave transformer 3-quarter-wave transformer

Tapered Transmission line transformer Impedance distribution Reflection Exponential Taper transformer Triangular Taper transformer Klofenstein Taper transformer where

Transmission line transformer Nonsynchronous transformer E.g matching Z1=75 W to Z2=50 W qT = 29.3o

Complex to Complex Impedance Matching