SmithChart & MatchingNetwork James Lu JamesJ.Q.Lu ECSE2100Fields&WavesI 1 NonmatchedImpedance(0) • ReflectionsleadtoZin variationswithline lengthandfrequency • Poweriswastedbecauseofreactivepower, whichcanalsodamageequipmentduring shortcircuit(forexample) • Onlypartialpowerisdeliveredtotheload. • SWR>1:therewillbevoltagemaximaon theline,voltagebreakdownathighpower levels • Noises(bouncesorechoes) JamesJ.Q.Lu ECSE2100Fields&WavesI 2 BenefitsofMatching(=0) • Zin =ZO, independentoflinelength, andfrequency(overthebandwidth ofthematchingnetwork) • Maximumpowertransfertotheload isachieved • SWR=1:novoltagepeaksontheline • Nobounces(echoes) JamesJ.Q.Lu ECSE2100Fields&WavesI 3 UQ LoadMatching What if the load cannot be made equal to Zo for some other reasons? Then, we need to build a matching network so that the source effectively sees a match load. Ps Z0 M ZL 0 Typically we only want to use lossless devices such as capacitors, inductors, transmission lines, in our matching network so that we do not dissipate any power in the network and deliver all the available power to the load. JamesJ.Q.Lu ECSE2100Fields&WavesI 4 NormalizedImpedance It will be easier if we normalize the load impedance to the characteristic impedance of the transmission line attached to the load. Z z r jx Zo 1 z 1 Since the impedance is a complex number, the reflection coefficient will be a complex number u jv 2 2 2v 1 u v x r 2 2 1 u 2 v2 1 u v JamesJ.Q.Lu ECSE2100Fields&WavesI 5 SmithCharts The impedance as a function of reflection coefficient can be re-written in the form: 2 2 C r S2 1 1 u v Du T v2 r 2 1 u 2 v2 E 1 r U 1 r 2v C S2 x 2 1 1 2 2 u 1 Dv T 1 u v E x U x2 These are equations for circles on the (u,v) plane 2 2 2 x xo (y yo ) a JamesJ.Q.Lu ECSE2100Fields&WavesI 6 SmithChart– RealCircles 1 Im 1 Circle 0.5 r=0 r=1/3 r=1 r=2.5 1 0.5 0 0.5 1 Re 0.5 1 JamesJ.Q.Lu ECSE2100Fields&WavesI 7 SmithChart– ImaginaryCircles Im1 x=1/3 x=1 x=2.5 0.5 1 0.5 0 0.5 1 Re x=-1/3 x=-1 x=-2.5 0.5 1 JamesJ.Q.Lu ECSE2100Fields&WavesI 8 z L1 L Impedances, voltages, SmithChart z L1 currents, etc. all repeat 1 every half wavelength zL r L jxL 1 1 l e j2 l zl l 1 l j 4 l Inductive e TG W R z=1+j 1 W S ro (atP max) S 1 z=1 ro Short P Open Pmin =0 Pmax (z=0) (z=) =1 =1 L y=1/(1+j) T W =0.5-j0.5 Capacitive Purely real impedances along Purely imaginary impedances the horizontal centre line along the periphery JamesJ.Q.Lu ECSE2100Fields&WavesI 9 SmithChartExample1 Given: - L 0.5 45 Z 50 o G WT What is ZL? ZL 501.39 j1.35 69.5 j67.5 L T W JamesJ.Q.Lu ECSE2100Fields&WavesI 10 SmithChartExample2 Given: ZL 15 j25 Zo 50 G What is L? WT 15 j25 z L 50 0.3 j0.5 L 0.6- 123 z1 = 2 + j z2 = 1.5 -j2 L T z3 = j4 W z4 = 3 z5 = infinity z6 = 0 z7 = 1 z8 = 3.68 -j18 JamesJ.Q.Lu ECSE2100Fields&WavesI 11 SmithChartExample3 Given: ZL 50 j50 Zo 50 6.78ns G WT What is Zin at 50 MHz? 50 j50 z L 50 1.0 j1.0 - L 0.445 64 e j2l e j4 l / e j2 in L L L L T 2 244 W l f 504106 46.78410 9 0.339 in 180 - in 0.445 180 Zin 500.38 j0.0 19 JamesJ.Q.Lu ECSE2100Fields&WavesI 12 Admittance A matching network is going to be a combination of elements connected in series AND parallel. Z1 Impedance is well suited when working with series configurations. For example: Z2 V ZI ZL Z1 Z2 Impedance is NOT well suited when working with parallel configurations. Z1 Z2 Z Z Z 1 2 L Z1 Z2 For parallel loads it is better to work with admittance. 1 Y1 Y2 Y1 I YV Z1 YL Y1 Y2 JamesJ.Q.Lu ECSE2100Fields&WavesI 13 NormalizedAdmittance Y y YZo g jb Yo 1 y 1 2 1 u2 v2 C g S 1 Du T v2 g D T 2 1 u 2 v2 E 1 g U 1 g 2v C S2 b 2 1 1 2 2 u 1 Dv T 1 u v E b U b2 These are equations for circles on the (u,v) plane JamesJ.Q.Lu ECSE2100Fields&WavesI 14 AdmittanceSmithChart Conductance Circles Susceptance Circles 1 1 Im Im b=-1 b=-1/3 b=-2.5 0.5 0.5 g=2.5 g=1 g=1/3 g=0 b= 1 0.5 0 0.5 1 1 g= 0.5 0 0.5Re 1 Re b=2.5 b=1/3 0.5 0.5 b=1 1 1 JamesJ.Q.Lu ECSE2100Fields&WavesI 15 ImpedanceandAdmittanceSmithCharts • For a matching network that contains elements connected in series and parallel, we will need two types of Smith charts Z impedance Smith chart Z admittance Smith Chart • The admittance Smith chart is the impedance Smith chart rotated 180 degrees. – We could use one Smith chart and flip the reflection coefficient vector 180 degrees when switching between a series configuration to a parallel configuration. JamesJ.Q.Lu ECSE2100Fields&WavesI 16 AdmittanceSmithChartExample1 Given: y 1 j1 Find z G WT • Procedure: Plot y •Plot 1+j1 on chart •vector =0.445-64 Flip 180 •Flip vector 180 degrees degrees 0.445- 116 Read L T & z z 0.5 j0.5 W JamesJ.Q.Lu ECSE2100Fields&WavesI 17 AdmittanceSmithChartExample2 Given: - 0.5 45 Zo 50 Find Y G • Procedure: WT •Plot Plot • Flip vector by 180 degrees • Read coordinate Read y Flip 180 degrees L y 0.38 j0.36 T W 1 Y 0.38 j0.36 50 7.6 j7.2 410 3 S JamesJ.Q.Lu ECSE2100Fields&WavesI 18 MatchingExample Ps Z0 50 M 100 0 Match 100 load to a 50 system at 100MHz A 100 resistor in parallel would do the trick, but ½ of the power would be dissipated in the matching network. We want to use only lossless elements such as inductors and capacitors so we don’t dissipate any power in the matching network JamesJ.Q.Lu ECSE2100Fields&WavesI 19 MatchingExample We need to go from z=2+j0 to z=1+j0 on the Smith chart We won’t get any G closer by adding WT series impedance so we will need to add something in parallel. We need to flip over to the admittance chart L T W Impedance Chart JamesJ.Q.Lu ECSE2100Fields&WavesI 20 MatchingExample y=0.5+j0 Before we add the admittance, add a mirror of the r=1 G circle as a guide. WT L T W Admittance Chart JamesJ.Q.Lu ECSE2100Fields&WavesI 21 MatchingExample y=0.5+j0 Before we add the admittance, add a mirror of the r=1 G circle as a guide WT Now add positive imaginary admittance. L T W Admittance Chart JamesJ.Q.Lu ECSE2100Fields&WavesI 22 MatchingExample y=0.5+j0 Before we add the admittance, add a mirror of the r=1 G circle as a guide WT Now add positive imaginary admittance jb = j0.5 jb j0.5 j0.5 j2100MHz C L 50 T C 16pF W 16pF 100 Admittance Chart JamesJ.Q.Lu ECSE2100Fields&WavesI 23 MatchingExample We will now add series impedance Flip to the impedance Smith G Chart WT We land at on the r=1 circle at x=-1, i.e. z = 1 – j1 L T W Impedance Chart JamesJ.Q.Lu ECSE2100Fields&WavesI 24 MatchingExample Add positive imaginary admittance to get to z=1+j0 G WT jx j1.0 j1.0 50 j2 100MHz L L 80nH 80nH L T W 16pF 100 Impedance Chart JamesJ.Q.Lu ECSE2100Fields&WavesI 25 MatchingExample This solution would have also worked G WT 32pF L T W 160nH 100 Impedance Chart JamesJ.Q.Lu ECSE2100Fields&WavesI 26 MatchingBandwidth 0 80 nH 5 16 pF 10 100 G T W 15 150 MHz 20 25 Reflection Coefficient (dB) 30 35 50 MHz L 40 T 50 60 70 80 90 100 110 120 130 140 150 W Frequency (MHz) Because the inductor and capacitor impedances change with frequency, the match works over a narrow frequency range Impedance Chart JamesJ.Q.Lu ECSE2100Fields&WavesI 27 SingleStubTuner yl1 l 1 yl2 Z Z0 Z0 L Z Goal: 0 0 z =1 (y =1) l in in 2 Open 1 1 l or yl Short zl 1 l 2 l = 1 l e yin = yl1 + yl2 = (gl1 + jbl1) + jbl2 gl1 = 1 (real-part condition) Stub length l = up=f bl1 = -bl2 (imaginary-part condition) Phase shift: =2l=2(2/)l=4(l/) (2 Degrees of freedom) Open: Z = -jZ cotl, or z = -jcotl in o l in l Short: Zin = jZotan , or zin = jtan JamesJ.Q.Lu ECSE2100Fields&WavesI 28 SingleStubTuner Match 100 load to a 50 system at 100MHz using two transmission lines connected in parallel Flip to Admittance G chart WT y=0.5+j0 Adding length to Cable 1 rotates the reflection coefficient L clockwise to g=1. T W l1 = 0.152 y1l=1+j0.72 Admittance Chart JamesJ.Q.Lu ECSE2100Fields&WavesI 29 SingleStubTuner The stub has to add a normalized admittance of -0.72 to bring the G trajectory to the WT center of the Smith Chart L T W Admittance Chart JamesJ.Q.Lu ECSE2100Fields&WavesI 30 SingleStubTuner An short stub of zero length has an admittance=j By adding enough G WT cable to the short stub, the admittance of the stub will reach to -0.72 L T W l2 = (0.401-0.25) = 0.151 Admittance Chart JamesJ.Q.Lu ECSE2100Fields&WavesI 31 SingleStubTuner 100 l1 = 0.152 0 G l2 = 0.151 WT L T W Admittance Chart JamesJ.Q.Lu ECSE2100Fields&WavesI 32 SingleStubTuner l1 = 0.347 100 0 G l2 = 0.097 WT This solution would have worked as well.