) Rotate L by 120 clockwise to obtain in!! anywhere between load and generator: rotation by 2d in clockwise direction. d=distance from the load to the

Transcription

1 3.1 Smith Chart Construction: Start with polar representation of. L ; in on lossless lines related by simple phase change ) Idea: polar plot going from L to in involves simple rotation. in jj 1 ) circle with radius=1 covers all possible s Special points O-C: = S-C: = = points: take e.g. Z L = 17:7 + j11:8 on Z 0 = 50 line; Other = =6 ) L = (Z L Z 0 )=(Z L + Z 0 ) = 0: l g. Find in = 6 L 150 2l = 0:5 6 ( ). See from 2l = 2 2 =6 = 2=3 = 120 change Phase Microwave Circuit Design I cb. Pejcinovic 1

2 ) Rotate L by 120 clockwise to obtain in!! anywhere between load and generator: rotation by 2d in clockwise direction. d=distance from the load to the point. Conversely if is known CCW rotation by 2d gives L. line ) spiral; no need to trace it just adjust the nal Lossy by e 2d. magnitude Simple graphical solution for transformation along tr. line. about impedance transformation? We'll use normalized What impedances. Microwave Circuit Design I cb. Pejcinovic 2

3 Direct relationship between and Z: 1 + Z = and Y = Special points: S-C O-C. What else can we learn? Z 0 is real if has angle 0 or 180 then Z is real as well Since horiz. axis in plot contains real Z-s starting with 0 on the ) left to 1 on the right. For = 0 Z=1 i.e. Z= Z 0. Origin of plot is Z = 1 in impedance plot. Fig each we can nd Z. Idea: connect the points with same For part of Z! same for imaginary parts. (contour plots) real Microwave Circuit Design I cb. Pejcinovic 3 (1)

4 real parts a.k.a const. resistance circles. Fig. Constant 3{19. all pass through R = 1 point outermost circle is R = 0 circle (jj = 1); all impedances on are purely reactive (0 to j1). it imaginary parts a.k.a. const. reactance lines Constant (circles) X = 0 = horizontal line through origin upper half ) postive reactance X ) inductive components lower half ) negative reactance ) capacitive components Microwave Circuit Design I cb. Pejcinovic 4

5 polar jj with Re(Z) and Im(Z) contours ) Smith chart Combine Drop the polar circles and calculate jj via SWR or the (S-chart). scale at bottom = 1 + jj SWR jj 1 jj = SWR 1 or + 1 SWR read of from the right part of horizontal axis (between Z=1 SWR Z = 1). Note: purely resistive loads have SWR = R. and show S-chart. We need compass and ruler. Microwave Circuit Design I cb. Pejcinovic 5 (2)

6 is not even necessary to calculate 2l (or 2d) since the scales It given in fractions of wavelength. are 2 (or 360 ) rotation corresponds to d = =2 since in that One 2l = 360 case Full circle = going along transmission line for a distance equal ) =2. to Impedance and standing wave patterns repeat every =2 distance. Microwave Circuit Design I cb. Pejcinovic 6

7 chart also used as admittance chart. Y = 1= Z; Y = Smith 0 = Y Z 0 but jj does not change with the change from Z Y=Y to Y. ) Y is 180 away from Z but on the same jj circle. conclusion by looking at eqs. 1 rotation of by 180 Same Z to Y. transforms Note on using S-chart as Y-chart: Y = 0 ) O-C; for Y = 1 ) S-C. resistance coord. ) conductance coord. reactance coord ) susceptance coord. positvie susceptance ) capacitive components negative susceptance ) inductive components (see chart) Microwave Circuit Design I cb. Pejcinovic 7

8 when in Y coordinates angle of re. coe. ( ) scale must rotated by 180 (or we have to go back to Z chart to look be at -s) Two scales: toward generator and toward load. generator: starting from load and moving along tr. line Toward generator i.e. getting closer to the input ) clockwise toward (CW) rotation. load: starting point in and we want to determine of Toward closer to the load ) CCW rotation. points Example 3-6 Microwave Circuit Design I cb. Pejcinovic 8

9 a) nd j L j; L ;SWR along tr. line calc. Z L = Z=Z 0 = 0:3 + j0:5; follow the lines until you 1. where Z L is. Move along const. resistance circle until nd proper imaginary comp. is found. circle through Z L with center at origin ) circle of constant 2. (SWR circle). jj Z L on this circle corresponds to L. j L j is obtained from 3. radius and angle by extending the line from origin the the point all the way to the outer circle (jj = 1) through reading o the angle ( L = 124 ). jj read o either and from the scale at the bottom or from SWR. Microwave Circuit Design I cb. Pejcinovic 9

10 Intersection of const. jj circle with right part of the horizontal 4. axis gives SWR. Z is real there and we know that SWR = R. (here: SWR = 4.2 ) jl j = (SWR 1)=(SWR +1) = 0:62) Finding in from known L and l means rotating along constant b) jj circle by 2l (lossless line). Once in is know Z in is read o S-chart. 1. Plot Z L and SWR circle (const. jj) draw radial line through Z L and read o the value of -s 2. generator ( = 0:078; no particular meaning) toward 0 = 40cm at 750MHz l=5.2cm ) l= = 5:2=40 = 3. Start from rotate CW by to :130. (= ) Microwave Circuit Design I cb. Pejcinovic 10

11 intersection of radial line with SWR circle give in and Z in. 4. is the same as before only angle changed to 30. jj 5. read Z in o the S-chart ( Z in = 2 + j2 Z in = j200. for l=2cm l= = 2=40 = 0:05; draw a radial line nd angle. c) with SWR circle gives (2cm) and Z(2cm). Z = Intersection 0:47 + j0:93 or Z = 47 + j93 Y L was given; procedure is the same: use S-chart as Suppose chart to nd position of Y L rest is the same. admittance Y L from Z L is easy: it's on the opposite side of the circle Finding values are read o ( Y L = 0:9j1:47; note on normalization where Y0 = 1=Z 0 = 0:01;Y L = 0:01 Y L ) To nd in from Y in rotate by 180 rst! Microwave Circuit Design I cb. Pejcinovic 11

12 chart is a very nice visual tool. It can easily show which Smith of and Z can be obtained and which cannot. Say Z in = values + j0:4 cannot be obtained in this case since that impedance 0:7 not on the SWR circle. is questions like nding a maximum reactive component attainable Also for this load can be answered easily. Z in 2:3 + j2:0; to get that Z in we need a line of ( ) )= 0:138 length Examples of calculations with S-chart. Microwave Circuit Design I cb. Pejcinovic 12

13 ex. 3{7. Finding Z in for a given circuit. Split it up in parts as we move on the line away from Z L Z in. towards pt. A: add the line delay (lossless) A! B: add j!l (j30) to total reactance; added in series. B! C: add j!c (j200) in shunt C! in: again phase delay along the line. Microwave Circuit Design I cb. Pejcinovic 13

14 Z L = 2 + j1:5; draw SWR circle through that pt. rotate by 0:12 to get to pt. B add j30=50 to impedance at pt. B looking towards Z L ; on a const. resistance circle (real comp. does not done addition is in series ) use resistances ) Z B = change); j1:3 + j0:6 = 1 j0:7 1 add C in shunt; impedances in shunt don't add up ) switch admittances; convert Z B to Y B by 180 rotation; Y = to 1=(j200) = j0:005 or normalized Y = j0: = j0:25 Microwave Circuit Design I cb. Pejcinovic 14

15 now use constant conductance circle to pt. YC = 0: j0:25 = 0:67 + j0:72 j0:47 phase change along the line: CW rotation on SWR circle for ) Y in = 1:3 + j1:1; Z in obtained by 180 rotation 0:06 Zin = 0:45 j0:38 nally de-normalize your Z-s and Y-s. What would happen if L and C were exchanged? you must remember what you are using: Z or Y Smith Note: chart! Microwave Circuit Design I cb. Pejcinovic 15

16 ex. 3-8: using lines with dierent Z 0 normalize Z L with Z 01 = 50! Z L = 0:4. CW rotation by l= = 5=20 = 0:25! l = 0:25. nd Z on the input terminals of Z 01 tr. line: ZA = 2:5! = 125 (note the transformation : Z in = Z 2 0=Z L or ZA Zin = 1= Z L ) normalize with Z 02 = 90! Z Z = 125=90 = 1:39 draw new SWR circle and rotate CW by 12:8=20 = 0:64. 0:5 is one full circle! Note: read o Z in = 0:9 j0:3 or Z in = 81 j27. last SWR is 1.39 ) jj = 0:163; angle from Z S-chart. Microwave Circuit Design I cb. Pejcinovic 16

17 What can we do with Smith chart? 1. plot Z or Y 2. calculate SWR jj; 6 3. impedance transformation due to tr. line 4. Z! Y and Y! Z 5. add series or shunt circuit elements at any pt. on the tr. line Microwave Circuit Design I cb. Pejcinovic 17

18 Variation of Z; Y with frequency reactive networks > 0 (gs. 3{14 Purely 16). 15 translates into requirement that Z and Y curves must have This CW trend with increasing frequency. Same conclusion holds for a R with X and G with B. see g. 3{23 for qualitative behavior. Arrows indicate direction of increasing f. tr. line with series RLC circuit (g. 3{24). Resonance Examine 1 GHz. Load is Z L = 0:5+j X and X = (!L 1=(!C))=50. at Microwave Circuit Design I cb. Pejcinovic 18

19 Calculate Z L at resonance above below: see table 3{1. CW change with freq. increase. What's the eect of tr. line? Each Z L has SWR circle; rotate according to table 3{1. (S-chart) rotation is frequency dependent! l= = 0:1; 0:133 Problem: ) proportional to f! (electrical length l = 2l= =!l=v) 0:167 Note: Z in vs. f is more spread out on S-chart than Z L itself! Zin covers 0:151 while Z L covers only 0:084. Di. of 0:067 is due to freq. change of tr. line. line (large f) ) larger impedance spread ) more dif- Longer to match impedance on f. cult Problem 3.51 for illustration. Microwave Circuit Design I cb. Pejcinovic 19

20 Standing wave patterns Smith Chart Use S-chart for this; idea from ^V = ^V + = 1 + ^I=^I + = 1 Magnitudes: V=V + = j1 + j I I + = j1 j are functions of distance; for lossless lines V + ;I + are not! V; vector addition and subtraction gives the ratios (normal- Simple ized rms voltage and current). 3{25 for Z L = 1 + j2. Remember Vector Fig. (subtraction) gives V=V + (I=I +. addition moves away from L on tr. line (SWR circle). Pick a few el. lengths and do vector summation and reconstruct SW specic pattern. Microwave Circuit Design I cb. Pejcinovic 20

21 Max. V when 6 = 0; (min. I); min. V (max I) when 6 = 180. position on tr. line by reading o the distance from \wavelengths Read toward generator" (need freq. and velocity). Fig. 3{26. Microwave Circuit Design I cb. Pejcinovic 21

22 Summary of Smith-chart: Smith Chart polar plot of = (Z Z 0 )=(Z + Z 0 ). Z 0 is the tr. line or reference (normalizing) impedance. characteristic Normalized impedances: z = Z=Z 0 ) = (z 1)=(z + 1) Normalized admittances: y = Y=Y 0 ) = (y 1)=(y + 1) Moving along tr. line ) CW circular motion on S-chart (lossless). Note the positive and negative reactances/susceptances Constant real and imaginary impedances are on circles. Unit circle is for pure reactances/susceptances. conversion from Z to Y done by 180 rotation. Microwave Circuit Design I cb. Pejcinovic 22

23 resistance results in jj > 1. Where to put it on S-chart? Negative choice is \compressed" Smith chart showing a fraction of the One negative resistance values on the horizontal axis. (g ) plot 1= but the resistance circles are now taken Alternative: be negative of what is read o the chart while the reactance to circles are as labeled (ex ). For S-chart calculations: ex and g ZY Smith Chart idea: Y is obtained by rotating Z in Smith chart by 180 Basic ZY chart has them both on one piece of paper. Note: degrees. half is for negative susceptance lower for positive (opposite upper the Z chart). Ex and g of Microwave Circuit Design I cb. Pejcinovic 23

24 Impedance matching networks Smith Chart matching? For ampliers: to deliver max. power ensure Why minimum noise etc. See g stability to do it? For lower freq. use lumped elements: el l circuits How (g ). Simple procedure: ex for eects of series/shunt inductor and capacitor. Add series reactance ) move on const.-resistance circle (g.) Add series susceptance ) move on const.-conductance circle. circuit design: moving from one impedance/admittance Matching another via const. resistance and conductance circles. Ex. to Not all ell circuits can be used for a given match: g Microwave Circuit Design I cb. Pejcinovic 24

25 factor Q: loaded Q dened as Q L =! 0 = BW. ell circuits Quality low- or high-pass lters. Dene node Q as: Q n = jx s j=r s either R s + jx s is an equivalent series input impedance at that where For admittance: Q n = jb p j=g p. Fig node. to nd Q? Need! 0 and BW. Not that simple since -3dB How are not dened. Instead use \equivalent" bandpass lter points replacing the \input" impedance Z B with its shunt equivalent by g.) Note: these are the same only at frequency! 0! (see ) resonant circuit loaded with 25 QL =! 0 =! 0R T C = jb Cj GT BW R T = C j = 25 jx Microwave Circuit Design I cb. Pejcinovic = 1 (3)

26 of the \equivalent" circuit can be calculated and it is the Gain as the original one at the resonant frequency. same question: what is the relationship between nodal and loaded Real In this case Q n = 2 and Q L = 1 ) Q L = Q n =2 same for Q? ell. Since the BW is not really dened for the original circuit all it by using BW f 0 =Q L = 500 MHz. Alternatively estimate it o the plot ) BW= 440 MHz which is close to the read Example estimate. to get high Q L high value of Q n is needed. (note on complex ) Problem: with el l circuits we cannot adjust Q! Solution: loads). use three elements i.e. T or networks. Microwave Circuit Design I cb. Pejcinovic 26

27 Q n and Q L not easily related. Normally we just use Problem: highest Q in a T or circuit. Useful tool: constant Q circles the Smith chart g Center of circles at (0; 1=Q n ) and on is (1 + 1=Q 2 n )1=2. radius example(s) Note that the nal Q is determined Design from frequency plot. Also there are many possible solutions. example (why?). note on ampliers: full circuit dc part ac part (g ). Final C-s and L-s (RFC-s) serve to decouple various part of the Various circuit. I/O matching done via el l circuits. Microwave Circuit Design I cb. Pejcinovic 27