ARRL Radio Designer and the Circles Utility 7KLV(cid:253)WZR(cid:240)SDUW(cid:253)DUWLFOH(cid:253)E\(cid:253):LOOLDP(cid:253)((cid:239)(cid:253)6DELQæ(cid:253):(cid:144),<+æ(cid:253)ZDV(cid:253)SXEOLVKHG(cid:253)LQ 4(;(cid:253)PDJD]LQH(cid:253)LQ(cid:253)6HSW(cid:238)2FW(cid:253)DQG(cid:253)1RY(cid:238)’HF(cid:253)(cid:236)(cid:228)(cid:228)(cid:229)(cid:239) 3DUW(cid:253)(cid:236)ª(cid:253)6PLWK(cid:253)&KDUW(cid:253)%DVLFV 3DUW(cid:253)ºª(cid:253)6PDOO(cid:240)6LJQDO(cid:253)$PSOLILHU(cid:253)’HVLJQ &RS\ULJKW(cid:253)(cid:139)(cid:253)(cid:236)(cid:228)(cid:228)(cid:229)æ(cid:253)7KH(cid:253)$PHULFDQ(cid:253)5DGLR(cid:253)5HOD\(cid:253)/HDJXHæ(cid:253),QF(cid:239) ARRL Radio Designer and the Circles Utility Part 1: Smith Chart Basics The Smith Chart is a venerable tool for graphic solution of RCL and transmission-line networks. The Circles Utility of ARRL’s Radio Designer software provides an “electronic” Smith Chart that is very useful. For those unfamiliar with the Circles Utility, we begin by exploring basic concepts and techniques of matching-network design. By William E. Sabin, WØIYH O ne of the interesting and gain-circle operations that are widely plified in Fig 1A, a resonant filter cir- useful features of the ARRL used in active-circuit design, espe- cuit that is also an impedance-match- Radio Designer program is the cially amplifiers, using S-parameter ing network. The schematic is shown Circles Utility. This two-part article equations. Input- and output-match- in Fig 1B. Start with the list in Fig 1A, will look at some of the ways of using ing networks can be evaluated. The and modify it later as required. I as- Circles. A brief overview of basic prin- stability of an active circuit can be sume the reader already knows how to ciples will be followed by some “walk- evaluated by plotting stability circles. use the Report Editor to get XY rect- through” examples that can be used as Noise-figure circles (circles of con- angular plots, tables and polar plots “templates” or guidance for future ref- stant noise-figure values) can be plot- versus frequency of such things as erence. The Circles Utility can do the ted. Figures of merit are calculated. MS11, MS21 etc, and how to set the following things: We can perform plots of certain quan- Terminations. An Optimization of the •Perform Smith Chart operations to tities over a frequency range. The circuit is also performed. design and analyze transmission-line result is a visual estimate of the per- Fig 2A shows how various lines of networks and LCR impedance-match- formance of the circuit. Part 2 of this constant resistance and constant reac- ing networks. Part 1 of this article article will deal with this. tance are plotted on the Smith Chart, deals with these topics. using the Circles Control Window (we •Using the Smith Chart, perform Smith Chart Basics will henceforth call it CCW). The vari- It is a good idea to bring up the ous entries that we type into the CCW Circles Utility and perform the follow- text window are just as shown (the ing operations as you read about them. “ = “ is optional and a “space” can be 1400 Harold Dr SE First, to enable Circles, we must used instead). In particular, pure re- Cedar Rapids, IA 52403 enter and Analyze (key F10) a circuit actances can only exist on the outer e-mail [email protected] listing of a two-port network as exem- circle of the chart; to move inside, some Sept/Oct 1998 3 resistance must be added. Positive reactance (+X, induc- where the vertical bars “| |” denote “magnitude.” From this tance) is in the upper half, negative reactance (–X, capaci- equation we see that a V circle is also a circle of constant tance) in the lower half. High resistance (R, low conduc- |RHO|, so the V circle is also called a “constant-reflection” tance G = 1/R) is toward the right. circle. Also of considerable interest is the return loss caYtiooun .c Taynp ael “sRo ”c orre a“Xte” ainn t Rhe o Cr CXW ci,r Eclxee caut tseo,m thee snp pelcaicfiec t lhoe- RL(dB)=- 20(cid:215) logRHO=- (cid:215)20 log(cid:230)Ł VV-+11(cid:246)ł dB (Eq 3) cursor at the desired location and press the left mouse but- ton. This creates an R or X circle and a label with some Return loss is a very sensitive measure of impedance specific value. match that is widely used in test equipment, such as net- To remove a single error in an entry, type “DEL” into the work analyzers. This term means “what fraction of the CCW, then Execute. To delete two entries, type “DEL 2,” power that is sent toward the load returns to the genera- and so forth. To get the best accuracy in all of the various tor?” In Radio Designer, RL (dB) is the same as MS11 (dB) operations, click on Settings/Display/Graphics/Line Width and MS22 (dB) at the input and output, respectively, of a and set the line width to “1,” the narrowest line. two-port network. The values of RHO and V can be found Fig 2B shows how various lines of constant-conductance for any combination of R and X or G and B, using the cursor and constant-susceptance values are plotted. This chart is method as described. a left-to-right and top-to-bottom reverse image of Fig 2A. The Smith Chart is basically a reflection-coefficient Positive susceptance (+B, capacitance) is below and nega- (RHO) chart. The distance from the center to the outer tive susceptance (–B, inductance) above the horizontal circle corresponds to |RHO| = 1.0, which is defined as axis. Both charts always show inductance above the axis “complete” reflection of a wave, which corresponds to a and capacitance below the axis. High conductance (G, low short-circuit load (R = 0), an open-circuit load (G = 0) or a resistance, R = 1/G) is toward the left. purely reactive load. The chart then assigns the R, X, G G and B circles at some cursor location are created by and B values in terms of the corresponding complex values typing “G” or “B,” Execute, then place the cursor at the of RHO according to the equation desired location and press the left mouse button. 1+RHO 1 The Circles Smith Chart is a YZ chart, which means that Z=Z0(cid:215) 1- RHO; Y= Z; RHOandZ0 possiblycomplex (Eq 4) Figs 2A and 2B and all four of the quantities shown there The method described for RHO, steps 1, 2, 3 and 4, can be can be plotted simultaneously on the same chart in two dif- ferent colors. The term “Z” refers to R and X in series and the term “Y” refers to G and B in parallel. It’s important to keep this distinction in mind, especially when switching between the two. This YZ capability is a powerful feature that we will use often. Also, the Z chart is “normalized” to 1 W at the origin. To normalize a 50 W system, divide all actual R and – X input values by 50. For other values of Z , such as 52 W , 0 450 W etc, use Z as a scaling factor. The Y chart is normal- 0 ized to 1 S (Siemens = 1 / W ) at the origin. For a 0.02 S (1 / (50 W )) system, multiply actual G and – B values by 50, or whatever Z is correct. It is very desirable to have a cal- 0 culator available to normalize values; the computations are (A) no problem once you get the “hang of it.” In the interest of simplicity, it is best (at least at the beginning) to use resis- * Smith chart example tive (not complex) values of Z . For a transmission line, a 0 resistive Z means a line with no attenuation (loss). BLK 0 Fig 2C shows various values of V (SWR) circles, entered IND 1 2 L=?1.94198UH? Q=250 F=7.15MHz as shown. At any point on a particular V circle, a value of CAP 2 0 C=?860.327PF? Q=10000 F=7.15MHz RHO, the reflection coefficient, can also be found. For ex- IND 2 3 L=?1.83567UH? Q=250 F=7.15MHz ample, at the intersection of R = 1 and V = 8, the reflection coefficient is 0.77 at 39.17(cid:176) (angle measured counterclock- TUNER:2POR 1 3 wise from the horizontal axis). To find this RHO, we use the END following procedure: FREQ •Enter the word “RHO” into the CCW and click the Ex- STEP 7.0MHZ 7.35MHZ 10 KHZ (B) ecute button. END •Place the cursor at the intersection of V = 8 and R = 1. •Press “M” on the keyboard. OPT •This places a mark at this location and also puts the TUNER R1=50 Z2=5 –50 MS11 complex (magnitude and angle) value of RHO in the CCW. The following equations give the relationships involved F=7.15MHZ MS11= –50 in the RHO and V operations: END RHO=((RR++jjXX))-+ZZ0 (Eq 1) ** CSeotm omutepnutst :load in Report Editor to 5 – j50 ohms 0 * Set generator in Report Editor to 50 ohms where R, X and Z are normalized to 1.0 as discussed pre- 0 viously and * Plot MS11 and MS21 1+ RHO V- 1 V=1- RHO ; RHO =V+1 (Eq 2) Ffoigr S1m—iAth i sC haa critr cauniatl ylissitsin. gB fiosr aa nsc ehxeammaptliec towf oth-peo critr ncuetitw.ork 4 QEX used to find values of Z and Y at some location. Type “Y” or •Create the appropriate R (or G) circle and an X (or B) “Z” in the CCW, then Execute, place the cursor at a location circle, as previously described. and press the “M” key. A marker appears on the chart and •Place the cursor exactly at the intersection and remove the R and X, or G and B, values appear in the CCW. If a your hand from the mouse. marker is not wanted, press the left mouse button instead. •Delete the two circles (“DEL 2”) if you do not want them This operation suggests an easy way to transform a series R to show. S and X to a parallel R and X because R = 1 / G and X = •Press the “M” key. This places a marker at the Z (or Y) S P P P P –1 / B at any selected location,. Working this backward, we location. can change parallel R to G (= 1 / R ) and X to B (= –1 / X ), Fig 2D shows arcs of constant Q values. Any values of R P P P P then to R and X by doing the Y to Z change. and +X that lie on the Q = +2 line correspond to X / R = Q S S To place a marker at a specific value of Z = R + jX (or Y = 2: for example, the intersection of the R = 1.5 circle and = G + jB) do the following: the X = +3 circle. For capacitive values of X, –Q is plotted Fig 2—A shows Smith X Chart circles of constant resistance, R, and reactance, X. Positive X is above the horizontal axis. B shows Smith Y Chart circles of constant conductance, G, and susceptance, B. Positive B is below the axis. C shows Smith Chart circles of constant SWR (V) and an example of a reflection-coefficient, RHO, location. D shows Smith Chart arcs of constant Q. Positive values of Q apply to inductive reactance or susceptance. Negative values of Q apply to capacitive reactance or susceptance. Sept/Oct 1998 5 below the horizontal axis. The inter- •Type DX into the CCW, then •Capacitive susceptance increases section of B = –3 and G = 1.5 is also at Execute. (capacitance increases) in a clockwise Q = B / G = 2. Q lines are useful in •Place the cursor first at Z and direction along a line of constant con- certain applications that are dis- press the left button. ductance. cussed in Part 2 of this article. •Move the cursor to Z1 and press •Conductance (G) increases in a The CCW contents and the complete the left button again. clockwise direction along a B line for B circle that we create can be saved to •The CCW displays the value of in- less than zero (above the horizontal disk by typing the SPLT command in ductance, L = DX / (2 p FREQ), that axis) and counterclockwise along a B the CCW. A name for the file is re- produces the change of reactance from line for B greater than zero (below the quested. This same file can be recalled Z to Z1 at the FREQ that was entered. horizontal axis). from disk by using the RPLT command. Instead of moving to Z1, we can move The next task is to move back and The name of the file is requested. from Z to Z2 along the R1 circle. This is forth between the Z chart and the Y a capacitive reactance of value –0.66. chart, using these steps: Navigating the Chart (Capacitive reactance increases —C •Starting at some Z point, move Z An important topic in Circles opera- decreases—in the counterclockwise di- to Z1 (Fig 3A). Then, using the CCW, tions concerns the ways that we rection.) The DX option now gives the find the value of Y1 at this point. modify the Z, Y, R, X, G and B quanti- needed value of capacitance, provided •Then move Y1 to Y2 (Fig 3B). ties from one chart location to another. that we place the cursor first at Z, then •Find the value of Z2 at Y2 and then Fig 3A shows an initial value of Z at at Z2. We can also move from Z1 to Z2 move Z2 to Z3, and so forth until the one location. We want to change Z to for a reactance change of –1.31. target is reached. any of the values Z1 through Z5. The To increase the value of R, from loca- Many impedance-matching prob- value of Z = R + jX is at the junction of tions Z, Z1 or Z2, move counterclock- lems start at a load impedance some- the R = 0.18 circle and the +X = 0.26 wise along a line of constant X, as what removed from the center of the circle, where +X is inductive. The fol- shown, if X is positive (above the hori- chart and work toward the chart cen- lowing rules are observed in Fig 3A: zontal axis). If X is negative, the mo- ter, R = 1, X = 0 or G = 1, B = 0. (We The impedance, Z, is at the junction tion is clockwise. The proper directions assume that the chart center is the of a constant-R circle and a constant- to reduce R are obvious in Fig 3A. impedance that the generator wants X circle. To increase X (make the Having arrived at Z3, Z4 or Z5, the to “see.” This process models adjust- inductive reactance, therefore the in- preceding operations can be repeated, ment of an antenna tuner. The target ductance, larger), move clockwise and in this manner we can travel impedance need not be at the chart along the R1 circle to point Z1. The around the chart. center. For example, if Z = 1.1 + j0.05 change in reactance is 0.91 – 0.26 = Fig 3B shows the travels for the Y or its complex conjugate, place a Z +0.65. (Inductive reactance increases chart. The rules are nearly the same marker there and make that the tar- in the clockwise direction.) as for the Z chart, and the directions get. It’s equally possible to place the The Circles Utility has another op- are as shown. The differences are: load at or near the chart center and the tion, called DX, which works as follows: •Inductive susceptance increases generator farther out. In that case, we •Enter a frequency in the CCW, (inductance decreases) in the counter- start at the center and work out to- for example FREQ = 7.15E6, then clockwise direction along a line of ward the generator. When we do this Execute. constant conductance. reversal of direction, a series inductor Fig 3—A shows directions of travel on a Z chart for X and R changes by adding or subtracting series XL, XC or R. B shows directions of travel on a Y chart for B changes by adding or subtracting shunt BL or BC. 6 QEX becomes a series capacitor, and a line that transforms impedance Z1 to Note also that the two Z points need shunt inductance becomes a shunt Z2. Z1 and Z2 can lie on the same V not be on the same V circle. If they are capacitance, and so forth. (SWR) circle as shown in Fig 4A, or on two different V circles (but not For example, a 50 W load is to be they may be on different V circles. Use V = ¥ ) the correct Z of the line and its 0 transformed to some complex imped- the following approach for the first length (in degrees) will be calculated. ance that a transistor collector (the option: on the same V circle: Two lines may be required. This valu- generator) wants to “see.” Similarly (a •Draw the V circle able feature is often used, especially little more difficult to visualize but •Locate Z1 and Z2 on the V circle in microstrip design. very important) a transistor input •Type DT in the CCW, then Execute Referring to Fig 2A, the difference looks backward toward a transformed •Place cursor at Z1 and click left between two points on a constant-re- 50 W (the load that the transistor input button, then place cursor at Z2 and sistance circle is a value of reactance. sees, looking back). In order to avoid click left button again. The clockwise After we mark two reactance points on confusion and errors the rule is: Start electrical angle is measured. Alterna- this circle (first the initial value, then at the load, wherever it is, and work tively, do Z2 first then Z1, in which the final value), the DXD operation toward the generator, wherever it is. case the angle is measured clockwise allows us to get this change of reac- The changing of impedance using from Z2 to Z1. The complete circle is tance with a section of transmission transmission lines and stubs is cov- 180(cid:176) electrical length. line connected as a stub in series with ered by three additional Circles func- The information appears in the CCW the line, as shown in Fig 4B. Quite tions as follows: text window. The physical length is: often two stubs—each having one deDgrTe efsin) dasn tdh ec healercatcrtiecrails tliecn gimthp e(idn- (meters)=(degrees)(cid:215) (velocityfactor)(cid:215) 0.833 hinasltfa olfl etdh,e o nneee dine de arecahc twainrcee o—f caa nb able- ( ) ance [Z (n) when normalized to 1.0, FREQ MHz anced transmission line as shown in 0 Z (u) for a 50 W line] of a transmission (Eq 5) Fig 4C. The stub(s) can be: 0 Fig 4—A shows the transmission line required to transform Z1 to Z2 along a line of constant SWR, where V = 3.0:1 (RHO = 0.5). Z0 (unnormalized) =49.53 W , Z0 (normalized) = 0.99. B (unbalanced) and C (balanced) show the DXD function; D shows the DBD function. Sept/Oct 1998 7 •A shorted stub (SS) less than l /4 for the components. MS21 tells the from 0.1 W to 0.9 W , where a V circle is an inductive reactance. circuit loss in decibels, about 0.35 dB has the low value 1.11. This transmis- •An open stub (OS) less than l /4 is (when MS11 dB is very large). The sion line should ideally have a Z of 0 a capacitive reactance. optimized L and C values are shown. 0.3 W . For a 50-W system, this would cor- The CCW asks for one of the follow- The last inductor “IND 2 3” could have respond to 15 W . Three small, parallel- ing: I (= Z , impedance of the transmis- a constant value (remove the question connected segments of 50 W coax would 0 sion line), D ( = degrees of electrical marks from the Netlist), and the other be okay. We can do some fine-tuning length) or W ( = wavelength). The CCW L and C are then optimized. with this method. The resistance added then tells which stub to use (OS or SS), The first coil from point A could be a by the inductor can be considered for a its reactance and other parameters. fixed value if it gets us beyond point B. more-accurate graphical answer. Use The DBD operation is just like DXD, This inductance could possibly be built the following approach: except that we put two susceptance right into the load device (antenna or •Get a first estimate of the normal- markers on a constant-conductance whatever). A length of transmission ized inductive reactance (= 1.0) as circle as in Fig 2B. We then get a par- line (Z = 1.0) could get us from, say described above. 0 allel stub (OS or SS) as shown in point C, along a constant V circle, over •Assume a value of Q for the in- L Fig 4D. The reactance of the stub and to the R = 1 circle. A simpler approach ductor. The R of the coil is then R = L L other parameters are displayed. This could use series inductance up to point X / Q . L L kind of stub is widely used. B and shunt capacitance from B to O. •Add R to the resistance R (= 0.1) L The combined usage of DT, DXD and Better yet, a series coil from A to C and of the load and relocate point A on the DBD in the Circles Utility is a very two low-loss capacitors, one in shunt Smith Chart accordingly. powerful way to match impedances us- from C to F and one in series from F to •On the chart, find a better value of ing transmission lines. By switching O. If the series inductive reactance the inductor needed to reach the hori- from one to another (and between Z and doesn’t go beyond point B at the lowest zontal axis. Y) we can travel the chart in transmis- frequency of interest, however, these •Further repetitions of this proce- sion-line segments and stubs in a very schemes won’t work. Another interest- dure are not necessary. elegant way. It is worthwhile to keep in ing idea would be a shunt coil from A to •The 1:9 transformer (assume it to mind that the reactances of transmis- K and a series coil from K to O. be loss-less) now takes us to a new sion-line segments vary over frequency Consider also an inductance from A value, which will still be close to the in somewhat different ways than ordi- up to the horizontal axis (the 0.1 mark) origin if Q is large. L nary lumped LC components. This can and a 9:1 (impedance ratio) ferrite- In this example a coil Q of 250 would L be a factor in designing a network. core transmission-line transformer mean a correction of 1.0 / 250 = 0.004 W , Problem Solving The art and gamesmanship of these matching exercises are to find the minimum number of components, es- pecially lossy inductors. Always keep the possibility of using transmission- line segments in mind because of their high Q (low loss) values. Sometimes a greater frequency-response band- width is achieved by using transmis- sion lines than lumped L and C. We also want values of L, C and lines that are realistic, efficient and economical. Fig 5 (drawn with fine lines to im- prove accuracy) illustrates the many different possibilities for converting a load impedance Z to R = 1, X = 0. We load will use, just as one example, the cir- cuit shown in Fig 1B. Z is at point load A, 0.10 – j1.0. The first component is a series inductor along the R = 0.10 line from A to C. A shunt capacitor moves the impedance along the G = 0.5 line from C to I, and a series inductor moves it along the R = 1.0 line from I to O (the origin). Fig 1B shows the values calculated in this manner. This circuit would have a very good low- pass filtering property, but requires three components, and at least two of them must be tunable. These values are placed in the circuit listing in Fig 1A and then optimized for best MS11 at 7.15 MHz, using realistic Q values Fig 5—Solution paths for the design example. 8 QEX which is negligible. In more extreme * Band tuning example situations it may be important. Keep in mind also the discussion BLK regarding the reversal of direction: for IND 1 2 L=?3.92257UH? Q=250 F=7.15MHZ example, starting at the origin (the CAP 2 0 C=?414.698PF? Q=1000 F=7.15MHZ load) and working outward to the gen- IND 2 3 L=2.3UH Q=250 F=7.15MHZ erator impedance or possibly its com- * Note that this L is held constant plex conjugate. ONE 3 0 ZDAT The main idea here is to illustrate TUNER:1POR 1 0 the power of the Smith Chart in visu- END alizing the myriad possible solutions, FREQ each of which has possible merits and STEP 6.9MHZ 7.4MHZ 10KHZ possible problems. With such a “shop- END ping list” the designer can make the best decisions. The Circles Utility can OPT then quickly and easily get values for TUNER R1=50 MS11 the various components. F=7.25MHZ MS11= -50 Also, a candidate circuit of lines/ END stubs and adjustable L and C should be tested over a frequency band for “tun- DATA ZDAT: Z RI ing range” of the components. A new * Freq Real Imag Designer file is created as shown in 7.00MHZ 3 -70 Fig 6 for this purpose. At any arbitrary 7.10MHZ 4 -60 frequency that is entered into the OPT 7.20MHZ 6 -40 block, the circuit values are optimized 7.30MHZ 8 -30 for maximum MS11. The load imped- END ance data is in the DATA block. This data is linked to the ONE circuit ele- * Comments: ment that has been placed in the main * Put the ONE element in the circuit block. * Change TUNER to a 1POR 1 0 (one port) as shown. circuit block. We can now see what * Set Report Editor to MS11 with TERM=50+j0 range of L and C values are needed to * Put load impedance data in the DATA block, real then tune the desired frequency range. imaginary. In the example of Fig 6, “IND 2 3” is * DATA is interpolated between freq entries. a fixed value (by assumption, it can’t * In the OPT block, insert any freq value in the range, be tuned) and it was necessary to in- one at a time. crease its fixed value (see Fig 1A) so * Analyze, then Optimize at the freq that is in the OPT that we could tune to the low end of the block. 7.0 to 7.3 MHz band with reasonable * Check to see the min and max values of tuning Ls and Cs values of the other components. This that are needed. is typical of problems that we might * Check to see that MS11 reaches a very large value. encounter. Fig 6—Net list for band tuning example. Another interesting exercise is to design a network that has a certain V (SWR) over a certain frequency band (without retuning). Using Eq 3, a V = 2 value corresponds to an MS11 (dB) cedure in this situation is to perform cations are available from your local ARRL of –9.5. A V = 1.5 value corresponds to DT in two segments. The first segment dealer or directly from the ARRL. Mail or- an MS11 (dB) of –14. By observing the begins “away from” the horizontal axis ders to Pub Sales Dept, ARRL, 225 Main plots of MS11 (dB) versus frequency and terminates “at” the horizontal St, Newington, CT 06111-1494. You can generated by Designer, the –9.5 dB call us toll-free at tel 888-277-5289; fax axis. The second segment begins “at” and –14 dB frequency ranges can be your order to 860-594-0303; or send e-mail the horizontal axis and proceeds to the easily observed. It is usually neces- to [email protected]. Check out the full desired finish point. The total electri- ARRL publications line on the World Wide sary to “Rescale” the graph, to see cal angle of the transmission line is Web at http://www.arrl.org/catalog. these levels more easily. For further then the sum of the lengths of the two Wilfred Caron, Antenna Impedance Match- information on the art and witchcraft segments. Some practice will make ing (Newington: 1989, ARRL), Order No. of broadband impedance matching, 2200. This book has very thorough cover- this a simple operation. Two RHO op- see References 2 and 3. age of matching methods using transmis- erations on a V circle also provide the sion lines and LC components. A Minor Bug DT electrical angle with no error. This W. E. Sabin, WØIYH, “Broadband HF An- problem does not occur often, but you tenna Matching With ARRL Radio De- The Circles Utility has a problem should be aware of it. signer,” QST, August 1995, p 33. when performing the DT (transmis- W. E. Sabin, WØIYH, “Computer Modeling of sion line) operation over a small region Additional Reading Coax Cable Circuits,” QEX, August 1996, of a V (SWR) circle close to the horizon- pp 3. R. Dean Straw, Editor, The ARRL Antenna tal (X = 0) axis (see Fig 4A). Incorrect W. E. Sabin, WØIYH, “Understanding the T- Book, 18th edition, Chapter 28, has an ex- tuner (C-L-C) Transmatch,” QEX, Decem- results may appear, due to a minor cellent general discussion of the Smith ber 1997, pp 13. glitch in the software. The correct pro- Chart. ARRL Order No. 6133. ARRL publi- Sept/Oct 1998 9 ARRL Radio Designer and the Circles Utility, Part 2: Small-Signal Amplifier Design Have you been intimidated by amplifier design? Maybe you want to step up to ARRL Radio Designer for the task. Either way, this will help you grasp this powerful design tool. By William E. Sabin, WØIYH Part 1 of this article was in- ticular features of the Radio Designer travels back to the generator (–I– is 1 tended to help the reader to program in this regard. correct because I– moves in phase 1 acquire greater familiarity with V– in a direction opposite to I+). with the Circles Utility of Radio De- S-Parameter Basics If we s1quare both sides of this equ1a- signer. We looked at passive network We begin with a brief overview of tion and divide both sides by Z01we get design using transmission lines and S-parameters, especially as they re- ( )2 LmCe thcoodmsp oonfe nntasv, iguastiningg ththee vaSrmioiuths ltawtoe- ptoo rRt andeitow Doreksi g(wneer w. Fililg c 1a lsl hito w“Ns ”a) V1– =(- I1–)2(cid:215) Z01= P1- (Eq 1) Z Chart. In this Part 2 we will consider that is connected to a generator and a 01 the interesting and very useful meth- load. The generator has a resistance This is a power wave that is re- ods of designing small-signal linear RG and is connected to the input port flected from N back to the generator. amplifiers using gain circles, stability through a transmission line having a We see that V+ and I+ are in-phase 1 1 circles, noise-figure circles and vari- characteristic impedance Z01. The out- (Z01 is a resistance) and V1– and –I1– ous figures of merit. These techniques put is connected to a load RL through are also in-phase, but V1+ and V1– (and are well known and widely used in in- a transmission line Z02. If t+he genera- I1+ and –I1–) may not be in-phase with dustry, and we will focus on the par- tor sends a voltage wave V toward N, each other. Both waves are in fact a current wave, I+ =V+/Z1, accompa- traveling simultaneously in opposite 1 1 01 nies it. The power in this forward wave directions (they are easily measured 1400 Harold Dr SE is P+ =V+I+ =(V+)2/Z =(I+)2Z . A independently using a directional cou- 1 1 1 1 01 1 01 Cedar Rapids, IA 52403 pler) but at any point the net voltage e-mail [email protected] reflected voltage wave, V1– =–I1–Z01, (or current) is the vector sum of the Nov/Dec 1998 3 two waves. The difference between P+ and P– is the power the effect of these re-reflections is that they create errors 1 1 that is accepted by N, and it is also the power that is deliv- in our measurements of the S-parameters (unless we com- ered by the generator. At the output of N, Fig 1 shows a pensate for them). In Radio Designer, we will always as- power wave traveling from the load toward N and a power sume that R = Z . An identical situation occurs on the L 02 wave from N to the load, and the same rules apply here. input side and we make R = Z . Because R = Z , it is G 01 L 02 The square roots of these four power waves are voltage true that a = 0 when the generator is connected at the 2 waves that are said to be normalized with respect to Z . input. Also, because R = Z , a = 0 when the generator is 0 G 01 1 + - + - connected to the output. Radio Designer takes care of this. V V V V a = 1 ; b = 1 ; a = 2 ; b = 2 We can now define the S-parameters, which we “custom- 1 1 2 2 Z Z Z Z arily,” although not necessarily, do in terms of voltage 01 01 02 02 waves as follows: (Eq 2a) They can also be expressed as normalized current waves - V 1 a1=I1+(cid:215) Z01 ; b1= I1- (cid:215) Z01 ; a2= I+2(cid:215) Z02 ; b=2 I-2(cid:215) Z02 S11= Z01 =V1- = b1 ; a =0 (Eq 2b) • V1+ V1+ a1 2 (Eq 3) Although Z and Z are often the same, especially in 01 02 Z network analyzers, they need not be the same. In Radio 01 Designer the Report Editor allows us to specify the Termi- This is the complex ratio of a reflected wave leaving port nations (Z0) separately. For example, if N is a loss-less 25:1 1 to a wave arriving at port 1 with no wave arriving at port impedance-ratio matching network from a 200 + j0 W gen- 2 (no reflection from R ,and the generator is at the left). erator to an 8 + j0 W load we can set the Input Termination It is also known as thLe reflection coefficient G and in at 200 + j0 W and the Output Termination at 8 + j0 W . Radio Radio Designer Circles Utility as RHO. Input SWINR can be Designer then tells us that this particular N has no mis- found from |S11| [SWR = (1 + |S11|) / (1 – |S11|)—Ed.]. match loss (the impedance matches at the input and out- S11 can be plotted on a Smith Chart. put are perfect). If the input impedance of N in this example is not 200 W , V- 2 the power available from the 200 W generator will not be - accepted by N. There is a mismatch loss. An exactly iden- S22= Z02 =V2 = b2 ; a =0 + + 1 tical statement is that the power not accepted by N is re- • V2 V2 a2 (Eq 4) flected and returns to the generator. The same is true at Z 02 the output. If we move the voltage generator to the right side (leave R = Z and R = Z where they are) the same This is the complex ratio of a reflected wave leaving port G 01 L 02 situation exists at the right-hand side of N. If the output 2 to a wave arriving at port 2 with no wave arriving at port impedance of N is not 8 W , some power is reflected at the 1 (no reflection from RG and the generator is at the right). output terminals. In practice, we routinely make this Also known as the reflection coefficient G OUT and in Radio switch of the generator so that we can measure the trans- Designer’s Circles Utility as RHO. Output SWR can be mission and reflection properties of N at the output side. found from |S22|. S22 can be plotted on a Smith Chart. Fig 1 shows a and a as normalized voltage waves rep- 1 2 - resenting the signals arriving at N and b1 and b2 as nor- V2 malized voltage waves leaving N. The other thing we see is Z b tthhae tl ethfte) piso rttriaonns omf ait1t eadcc, eapftteerd ibmy pNe d(farnomce tthrea ngsefnoerrmaatotrio onn, •S21= V1+02 = a21 ; a2=0 (Eq 5) to the output terminals. A portion of a2 is transmitted (from Z01 generator on the right), after impedance transformation, This is the complex ratio (gain or attenuation) of a wave from output back to input. leaving port 2 to a wave arriving at port 1 with no wave We need one other thing to complete the picture. If the arriving at port 2 (no reflection from R ). S21 is not an load R = Z then a wave b will be completely absorbed L L 02 2 impedance or a reflection factor, so it is plotted on a polar by R . If R = Z a wave b will be absorbed by R . If this L G 01 1 G graph. The quantity |S21|2 is referred to in transistor data is not true, re-reflections occur at R or R , and the situa- L G sheets as the “intrinsic” (same as “transducer”) power gain, tion becomes a little more complicated. In test equipment, Fig 1—Definitions of terms used in S-parameter analysis and design. 4 QEX