ARRL Radio Designer and the Circles Utility, Part I: Smith Chart Basics

17 Pages·1998·0.53 MB·English

ARRL Radio Designer and the Circles Utility
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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 [phone]; 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