Fields and Waves I Lecture 6 K. A. Connor Impedance Matching and Smith Charts K. A. Connor Electrical, Computer, and Systems Engineering Department Rensselaer Polytechnic Institute, Troy, NY Welcome to Fields and Waves I Before I start, can those of you with pagers and cell phones please turn them off? Thanks.

J. Darryl Michael – GE Global Research Center, Niskayuna, NY These Slides Were Prepared by Prof. Kenneth A. Connor Using Original Materials Written Mostly by the Following: Kenneth A. Connor – ECSE Department, Rensselaer Polytechnic Institute, Troy, NY J. Darryl Michael – GE Global Research Center, Niskayuna, NY Thomas P. Crowley – National Institute of Standards and Technology, Boulder, CO Sheppard J. Salon – ECSE Department, Rensselaer Polytechnic Institute, Troy, NY Lale Ergene – ITU Informatics Institute, Istanbul, Turkey Jeffrey Braunstein – Chung-Ang University, Seoul, Korea Materials from other sources are referenced where they are used. Those listed as Ulaby are figures from Ulaby’s textbook. 7 November 2018 Fields and Waves I

Ulaby 7 November 2018 Fields and Waves I

Impedance Matching Game http://contact.tm.agilent.com/Agilent/tmo/an-95-1/classes/imatch_popup.html 7 November 2018 Fields and Waves I

Impedance Match Designer http://www.rfengineer.cc/match1.htm 7 November 2018 Fields and Waves I

The Smith Chart Philip Smith, Electrical Engineer, Interview: http://www.ieee.org/organizations/history_center/oral_histories/abstracts/smithab.html http://my.ece.ucsb.edu/sanabria/tattoo.html http://www.ece.utah.edu/~cfurse/ http://flickr.com/photos/jonsphotos/218609214/ 7 November 2018 Fields and Waves I

Smith Chart 7 November 2018 Fields and Waves I http://n2pk.com/ http://www.rfcafe.com/references/electrical/smith-chart-artwork.htm 7 November 2018 Fields and Waves I http://www.tmworld.com/article/CA187342.html http://foe.mmu.edu.my/software/fksmith/

Impedance Matching Examples Why impedance match? Matching methods Overview Impedance Matching Examples Why impedance match? Matching methods The quarter-wave transformer The single stub match The double stub match 7 November 2018 Fields and Waves I

Load Matching Traditional application of Transmission Lines connect radio transmitter and antenna Antenna Impedance: Real Part - Radiation; Radiative Power looks like Resistive Loss to rest of circuit Imaginary Part - Reactive Power - not Radiated Function of Frequency Good antennas have real impedance - 1/2 wave dipole has ZL ~ 73W, but non-ideal properties affect this, e.g. nearby metal, people, ground Need to adjust for individual setup 7 November 2018 Fields and Waves I

Load Matching Objective in Load Matching: 1. Keep Reactive Power Small waste of power can damage equipment during short circuit (for example) Why? 2. Maximize Power to the Load can be achieved by minimizing G, by minimizing ZL - Zo can also improve coupling if Gs = 0 that’s why Function Generator has 50W output impedance also to reduce bounces 7 November 2018 Fields and Waves I

Load Matching - Example First, a simple example If ZL is complex, eliminating the imaginary part will decrease G 7 November 2018 Fields and Waves I

Load Matching - Example Characteristic Impedance Reflection Coefficient Voltage Standing Wave Ratio For matching, will need reactive components At high frequencies, use transmission lines to make L & C - standard components often will not work 7 November 2018 Fields and Waves I

Load Matching - Example Add Open Circuit Stub to Load to Cancel Out the Imaginary Impedance For matching, will need reactive components At high frequencies, use transmission lines to make L & C - standard components often will not work 7 November 2018 Fields and Waves I

Load Matching - Example Convert Combo to Admittance The Line Admittance and Impedance Should Be: For an Open Circuited Line The Total Load 7 November 2018 Fields and Waves I

Load Matching - Example The Total Load The New Reflection Coefficient The New VSWR Significant Improvement 7 November 2018 Fields and Waves I

Much of the following slides were written by Prof Much of the following slides were written by Prof. Nick Shuley from the University of Queensland (UQ). The books he uses the most are: Cheng Ulaby Pozar 7 November 2018 Fields and Waves I UQ

Why Impedance Match? Reflections lead to variations in the input impedance of the line. The input impedance changes with line length and frequency. Power is wasted. An impedance match provides maximum power transfer to the load. A VSWR > 1 means there will be voltage maxima on the line. These can lead to voltage breakdown at high power levels. 7 November 2018 Fields and Waves I UQ

VSWR = 1. Therefore there are no voltage peaks on the line. Benefits of Matching The input impedance remains constant at the value ZO. Therefore, the input impedance is independent of line length, and frequency (over the bandwidth of the matching network). VSWR = 1. Therefore there are no voltage peaks on the line. Maximum power transfer to the load is achieved. 7 November 2018 Fields and Waves I UQ

Matching Techniques We will now investigate two matching techniques which use sections of transmission line as circuit elements. The quarter-wave transformer The single stub matching network An excellent description of various other matching techniques is covered in Pozar. 7 November 2018 Fields and Waves I UQ

Quarter-Wave Transformer Consider a lossless quarter-wave length of line terminated by a resistance RL: 7 November 2018 Fields and Waves I UQ

Quarter-Wave Transformer (2) Assuming the line is lossless: Note that ZS is purely real, so the line allows us to transform one resistance value to another resistance. 7 November 2018 Fields and Waves I UQ

Key Properties of /4 Lines Impedance inversion: We can therefore convert an open circuit to a short circuit, and vice versa: short circuit termination: Zin, sc =   open circuit termination: Zin, oc = 0  7 November 2018 Fields and Waves I UQ

Example - Resistive Load A mismatched load can be matched to a transmission line using a quarter-wave transformer of suitable characteristic impedance. e.g.: match a 100  resistor to a 50  line.  the transformer characteristic impedance ZOT must be: 7 November 2018 Fields and Waves I UQ

Arbitrary Load If ZL is not real, a length of line (with characteristic impedance ZO) may be used to transform ZL to a real impedance, which can then be converted to ZO by the quarter-wave transformer, of characteristic impedance ZOT. 7 November 2018 Fields and Waves I UQ

Impedances, voltages, currents, etc. all repeat every half wavelength Smith Chart Impedances, voltages, currents, etc. all repeat every half wavelength The magnitude of the reflection coefficient, the standing wave ratio (SWR) do not change, so they characterize the voltage & current patterns on the line If the load impedance is normalized by the characteristic impedance of the line, the voltages, currents, impedances, etc. all still have the same properties, but the results can be generalized to any line with the same normalized impedances 7 November 2018 Fields and Waves I

The Smith Chart is a clever tool for analyzing transmission lines The outside of the chart shows location on the line in wavelengths The combination of intersecting circles inside the chart allow us to locate the normalized impedance and then to find the impedance anywhere on the line 7 November 2018 Fields and Waves I

Complex reflection coefficients Recall where we found that the reflection coefficient was a complex quantity. We can thus plot reflection coefficients in the complex  plane. The components are: 7 November 2018 Fields and Waves I UQ

Complex -plane Ulaby 7 November 2018 Fields and Waves I UQ

The Smith Chart We need to relate impedances to reflection coefficients: First, we normalize all impedances with respect to the characteristic impedance of the line: For an impedance of ZR becomes: 7 November 2018 Fields and Waves I UQ

Derivation Now since the normalized impedance can be written as: we set (6.3) equal to (6.2) using the real and imaginary parts of (6.1). This gives: We can then solve for the rR and xR in terms of . Graphical families of all possible solutions to this equation constitute the Smith Chart. 7 November 2018 Fields and Waves I UQ

Smith Chart Structure A Smith chart is This allows easy conversion therefore a polar plot of , with contours of real and imaginary parts of z superimposed on top. This allows easy conversion between normalized impedance z and  Ulaby 7 November 2018 Fields and Waves I UQ

Smith.pdf Radial scales around the outside are in wavelengths. These are used to determine lengths of line. Some Smith charts have a number of scales at the bottom of the chart but these are usually not necessary. Used very commonly in the literature to plot impedances as a function of frequency Ulaby 7 November 2018 Fields and Waves I UQ

A tour of the Smith chart two scales on the periphery (in wavelengths) 1 towards generator (clockwise) 1 towards load (anticlock) Note also that once around the whole chart is a total length of /2 Ulaby 7 November 2018 Fields and Waves I UQ

location of points All impedances in the top half are inductive e.g. 1+j 1+j All impedances in the bottom half are capacitive e.g. 1-2j 1-2j Ulaby 7 November 2018 Fields and Waves I UQ

location of points (2) 1.2j purely real impedances are along the horizontal centre line 1.2j 0.1 2 purely imaginary impedances are along the periphery -0.8j Ulaby 7 November 2018 Fields and Waves I UQ

A few special points open circuit point (infinite impedance) inductive unity impedance z =1 (match point) short circuit point (zero impedance) capacitive Ulaby 7 November 2018 Fields and Waves I UQ

Admittance  impedance Any point reflected through the centre point converts admittance to impedance and vice versa. z=2+3j Top Half: inductive reactance or capacitive susceptance usually marked on charts y=0.15-0.23j = 1/(2+3j) Bottom Half: capacitive reactance or inductive susceptance Ulaby 7 November 2018 Fields and Waves I UQ

Reflection coefficient The reflection coefficient is proportional to the length of the radial vector on the chart. The length of the vector to the periphery corresponds to  =1. The phase angle of the reflection coefficient is measured from the positive direction of the horizontal axis   Ulaby 7 November 2018 Fields and Waves I UQ

VSWR The VSWR corresponds to where the rotation of the reflection coefficient cuts the +ve real axis.  VSWR The VSWR must be always 1 Ulaby 7 November 2018 Fields and Waves I UQ

Example (1) Given a (normalised) load impedance of ZL = 2+3j Find the reflection coefficient and VSWR at the load. Analytically: 7 November 2018 Fields and Waves I UQ

By Smith chart locate the point (2+3j) measure the length of the line measure the length of the radius ratio of (2):(3) gives = 0.745 measure angle of (2+3j) gives 26.5 deg rotate (2+3j) on circle to +ve real axis, read VSWR = 6.9 r 2+3j Ulaby 7 November 2018 Fields and Waves I UQ

input impedance with a complex load Find the input impedance of a lossless transmission line given the following parameters: Z0=100. ZL=100+j100, line length =0.676 Z0=50 Zin = ? ZL=100+j100 7 November 2018 Fields and Waves I UQ

Solution Note that the normalised length is > than /2. Since once around the Smith Chart is half a wavelength, the input impedance repeats itself every /2. We can write  = 0.676  = 0.5+0.176 Next, normalise the load impedance Locate this point on the chart and then move this point, towards the generator (clockwise) 0.167, the impedance at the new point is zin=1-j1, denormalising, Zin=100-j100 7 November 2018 Fields and Waves I UQ

Solution 0.166λ 0.166λ+0.176λ = 0.340λ 1+1j 0.25 1-1j VSWR = 2.6 Ulaby 7 November 2018 Fields and Waves I UQ

The Chart gives direct conversion between  and Z. SUMMARY The Smith Chart allows the graphical solution of the transmission line equation for Z. The Chart gives direct conversion between  and Z. It may also be used to convert impedance to admittance and vice versa. 7 November 2018 Fields and Waves I UQ

Imaginary Impedance Axis Real Impedance Axis http://www.ife.ee.ethz.ch/~ichsc/smith_charts/Black_Magic.pdf 7 November 2018 Fields and Waves I

Smith Chart Constant Imaginary Impedance Lines Impedance Z=R+jX Normalized z=2+j for Zo=50 Constant Real Impedance Circles 7 November 2018 Fields and Waves I

Smith Chart from MAXIM Impedance divided by line impedance (50 Ohms) Z1 = 100 + j50 Z2 = 75 -j100 Z3 = j200 Z4 = 150 Z5 = infinity (an open circuit) Z6 = 0 (a short circuit) Z7 = 50 Z8 = 184 -j900 Then, normalize and plot. The points are plotted as follows: z1 = 2 + j z2 = 1.5 -j2 z3 = j4 z4 = 3 z5 = infinity z6 = 0 z7 = 1 z8 = 3.68 -j18S 7 November 2018 Fields and Waves I

Smith Chart Thus, the first step in analyzing a transmission line is to locate the normalized load impedance on the chart Next, a circle is drawn that represents the reflection coefficient or SWR. The center of the circle is the center of the chart. The circle passes through the normalized load impedance Any point on the line is found on this circle. Rotate clockwise to move toward the generator (away from the load) The distance moved on the line is indicated on the outside of the chart in wavelengths 7 November 2018 Fields and Waves I

Constant Reflection Coefficient Circle Toward Generator Constant Reflection Coefficient Circle Away From Generator Scale in Wavelengths Full Circle is One Half Wavelength Since Everything Repeats 7 November 2018 Fields and Waves I

Smith Chart References http://www.maxim-ic.com/appnotes.cfm/appnote_number/742/ http://www.sss-mag.com/smith.html http://www.educatorscorner.com/index.cgi?CONTENT_ID=2482 to download applet http://www.web-ee.com/primers/files/SmithCharts/smith_charts.htm http://www.amanogawa.com/index.html Two examples from this page are shown in the following slides 7 November 2018 Fields and Waves I

Smith Chart Problem http://cnx.rice.edu/content/m1059/latest/ Does not work with IE 7 November 2018 Fields and Waves I

First, locate the normalized impedance on the chart for ZL = 50 + j100 Smith Chart Example First, locate the normalized impedance on the chart for ZL = 50 + j100 Then draw the circle through the point The circle gives us the reflection coefficient (the radius of the circle) which can be read from the scale at the bottom of most charts Also note that exactly opposite to the normalized load is its admittance. Thus, the chart can also be used to find the admittance. We use this fact in stub matching 7 November 2018 Fields and Waves I

7 November 2018 Fields and Waves I

Note – the cursor is at the load location 7 November 2018 Fields and Waves I

Load of 100 + j100 Ohms on 50 Ohm Transmission Line Single Stub Matching Load of 100 + j100 Ohms on 50 Ohm Transmission Line The frequency is 1 GHz = 1x109 Hz Want to place an open circuit stub somewhere on the line to match the load to the line, at least as well as possible. The steps are well described at http://www.amanogawa.com/index.html First the line and load are specified. Then the step by step procedure is followed to locate the open circuit stub to match the line to the load 7 November 2018 Fields and Waves I

7 November 2018 Fields and Waves I

7 November 2018 Fields and Waves I

7 November 2018 Fields and Waves I

7 November 2018 Fields and Waves I

7 November 2018 Fields and Waves I

7 November 2018 Fields and Waves I

7 November 2018 Fields and Waves I

Smith Chart Construction 7 November 2018 Fields and Waves I UQ

/4 Transformer Design Example Design a quarter-wave transformer to match a load ZL = 30  j100  to a 50  line. The transformer is to be placed as close as possible to the load. 7 November 2018 Fields and Waves I UQ

Plot the load on the Smith Chart Rotate around to Vmin Solution Plot the load on the Smith Chart Rotate around to Vmin Determine the distance to the transformer (in wavelengths) Read off R2 Calculate the ZOT of the transformer 7 November 2018 Fields and Waves I UQ

Single-stub Matching Networks (can also use an open circuited stub) Ulaby 7 November 2018 Fields and Waves I UQ

Single-stub Equivalent Circuit Ulaby 7 November 2018 Fields and Waves I UQ

Single-stub Design Method Convert the load impedance ZL to an equivalent admittance YL = 1/ZL. Use a length of line of characteristic impedance Zo to transform YL to Yd = Yo + jB. Note Yo = 1/Zo Combine a stub in parallel which has an input admittance Ys = -jB. Therefore, the total admittance at MM’ is: i.e. we have an impedance match! 7 November 2018 Fields and Waves I UQ

Single-stub Design Example We wish to match a load impedance of ZL = (25 - j50)  to a 50  transmission line. convert the load to a normalised impedance convert the load impedance to an admittance using the Smith Chart (transform point A to point B) 7 November 2018 Fields and Waves I UQ

B C E A F 7 November 2018 Fields and Waves I UQ

Single-stub Design Example In the admittance domain, constant resistance circles (r) become constant conductance circles (g). Rotate towards the generator until the VSWR circle cuts the g = 1 circle (this means the real part of Y is equal to Yo). (This point is labelled C on the Smith Chart). Note the distance travelled (d), and the admittance at C (yd): d = (0.178-0.115) = 0.063 yd = 1 + j1.58 7 November 2018 Fields and Waves I UQ

Single-stub Design Example Looking from the generator toward the parallel combination of the line connected to the load and the stub, the normalised input admittance at the junction is which must be equal to to ensure that we have an impedance match to Zo. 7 November 2018 Fields and Waves I UQ

Single-stub Design Example Therefore For a short circuited stub, the normalised admittance of a short circuit is  and is located at point E on the Smith Chart. We need to rotate this point towards the generator to obtain the desired input admittance of -j1.58, which is located at point F on the Smith Chart. 7 November 2018 Fields and Waves I UQ

Single-stub Design Example The distance travelled (and hence the length of the stub) is: The design is now complete, as we have the length of the stub and the length of the line connecting the load to the stub. Note that the transmission lines all have a characteristic impedance equal to Zo, the characteristic impedance of the line we are matching to. 7 November 2018 Fields and Waves I UQ

Single-stub Solutions A single-stub matching network design can lead to 4 possible solutions. In the example just completed, we could have selected: yd = 1 + j1.58 OR yd = 1 - j1.58 a short circuit terminated stub OR an open circuit terminated stub 7 November 2018 Fields and Waves I UQ

Single-stub Solutions Which you choose depends upon practical considerations: Can I realise open or short circuit terminations in the transmission line I am using? Does it matter if there is a voltage maximum on the line between the stub and the load termination? Is the physical length of line between the stub and the load termination too short/long? As an engineer, these are the decisions you must be able to make! 7 November 2018 Fields and Waves I UQ

Impedance matching is necessary to: SUMMARY Impedance matching is necessary to: reduce VSWR obtain maximum power transfer A quarter-wave line can be used to transform resistance values, and act as an impedance inverter. 7 November 2018 Fields and Waves I UQ

A single stub matching network can also be used. SUMMARY Combined with a series length of line, a quarter-wave transformer can match complex loads to a resistive Zo. A single stub matching network can also be used. Both matching network types are narrow-band: they are designed to operate at a single frequency only. 7 November 2018 Fields and Waves I UQ

disadvantage of single stubs Single stub placement means that we have no choice in the position (distance from the load) of the stub. This not might be practical. Furthermore, not all load impedances can be matched! A double stub tuner does not have this problem, the distance from the load is more or less arbitrary. However, it still cannot match all impedances. 7 November 2018 Fields and Waves I UQ

sample arrangement determine stub lengths 1, 2 given position d1 could be anything, but usually fixed. λ/8 d1 ZL 2 1 short circuited stubs (could be also open circuited) matched s =1 7 November 2018 Fields and Waves I UQ

rotate the unit circle rotate the unit circle λ/8 (1/4 turn) towards the load displaced circle Strategy: A way of interpreting this is that all impedances on the blue circle, when rotated λ/8 towards the generator will end up on the red circle. The load stub can then take out any remaining reactance. 7 November 2018 Fields and Waves I UQ

start from the load impedance ZL 7 November 2018 Fields and Waves I UQ

invert to find admittance YL we have to work in admittance because the stubs are in parallel with the main line and therefore admittances add! 7 November 2018 Fields and Waves I UQ

we are up to here d1 ZL YL 7 November 2018 Fields and Waves I UQ

step 4 rotate the admittance d1 we are rotating (towards the generator) along a circle whose radius is determined by the VSWR in the load section. YL Y'L 7 November 2018 Fields and Waves I UQ

up to here now d1 ZL Y'L 7 November 2018 Fields and Waves I UQ

move along a constant conductance line Now, move along a constant conductance circle so as to end up on the blue circle. YL Y'L Moving along a constant conductance line does not change the real part of the admittance, only the imaginary part. 7 November 2018 Fields and Waves I UQ

determine the load stub length From the previous slide: N.B. The signs on the imaginary component are generally . The signs here correspond to the example. in moving from Y'L to Y"L we have to add j(B1+B2), this susceptance will be provided by the stub. The load stub is now to provide +j(B1+B2); Note that the stub can only provide an imaginary contribution. 7 November 2018 Fields and Waves I UQ

load stub length generator stub length in wavelengths 1 +j(B1+B2) s/c admittance point generator stub length in wavelengths 1 7 November 2018 Fields and Waves I UQ

Up to here d1 ZL 2 1 Y'L 7 November 2018 Fields and Waves I UQ

Generator stub We can now rotate the point Y"L on the blue circle λ/8 to the red circle to locate point YL"'. YL"' This is the whole point of displacing the unit circle by λ/8! YL"' has admittance of 1+jB3; the imaginary part can be taken out with the second stub. 7 November 2018 Fields and Waves I UQ

completed d1 d1 ZL ZL 2 2 1 Y"'L Y'L admittances before addition of stubs 7 November 2018 Fields and Waves I UQ

generator stub length -jB3 generator stub length in wavelengths 2 s/c admittance point -jB3 generator stub length in wavelengths 2 7 November 2018 Fields and Waves I UQ

Alternative solutions Note in slide 10 that we could have intersected the transposed circle in two places. By choosing the second intersection, we can have a second solution. × Y"L × This would then rotate to the bottom half of the red unit circle on rotation through λ/8. Y'L 7 November 2018 Fields and Waves I UQ

Try it yourself! Given a load impedance of (65-j50)  with two stubs λ/8 apart and the load stub is 0.11λ from the load. design a double stub matching circuit. One solution is: 1 = 0.128λ 2 = 0.289λ Usually we choose the shortest stubs The other is: 1 = 0.377λ 2 = 0.417λ 7 November 2018 Fields and Waves I UQ

Added complications We have assumed that the characteristic impedances to be all the same (50). We do not have to have this restriction. For example the stubs could be 70 and the main lines 50. These problems are usually handled by repeatedly normalizing and de-normalizing with respect to the different characteristic impedances. 7 November 2018 Fields and Waves I UQ

Other Constructions The method outlined in this lecture is only one method of solving the double stub matching problem. There are other graphical methods as well as computer programs that carry out the calculations instead. 7 November 2018 Fields and Waves I UQ

Online Aids http://www.amanogawa.com/archive/transmissionA.html 7 November 2018 Fields and Waves I