Chapter 5 Impedance Matching and Tuning

Transcription

1 3/25/29 section 5_1 Match with umped Elements 1/3 Chapter 5 Impedance Match and Tun One of the most important and fundamental two-port networks that microwave eneers des is a lossless match network (otherwise known as an impedance transformer). HO: MATCHING NETWORKS Q: In microwave circuits, a source and load are connected by a transmission le. Can we implement match networks transmission le circuits? A: HO: MATCHING NETWORKS AND TRANSMISSION INES Q: These match networks seem too ood to be true can we really des and construct them to provide a perfect match? A: We can easily provide a near perfect match at precisely one frequency. But, sce lossless match and transmission les are made of entirely reactive elements (not to mention the reactive components of source and load impedance), we fd that chan the frequency will typically unmatch our circuit!

2 3/25/29 section 5_1 Match with umped Elements 2/3 Thus, a difficult challene for any microwave des eneer is to des a wideband match network a match network that provides an adequate match over a wide rane of frequencies. Generally speak, match network des requires a tradeoff between these for desirable attributes: 1. Bandwidth 2. Complexity 3. Implementation 4. Adjustability 5.1 Match with umped Elements Read Assment: pp Now let s be to exame how match networks are built! We be with the simplest solution: An -network, consist of a sle capacitor and a sle ductor. Q: Just two elements! That seems simple enouh. Do we always use these -networks when construct lossless match networks?

3 3/25/29 section 5_1 Match with umped Elements 3/3 A: Nope. -networks have two major drawbacks: 1. They are narrow-band. 2. Capacitors and ductors are difficult to make at microwave frequencies! Now, let s see how these -networks actually work: HO: -NETWORK ANAYSIS

4 3/12/27 Match Networks 1/9 Match Networks Consider aa the problem where a passive load is attached to an active source: V = R + jx The load will absorb power power that is delivered to it by the source. 1 P = Re{ VI} 2 1 V = Re V Re { } = V = V 2 2 R Recall that the power delivered to the load will be maximized (for a iven V and ) if the load impedance is equal to the complex conjuate of the source impedance ( = ).

5 3/12/27 Match Networks 2/9 We call this maximum power the available power P avl of the source it is, after all, the larest amount of power that the source can ever deliver! P max P avl 1 = V R = V 2 2R R = V 2 8 R * Note the available power of the source is dependent on source parameters only (i.e., V and R ). This makes sense! Do you see why? * Thus, we can say that to take full advantae of all the available power of the source, we must to make the load impedance the complex conjuate of the source impedance. * Otherwise, the power delivered to the load will be less than power made available by the source! In other words : P P avl

6 3/12/27 Match Networks 3/9 Q: But, you said that the load impedance typically models the put impedance of some useful device. We don t typically et to select or adjust this impedance it is what it is. Must we then simply accept the fact that the delivered power will be less than the available power? A: NO! We can fact modify our circuit such that all available source power is delivered to the load without any way alter the impedance value of that load! To accomplish this, we must sert a match network between the source and the load: = R + jx I V I + V Match Network + V = R + jx The sole purpose of this match network is to transform the load impedance to an put impedance that is conjuate matched to the source! I.E.:

7 3/12/27 Match Networks 4/9 V = = I * = * Match Network = R + jx Because of this, all available source power is delivered to the put of the match network (i.e., delivered to ): P = P avl Q: Wait just one second! The match network ensures that all available power is delivered to the put of the match network, but that does not mean (necessarily) that this power will be delivered to the load. The power delivered to the load could still be much less than the available power!

8 3/12/27 Match Networks 5/9 A: True! To ensure that the available power delivered to the put of the match network is entirely delivered to the load, we must construct our match network such that it cannot absorb any power the match network must be lossless! We must construct our match network entirely with reactive elements! Examples of reactive elements clude ductors, capacitors, transformers, as well as lenths of lossless transmission les. Thus, construct a proper lossless match network will lead to the happy condition where: P = P = Pavl * Note that the des and construction of this lossless network will depend on both the value of source impedance and load impedance. * However, the match network does not physically alter the values of either of these two quantities the source and load are left physically unchaned! Now, let s consider the match network from a different perspective. Instead of def it terms of its put impedance when attached the load, let s describe it terms of its output impedance when attached to the source:

9 3/12/27 Match Networks 6/9 = R + jx I out V Match Network + V out This new source (i.e., the orial source with the match network attached) can be expressed terms of its Theven s equivalent circuit: = R + jx out out out V s This equivalent circuit can be determed by first evaluat oc (or measur) the open-circuit output voltae V : out = R + jx I out = V Match Network + oc V out

10 3/12/27 Match Networks 7/9 And likewise evaluat (or measur) the short-circuit output current I sc : out = R + jx sc I out V Match Network + V out = oc sc From these two values ( V and I ) we can determe the out Theven s equivalent source: out V oc oc out s = out out = oc Iout V V Note that eneral that V s V and the match out network transforms both the values of both the impedance and the voltae source. Q: Arrr! Doesn t that mean that the available power of this transformed source will be different from the orial?

11 3/12/27 Match Networks 8/9 A: Nope. If the match network is lossless, the available power of this equivalent source is identical to the available power of the orial source the lossless match network does not alter the available power! Now, for a properly desed, lossless match network, it turns out that (as you miht have expected!) the output impedance is equal to the complex conjuate of the load out impedance. I.E.: = out The source and load are aa matched! Thus, we can look at the match network two equivalent ways: = R + jx V Match Network = R + jx 1. As a network attached to a load, one that transforms its impedance to a value matched to the source impedance :

12 3/12/27 Match Networks 9/9 V = 2. Or, as network attached to a source, one that transforms its impedance to out a value matched to the load impedance : out = V V s Either way, the source and load impedance are conjuate matched all the available power is delivered to the load!

13 3/12/27 Match Networks and Transmission es 1/7 Match Networks and Transmission es Recall that a primary purpose of a transmission le is to allow the transfer of power from a source to a load. V Q: So, say we directly connect an arbitrary source to an arbitrary load via a lenth of transmission le. Will the power delivered to the load be equal to the available power of the source? A: Not likely! Remember we determed earlier that the efficacy of power transfer depends on: 1. the source impedance. 2. load impedance. 3. the transmission le characteristic impedance.

14 3/12/27 Match Networks and Transmission es 2/7 4. the transmission le lenth. Recall that maximum power transfer occurred only when these four parameters resulted the put impedance of the transmission le be equal to the complex conjuate of the source impedance (i.e., = ). It is of course unlikely that the very specific conditions of a conjuate match will occur if we simply connect a lenth of transmission le between an arbitrary source and load, and thus the power delivered to the load will enerally be less than the available power of the source. Q: Is there any way to use a match network to fix this problem? Can the power delivered to the load be creased to equal the available power of the source if there is a transmission le connect them? A: There sure is! We can likewise construct a match network for the case where the source and load are connected by a transmission le. For example, we can construct a network to transform the put impedance of the transmission le to the complex conjuate of the source impedance:

15 3/12/27 Match Networks and Transmission es 3/7 V Match Network Q: But, do we have to place the match network between the source and the transmission le? A: Nope! We could also place a (different) match network between the transmission le and the load. V Match Network In either case, we fd that at any and all pots alon this matched circuit, the output impedance of the equivalent source (i.e., look left) will be equal to the complex conjuate of the put impedance (i.e., look riht).

16 3/12/27 Match Networks and Transmission es 4/7 out = V V s = out Q: So which method should we chose? Do eneers typically place the match network between the source and the transmission le, or place it between the transmission le and the load? A: Actually, the typical solution is to do both! We fd that often there is a match network between the a source and the transmission le, and between the le and the load. V Match Network Match Network

17 3/12/27 Match Networks and Transmission es 5/7 The first network matches the source to the transmission le other words, it transforms the output impedance of the equivalent source to a value numerically equal to characteristic impedance : out = V V s The second network matches the load to the transmission le other words it transforms the load impedance to a value numerically equal to characteristic impedance : = Q: Yikes! Why would we want to build two separate match networks, stead of just one?

18 3/12/27 Match Networks and Transmission es 6/7 A: By us two separate match networks, we can decouple the des problem. Recall aa that the des of a sle match network solution would depend on four separate parameters: 1. the source impedance. 2. load impedance. 3. the transmission le characteristic impedance. 4. the transmission le lenth. Alternatively, the des of the network match the source and transmission le depends on only: 1. the source impedance. 2. the transmission le characteristic impedance. Whereas, the des of the network match the load and transmission le depends on only: 1. the source impedance. 2. the transmission le characteristic impedance. Note that neither des depends on the transmission le lenth!

19 3/12/27 Match Networks and Transmission es 7/7 Q: How is that possible? A: Remember the case where = =. For that special case, we found that a conjuate match was the result reardless of the transmission le lenth. Thus, by match the source to le impedance and likewise match the load to the le impedance, a conjuate match is assured but the lenth of the transmission le does not matter! In fact, the typically problem for microwave eneers is to match a load (e.., device put impedance) to a standard transmission le impedance (typically = 5Ω); or to dependently match a source (e.., device output impedance) to a standard le impedance. A conjuate match is thus obtaed by connect the two with a transmission le of any lenth! V s

20 3/25/29 Network Analysis 1/1 -Network Analysis Consider the first match -network, which we shall denote as match network (A): = jx, β Γ ( ) ( ) = Γ = Y = jb z = Note that this match network consists of just two lumped elements, which must be purely reactive other words, a capacitor and an ductor! To make Γ =, the put impedance of the network must be: = Note that us basic circuit analysis we fd that this put impedance is: 1 jb = jx jb = jx jb

21 3/25/29 Network Analysis 2/1 Note that a matched network, with =, means that: Re{ } = AND Im{ } = Note that there are two equations. This works out well, sce we have two unknowns (B and X)! Essentially, the -network match network can be viewed as consist of two distct parts, each attempt to satisfy a specific requirement. Part 1: Select Y = jb Sce the shunt element Y and are parallel, we can combe them to one element that we shall call Y 1 : 1 Y1 Y + = jb + Y The impedance of this element is therefore: = = = + + Y 1 jb Y j B, β Γ ( ) ( ) = Γ = = jx 1 = + j B z =

22 3/25/29 Network Analysis 3/1 To achieve a perfect match, we must set the value of susceptance B such that: Re { } = Re = 1 + j B Thus, if B is properly selected: = jx, β Γ ( ) ( ) = Γ = 1 = + j X1 z = Hopefully, the second part of the match is now very obvious to you! Part 2: Select = jx Note that the impedance 1 1 = jb has the ideal real value of. However, it likewise posses an annoy imaary part of: = { } = X1 Im 1 Im + j B However, this imaary component is easily removed by sett the series element = j X to its equal but opposite value! I.E.,:

23 3/25/29 Network Analysis 4/1 = = X X1 Im + j B = jx 1, β Γ ( ) ( ) = Γ = 1 = + j X1 Thus, we fd that: z = = + 1 = jx + + jx = 1 1 We have created a perfect match! Go throuh this complex alebra, we can solve for the required values X and B to satisfy these two equations to create a matched network! and, B = X ± R R + X R R + X X 1 X = + B R B R

24 3/25/29 Network Analysis 5/1 where = R + jx. Note: 1) Because of the ±, there are two solutions for B (and thus X). 2) For jb to be purely imaary (i.e., reactive), B must be real. From the term: R + X R 2 2 the expression for B, we note that R must be reater than ( R > ) to sure that B and thus X is real. In other words, this match network can only be used when R >. Notice that this condition means that the normalized load z lies side the r = 1 circle on the Smith Chart! Now let s consider the second of the two -networks, which we shall call network (B). Note it also is formed with just two lumped elements. = jx, β Γ ( ) ( ) = Γ = Y = jb z =

25 3/25/29 Network Analysis 6/1 To make Γ =, the put admittance of the network must be: Y = Y Note from circuit theory that the put admittance for this network is: 1 Y = jb + jx + Therefore a matched network, with Y = Y, is described as: Re{Y } = Y AND Im{Y } = For this des, we set the value of = jx such that the admittance Y 1 : 1 1 Y1 = + jx + has a real part equal to Y : 1 = { } = + Y Re Y Re 1 jx, β Γ ( ) ( ) = Γ = Y = jb Y1 = Y + jb1 z =

26 3/25/29 Network Analysis 7/1 Now, it is evident that a perfect match will occur if the shunt element Y = jb is set to cancel the reactive component of Y 1 : So that we fd: 1 B = Im{ Y} 1 = Im jx + A perfect match! ( ) Y = Y + Y = jb + Y + jb = Y 1 1 1, β ( ) ( ) Γ =Γ = Y = Y z = With these two equations, we can directly solve for the required values X and B for a matched network: and, ( ) X = ± R R X where = R + jx. B =± ( ) R R

27 3/25/29 Network Analysis 8/1 Note: 1) Because of the ±, there are two solutions for B (and thus X). 2) For jb and jx to be purely imaary (i.e., reactive), B and X must be real. We note from the term: ( R ) that R must be less than ( R < ) to sure that B and thus X are real. In other words, this match network can only be used when R <. Notice that this condition means that the normalized load z lies outside the r = 1 circle on the Smith Chart! (B) (A)

28 3/25/29 Network Analysis 9/1 Once the values of X and B are found, we can determe the required values of ductance and/or capacitance C, for the sal frequency ω! Recall that: and that: ω if X > X = 1 if X < ω C ω C if B > B = 1 if B < ω Make sure that you see and know why these equations are true. As a result, we see that the reactance or susceptance of the elements of our -network will have the proper values for match at precisely one and only one frequency! And this frequency better be the sal frequency ω! If the sal frequency chanes from this des frequency, the reactance and susceptance of the match network ductors and capacitors will likewise chane. The circuit will no loner be matched.

29 3/25/29 Network Analysis 1/1 This match network has a narrow bandwidth! An -Network Des Example One other problem; it becomes very difficult to build quality lumped elements with useful values past 1 or 2 GHz. Thus, - Network solutions are enerally applicable only the RF reion (i.e., < 2GHz).