9/4/2007 Matching Networks Notes 1/2 Jim Stiles The Univ. of Kansas Dept. of EECS D. Matching Networks Q: Yikes! The signal source is generally a Thevenin’s equivalent of the output of some useful device, while the load impedance is generally the input impedance of some other useful device. I do not want to—nor typically can I—change these devices or alter their characteristics. Must I then just accept the fact that I will achieve suboptimum power transfer? A: NOPE! All of the available power of the source can be delivered to the load—if we properly construct a matching network. HO: Matching Networks Q: But in microwave circuits, a source and load are connected by a transmission line. Can we implement matching networks in transmission line circuits? A: HO: Matching Networks and Transmission Lines Q: Matching networks seem almost too good to be true; can we really design and construct them to provide a perfect match? A: It is relatively easy to provide a near perfect match at precisely one frequency!9/4/2007 Matching Networks Notes 2/2 Jim Stiles The Univ. of Kansas Dept. of EECS But, since lossless matching networks are made entirely of reactive elements (not to mention the reactive components of the source and load impedance), we find that changing the signal frequency will typically “mismatch” our circuit! Thus a difficult challenge for any microwave component designer is to provide a wideband match to a transmission line with characteristic impedance Z0. Æ All microwave components thus have a finite operating bandwidth!9/4/2007 Matching Networks 1/9 Jim Stiles The Univ. of Kansas Dept. of EECS Matching Networks Consider again the problem where a passive load is attached to an active source: The load will absorb power—power that is delivered to it by the source. {}{}222212121212LLLgLggL gLLggLLggLPReVIVZRe VZZ ZZRe ZVZZRVZZ∗∗=⎧⎫⎛⎞⎛⎞⎪⎪=⎜⎟⎜⎟⎨⎬⎜⎟⎜⎟++⎪⎪⎝⎠⎝⎠⎩⎭=+=+ Recall that the power delivered to the load will be maximized (for a given gV and gZ) if the load impedance is equal to the complex conjugate of the source impedance (LgZZ∗= ). Vg gZ LL LZRjX=+9/4/2007 Matching Networks 2/9 Jim Stiles The Univ. of Kansas Dept. of EECS We call this maximum power the available power avlP of the source—it is, after all, the largest amount of power that the source can ever deliver! 22222121228maxLavlgggggggggPPRVZZRVRVR∗=+== * Note the available power of the source is dependent on source parameters only (i.e., gV and gR). This makes sense! Do you see why? * Thus, we can say that to “take full advantage” of all the available power of the source, we must to make the load impedance the complex conjugate of the source impedance. * Otherwise, the power delivered to the load will be less than power made available by the source! In other “words”: LavlPP≤9/4/2007 Matching Networks 3/9 Jim Stiles The Univ. of Kansas Dept. of EECS A: NO! We can in fact modify our circuit such that all available source power is delivered to the load—without in any way altering the impedance value of that load! To accomplish this, we must insert a matching network between the source and the load: The sole purpose of this matching network is to “transform” the load impedance into an input impedance that is conjugate matched to the source! I.E.: LL LZRjX=+ Vg gggZRjX=+ Matching Network inV+− inI LV+− LI Q: But, you said that the load impedance typically models the input impedance of some useful device. We don’t typically get 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?9/4/2007 Matching Networks 4/9 Jim Stiles The Univ. of Kansas Dept. of EECS *inin ginVZZI== Because of this, all available source power is delivered to the input of the matching network (i.e., delivered to inZ): inavlPP= LL LZRjX=+ Matching Network *in gZZ= Q: Wait just one second! The matching network ensures that all available power is delivered to the input of the matching network, but that does not mean (necessarily) that this power will be delivered to the load LZ . The power delivered to the load could still be much less than the available power!9/4/2007 Matching Networks 5/9 Jim Stiles The Univ. of Kansas Dept. of EECS A: True! To ensure that the available power delivered to the input of the matching network is entirely delivered to the load, we must construct our matching network such that it cannot absorb any power—the matching network must be lossless! We must construct our matching network entirely with reactive elements! Examples of reactive elements include inductors, capacitors, transformers, as well as lengths of lossless transmission lines. Thus, constructing a proper lossless matching network will lead to the happy condition where: LinavlPP P== * Note that the design and construction of this lossless network will depend on both the value of source impedance gZand load impedance LZ. * However, the matching network does not physically alter the values of either of these two quantities—the source and load are left physically unchanged! Now, let’s consider the matching network from a different perspective. Instead of defining it in terms of its input impedance when attached the load, let’s describe it in terms of its output impedance when attached to the source:9/4/2007 Matching Networks 6/9 Jim Stiles The Univ. of Kansas Dept. of EECS This “new” source (i.e., the original source with the matching network attached) can be expressed in terms of its Thevenin’s equivalent circuit: This equivalent circuit can be determined by first evaluating (or measuring) the open-circuit output voltage ocoutV: outV+− Vg gggZRjX=+ Matching Network outI Vs out out outZRjX=+ ocoutV+− Vg gggZRjX=+ Matching Network 0outI=9/4/2007 Matching Networks 7/9 Jim Stiles The Univ. of Kansas Dept. of EECS And likewise evaluating (or measuring) the short-circuit output current scoutI: From these two values ( and oc scout outVI) we can determine the Thevenin’s equivalent source: ococoutsout outscoutVVVZI== Note that in general that sgVV≠ and out gZZ≠—the matching network “transforms” both the values of both the impedance and the voltage source. 0outV+−= Vg