EECS 242 Two Port Gain and Stability Prof Niknejad University of California Berkeley University of California Berkeley EECS 242 p 1 33 Input Output Admittance The input and output impedance of a two port will play an important role in our discussions The stability and power gain of the two port is determined by these quantities In terms of y parameters Yin V2 I1 Y11 V1 Y12 V2 Y11 Y12 V1 V1 V1 The voltage gain of the two port is given by solving the following equations I2 V2 YL Y21 V1 V2 Y22 Y21 V2 V1 YL Y22 Note that for a simple transistor Y21 gm and so the above reduces to the familiar gm Ro RL University of California Berkeley EECS 242 p 2 33 Input Output Admittance cont We can now solve for the input and output admittance Yin Y11 Y12 Y21 YL Y22 Yout Y22 Y12 Y21 YS Y11 Note that if Y12 0 then the input and output impedance are de coupled Yin Y11 Yout Y22 But in general they are coupled and changing the load will change the input admittance It s interesting to note the same formula derived above also works for the input output impedance Z12 Z21 Zin Z11 ZL Z22 The same is true for the hybrid and inverse hybrid matrices University of California Berkeley EECS 242 p 3 33 Power Gain Pin PL YS vs y11 y21 y12 y22 Pav S YL Pav L We can define power gain in many different ways The power gain Gp is defined as follows PL f YL Yij 6 f YS Gp Pin We note that this power gain is a function of the load admittance YL and the two port parameters Yij The available power gain is defined as follows Ga Pav L f YS Yij 6 f YL Pav S The available power from the two port is denoted Pav L whereas the power available from the source is Pav S University of California Berkeley EECS 242 p 4 33 Power Gain cont Pin PL YS vs y11 y21 Pav S YL Pav L Finally the transducer gain is defined by GT y12 y22 PL f YL YS Yij Pav S This is a measure of the efficacy of the two port as it compares the power at the load to a simple conjugate match University of California Berkeley EECS 242 p 5 33 Derivation of Power Gain The power gain is readily calculated from the input admittance and voltage gain Pin V1 2 Yin 2 V2 2 PL YL 2 2 V2 YL Gp V1 Yin Y21 2 YL Gp YL Y22 2 Yin University of California Berkeley EECS 242 p 6 33 Derivation of Available Gain IS YS Y11 Y21 Y12 Y22 Ieq Yeq To derive the available power gain consider a Norton equivalent for the two port where Y21 IS Ieq I2 Y21 V1 Y11 YS The Norton equivalent admittance is simply the output admittance of the two port Yeq Y22 Y21 Y12 Y11 YS The available power at the source and load are given by Pav S IS 2 8 YS Pav L Ieq 2 8 Yeq Y21 2 YS Ga Y11 YS 2 Yeq University of California Berkeley EECS 242 p 7 33 Transducer Gain Derivation The transducer gain is given by GT PL Pav S 1 YL V2 2 2 IS 2 8 YS 2 V2 4 YL YS IS We need to find the output voltage in terms of the source current Using the voltage gain we have and input admittance we have V2 Y21 V Y Y 1 22 L IS V YS Yin V2 Y21 1 I Y Y Y Y 22 in S L S Y Y 12 21 YS Yin YS Y11 YL Y22 University of California Berkeley EECS 242 p 8 33 Transducer Gain cont We can now express the output voltage as a function of source current as 2 V2 Y21 2 I YS Y11 YL Y22 Y12 Y21 2 S And thus the transducer gain GT 4 YL YS Y21 2 YS Y11 YL Y22 Y12 Y21 2 It s interesting to note that all of the gain expression we have derived are in the exact same form for the impedance hybrid and inverse hybrid matrices University of California Berkeley EECS 242 p 9 33 Comparison of Power Gains In general PL Pav L with equality for a matched load Thus we can say that GT Ga The maximum transducer gain as a function of the load impedance thus occurs when the load is conjugately matched to the two port output impedance GT max L PL YL Yout Ga Pav S Likewise since Pin Pav S again with equality when the the two port is conjugately matched to the source we have GT Gp The transducer gain is maximized with respect to the source when GT max S GT Yin YS Gp University of California Berkeley EECS 242 p 10 33 Bi Conjugate Match When the input and output are simultaneously conjugately matched or a bi conjugate match has been established we find that the transducer gain is maximized with respect to the source and load impedance GT max Gp max Ga max This is thus the recipe for calculating the optimal source and load impedance in to maximize gain Y12 Y21 Yin Y11 YS YL Y22 Yout Y22 Y12 Y21 YL YS Y11 Solution of the above four equations real imag results in the optimal YS opt and YL opt University of California Berkeley EECS 242 p 11 33 Calculation of Optimal Source Load Another approach is to simply equate the partial derivatives of GT with respect to the source load admittance to find the maximum point GT 0 GS GT 0 GL GT 0 BS GT 0 BL Again we have four equations But we should be smarter about this and recall that the maximum gains are all equal Since Ga and Gp are only a function of the source or load we can get away with only solving two equations For instance Ga 0 BS Ga 0 GS we can find the Y This yields YS opt and by setting YL Yout L opt Likewise we can also solve Gp 0 GL Gp 0 BL And now use YS opt Yin University of California Berkeley EECS 242 p 12 33 Optimal Power Gain Derivation Let s outline the procedure for the optimal power gain We ll use the power gain Gp and take partials with respect to the load Let Yjk mjk jnjk YL GL jXL Y12 Y21 P jQ Lej Y21 2 Gp GL D Y12 Y21 YL Y22 Y12 Y21 m11 Y11 YL Y22 YL Y22 2 D m11 YL Y22 2 P GL m22 Q BL n22 Gp Y21 2 GL D 0 BL D2 BL University of California Berkeley EECS 242 p 13 33 Optimal Load cont Solving the above equation we arrive at the following solution BL opt In a similar fashion solving for the optimal load conductance GL opt Q n22 2m11 1 2m11 q 2m11 m22 P 2 L2 If we substitute these values into the equation …