Terminated Two-Port Circuits Zin ZTh In a typical application, the two-port circuit is driven at port 1 and loaded at port 2. From the two-port equations, and including relationships at the input and output, a number of characteristics of terminated twoport circuits, such as Thevenin impedance, Thevenin voltage, input impedance, current gain, voltage gain, etc, can be derived. 69 Terminated Two-Port Circuits Assuming the z parameters of the network are given, find the input impedance. Z in = V1 I1 V1 = z11 I1 + z12 I 2 Zin ZTh V2 = -Z L I 2 = z21 I1 + z22 I 2 V2 = z21 I1 + z22 I 2 z21 I1 I2 = z22 + Z L V1 = z11 I1 + z12 I 2 = z11 I1 - z12 z21 I1 z22 + Z L V1 z12 z21 Z in = = z11 I1 z22 + Z L Note: Z in = z11 only if Z L ¥, which represents the open-circuit condition for z parameters. 70 Terminated Two-Port Equations 71 Terminated Two-Port Equations 72 Terminated Two-Port Equations 73 Example 11 The b parameters of the network are given. Find a) voltage V2 , b) the average power delivered to the load. 20 3k b = 2 mS 0.2 a) From table: V2 DbZ L = Vg b12 + b11Z g + b22 Z L + b21Z g Z L (20 ⋅ 0.2 - 3⋅103 ⋅ 2 ⋅10-3 )⋅ 5⋅103 10 = = 3 3 3 -3 19 -3⋅10 - 20 ⋅ 500 - 0.2 ⋅ 5 ⋅10 - 2 ⋅10 ⋅ 500 ⋅ 5 ⋅10 V2 = b) Average power: 10 10 Vg = 500 V = 263.16 V 19 19 P2 = V22 2 RL 2 = (263.16 V) 2 ⋅ 5 k = 6.93 W 74 Example 12 Given the a parameters and Vg =50mV, Zg =100, ZL =5k, find a) the average power delivered to the load, b) the load resistance for maximum power transfer. 5 104 a = 1μS a) From table:I 2 = 10 0.03 -Vg a11Z L + a12 + a21Z g Z L + a22 Z g = (5⋅10 -4 -50 ⋅10-3 V 3 -6 ⋅ 5 ⋅10 + 10 + 10 3 ) ⋅100 ⋅ 5 ⋅10 - 0.03⋅100 = -5 mA 1 2 1 2 PL = I 2 Z L = (-5 mA) 5 k = 62.5 mW 2 2 75 Example 12 cont’d b) Load resistance for maximum power transfer: Vg =50mV, Zg =100, ZL =5k 5 104 a = 1μS From table: Z = a12 + a22 Z g = 10 - 0.03⋅100 Th -4 -6 a11 + a21Z g 5 ⋅10 +10 10 0.03 S ⋅100 = 11.67 k * Z L = ZTh = 11.67 k 76 Example 13 Two identical amplifiers are cascaded; their h parameters are given. Find the voltage gain V2 /Vg . 1k 0.0015 h = 100 100μS Convert to a parameters: h (0.05) a A11 5 103 , 100 h21 a A21 h22 100μS 1μS, 100 h21 5 103 a aA aA 1μS 10 5 103 0.01 1μS a A12 h11 1k 10 100 h21 a A22 1 1 0.01 h21 100 10 10.25 106 0.095 0.01 9.5 109 S 1.1104 77 Example 13 cont’d 10.25 106 0.095 a aA1 aA2 9 4 9.5 10 S 1.1 10 V2 ZL = a) From table: Vg (a11 + a21Z g ) Z L + a12 + a22 Z g = 104 (10.25⋅10-6 + 9.5⋅10-9 ⋅ 500)⋅104 + 0.095 +1.1⋅10-4 ⋅ 500 104 = = 33,333.33 0.15 + 0.095 + 0.055 78 Example 14 Evaluate V2/Vs in the following circuit. Series connection: add z parameters. z parameters of 10 resistor: z11 z12 z22 z12 z12 z12 10 z11 z12 0 z11 z12 10 z22 z12 0 z22 z12 10 Add z matrices 12 8 10 10 z 8 20 10 10 22 18 18 30 79 Example 14 cont’d Find V2/Vs from table V2 z21Z L ( z11 Z g )( z22 Z L ) z12 z21 Vs z 22 18 18 30 Z g 5 , Z L 20 V2 18 20 Vs (22 5)(30 20) 182 0.351 80 Ladder Network Synthesis LC ladder networks for low-pass filters of (a) odd order, (b) even order 81 Ladder Network Synthesis To synthesize the transfer function of the LC ladder To make the design less complicated, assume Zs=0. 82 Ladder Network Synthesis 83 Example 15 Design the LC ladder network terminated with a 1 resistor that has the following normalized transfer function 84 Example 15 cont’d 85 Example 15 cont’d Note that any realization of y22 will automatically realize y21, since y22 is the output driving-point admittance, that is, the output admittance of the network with the input port shortcircuited. 86 Example 16 Realize the following transfer function using an LC ladder network terminated in a 1 resistor: 87 Example 16 cont’d 88 Example 16 cont’d 89