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1 TWO PORT NETWORKS CHAPTER 6

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2 OBJECTIVES To understand about two – port networks and its functions. To understand the different between z- parameter, y-parameter, ABCD- parameter and terminated two port networks. To investigate and analysis the behavior of two – port networks.

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3 SUB - TOPICS 6-1 Z – PARAMETER 6-2 Y – PARAMETER 6-3 ABCD – PARAMETER 6-4 TERMINATED TWO PORT NETWORKS

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4 TWO – PORT NETWORKS A pair of terminals through which a current may enter or leave a network is known as a port. Two terminal devices or elements (such as resistors, capacitors, and inductors) results in one – port network. Most of the circuits we have dealt with so far are two – terminal or one – port circuits. (Fig. a) A two – port network is an electrical network with two separate ports for input and output. It has two terminal pairs acting as access points. The current entering one terminal of a pair leaves the other terminal in the pair. (Fig. b)

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5 One port or two terminal circuit Two port or four terminal circuit It is an electrical network with two separate ports for input and output. No independent sources.

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6 6-1 Z – PARAMETER Z – parameter also called as impedance parameter and the units is ohm (Ω) The “black box” is replace with Z-parameter is as shown below. +V1-+V1- I1I1 I2I2 +V2-+V2- Z 11 Z 21 Z 12 Z 22 where the z terms are called the impedance parameters, or simply z parameters, and have units of ohms.

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7 z 11 = Open-circuit input impedance z 21 = Open-circuit transfer impedance from port 1 to port 2 z 12 = Open-circuit transfer impedance from port 2 to port 1 z 22 = Open-circuit output impedance

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8 Example 1 Find the Z – parameter of the circuit below. 40Ω 240Ω 120Ω +V1_+V1_ +V2_+V2_ I1I1 I2I2

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9 Solution 40Ω 240Ω 120Ω +V1_+V1_ +V2_+V2_ I1I1 IaIa IbIb When I 2 = 0(open circuit port 2). Redraw the circuit.

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10 40Ω 240Ω 120Ω +V1_+V1_ +V2_+V2_ IyIy I2I2 IxIx When I 1 = 0 (open circuit port 1). Redraw the circuit. In matrix form:

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11 Example 2 Find the Z – parameter of the circuit below +_+_ +V1_+V1_ +V2_+V2_ -j20Ω 10Ω j4Ω 2Ω2Ω 10I 2 I2I2 I1I1

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12Solution i) I 2 = 0 (open circuit port 2). Redraw the circuit. +V1_+V1_ j4Ω 2Ω2ΩI1I1 +V2_+V2_ I 2 = 0

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13 ii) I 1 = 0 (open circuit port 1). Redraw the circuit. +_+_ +V1_+V1_ +V2_+V2_ -j20Ω 10Ω 10I 2 I2I2 I 1 = 0

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Y - PARAMETER Y – parameter also called admittance parameter and the units is siemens (S). The “black box” that we want to replace with the Y- parameter is shown below. +V1-+V1- I1I1 I2I2 +V2-+V2- Y 11 Y 21 Y 12 Y 22 where the y terms are called the admittance parameters, or simply y parameters, and they have units of Siemens.

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15 y 11 = Short-circuit input admittance y 21 = Short-circuit transfer admittance from port 1 to port 2 y 12 = Short-circuit transfer admittance from port 2 to port 1 y 22 = Short-circuit output admittance

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16 Example 3 Find the Y – parameter of the circuit shown below. 5Ω5Ω 15Ω 20Ω +V1_+V1_ +V2_+V2_ I1I1 I2I2

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17 Solution i)V 2 = 0 5Ω5Ω 20Ω +V1_+V1_ I1I1 I2I2 IaIa

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18 ii) V 1 = 0 In matrix form; 5Ω5Ω 15Ω +V2_+V2_ I1I1 I2I2 IxIx

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19 Example 4 Find the Y – parameters of the circuit shown. +_+_ +V1_+V1_ +V2_+V2_ -j20Ω 10Ω j4Ω 2Ω2Ω 10I 2 I2I2 I1I1

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20Solution i) V 2 = 0 (short – circuit port 2). Redraw the circuit. +_+_ +V1_+V1_ 10Ω j4Ω 2Ω2Ω 10I 2 I2I2 I1I1

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21 ii) V 1 = 0 (short – circuit port 1). Redraw the circuit. +_+_ +V2_+V2_ -j20Ω 10Ω j4Ω 2Ω2Ω 10I 2 I2I2 I1I1

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T (ABCD) PARAMETER T – parameter or also ABCD – parameter is a another set of parameters relates the variables at the input port to those at the output port. T – parameter also called transmission parameters because this parameter are useful in the analysis of transmission lines because they express sending – end variables (V 1 and I 1 ) in terms of the receiving – end variables (V 2 and -I 2 ). The “black box” that we want to replace with T – parameter is as shown below. +V1-+V1- I1I1 I2I2 +V2-+V2- A 11 C 21 B 12 D 22

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23 where the T terms are called the transmission parameters, or simply T or ABCD parameters, and each parameter has different units. A=open-circuit voltage ratio C= open-circuit transfer admittance (S) B= negative short- circuit transfer impedance ( ) D=negative short- circuit current ratio

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24 Example 5 Find the ABCD – parameter of the circuit shown below. 2Ω2Ω 10Ω +V2_+V2_ I1I1 I2I2 +V1_+V1_ 4Ω4Ω

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25 Solution i) I 2 = 0, 2Ω2Ω 10Ω +V2_+V2_ I1I1 +V1_+V1_

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26 ii) V 2 = 0, 2Ω2Ω 10Ω I1I1 I2I2 +V1_+V1_ 4Ω4Ω I 1 + I 2 In matrix form;

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TERMINATED TWO – PORT NETWORKS In typical application of two port network, the circuit is driven at port 1 and loaded at port 2. Figure below shows the typical terminated 2 port model. +V1-+V1- I1I1 I2I2 +V2-+V2- ++ ZgZg ZLZL VgVg Two – port network Terminated two-port parameter can be implement to z-parameter, Y-parameter and ABCD-parameter.

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28 Z g represents the internal impedance of the source and V g is the internal voltage of the source and Z L is the load impedance. There are a few characteristics of the terminated two- port network and some of them are;

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29 The derivation of any one of the desired expression involves the algebraic manipulation of the two – port equation. The equation are: 1) the two-port parameter equation either Z or Y or ABCD. For example, Z-parameter, 2) KVL at input, 3) KVL at the output, From these equations, all the characteristic can be obtained.

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30 Example 6 For the two-port shown below, obtain the suitable value of R s such that maximum power is available at the input terminal. The Z-parameter of the two-port network is given as With R s = 5Ω,what would be the value of +V1-+V1- I1I1 I2I2 +V2-+V2- ++ RsRs 4Ω4ΩVsVs Z

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31 Solution Z-parameter equation becomes; KVL at the output; Subs. (3) into (2); Subs. (4) into (1); For the circuit to have maximum power, Hence, the input impedance;

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32 To find at max. power transfer, voltage drop at Z 1 is half of Vs From equations (3), (4), (5) & (6), Overall voltage gain,

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33 Example 7 The ABCD parameter of two – port network shown below are. The output port is connected to a variable load for a maximum power transfer. Find R L and the maximum power transferred.

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34 ABCD parameter equation becomes V 1 = 4V 2 – 20I 2 I 1 = 0.1V 2 – 2I 2 At the input port, V 1 = -10I 1 (1) (2) (3) Solution

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35 (4)(5) (6) (3) Into (1) -10I 1 = 4V 2 – 20I 2 I 1 = -0.4V 2 + 2I 2 (4) Into (2) -0.4V 2 + 2I 2 = 0.1V 2 – 2I 2 0.5V 2 = 4I 2 From (5); Z TH = V 2 /I 2 = 8Ω But from Figure (b), we know that V 1 = 50 – 10I 1 and I 2 =0 Sub. these into (1) and (2) 50 – 10I 1 = 4V 2 I 1 = 0.1V 2 (8) (7) Sub (8) into (7); V 2 = 10 Thus, V TH = V 2 = 10V R L for maximum power transfer, R L = Z TH = 8Ω The maximum power; P = I 2 R L = (V TH /2R L ) 2 x R L = V 2 TH /4R L = 3.125W

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