Objectives: Introduction Notes for course EE1.1 Circuit Analysis 4-5 Re-examination of 1-port sub-circuits Admittance parameters for -port circuits TOPIC 1 -PORT CIRCUITS Gain and port impedance from -port admittance parameters Impedance parameters for -port circuits Hybrid parameters for -port circuits 1 INTRODUCTION Amplifier circuits are found in a large number of appliances, including radios, TV, video, audio, telephony (mobile and fixed), communications and instrumentation. The applications of amplifiers are practically unlimited. It is clear that an amplifier circuit has an input signal and an output signal. This configuration is represented by a device called a -port circuit, which has an input port for the input signal and an output port for the output signal. In this topic, we look at circuits from this point of view. To develop the idea of -port circuits, consider a general 1-port sub-circuit of the type we are very familiar with: The basic passive elements, resistor, inductor and capacitor, and the independent sources, are the simplest 1-port sub-circuits. A more general 1-port sub-circuit contains any number of interconnected resistors, capacitors, inductors, and sources and could have many nodes. Sometimes, perhaps when we have finished designing a circuit, we become less interested in the detail of the elements interconnected in the circuit and are happy to represent the 1-port circuit by how it behaves at its terminals. This is achieved by use of circuit analysis to determine the relationship between the voltage across the terminals and the current flowing through the terminals which leads to a Thevenin or Norton equivalent circuit, which can be used as a simpler replacement for the original complex circuit. Consider now the -port circuit. A -port is a network having two pairs of terminals: Some elements intrinsically have more than the two terminals; examples are the bipolar junction transistor or the MOSFET. Such circuits cannot be represented as a 1-port circuit and the -port is the simplest description available. -port circuits also have the same application as 1-port circuits, which is that they enable us to describe the input-output behaviour of a circuit without worrying about circuit details. As an example, consider the BJT amplifier:
Topic 1 -port Circuits This circuit has 6 nodes, 7 -terminal elements and one 3-terminal element. Once the circuit has been designed we may be happy to describe its behaviour between the input port and the output port, perhaps using parameters like gain. A -port description of the circuit allows us to do this systematically. In practice, -port circuits often represent devices in which a source delivers energy to a load through the -port network. For example, stereo amplifiers take a low power audio signal and increase its power so that it will drive a speaker system. Determining and knowing ratios such as voltage gain, current gain, and power gain of a -port circuit is very important when dealing with a source that delivers power through a -port to a load. This topic deals with precisely this need. Before proceeding to look at -port circuits, we re-examine 1-port circuits, and in particular the derivation of Thevenin and Norton equivalent circuits, in a more systematic way than hitherto. RE-EXAMINATION OF 1-PORT SUB-CIRCUITS In this section, we look again at deriving Thevenin and Norton equivalents for 1-port sub-circuits and introduce systematic ways of deriving the equivalents based on nodal analysis. Consider first the following example: Example: Evaluate the Thevenin and Norton equivalents for the following circuit when viewed from terminals 1 and : The Thevenin voltage is the open-circuit voltage at node 1; using current division and Ohm's law, we have: V oc = 1 3 1 1 + 1 + 1 3 4 = 1 3 6 6 + 3 + 4 = 8 11 V The Thevenin impedance is obtained by de-activating the current source and determining the equivalent impedance seen between terminals 1 and. R eq = 1 3 3 1 3 + 3 = 1 + 9 6 = 3 11 Ω
Topic 1 -port Circuits Hence, we have the Thevenin equivalent circuit and, by calculating I sc = V oc /R eq, the Norton equivalent circuit: Thevenin and Norton equivalent circuits are simplifications of a circuit; however large the number of nodes in the original circuit, the Norton equivalent has just two nodes (1 and ). The simplification can be viewed as a process of eliminating nodes in the original circuit which are not port nodes (node in the above example). This process can be viewed as a simplification of equations obtained by systematic nodal analysis of the circuit. We now apply this systematic method to the above circuit. We start by labelling the reference node as zero and the port node as 1; the remaining nodes are numbered sequentially from onwards. At the port for which we require the Thevenin or Norton equivalent, i.e. port 1, we apply a test current source which we label : We now perform by-inspection nodal analysis: 5 3 3 = 4 Node is not a port node and is therefore an internal node, whose voltage must be eliminated; we use Gaussian elimination: 5 3 V = I 1 4 5 ( ) ( ) 3 = ( )4 11 3 3 V 1 = + 8 3 The resulting equation leads directly to the Norton and Thevenin equivalent circuits, derived above: 11 3 = + 8 3 = 11 3 8 3 = 3 11 + 8 11 If the original circuit has no independent sources, then the above process leads just to the equivalent Thevenin/Norton equivalent impedance. We consider a more complex example: Example: Determine the equivalent resistance between nodes 1 and of the following circuit (note that is a test current source):
Topic 1 -port Circuits 4 We first perform by-inspection nodal analysis: 1 + 1 1 + 1 + 1 1+ 1 V 3 = 3 5 V 3 = The partition lines show that although this circuit has 3 nodes, we are treating it as a 1-port circuit, i.e. we are only interested in the relationship between and. Nodes and 3 are not port nodes and therefore and V 3 must be eliminated; we use Gaussian elimination: 3 5 V 3 = 3 5 ( ) ( ) = 3 = 3 = 3 ( ) ( ) = 1 = The resulting equation leads directly to the relationship between and : = Hence, the equivalent sub-circuit is a 1 Ω resistor connected between node 1 and node. This approach is applicable to a circuit with any number of nodes, although a computer method of reducing the matrix would be needed for larger circuits. Having shown that an n-node circuit may be reduced to a 1-port description systematically, we are now ready to consider -port circuits. 3 -PORT ADMITTANCE PARAMETERS 3.1 Definition Rather than always having to deal with all the internal variables of a -port circuit, it is often more convenient to deal only with the terminal voltages and currents:
Topic 1 -port Circuits We assume that all excitations are external to the -port; hence the -port has no internal independent sources. We also assume that all dynamic elements (inductors and capacitors) are initially relaxed, i.e., have zero initial conditions, i.e. capacitor voltages and inductor currents are initially zero. Port currents are defined positive into the circuit. Under these assumptions, the admittance equations for a -port are expressions for the terminal currents and I in terms of the port voltages and, i.e.: = y 11 + y 1 I = y 1 + y where y 11, y 1, y 1 and y are called admittance parameters or y-parameters. The unit for all the admittance parameters is the Siemens (S). If the circuit has no connection between port 1 and port, then each port can be described by Ohm's law: = y 11 I = y where y 11 and y are the effective admittances at port 1 and, respectively. The remaining parameters y 1 and y 1 describe the effect of on and the effect of on I ; these terms are necessary when there are elements which link port 1 to port including where the -port is an active device, such as a dependent source or transistor The -port equations may be written more compactly in matrix notation: Example I = y 11 y 1 y 1 y Compute the admittance parameters of the following -port circuit: Solution Write the nodal admittance matrix by inspection; tape the VCCS to i c : G 1 + G 3 G 3 G 3 G + G 3 G 1, G and G 3 are conductances of the resistors. Un-tape the source i c = g m : 5 = I i c
Topic 1 -port Circuits G 1 + G 3 G 3 G 3 G + G 3 = I g m V 1 Move the VCCS term to the LHS (to column 1 because it depends on ): G 1 + G 3 G 3 g m G 3 G + G 3 = I Since these equations are in the standard form for -port admittance matrices, we may easily identify the y-parameters: y 11 = G 1 + G 3 y 1 = G 3 y 1 = g m G 3 y = G + G 3 This example was straightforward because the circuit has only two nodes. In general a -port circuit will have more than nodes. We now consider this more general case. 3. General -port analysis Consider a general circuit with n nodes consisting of resistors, capacitors, inductors and dependent voltage and current sources: The reference node is labelled node, and the remaining nodes are numbered sequentially from 1 to n; the nodal voltages are,,... V n and the currents injected at each node are, I,... I n. Nodal analysis would lead to following set of nodal equations: y' 11 y' 1.. y' 1,n y' 1n y' 1 y'.. y',n y' n V : :. : : : y' n,1 y' n,.. y' n,n y' n,n y' n1 y' n1.. y' n,n y' nn V n V n I = : where the y' ij are general representations of the elements of the nodal admittance matrix. Consider now the case where two of the nodes (designated 1 and ) are taken to form the ports of a -port circuit: I n I n 6
Topic 1 -port Circuits Consider the effect of this on the nodal equations. Voltages and, which were just nodal voltages are now the port voltages and and I become the currents at the external port nodes; since nodes 3, 4,... n are now internal nodes, the injected current for these nodes is zero: y' 11 y' 1 y' 13.. y' 1n y' 1 y' y' 3.. y' n V I y' 31 y' 3 y' 33 : y' 3n V 3 = : :... : : : y' n1 y' n1 y' n3.. y' nn V n We use partition lines to show that the n n nodal matrix is that of a -port circuit which can be described by a port admittance matrix. The port admittance matrix has the following form: or y 11 y 1.. y 1 y.. V I : = : :... : : :.. I = y 11 y 1 y 1 y Obtaining the port matrix from the n n nodal matrix is essentially one of eliminating the internal voltage variables V 3, V 4,... V n in an equation solving process. This can be done by a variety of methods; a method that generally works reasonably well for circuit equations is Gaussian elimination. Example: Determine the y-parameters for the following -port, 3 node circuit: Solution Determination of the -port y-parameters starts from nodal analysis. Writing the node equations by inspection leads to the following: 7
Topic 1 -port Circuits jω + 3 jω jω jω + 5 V 3 = I where jω is the admittance of the 1 F capacitor; the nodal matrix is symmetrical because the RLC network contains no dependent sources. We will use Gaussian elimination for matrix reduction. Whereas in nodal analysis we carried on the reduction until we had a single variable, in this case we only need to eliminate row 3 and column 3 to obtain the matrix; the pivot is y' 33 = 5; the matrix becomes: ( )( ) ( )( ) jω + 3 jω 5 5 jω ( ) ( ) jω + ( ) ( ) 5 5 jω + 3.8 jω.4 jω.4 jω +. jω +. jω.4 jω.4 jω + 1.8 Hence, we have the port admittance equations: I = jω +. jω.4 jω.4 jω + 1.8 The port admittance matrix like the nodal admittance matrix is symmetrical. 3.3 Equivalent Circuit for -port admittance equations It can be helpful to visualise -port admittance equations by means of an equivalent circuit. Consider the first admittance equation: = y 11 + y 1 A valid interpretation of this equation is that the port current is equal to the port voltage times an admittance y 11 in parallel with a voltage-controlled current source y 1. A similar interpretation is possible for the second equation: I = y 1 + y This yields an admittance branch y in parallel with a voltage-controlled current source y 1. These interpretations lead to the following equivalent circuit for a -port circuit: where the resistance symbol designates a general impedance or admittance. However complex the circuit and however high the number of nodes, this equivalent circuit with four admittances always exists. 8
Topic 1 -port Circuits Each port consists of an admittance connected in parallel with a dependent current source; this may be regarded as a generalisation of the Norton equivalent circuit for 1-port circuits to -port circuits. 3.4 Determining -port parmeters by port tests It is possible to avoid a complete nodal analysis of a circuit in order to determine the port admittance parameters. This can be achieved instead by carrying out tests on the ports of the circuit. This method may be used in three ways, 1) in the laboratory, ) in computer simulation or 3) for hand analysis. Consider a circuit which has the general -port admittance equations: = y 11 + y 1 I = y 1 + y Let us apply a short-circuit to port such that = ; we then have: This leads to: = y 11 I = y 1 y 11 = = y 1 = I = We can envisage a short-circuit on port and application of a known voltage source to port 1; Determining the port currents and I yields the two admittance parameters. Referring to the original circuit and its equations, let us now apply a short-circuit to port 1 such that = ; we then have: This leads to: = y 1 I = y y 1 = = y = I = We can envisage a short-circuit on port 1 and application of a known voltage source to port ; Again, determining the port currents and I yields the two admittance parameters. Since each admittance parameter is defined with regard to a short-circuited terminal voltage the admittance parameters are sometimes called short-circuit admittance parameters. 9
Topic 1 -port Circuits An example illustrates derivation of admittance parameters by port tests. Example: We take the previous example of a -port circuit with nodes: We begin by placing a short-circuit on port and voltage source on port 1; in order to obtain y 11 and y 1, we determine and I : We then place a short-circuit on port 1 and voltage source on port, and in order to obtain y 1 and y, we again determine and I : The y-parameters obtained are identical with those obtained previously by nodal analysis of the circuit. 4 GAIN AND IMPEDANCE FROM -PORT ADMITTANCE PARAMETERS 4.1 General We have defined admittance parameters and shown that they may be obtained by nodal analysis and equation reduction or by port tests. One might ask what have we gained by this? We hope to answer this question in this section where we derive expressions for gain and input and output impedance of a circuit in terms of its -port admittance parameters. Consider a -port circuit in isolation: The circuit is described by: = y 11 + y 1 I = y 1 + y where, I, and are port variables. 1
Topic 1 -port Circuits The impedance and gain expressions we are interested in are ratios of port variables. Ratios of variables at the same port, such as /, I / are termed driving point functions. Ratios of variables at different ports, such as /, /I, I /, /I, are termed transfer functions. Driving point and transfer functions together are termed network functions. Note that transfer functions may be dimensionless, e.g. /, or have dimensions of impedance, e.g. /, or admittance, e.g. I / Consider the constraints acting on the variables. We have 4 port variables,,,, I, and therefore there are just 3 independent network functions which can be determined. The port admittance equations clearly provide two constraints. It follows that in order to define the network functions, a third constraint must be provided. This constraint is provided by the terminations or embedding of the circuit and it may be called the termination constraint. We will consider various termination constraints in due course. Once the termination constraint is in place and all network functions are defined, then a single source is sufficient to determine all voltages and currents. 4. Gain with simple terminations Consider a -port circuit with port terminated in admittance Y L : The equations for the -port circuit alone are: The termination constraint is: = y 11 + y 1 I = y 1 + y I = Y L The minus sign is due to the voltage and current reference directions. Substitution of the termination constraint into the -port equation for I gives: Y L = y 1 + y I = Y L = 11 y 1 y + Y L It is good practice to state the termination constraint along with the network function as we have done here. This expression may be used to determine voltage gain for any load admittance; it is valid independent of the nature of the source at port 1, so port 1 does not have to be driven by a voltage source.
Topic 1 -port Circuits If we terminate port in an open-circuit, we have: A v( o/c) = = y 1 y I = If is known, then this equation may be used to determine for the case when port is terminated in an open-circuit; the value of obtained is actually the voltage of the equivalent Thevenin voltage source representing the output port (port ). Termination of port in a short-circuit leads to =, so voltage gain is not useful in this case. Now consider current gain; terminating port in an open-circuit is not valid because it forces I to be zero. In the case of a short-circuit termination, we can set = in the -port equations: We can now determine the current gain: Example = y 11 I = y 1 A i( s/c) = I = y 1 y = 11 For the simple circuit of the previous example determine from the admittance matrix (i) the voltage gain with a load resistance of 1 Ω and (ii) the current gain with a short-circuit at port ; the element values are G 1 = 1 S, G =S, G 3 = 3 S, g m = 1 S: The previously determined admittance matrix is: We have: and Y = y 11 y 1 y 1 y = G 1 + G 3 G 3 g m G 3 G + G 3 = 4 3 18 5 A v = y = 1 = 18 y Y L =.5 + Y L 5 + 1 = 3 A i( s/c) = I = y 1 = 18 y = 11 4 = 4.5 Having considered circuit gain, we now consider circuit input and output admittances. 4.3 Input and Output Admittance with simple terminations The input and output admittances of a terminated -port are important quantities, especially when determining voltage gain, current gain and power gain. 1
Topic 1 -port Circuits We are interested in the input admittance of a terminated -port circuit because this affects the load it presents to a source: The circuit without the load admittance Y L can be represented by its admittance parameters: I = y 11 y 1 y 1 y The load admittance Y L imposes the termination constraint: We now have: I = Y L Y L V = y 11 y 1 y 1 y We can move the Y L term across the to the RHS and into column, since it contains : = y 11 y 1 y 1 y + Y L We can solve the equations to eliminate by a number of methods including simple substitution. We use Gaussian elimination: = y 11 y 1 y 1 y + Y L Hence, the input admittance of the -port circuit is: = y 11 y 1y 1 y + Y L Y in = = y 11 y 1y 1 y + Y L In the case where port is terminated in an open-circuit, we have Y L = : Y in( o/c) = = y 11 y 1y 1 y I = Note that Y L = is equivalent to the constraint I =. We have stated the termination constraint along with the network function, which is good practice. In the case where there is a short-circuit termination at port, we have Y L ; the general expression now reduces to: Y in( s/c) = = y 11 = 13
Topic 1 -port Circuits Note that Y L is equivalent to the constraint =. Consider now the -port circuit with voltage source V S and admittance Y S at port 1: We are interested in the output impedance at port, as this determines the effect of a load at port. The output impedance is actually the Thevenin equivalent impedance seen looking into port for the combination of the -port circuit and the source. Hence, we define output admittance as being obtained with the input voltage source de-activated: The circuit is identical with the circuit with load admittance at port of Y L, except that port 1 and port are interchanged and Y L is replaced by Y S. Therefore, we can use the previous expression for Y in to determine Y out : Y in = = y 11 y 1y 1 y + Y L Y out = I = y y 1y 1 y 11 + Y S In the case where port 1 is terminated in an open-circuit, we have Y S = : Y out( o/c) = I = y y 1y 1 y = 11 This corresponds to the case where port 1 is driven by a current source alone. In the case where there is a short-circuit termination at port 1, we have Y S ; the general expression now reduces to: Y out( s/c) = I = y = This corresponds to the case where port 1 is driven by a voltage source alone. Examples For the simple circuit of the previous example determine from the admittance matrix (i) the input admittance at port 1 with a with a port load resistance of 1 Ω and (ii) the output admittance at port when (a) the input at port 1 is a voltage source and (b) is a current source; the element values are G 1 = 1 S, G =S, G 3 = 3 S, g m = 1 S: 14
Topic 1 -port Circuits The previously determined admittance matrix is: Y = y 11 y 1 y 1 y = G 1 + G 3 G 3 g m G 3 G + G 3 = 4 3 18 5 i: For the input admittance at port 1 with Y L = 1 S, we have: Y in = = y 11 y 1y 1 = 4 ( 3)18 y I = Y L + Y L 5 + 1 = 13 S The input resistance, if required is: Z in = 1 Y in = 1 13 Ω ii The output admittance at port with source admittance at node 1 of Y S is given by: Y out = I = Y S = y y 1y 1 y 11 + Y S (a) When the input excitation is a voltage source, de-activation of the source gives Y S = : Y out( s/c) = I = y = 5 S = (b) When the input excitation is a current source, de-activation of the source gives Y S = : Y out( o/c) = I = y y 1y 1 y = 11 4.4 Circuits with non-existent admittance description = 5 ( 3)18 = 5 + 1 4 5 5 = 1 4 5 = 15.8 S Consider the following -port circuit for which we wish to determine the admittance parameters: Use of KCL leads to the following two equations: = 1 R I I = 1 R It turns out that it is impossible to eliminate the current terms on the RHS in favour of voltages (try it!); it follows that it is impossible to describe this circuit using the admittance matrix, even though the circuit is a perfectly valid one. The problem is that the short-circuit between the ports has an infinite admittance. Apart from short-circuits, admittance matrices also cannot describe dependent voltage sources and ideal op-amps (nullors). 15
Topic 1 -port Circuits These problems can be solved using a number of methods, of which a very important one is the modified nodal admittance (MNA) matrix approach which is used in SPICE. However, it is easy to see that the equations for the above example can easily be arranged in an alternative form with voltages in terms of currents: = R + RI = R + RI This format is the dual of the admittance description and is called the impedance description. The impedance approach is not so important as the admittance approach because the impedance equations are not naturally produced by nodal analysis of a circuit. However, we shall summarise its main features. 5 IMPEDANCE PARAMETERS FOR -PORT CIRCUITS Consider again the general -port circuit: The impedance parameters, or z-parameters, relate the port currents to the port voltages according to the matrix equation: = z 11 z 1 z 1 z As for the admittance parameters, it is possible to obtain impedance parameters by carrying out tests on the ports of the circuit. We can write the -port circuit impedance equations: = z 11 + z 1 I = z 1 + z I Let us apply an open-circuit to port such that I = ; we then have: This leads to: = z 11 = z 1 I z 11 = I = z 1 = I = We can envisage an open-circuit on port and application of a known current source to port 1; the port voltages and yield admittance parameters z 11 and z 1. We now apply an open-circuit to port 1 such that = ; we then have: This leads to: = z 1 I = z I 16
Topic 1 -port Circuits z 1 = I = z = I = We can envisage an open-circuit on port 1 and application of a known current source I to port ; the port voltages and yield the two admittance parameters z 1 and z. Since each impedance parameter is defined by applying an open-circuit to a port, the impedance parameters are sometimes called open-circuit impedance parameters. As with the y-parameters, the z-parameters have a two-dependent source equivalent circuit. Consider the first impedance equation: = z 11 + z 1 I This equation can be interpreted as an application of KVL around the left hand loop of the following circuit: Consider the second impedance equation: = z 1 + z I It has a similar interpretation as application of KVL to the right-hand loop in the equivalent circuit. Thus a set of z-parameters for a -port circuit may always be represented by an equivalent circuit containing the four z-parameters. Notice that the elements in the equivalent circuit are connected in series, in contrast to the equivalent circuit for the admittance parameters where they were connected in parallel. This equivalent circuit may be regarded as an extension of the Thevenin equivalent circuit to the - port case. Given a circuit with a set of z-parameters, it is possible to determine voltage and current gain as well as input and output impedance for arbitrary source and load impedances at ports 1 and ; it is similar to the process using y-parameters, so we will not do this here. Since the z-parameters relate port voltages to port currents and the y-parameters relate port currents to port voltages, one might expect that the z-parameter matrix and the y-parameter matrix are related to each other in a precise mathematical way. Consider the impedance description for a -port circuit: = z 11 z 1 z 1 z Using matrix algebra, we may invert the z-matrix and use it to pre-multiply both sides of the impedance equation set: z 11 z 1 z 1 z V1 = z 11 z 1 z 1 z I z11 z 1 z 1 z I 17
Topic 1 -port Circuits where a superscript on a matrix of 1 represents inversion. It is known from matrix algebra that the inverse of a matrix times the matrix is the unit matrix which may be dropped when it multiplies a column vector. Hence, we have: z 11 z 1 z 1 z V1 = I The voltage and current vector framework is now that for an admittance description. Hence, the inverse of the z-matrix is the admittance matrix and vice versa: z 11 z 1 z 1 z = y 11 y 1 y 11 y 1 y 1 y y 1 y = z 11 z 1 z 1 z This is a reasonably efficient method for transforming between z- and y-matrices for -port circuits, e.g.: a c b d = 1 ad bc d c b a The term ad bc is the determinant Δ of the first matrix; if Δ =, the inverse does not exist, meaning that the alternative representation is not possible. Example: For the following circuit, (i) write the y-matrix by inspection, (ii) invert the y-matrix to obtain the z-matrix and (iii) check that the z-matrix describes the circuit: By inspection, the y-matrix is: The corresponding admittance equations are: Invert the y-matrix to obtain the z-matrix: Z = Y = G 1 G 1 G 1 G 1 + G Y = G 1 G 1 G 1 G 1 + G = G 1 G 1 I = G 1 + ( G 1 + G ) 1 = G 1 G 1 + G ( ) G 1 G 1 + G G 1 G 1 G 1 1 + 1 1 1 G 1 + G G 1 = G 1 G G 1 G 1 = G G 1 G = R 1 + R R 1 1 R R G G We can now write the impedance equations: 18
Topic 1 -port Circuits = ( R 1 + R ) + R I = R + R I or ( ) R + R 1 ( ) R = + I = + I We can see that these equations correctly describe the circuit. 6 HYBRID PARAMETERS FOR -PORT CIRCUITS In the z- and y-parameter descriptions, the port variables are classified according to variable type; thus for the admittance description, we express currents (port 1 and port ) in terms of voltages (port 1 and port ); and for the impedance description, we express voltages (port 1 and port ) in terms of currents (port 1 and port ). Hybrid parameters, as their name implies, are a cross between y- and z-parameters; we take a voltage and a current on the LHS and the remaining voltage and remaining current on the RHS. Specifically, a -port circuit containing no internal independent sources and with no initial stored energy, can be defined by the matrix equation: = h 11 + h 1 I = h 1 + h where h 11, h 1, h 1 and h are the hybrid parameters for the -port circuit. This leads to the matrix description: I = h 11 h 1 h 1 h This combination of variables naturally arises when the model of a common emitter transistor is simplified; it is therefore encountered quite frequently. Unlike y- and z-parameters, the h-parameters have different units; h 11 has units of Ohms, h 1 and h 1 are dimensionless, and h has units of Siemens. As with both y- and z-parameters, we can interpret the hybrid equation set as a two-dependent source equivalent circuit: Notice that at port 1 the elements are connected in series and at port they are connected in parallel. The h-parameters may be determined by tests on each port. Initially, we apply a short-circuit to port ; we have: = h 11 I = h 1 19
Topic 1 -port Circuits This leads to: h 11 = = h 1 = I = We can envisage a short-circuit on port and application of a known current source to port 1; the port variables and I yield hybrid parameters h 11 and h 1. We now apply an open-circuit to port 1 such that = ; we then have: This leads to: = h 1 I = h h 1 = = h = I = We can envisage an open-circuit on port 1 and application of a known voltage source to port ; the port variables and I yield the two hybrid parameters h 1 and h. Just as for the other parameter sets, we can determine gain and input and output impedance of a - port circuit in terms of its h-parameters; we will not do that here. A further set of parameters, called the hybrid-g parameters are similar to the hybrid h-parameters but with the dependent and independent variable sets interchanged: = g 11 g 1 g 1 g Further parameter sets classify variables according to port location. For example, the transmission matrix expresses the port 1 variables, and, in terms of the port variables, and I. These forms are left for work in other courses. We end with an example where we describe the bipolar junction transistor model using h- parameters: Example: Obtain a hybrid description for a bipolar transistor with current gain β = 1, baseemitter resistance r π =1 kω and collector-emitter resistance of 1 MΩ: Solution: An equivalent circuit for the transistor is as follows: I Comparing this with the -dependent source model for the h-parameter description, we identify: h 11 = r π = 1 Ω h 1 = h 1 = β = 1 h = 1 = 1 6 S r o
Topic 1 -port Circuits Hence, the h-parameter equation set is: I = r π β r o = 13 1 1 6 Because of its suitability for describing the common-emitter bipolar-junction transistor, the h- parameter subscripts are sometimes altered as follows: I = hie h re h fe h oe V In this scheme, the second subscript 'e' denotes that the parameters are applicable to a commonemitter transistor. The first subscripts have the following meanings: 'i' input (r π ); 'r' reverse; 'f' forward (β); 'o' output (1/r o ). Note it is only for the admittance matrix that by-inspection writing of the matrix from the circuit is possible. This makes the admittance matrix by far the most important of the circuit matrix formulations. 7 CONCLUSIONS We began this study of -port circuits by looking again at 1-port sub-circuits and developing a systematic method based on nodal analysis and Gaussian elimination for deriving their Thevenin and Norton equivalent circuits. This led to a definition of admittance parameters for -port circuits and a similar method for obtaining the parameters by nodal analysis of a circuit followed by reduction. We then showed that circuit gain and port impedance can be obtained from the -port admittance parameters once the termination of the -port circuit is specified. Finally, we looked briefly into impedance and hybrid parameters for -port circuits. 1