Notes for course EE1.1 Circuit Analysis 2004-05 TOPIC 10 ...cas.ee.ic.ac.uk/people/dhaigh/EE1.1...

Notes for course EE1.1 Circuit Analysis 2004-05

TOPIC 10 – 2-PORT CIRCUITS

Objectives:

Introduction

Re-examination of 1-port sub-circuits

Admittance parameters for 2-port circuits

Gain and port impedance from 2-port admittance parameters

Impedance parameters for 2-port circuits

Hybrid parameters for 2-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 amplifiersare practically unlimited. It is clear that an amplifier circuit has an input signal and an outputsignal. This configuration is represented by a device called a 2-port circuit, which has an input portfor the input signal and an output port for the output signal. In this topic, we look at circuits fromthis point of view.

To develop the idea of 2-port circuits, consider a general 1-port sub-circuit of the type we are veryfamiliar with:

The basic passive elements, resistor, inductor and capacitor, and the independent sources, are thesimplest 1-port sub-circuits. A more general 1-port sub-circuit contains any number ofinterconnected resistors, capacitors, inductors, and sources and could have many nodes.Sometimes, perhaps when we have finished designing a circuit, we become less interested in thedetail of the elements interconnected in the circuit and are happy to represent the 1-port circuit byhow it behaves at its terminals. This is achieved by use of circuit analysis to determine therelationship between the voltage V1 across the terminals and the current I1 flowing through theterminals which leads to a Thevenin or Norton equivalent circuit, which can be used as a simplerreplacement for the original complex circuit.

Consider now the 2-port circuit. A 2-port is a network having two pairs of terminals:

Some elements intrinsically have more than the two terminals; examples are the bipolar junctiontransistor or the MOSFET. Such circuits cannot be represented as a 1-port circuit and the 2-port isthe simplest description available. 2-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 worryingabout circuit details. As an example, consider the BJT amplifier:

Topic 10 – 2-port Circuits

2

This circuit has 6 nodes, 7 2-terminal elements and one 3-terminal element. Once the circuit hasbeen designed we may be happy to describe its behaviour between the input port and the outputport, perhaps using parameters like gain. A 2-port description of the circuit allows us to do thissystematically.

In practice, 2-port circuits often represent devices in which a source delivers energy to a loadthrough the 2-port network. For example, stereo amplifiers take a low power audio signal andincrease its power so that it will drive a speaker system. Determining and knowing ratios such asvoltage gain, current gain, and power gain of a 2-port circuit is very important when dealing with asource that delivers power through a 2-port to a load. This topic deals with precisely this need.

Before proceeding to look at 2-port circuits, we re-examine 1-port circuits, and in particular thederivation of Thevenin and Norton equivalent circuits, in a more systematic way than hitherto.

2 RE-EXAMINATION OF 1-PORT SUB-CIRCUITS

In this section, we look again at deriving Thevenin and Norton equivalents for 1-port sub-circuitsand 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 viewedfrom terminals 1 and 0:

The Thevenin voltage is the open-circuit voltage at node 1; using current division and Ohm's law,we have:

Voc =13×

1

1+ 12+ 1

3

× 4 =13×

66 + 3+ 2

× 4 =811

V

The Thevenin impedance is obtained by de-activating the current source and determining theequivalent impedance seen between terminals 1 and 0.

Req =

13× 3

213+ 3

2

=

12

2 + 96

=3

11 Ω


3

Hence, we have the Thevenin equivalent circuit and, by calculating Isc = Voc/Req, the Nortonequivalent circuit:

Thevenin and Norton equivalent circuits are simplifications of a circuit; however large the numberof nodes in the original circuit, the Norton equivalent has just two nodes (1 and 0).

The simplification can be viewed as a process of eliminating nodes in the original circuit which arenot port nodes (node 2 in the above example).

This process can be viewed as a simplification of equations obtained by systematic nodal analysisof 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 arenumbered sequentially from 2 onwards.

At the port for which we require the Thevenin or Norton equivalent, i.e. port 1, we apply a testcurrent source which we label I1:

We now perform by-inspection nodal analysis:

5 −2−2 3⎡

⎣⎢

⎤

⎦⎥V1V2⎡

⎣⎢

⎤

⎦⎥ =

I14

⎡

⎣⎢

⎤

⎦⎥

Node 2 is not a port node and is therefore an internal node, whose voltage V2 must be eliminated;we use Gaussian elimination:

5 −2−2 3⎡

⎣⎢

⎤

⎦⎥V1V2⎡

⎣⎢

⎤

⎦⎥ =

I14

⎡

⎣⎢

⎤

⎦⎥ ≡

5 − −2( ) −2( )3

0

0 0

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

V10

⎡

⎣⎢

⎤

⎦⎥ =

I1 −−2( )43

0

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥≡113

0

0 0

⎡

⎣

⎢⎢

⎤

⎦

⎥⎥V10

⎡

⎣⎢

⎤

⎦⎥ =

I1 +83

0

⎡

⎣

⎢⎢

⎤

⎦

⎥⎥

The resulting equation leads directly to the Norton and Thevenin equivalent circuits, derived above:

113V1 = I1 +

83

I1 =113V1 −

83

V1 =3

11I1 +

811

If the original circuit has no independent sources, then the above process leads just to the equivalentThevenin/Norton equivalent impedance.

We consider a more complex example:

Example: Determine the equivalent resistance between nodes 1 and 0 of the following circuit (notethat I1 is a test current source):


4

We first perform by-inspection nodal analysis:

12+1 −1 0

−1 1+ 12+1 −1

0 −1 1+1

⎡

⎣

⎢⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥⎥

V1V2V3

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥=

I100

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

32

−1 0

−1 52

−1

0 −1 2

⎡

⎣

⎢⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥⎥

V1V2V3

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥=

I100

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

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 I1 and V1.

Nodes 2 and 3 are not port nodes and therefore V2 and V3 must be eliminated; we use Gaussianelimination:

32

−1 0

−1 52

−1

0 −1 2

⎡

⎣

⎢⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥⎥

V1V2V3

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥=

I100

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

32

−1 0

−1 52−

−1( ) −1( )2

0

0 0 0

⎡

⎣

⎢⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥⎥

V1V20

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥=

I100

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

32

−1 0

−1 2 00 0 0

⎡

⎣

⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥

V1V20

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥=

I100

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

32

−1

−1 2

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

V1V2

⎡

⎣⎢

⎤

⎦⎥ =

I10

⎡

⎣⎢

⎤

⎦⎥

32−

−1( ) −1( )2

0

0 0

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

V10

⎡

⎣⎢

⎤

⎦⎥ =

I10

⎡

⎣⎢

⎤

⎦⎥

1 00 0⎡

⎣⎢

⎤

⎦⎥V10

⎡

⎣⎢

⎤

⎦⎥ =

I10

⎡

⎣⎢

⎤

⎦⎥

The resulting equation leads directly to the relationship between V1 and I1:

V1 = I1Hence, the equivalent sub-circuit is a 1 Ω resistor connected between node 1 and node 0.

This approach is applicable to a circuit with any number of nodes, although a computer method ofreducing 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 arenow ready to consider 2-port circuits.

3 2-PORT ADMITTANCE PARAMETERS

3.1 Definition

Rather than always having to deal with all the internal variables of a 2-port circuit, it is often moreconvenient to deal only with the terminal voltages and currents:


5

We assume that all excitations are external to the 2-port; hence the 2-port has no internalindependent sources.

We also assume that all dynamic elements (inductors and capacitors) are initially relaxed, i.e., havezero 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 2-port are expressions for the terminalcurrents I1 and I2 in terms of the port voltages V1 and V2, i.e.:

I1 = y11V1 + y12V2I2 = y21V1 + y22V2

where y11, y12, y21 and y22 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 2, then each port can be described by Ohm'slaw:

I1 = y11V1I2 = y22V2

where y11 and y22 are the effective admittances at port 1 and 2, respectively.

The remaining parameters y12 and y21 describe the effect of V2 on I1 and the effect of V1 on I2; theseterms are necessary when there are elements which link port 1 to port 2 including where the 2-portis an active device, such as a dependent source or transistor

The 2-port equations may be written more compactly in matrix notation:

I1I2⎡

⎣⎢

⎤

⎦⎥ =

y11 y12y21 y22⎡

⎣⎢

⎤

⎦⎥V1V2⎡

⎣⎢

⎤

⎦⎥

Example

Compute the admittance parameters of the following 2-port circuit:

Solution

Write the nodal admittance matrix by inspection; tape the VCCS to ic:

G1 +G3 −G3−G3 G2 +G3

⎡

⎣⎢

⎤

⎦⎥V1V2⎡

⎣⎢

⎤

⎦⎥ =

I1I2 − ic⎡

⎣⎢

⎤

⎦⎥

G1, G2 and G3 are conductances of the resistors.

Un-tape the source ic = gmV1:


6

G1 +G3 −G3−G3 G2 +G3

⎡

⎣⎢

⎤

⎦⎥V1V2⎡

⎣⎢

⎤

⎦⎥ =

I1I2 − gmV1⎡

⎣⎢

⎤

⎦⎥

Move the VCCS term to the LHS (to column 1 because it depends on V1):

G1 +G3 −G3gm −G3 G2 +G3⎡

⎣⎢

⎤

⎦⎥V1V2⎡

⎣⎢

⎤

⎦⎥ =

I1I2⎡

⎣⎢

⎤

⎦⎥

Since these equations are in the standard form for 2-port admittance matrices, we may easilyidentify the y-parameters:

y11 = G1 +G3 y12 = −G3 y21 = gm −G3 y22 = G2 +G3

This example was straightforward because the circuit has only two nodes. In general a 2-portcircuit will have more than 2 nodes. We now consider this more general case.

3.2 General 2-port analysis

Consider a general circuit with n nodes consisting of resistors, capacitors, inductors and dependentvoltage and current sources:

The reference node is labelled node 0, and the remaining nodes are numbered sequentially from 1 ton; the nodal voltages are V1, V2, ... Vn and the currents injected at each node are I1, I2, ... In.

Nodal analysis would lead to following set of nodal equations:

y '11 y '12 .. y '1,n−1 y '1ny '21 y '22 .. y '2,n−1 y '2n: : . : :

y 'n−1,1 y 'n−1,2 .. y 'n−1,n−1 y 'n−1,ny 'n1 y 'n1 .. y 'n,n−1 y 'nn

⎡

⎣

⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥

V1V2:

Vn−1Vn

⎡

⎣

⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥

=

I1I2:

In−1In

⎡

⎣

⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥

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 2) are taken to form the ports of a2-port circuit:


7

Consider the effect of this on the nodal equations. Voltages V1 and V2, which were just nodalvoltages are now the port voltages and I1 and I2 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 '12 y '13 .. y '1ny '21 y '22 y '23 .. y '2ny '31 y '32 y '33 : y '3n: : .. . :

y 'n1 y 'n1 y 'n3 .. y 'nn

⎡

⎣

⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥

V1V2V3:Vn

⎡

⎣

⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥

=

I1I20:0

⎡

⎣

⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥

We use partition lines to show that the n × n nodal matrix is that of a 2-port circuit which can bedescribed by a 2 × 2 port admittance matrix.

The 2 × 2 port admittance matrix has the following form:

y11 y12 0 .. 0y21 y22 0 .. 00 0 0 : 0: : .. . :0 0 0 .. 0

⎡

⎣

⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥

V1V20:0

⎡

⎣

⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥

=

I1I20:0

⎡

⎣

⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥

or

I1I2⎡

⎣⎢

⎤

⎦⎥ =

y11 y12y21 y22⎡

⎣⎢

⎤

⎦⎥V1V2⎡

⎣⎢

⎤

⎦⎥

Obtaining the 2 × 2 port matrix from the n × n nodal matrix is essentially one of eliminating theinternal voltage variables V3, V4, ... Vn in an equation solving process.

This can be done by a variety of methods; a method that generally works reasonably well for circuitequations is Gaussian elimination.

Example: Determine the y-parameters for the following 2-port, 3 node circuit:

Solution

Determination of the 2-port y-parameters starts from nodal analysis. Writing the node equations byinspection leads to the following:


8

jω + 3 − jω −2− jω jω + 2 −1−2 −1 5

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

V1V2V3

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥=

I1I20

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

where jω is the admittance of the 1 F capacitor; the nodal matrix is symmetrical because the RLCnetwork contains no dependent sources.

We will use Gaussian elimination for matrix reduction. Whereas in nodal analysis we carried onthe reduction until we had a single variable, in this case we only need to eliminate row 3 andcolumn 3 to obtain the 2 × 2 matrix; the pivot is y'33 = 5; the matrix becomes:

jω + 3− −2( ) −2( )5

− jω −−1( ) −2( )5

0

− jω −−2( ) −1( )5

jω + 2 − −1( ) −1( )5

0

0 0 0

⎡

⎣

⎢⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥⎥

≡jω + 3− 0.8 − jω − 0.4− jω − 0.4 jω + 2 − 0.2

⎡

⎣⎢

⎤

⎦⎥ ≡

jω + 2.2 − jω − 0.4− jω − 0.4 jω +1.8⎡

⎣⎢

⎤

⎦⎥

Hence, we have the 2 × 2 port admittance equations:

I1I2⎡

⎣⎢

⎤

⎦⎥ =

jω + 2.2 − jω − 0.4− jω − 0.4 jω +1.8⎡

⎣⎢

⎤

⎦⎥V1V2⎡

⎣⎢

⎤

⎦⎥

The port admittance matrix like the nodal admittance matrix is symmetrical.

3.3 Equivalent Circuit for 2-port admittance equations

It can be helpful to visualise 2-port admittance equations by means of an equivalent circuit.

Consider the first admittance equation:

I1 = y11V1 + y12V2A valid interpretation of this equation is that the port current I1 is equal to the port voltage V1 timesan admittance y11 in parallel with a voltage-controlled current source y12V2.

A similar interpretation is possible for the second equation:

I2 = y21V1 + y22V2This yields an admittance branch y22 in parallel with a voltage-controlled current source y21V1.

These interpretations lead to the following equivalent circuit for a 2-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 withfour admittances always exists.


9

Each port consists of an admittance connected in parallel with a dependent current source; this maybe regarded as a generalisation of the Norton equivalent circuit for 1-port circuits to 2-port circuits.

3.4 Determining 2-port parmeters by port tests

It is possible to avoid a complete nodal analysis of a circuit in order to determine the portadmittance parameters. This can be achieved instead by carrying out tests on the ports of thecircuit.

This method may be used in three ways, 1) in the laboratory, 2) in computer simulation or 3) forhand analysis.

Consider a circuit which has the general 2-port admittance equations:

I1 = y11V1 + y12V2I2 = y21V1 + y22V2

Let us apply a short-circuit to port 2 such that V2 = 0; we then have:

I1 = y11V1I2 = y21V1

This leads to:

y11 =I1V1 V2 =0

y21 =I2V1 V2 =0

We can envisage a short-circuit on port 2 and application of a known voltage source V1 to port 1;

Determining the port currents I1 and I2 yields the two admittance parameters.

Referring to the original circuit and its equations, let us now apply a short-circuit to port 1 such thatV1 = 0; we then have:

I1 = y12V2I2 = y22V2

This leads to:

y12 =I1V2 V1=0

y22 =I2V2 V1=0

We can envisage a short-circuit on port 1 and application of a known voltage source V2 to port 2;

Again, determining the port currents I1 and I2 yields the two admittance parameters.

Since each admittance parameter is defined with regard to a short-circuited terminal voltage theadmittance parameters are sometimes called short-circuit admittance parameters.


10

An example illustrates derivation of admittance parameters by port tests.

Example:

We take the previous example of a 2-port circuit with 2 nodes:

We begin by placing a short-circuit on port 2 and voltage source on port 1; in order to obtain y11

and y21, we determine I1 and I2:

We then place a short-circuit on port 1 and voltage source on port 2, and in order to obtain y12 andy22, we again determine I1 and I2:

The y-parameters obtained are identical with those obtained previously by nodal analysis of thecircuit.

4 GAIN AND IMPEDANCE FROM 2-PORT ADMITTANCE PARAMETERS

4.1 General

We have defined admittance parameters and shown that they may be obtained by nodal analysis andequation 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 andoutput impedance of a circuit in terms of its 2-port admittance parameters.

Consider a 2-port circuit in isolation:

The circuit is described by:

I1 = y11V1 + y12V2I2 = y21V1 + y22V2

where I1, I2, V1 and V2 are port variables.


11

The impedance and gain expressions we are interested in are ratios of port variables.

Ratios of variables at the same port, such as V1/I1, I2/V2 are termed driving point functions.

Ratios of variables at different ports, such as V2/V1, I1/I2, I2/V1, V1/I2, are termed transfer functions.

Driving point and transfer functions together are termed network functions.

Note that transfer functions may be dimensionless, e.g. V2/V1, or have dimensions of impedance,e.g. V2/I1, or admittance, e.g. I2/V1

Consider the constraints acting on the variables.

We have 4 port variables, V1, V2, I1, I2, and therefore there are just 3 independent network functionswhich 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 thetermination 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 singlesource is sufficient to determine all voltages and currents.

4.2 Gain with simple terminations

Consider a 2-port circuit with port 2 terminated in admittance YL:

The equations for the 2-port circuit alone are:

I1 = y11V1 + y12V2I2 = y21V1 + y22V2

The termination constraint is:

I2 = −YLV2The minus sign is due to the voltage and current reference directions.

Substitution of the termination constraint into the 2-port equation for I2 gives:

−YLV2 = y21V1 + y22V2V2V1 I2 =−YLV2

= −y21

y22 +YL

It is good practice to state the termination constraint along with the network function as we havedone here.

This expression may be used to determine voltage gain for any load admittance; it is validindependent of the nature of the source at port 1, so port 1 does not have to be driven by a voltagesource.


12

If we terminate port 2 in an open-circuit, we have:

Av o/c( ) =V2V1 I2 =0

= −y21y22

If V1 is known, then this equation may be used to determine V2 for the case when port 2 isterminated in an open-circuit; the value of V2 obtained is actually the voltage of the equivalentThevenin voltage source representing the output port (port 2).

Termination of port 2 in a short-circuit leads to V2 = 0, so voltage gain is not useful in this case.

Now consider current gain; terminating port 2 in an open-circuit is not valid because it forces I2 tobe zero.

In the case of a short-circuit termination, we can set V2 = 0 in the 2-port equations:

I1 = y11V1I2 = y21V1

We can now determine the current gain:

Ai s /c( ) =I2I1 V2 =0

=y21y11

Example

For the simple circuit of the previous example determine from the admittance matrix (i) the voltagegain with a load resistance of 1 Ω and (ii) the current gain with a short-circuit at port 2; the elementvalues are G1 = 1 S, G2 =2S, G3 = 3 S, gm = 21 S:

The previously determined admittance matrix is:

Y =y11 y12y21 y22⎡

⎣⎢

⎤

⎦⎥ =

G1 +G3 −G3gm −G3 G2 +G3⎡

⎣⎢

⎤

⎦⎥ =

4 −318 5⎡

⎣⎢

⎤

⎦⎥

We have:

Av =V2V1 YL =0.5

= −y21

y22 +YL= −

185 +1

= −3

and

Ai s /c( ) =I2I1 V2 =0

=y21y11

=184

= 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 2-port are important quantities, especially whendetermining voltage gain, current gain and power gain.


13

We are interested in the input admittance of a terminated 2-port circuit because this affects the loadit presents to a source:

The circuit without the load admittance YL can be represented by its admittance parameters:

I1I2⎡

⎣⎢

⎤

⎦⎥ =

y11 y12y21 y22⎡

⎣⎢

⎤

⎦⎥V1V2⎡

⎣⎢

⎤

⎦⎥

The load admittance YL imposes the termination constraint:

I2 = −YLV2We now have:

I1−YLV2⎡

⎣⎢

⎤

⎦⎥ =

y11 y12y21 y22⎡

⎣⎢

⎤

⎦⎥V1V2⎡

⎣⎢

⎤

⎦⎥

We can move the –YLV2 term across the to the RHS and into column 2, since it contains V2:

I10

⎡

⎣⎢

⎤

⎦⎥ =

y11 y12y21 y22 +YL⎡

⎣⎢

⎤

⎦⎥V1V2⎡

⎣⎢

⎤

⎦⎥

We can solve the equations to eliminate V2 by a number of methods including simple substitution.We use Gaussian elimination:

I10

⎡

⎣⎢

⎤

⎦⎥ =

y11 y12y21 y22 +YL⎡

⎣⎢

⎤

⎦⎥V1V2⎡

⎣⎢

⎤

⎦⎥ =

y11 −y12y21y22 +YL

0

0 0

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

V10

⎡

⎣⎢

⎤

⎦⎥

Hence, the input admittance of the 2-port circuit is:

Yin =I1V1

= y11 −y12y21y22 +YL

In the case where port 2 is terminated in an open-circuit, we have YL = 0:

Yin o/c( ) =I1V1 I2 =0

= y11 −y12y21y22

Note that YL = 0 is equivalent to the constraint I2 = 0.

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 2, we have YL → ∞; the generalexpression now reduces to:

Yin s /c( ) =I1V1 V2 =0

= y11


14

Note that YL → ∞ is equivalent to the constraint V2 = 0.

Consider now the 2-port circuit with voltage source VS and admittance YS at port 1:

We are interested in the output impedance at port 2, as this determines the effect of a load at port 2.

The output impedance is actually the Thevenin equivalent impedance seen looking into port 2 forthe combination of the 2-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 2 of YL, except that port 1 andport 2 are interchanged and YL is replaced by YS.

Therefore, we can use the previous expression for Yin to determine Yout:

Yin =I1V1

= y11 −y12y21y22 +YL

⇒ Yout =I2V2

= y22 −y12y21y11 +YS

In the case where port 1 is terminated in an open-circuit, we have YS = 0:

Yout o/c( ) =I2V2 I1=0

= y22 −y12y21y11

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 YS → ∞; the generalexpression now reduces to:

Yout s /c( ) =I2V2 V1=0

= y22

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 inputadmittance at port 1 with a with a port 2 load resistance of 1 Ω and (ii) the output admittance at port2 when (a) the input at port 1 is a voltage source and (b) is a current source; the element values areG1 = 1 S, G2 =2S, G3 = 3 S, gm = 21 S:


15

The previously determined admittance matrix is:

Y =y11 y12y21 y22⎡

⎣⎢

⎤

⎦⎥ =

G1 +G3 −G3gm −G3 G2 +G3⎡

⎣⎢

⎤

⎦⎥ =

4 −318 5⎡

⎣⎢

⎤

⎦⎥

i: For the input admittance at port 1 with YL = 1 S, we have:

Yin =I1V1 I2 =−YLV2

= y11 −y12y21y22 +YL

= 4 −−3( )185 +1

= 13 S

The input resistance, if required is:

Zin =1Yin

=1

13 Ω

ii The output admittance at port 2 with source admittance at node 1 of YS is given by:

Yout =I2V2 I1=−YSV1

= y22 −y12y21y11 +YS

(a) When the input excitation is a voltage source, de-activation of the source gives YS = ∞:

Yout s /c( ) =I2V2 V1=0

= y22 = 5 S

(b) When the input excitation is a current source, de-activation of the source gives YS = 0:

Yout o/c( ) =I2V2 V1=0

= y22 −y12y21y11

= 5 −−3( )18

5= 5 +10 4

5= 10 4

5= 15.8 S

4.4 Circuits with non-existent admittance description

Consider the following 2-port circuit for which we wish to determine the admittance parameters:

Use of KCL leads to the following two equations:

I1 =1RV1 − I2

I2 =1RV2 − I1

It turns out that it is impossible to eliminate the current terms on the RHS in favour of voltages (tryit!); it follows that it is impossible to describe this circuit using the admittance matrix, even thoughthe 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 andideal op-amps (nullors).


16

These problems can be solved using a number of methods, of which a very important one is themodified 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 analternative form with voltages in terms of currents:

V1 = RI1 + RI2V2 = RI1 + RI2

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 impedanceequations are not naturally produced by nodal analysis of a circuit.

However, we shall summarise its main features.

5 IMPEDANCE PARAMETERS FOR 2-PORT CIRCUITS

Consider again the general 2-port circuit:

The impedance parameters, or z-parameters, relate the port currents to the port voltages accordingto the matrix equation:

V1V2⎡

⎣⎢

⎤

⎦⎥ =

z11 z12z21 z22⎡

⎣⎢

⎤

⎦⎥I1I2⎡

⎣⎢

⎤

⎦⎥

As for the admittance parameters, it is possible to obtain impedance parameters by carrying outtests on the ports of the circuit.

We can write the 2-port circuit impedance equations:

V1 = z11I1 + z12I2V2 = z21I1 + z22I2

Let us apply an open-circuit to port 2 such that I2 = 0; we then have:

V1 = z11I1V2 = z21I1

This leads to:

z11 =V1I1 I2 =0

z21 =V2I1 I2 =0

We can envisage an open-circuit on port 2 and application of a known current source I1 to port 1;the port voltages V1 and V2 yield admittance parameters z11 and z21.

We now apply an open-circuit to port 1 such that I1 = 0; we then have:

V1 = z12I2V2 = z22I2

This leads to:


17

z12 =V1I2 I1=0

z22 =V2I2 I1=0

We can envisage an open-circuit on port 1 and application of a known current source I2 to port 2;the port voltages V1 and V2 yield the two admittance parameters z12 and z22.

Since each impedance parameter is defined by applying an open-circuit to a port, the impedanceparameters 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:

V1 = z11I1 + z12I2This equation can be interpreted as an application of KVL around the left hand loop of thefollowing circuit:

Consider the second impedance equation:

V2 = z21I1 + z22I2It 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 2-port circuit may always be represented by an equivalent circuitcontaining the four z-parameters.

Notice that the elements in the equivalent circuit are connected in series, in contrast to theequivalent 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 2-port case.

Given a circuit with a set of z-parameters, it is possible to determine voltage and current gain aswell as input and output impedance for arbitrary source and load impedances at ports 1 and 2; it issimilar 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 currentsto port voltages, one might expect that the z-parameter matrix and the y-parameter matrix arerelated to each other in a precise mathematical way.

Consider the impedance description for a 2-port circuit:

V1V2⎡

⎣⎢

⎤

⎦⎥ =

z11 z12z21 z22⎡

⎣⎢

⎤

⎦⎥I1I2⎡

⎣⎢

⎤

⎦⎥

Using matrix algebra, we may invert the z-matrix and use it to pre-multiply both sides of theimpedance equation set:

z11 z12z21 z22⎡

⎣⎢

⎤

⎦⎥−1 V1

V2⎡

⎣⎢

⎤

⎦⎥ =

z11 z12z21 z22⎡

⎣⎢

⎤

⎦⎥−1 z11 z12

z21 z22⎡

⎣⎢

⎤

⎦⎥I1I2⎡

⎣⎢

⎤

⎦⎥


18

where a superscript on a matrix of –1 represents inversion. It is known from matrix algebra that theinverse of a matrix times the matrix is the unit matrix which may be dropped when it multiplies acolumn vector. Hence, we have:

z11 z12z21 z22⎡

⎣⎢

⎤

⎦⎥−1 V1

V2⎡

⎣⎢

⎤

⎦⎥ =

I1I2⎡

⎣⎢

⎤

⎦⎥

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:

z11 z12z21 z22

⎡

⎣⎢

⎤

⎦⎥−1

=y11 y12y21 y22

⎡

⎣⎢

⎤

⎦⎥

y11 y12y21 y22

⎡

⎣⎢

⎤

⎦⎥−1

=z11 z12z21 z22

⎡

⎣⎢

⎤

⎦⎥

This is a reasonably efficient method for transforming between z- and y-matrices for 2-port circuits,e.g.:

a bc d⎡

⎣⎢

⎤

⎦⎥−1

=1

ad − bcd −b−c a⎡

⎣⎢

⎤

⎦⎥

The term ad – bc is the determinant Δ of the first matrix; if Δ = 0, 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 toobtain the z-matrix and (iii) check that the z-matrix describes the circuit:

By inspection, the y-matrix is:

Y =G1 −G1−G1 G1 +G2⎡

⎣⎢

⎤

⎦⎥

The corresponding admittance equations are:

I1 = G1V1 −G1V2I2 = −G1V1 + G1 +G2( )V2

Invert the y-matrix to obtain the z-matrix:

Z = Y −1 =G1 −G1−G1 G1 +G2⎡

⎣⎢

⎤

⎦⎥−1

=1

G1 G1 +G2( ) −G12G1 +G2 G1G1 G1

⎡

⎣⎢

⎤

⎦⎥

=1

G1G2

G1 +G2 G1G1 G1

⎡

⎣⎢

⎤

⎦⎥ =

1G2

+1G1

1G2

1G2

1G2

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥

=R1 + R2 R2R2 R2

⎡

⎣⎢

⎤

⎦⎥

We can now write the impedance equations:


19

V1 = R1 + R2( ) I1 + R2I2V2 = R2I1 + R2I2

or

V1 = I1 + I2( )R2 + I1R1V2 = I1 + I2( )R2

We can see that these equations correctly describe the circuit.

6 HYBRID PARAMETERS FOR 2-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 2) in terms of voltages(port 1 and port 2); and for the impedance description, we express voltages (port 1 and port 2) interms of currents (port 1 and port 2).

Hybrid parameters, as their name implies, are a cross between y- and z-parameters; we take avoltage and a current on the LHS and the remaining voltage and remaining current on the RHS.

Specifically, a 2-port circuit containing no internal independent sources and with no initial storedenergy, can be defined by the matrix equation:

V1 = h11I1 + h12V2I2 = h21I1 + h22V2

where h11, h12, h21 and h22 are the hybrid parameters for the 2-port circuit.

This leads to the matrix description:

V1I2⎡

⎣⎢

⎤

⎦⎥ =

h11 h12h21 h22⎡

⎣⎢

⎤

⎦⎥I1V2⎡

⎣⎢

⎤

⎦⎥

This combination of variables naturally arises when the model of a common emitter transistor issimplified; it is therefore encountered quite frequently.

Unlike y- and z-parameters, the h-parameters have different units; h11 has units of Ohms, h12 and h21

are dimensionless, and h22 has units of Siemens.

As with both y- and z-parameters, we can interpret the hybrid equation set as a two-dependentsource equivalent circuit:

Notice that at port 1 the elements are connected in series and at port 2 they are connected inparallel.

The h-parameters may be determined by tests on each port.

Initially, we apply a short-circuit to port 2; we have:

V1 = h11I1I2 = h21I1


20

This leads to:

h11 =V1I1 V2 =0

h21 =I2I1 V2 =0

We can envisage a short-circuit on port 2 and application of a known current source I1 to port 1; theport variables V1 and I2 yield hybrid parameters h11 and h21.

We now apply an open-circuit to port 1 such that I1 = 0; we then have:

V1 = h12V2I2 = h22V2

This leads to:

h12 =V1V2 I1=0

h22 =I2V2 I1=0

We can envisage an open-circuit on port 1 and application of a known voltage source V2 to port 2;the port variables V1 and I2 yield the two hybrid parameters h12 and h22.

Just as for the other parameter sets, we can determine gain and input and output impedance of a 2-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-parametersbut with the dependent and independent variable sets interchanged:

I1V2⎡

⎣⎢

⎤

⎦⎥ =

g11 g12g21 g22⎡

⎣⎢

⎤

⎦⎥V1I2⎡

⎣⎢

⎤

⎦⎥

Further parameter sets classify variables according to port location. For example, the transmissionmatrix expresses the port 1 variables, V1 and I1, in terms of the port 2 variables, V2 and I2.

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 = 100, base-emitter resistance rπ =1 kΩ and collector-emitter resistance of 1 MΩ:

Solution: An equivalent circuit for the transistor is as follows:

Comparing this with the 2-dependent source model for the h-parameter description, we identify:

h11 = rπ = 1000 Ωh12 = 0h21 = β = 100

h22 =1ro

= 10−6 S


21

Hence, the h-parameter equation set is:

V1I2⎡

⎣⎢

⎤

⎦⎥ =

rπ 0

β ro−1

⎡

⎣⎢⎢

⎤

⎦⎥⎥

I1V2⎡

⎣⎢

⎤

⎦⎥ =

103 0

100 10−6⎡

⎣⎢⎢

⎤

⎦⎥⎥

I1V2⎡

⎣⎢

⎤

⎦⎥

Because of its suitability for describing the common-emitter bipolar-junction transistor, the h-parameter subscripts are sometimes altered as follows:

V1I2⎡

⎣⎢

⎤

⎦⎥ =

hie hreh fe hoe⎡

⎣⎢

⎤

⎦⎥I1V2⎡

⎣⎢

⎤

⎦⎥

In this scheme, the second subscript 'e' denotes that the parameters are applicable to a common-emitter transistor. The first subscripts have the following meanings: 'i' input (rπ); 'r' reverse; 'f'forward (β); 'o' output (1/ro).

Note it is only for the admittance matrix that by-inspection writing of the matrix from the circuit ispossible. This makes the admittance matrix by–far the most important of the circuit matrixformulations.

7 CONCLUSIONS

We began this study of 2-port circuits by looking again at 1-port sub-circuits and developing asystematic method based on nodal analysis and Gaussian elimination for deriving their Theveninand Norton equivalent circuits. This led to a definition of admittance parameters for 2-port circuitsand a similar method for obtaining the parameters by nodal analysis of a circuit followed byreduction. We then showed that circuit gain and port impedance can be obtained from the 2-portadmittance parameters once the termination of the 2-port circuit is specified. Finally, we lookedbriefly into impedance and hybrid parameters for 2-port circuits.

Notes for course EE1.1 Circuit Analysis 2004-05 TOPIC 10 ...cas.ee.ic.ac.uk/people/dhaigh/EE1.1...

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Transcript of Notes for course EE1.1 Circuit Analysis 2004-05 TOPIC 10 ...cas.ee.ic.ac.uk/people/dhaigh/EE1.1...