Network Notes IMP

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Network A Netwov

i

Thevenin's Theorem and Norton's Theore

Two networks to describe Thevenin's the

LOADTo be replaced

P

Q



Network A

Thevenin's Theorem : Characteristics of Net

To be replaced1. Linear eleme

2. Sources :

dependent (con

or independent3. Initial condi

passive elemen4. No magnetic

controlled-sou

coupling to net

P

Q



hevenin's Theorem : Characteristics of Netw

Nonlinear ElemeLinear Elements

L R C D T M

Yes No

i

i

v1 i1

Independent Sources Controlled Sou

v

v

Yes Yes



Vg

L1 L2

M

C1R

C2

Vo

(1-D)(Vg-Vo) -D(Vg-Vo)

Vg

L1

M

L2

M

C1

C2

(1-D)(Vg-Vo) -D(Vg-

Network ANetw

NO


P

Q



Vg

L1 L2C1R

C2

Vo

(1-D)(Vg-Vo) -D(Vg-Vo)

Vg

L1

M

L2

M

C1

C2

(1-D)(Vg-Vo)-D(Vg-Vo)

Network ANetw

NO


P

Q



Thevenin's Theorem: Network A is replac

and Vth

Netwov

i

LOAD

Network C

Vth

Network A

P

Q

M



Thevenin's Theorem: How to find Vth?

Netw

LOAD

VthNetwork A

Disconnect

P

Q

Open Ckt. Voltage



Thevenin's Theorem: How to determine C

Network A

P

Q

1. Set Initial Conditions to

2. Independent Sources TuOff

3. Dependent Sources are O

Network C

P

M



Determination of Vth : An Example

Vcc

R1

R2

Rc

Re

IB

Ic

IB+IC

Simplification of the base circuit

Vcc

R1

R2

R

R

I

N

B

B

B

Network A Net

Vth

Vth=Vcc[R2/(R



hevenin's Theorem : Characteristics of Netw

Nonlinear ElemeLinear Elements

L R C D T M

Yes

i

i

v1 i1

Independent Sources Controlled Sou

v

v

Yes Yes

Yes



Determination of Network C : Example

Sim lification of the base circuit

Vcc

R1

R2

B

Network A

R1

R2

B

Network C

N

Shorted

B

Rb=(R

Note : No initial condition and No dependent

M

M



Vcc

R1

R2

Rc

Re

IB

Ic

IB+IC


R

R

I

N

B

B

Net

Vth=Vcc[R2/(R1+R2)]

Application of Thevenin's Theorem: An Ex

Vth

Rb

Rb=(R1||R2)

M



Vcc

R1

R2

Rc

Re

IB

Ic

IB+IC


R

RI

N

B

B

Net

Vth=Vcc[R2/(R1+R2)]

Application of Thevenin's Theorem: An Ex

VthRb

Rb=(R1||R2)

IB IB

Linear Model



Application of Thevenin's Theorem: Exa

Common Source Amplifier : Small Signal Gai

Vdd

R1

R2

RL

Rs

Id

Is

N

GCb

Vo=VOvi

D

S

Av=



Application of Thevenin's Theorem: Examp

R1R2

rd

Rs

N

G

Cb

vo

viRL

S

D

gm vgs

Linear Small Signal Equivalent Circuit of the CS A



Application of Thevenin's Theorem: Exampl

Simplified Small Signal Equivalent Circuit of CS Ampl

rd

Rs

N

G

vo

viRL

S

D

gm vgs

Network BNetwork A

Netw

Vth

M




rd

Rs

N

G

viS

D

gm.vgs

Network A

Vth=-(gm.vg

Determination of Vth

=-(gm.rd




rd

Rs

G

S

D

gm.vgs

Network C

Determination of Netwok C

Shorted

M

M




Determination of Rth

rd

Rs

G,M

S

D

gm.vgs

vgs

vx

ixvx=(ix-gm.vg

vgs=-ix.Rs

Therefore,

vx=ix.rd+gm+ix.R

Rth=vx

ix

=rd+Rs(1+




Vth

DM

N

The equivalent circuit of the CS amp

RL

rd Rs(1+ )

Rth

=gm

vivo



Application of Thevenin's Theorem: Example

vi

vo

L C

R RL

RLC Network : Zero Initial Conditions on L and

vi

Network A Network B L C

R

L C

RZth = 1/sC + sL||R

Vth=vi RR+sL



Norton's Theorem:Network A is replacedand Ith

Netwv

i

LOA

Network CIth

Network A

P

Q

M

Network A Ith

Network C as in Theveni

=>Norton's Theorem

Thevenin's Theore

P

Q



− vg + v1 + v2 + v3 = 0

v1 + v2 + v3 = vg

VgLoop

Direction

B

A

v1 v2

v3

R1 R2

R3

C D

One-loop resistive network towhich Kirchoff's Voltage Law is applied

Fig. 1

Fig.2



v1

R11

vg

1

1

vg

1

Req

ib

ia

i1 i2

i4i3

b

c

d

a

Fig. 3

Fig. 4(a)

Fig. 4(b)

i1+i2=i3+i4

Equivalent Network



Req = R1 + R2 + ........... + Rn = j=1n R j

Series Connected Resistive Network

R1 R21

vg

1

ia

Fig. 5

Rn



Fig. 5(a)

1

1

ia

L1 L2va

i1 i2

Fig. 5(b)

vg

ib

Equivalent Inductive One-Port Network

Leq= 1/L1+1/L2



Number of Network Equations

i1

i3

i2

ef'

f

3i1 + 2i2 - i3=4-i1 + 5i2 + 3i3 =-2i1 + 12 i2 + 5i3 = 0

* Equations must be linearly independent

Ax=b

UniqueSolution

gSolve:

* Formulate the network using minimum no of variabl

Formulation:(1)

(2)(3)

Note: 2*(2)+(1)=(3)



How to formulate the network using linearly independ

=> How many unknowns in a network with 'n' nodes a

=> 2*b unknown(branch voltage a

=> effectively b u

example : if voltacapacitor is knowbe found out

CdVc/dt = Ic

from

n=5 and b=8

0 : Ref Node

12

4

3

C1 C3

R1

C2 R3RL

b1

b2 b3

b4

b5

b6 b7

R2

b8



1

How many KCLs and how many KVLs?

(1) (n-1) nodes => (n-1(2) b-(n-1) KVL equat

example : n-1=4 => 4

b-(n-1)=8-4

0 : Ref Node

12

4

3

C1 C3

R1

C2 R3RL

b1

b2 b3

b4b5

b6 b7

R2

b8

note: easy to apply KC

apply KVL because thloops

V

(b4,b1,b2), (b2,b5,b8),(b4,b1,b5,b8), (b4,b6,b(b1,b6,b7,b5)

Problem: With a wrong choice the equations will be line



1

How to define voltages and currents in KCL an

0 : Ref Node

12

4

3

C1 C3

R1

C2 R3RL

b1

b2 b3

b4b5

b6 b7

R2

b8

Node Voltage : Voltwith respect to the r

Example : V4 (i.e. vbranch b3)

Branch Voltage : Vobranch of the netwo

Example : VC3 (i.e.

branch b7)

Note: Vc3 = V4-V3It is enough if we know the node voltages and v

VC3



How to define voltages and currents in KCL an

0 : Ref Node

12

4

3

C1 C3

R1

C2 R3RL

b1

b2 b3

b4b5

b6 b7

R2

b8

Loop Current : Currof the network

Example : i2 (i.e. cuconsisting of branch

Branch Current : Cubranch of the netwo

Example : Ib2 (i.e. c

branch b2, through c

Note: Ib2 = i1-i2 =>It is enough if we know the loop currents and v

i2

VIb2

i1



RL

0 : Ref Node

12

4

3

C1 C3

R1

C2 R3

b1

b2 b3

b4b5

b6 b7R2

b8V

Formulate network equations with explicit applic

Independent Equations: => Draw a graph : Figure with

information

0 :

12

b1

bb4

b6

Choice of variables:

Loop Currents and Branch Voltages

Graph of




0 : Ref Node

12

43

b1

b2

b3

b4

b5

b6 b7

b8

Graph of the network

0 : R

Construct a tree => A connected graph spanning all nodhave a circuit. A tree has n nodes an

A tree of th

=> Graph has more than one tree

12

b1

b2b

b4

b6

(b1,b2,b5,b7)Tree:




Tree of the network

0 : Ref Node

12

4

3

b1

b2b3

b4

b5

b6 b7

b8

(b1,b2,b5,b7)Tree: Co-tree:

(b4,b6,b3,b8)

=>The co-tree h

Every branch of(known as chordformation of a lo

is known as fund

Example : (b6 , fundamental circ

No of fundamen

no of chords = b

If we apply KVL to these fundamental circuits then thelinearly independent




0 : Ref Node0

12

C1

R1

C2

b1

b

b4

b6

V

Original N

12

4

3

b1

b2b3

b4

b5

b6 b7

b8


(b4,b6,b3,b8)

V=Vb1+Vb2

Vb2=Vb5+Vb8Vb6=Vb1+Vb5-Vb7Vb7=-Vb3-Vb2+Vb5

=>(b4: chord)

=>(b8: chord)=>(b6: chord)=>(b3: chord)

KVLs:

i1i2

i3 i4

7 voltage

are unkn




0 : Ref Node

12

4

3

C1 C3

R1

C2 R3

b1

b2 b3

b4b5

b6 b7R2

b8V

Original Network

V-I relationship

12

4

3

b1

b2 b3b4

b5

b6 b7

b8

0

I1

I3

I2

I4

Vb1=Ib1*R1

Vb2=1/C2 Ib2 dt

Vb3=Ib3*R3Vb4=V

Vb5=Ib5*R2

Vb6=1/C1 Ib6 dt

Vb7=1/C3 Ib7 dt

Vb8=R4*Ib8

R4

Expressed in terms of 8 (b



Formulate network equations using KVL

Appl

Ib1=

12

4

3

b1

b2b3

b4

b5

b6 b7

b8

0

I1

I3

I2

I4Appl

Ib2=

Appl

Ib5=

Apply

Ib7=I

Note: Ib4=-I1 ; Ib6=I3; Ib8=I2; Ib3=I4

Ib4

=> enouloop cur



Formulate network equations using KVL:

Ib1= (I1-I3)

Ib2=I1-I2+I4

Ib5=I2-I3-I4

Ib7=I4+I3

Ib4=-I1 ; Ib6=I3; Ib8=I2; Ib3=I4

V=Vb1+Vb2

Vb2=Vb5+Vb8Vb6=Vb1+Vb5-Vb7Vb7=-Vb3-Vb2+Vb5

=>(b4: chord)

=>(b8: chord)=>(b6: chord)=>(b3: chord)

Vb1=

Vb2=

Vb3=

Vb4

Vb5=

Vb6=

Vb7=

All the

Vb8

KVLs:

KCLs: Eleme



Network equations using KV

Vb6=Vb1+Vb5-Vb7

Vb7=-Vb3-Vb2+Vb5 =>

V=Vb1+Vb2 => V = (I1-I3)R1+ 1/C2

Vb2=Vb5+Vb8 => 1/C2 (I1-I2+I4)dt = (I2-I

1/C3 (I4+I3)dt=-I4*R3-1/C2 (I1-I2+I4)dt+(I

I3dt = (I1-I3)R1 + (I2-I3-I4)R2 -1/C3 (1/C1

Unknowns:Loop Currents: I1,I2,I3 and I4

Loop Variable Ana

Therefore the analy



Loop Variable Analysis: Matrix Representation

-1/C2 dt -R1

-1/C2 dt 1/C2 dt

+R2+R4

-1-R2 -R2

-R1 -R21/C3 dt +1/C1dt

+R1+R2 +1

1/C2 dt -1/C2 dt -R2 1/C2 dt +R2

1/C3 dt

+R

0

0

0

1/C2 dtR1+V

=

-1/C



Loop Variable Analysis: Generali

Graph of a network with L

independent loop currents

I1

I2

I3

I4

IL

Voltage drop in loop k produced by current Ij

Rkj*Ij+Lkj*dIj/dt +1/Ckj Ij dt = (Rkj+Lkj d/dt

=akjIj

akj: symbol to summarize the operation on Ij

Rkj = totalcom

k anLkj =

Ckj =




The general form of KVL

akjIj=Vk j=1

L

General form of KVL for L loop

akjIj=Vk , k =1,....,L j=1

L

a11 a12 a13

aL1 --- ---

--- --- ---

Operator onI1 I2 I3

V1

V2

VL

, Vk is the active voltage source in

Loop 1 =>

Loop 2 =>

Loop L =>

-=Vector Equation:



Loop Variable Analysis: Matrix Representation

-1/C2 dt -R1

-1/C2 dt 1/C2 dt

+R2+R4

-1/C2 d-R2 -R2

-R1 -R21/C3 dt +1/C1 dt

+R1+R2

R2

+1/C3 d

1/C2 dt -1/C2 dt -R2 1/C2 dt +R21/C3 dt +1/C

+R3+R2

0

0

0

1/C2 dtR1+V

=

=a11 =a12 =a13 =a14

=a21

-1/C2 dt

Solution : Find the inverse of the matrix (square: b-(n-1)*b-(n-1))

May use Gauss Elimination Method : A process called trangulariz

=a31

=a41

=a22

=a23 =a24

=a32



RL

0 : Ref Node

12

4

3

C1 C3

R1

C2 R3

b1

b2 b3

b4b5

b6 b7R2

b8V


Independent Equations: => Draw a graph : Figure with

information

0 :

12

b1

bb4

b6

Choice of variables:

Node Voltages and Branch Currents

Graph of




0 : Ref Node

12

43

b1

b2

b3

b4

b5

b6 b7

b8

Graph of the network

0 : R

Construct a tree => A connected graph spanning all nodhave a circuit. A tree has n nodes an

A tree of th

=> Graph has more than one tree

12

b1

b2b

b4

b6

(b1,b2,b5,b7)Tree:




Tree of the network

0 : Ref Node

12

4

3

b1

b2b3

b4

b5

b6 b7

b8


(b4,b6,b3,b8)

=>The tree has (

Every branch ofis responsible foformation of a c

is known as fund

Example : (b1 ,b4,

No of fundamentalno of branches in a

If we apply KCL to these fundamental cut-sets then thelinearly independent

A cut-set is the whose removal dis



Identify the fundamental cut-sets to apply KCL

2

4

3

b1

b2b4

b7

1

1 b1

b4

b6

2

0

4

b12

b6

b2 b3b4

0

4

b3

b5

b8

b6

(b5,b3,b8,b7)

b7

3

4b6

1

b3

0

(b7,b3,b6)

0

0

1

1

4

2 3This nodeis disconnected

This is afundamentalcut-set



RL

0 : Ref Node

12

4

3

C1 C3

R1

C2 R3

b1

b2 b3

b4b5

b6 b7R2

b8V

Formulate network equations with application of

b3

b6

21

0

Ib6+Ib3+Ib

-Ib4-Ib1-Ib

Ib6-Ib7+Ib

Ib2+Ib4-Ib

Cut-set of b1:

Cut-set of b7:

Cut-set of b2:

Cut-set of b5:

A BI1

In

I1+ I2 + ... + In =0

Conservation of Charges



0 : Ref Node

12

4

3

C1 C3

R1

C2 R3

b1

b2 b3

b4b5

b6 b7R2

b8V

Original Network

V-I relationship of the elements

Ib1=Vb1/R1

Ib2=C2 dVb2/dt

Ib3=Vb3/R3Vb4=V ; Ib4=? Ideal

Ib5=Vb5/R2

Ib6=C1dVb6/dt

Ib7=C3dVb7/dt

Ib8=Vb8/R4

R4

Expressed in terms of 8 (b

In terms of branch voltage an



Formulate network equations using KC

Apply KV

Vb1=

12

4

3

b1

b2b3

b4

b5

b6 b7

b8

0

Vb2=

Vb5

Vb7

=> enounode vo

Ib8

Vb3

Vb4

Vb6

Vb

Ib5

Ib4

Ib1

Ib6 Ib7

Ib3



Formulate network equations using KCL: All

Elements:KCLs:

Ib6+Ib3+Ib5-Ib8=0

-Ib4-Ib1-Ib6=0

Ib6-Ib7+Ib3=0

Ib2+Ib4-Ib3+Ib8=0

V

V

Ib1=Vb1/R1

Ib2=C2 dVb2/dt

Ib3=Vb3/R3

Vb4=V ;

Ib5=Vb5/R2

Ib6=C1dVb6/dt

Ib7=C3dVb7/dt

Ib8=Vb8/R4

V2,V3,V4 and Ib4

Note: V1 is given (V1=V)

However, Ib4 is unknown

Note: You can write KCLs at all nodes except the refereand obtain (n-1) linearly independent equations



Network equations using KC

Unknowns:Node Voltages: V2,V3, V4 and Ib4

Node Variable

Therefore the an

No. of Unknowns: n-1 =4

C1d/dt (V-V4) + V4/R3 + V2-V3/R2 - V3/R4 =

-Ib4 - (V1-V2))/R1 -C1d/dt (V-V4) = 0

C1d/dt (V-V4) -C3d/dt (V4-V3) + V4/R3 = 0

C2d/dt (V2-V3) + Ib4 - V4/R3 + V3/R4 = 0

Solve a system of linear integro-differential equ

Note: If (n-1) < b- (n-1) then it is advantageous to use n

as we need to solve less number of linearly indep



Node Variable Analysis:Matrix Representation

=

C1dV/dt 0 1/R2 -1/R2-1/R4 -C1d/dt +1/R

-1 1/R1 0 C1d/dtV/R1 +

C1dV/dt

0 0 C3d/dt 1/R3 + C3d/dC1dV/dt

1 C2d/dt 1/R4 + C2d/dt -1/R30

The voltage source V is equivalent to a current source Ib4 (u

voltage V (known)

Note : There is no active current source.



Node Variable Analysis: Generali

Current flowing out of node k due to voltage in

Vj*(1/R1 + 1/R2 + ...) + (1/L1+1/L2 + ...) Vjdt +

= (1/Rkj + 1/Lkjbkj dt + Ckjd/dt)Vj =

bkj: symbol to summarize the operation on Vj

C1C2

R1

R2

L1

L2

node jnode k

Elements Connecting Nodes j and k

1/Rkj = (1/R

1/Lkj = (1/L

Ckj = C1+C2

Ikj

node xIkx

Ik (Active Source)




The general form of KCL

bkjVj=Ik j=1

L

General form of KCL for N nodes

bkjVj=Ik , k =1,....,N j=1

L

b11 b12 b13

bL1 --- ---

--- --- ---

Operator of I1 I2 I3

I1

I2

IN

, Ik is the active current source con

Node 1 =>

Node 2 =>

Node N =>

-=Vector Equation:



Node Variable Analysis:Identify the operators

=

-1 1/R1 0 C1d/dV/R1 +C1dV/dt

0 1/R2 -1/R2-1/R4-C3d/dt C3d/d

-V/R1

0 0 C3d/dt -1/R3-(C3+C1)d-C1d/dt

Note : There is no active current source.

0 -1/R2-1/R1-C2d/dt 1/R2 0

0

The voltage source V is equivalent to a current source Ib4 (voltage V (known)

=b12 =b13 =b14

=b22 =b23 =b24

=b32 =b33 =b34

=b42 =b43 =b44



Apply KCL at each node: Formulation by observatio

0 : Ref Node

12

4

3

C1 C3

R1

C2 R3

b1

b2 b3

b4b5

b6 b7

R2

b8V

-Ib4 - (V-V2)/R1-C1d/dt KCL at Node 1:

KCL at Node 2 :

(V-V2)/R1 - C2d/dt (V2)

KCL at Node 3:

(V2-V3)/R2 -V3/R4 + C3

KCL at Node 4:

C1d/dt(V-V4) - V4/R3 - C

R4

Note: For a DC source dV/dt = 0

Even without identificationof the trees it will bepossible to formulatethe network equations



v1

v2

v1+v2=v

i1+i2=ii1 i2

Sources can be combined



i

v vR

R

i

No Difference



v(t)

R v1(t)

i1(t)

i1(t)

Rv/R

Source Transformation



v(t)

v1(t)i1(t)

L

vdti=1/L

v1(t)

v(t)

v1(t)i1(t)

v1(t)

i=Cdv/dt

C

C



Source Transformation

Equivalent Networks

v1

R1

R2

L1

v1

R1 L1

v1



Network Functions

Port : A pair of termi

current into one term

current out of otherOne-Port

Network V

I

I

A one-port network ispecified if voltage-c

at the terminals of the

1

I

I5

I

I



Network Functions

Two-Port Network :=>Defined by two pairs of

=>The variables are V1,I1,V2,I2.

Two-Port

Network V2

I2

V1

I1

voltage-current relationships

=>Two dependent variablesTwo independent variables

Depende

1. V1,V2

2. I1,I2

3. V1,I2

4. V1,I1

z par

y par

h (hy

A,B,



Network Functions

Mathematical description using dependent and indepe

V1=z11*I1+z12*I2V2=z21*I1 + z22*I2

z11 =V1

I1I2=0

z12 =V1

I2I1=0

Definition of z parameters:

z21 =V2

I1I2=0

z22 =V2

I2I1=0

I1=y11*V1+I2=y21*V1

Definition of y pa

z21 =V2

I1I2=0

z22 =V2

I2I1=0

Open-Circuit Parameters

y11 =I1

V1V2=0

y21 =I2

V1V2=0

Short- Circuit P

Short-Ckt. AdmittOpen-Ckt. Impedance



Network Functions

Mathematical description using dependent and indepe

V1=h11*I1+h12*V2I2= h21*I1 + h22*V2

h11 =V1

I1V2=0

h12 =V1

V2I1=0

Definition of h (hybrid) parameters:

h21 =I2

I1V2=0

h22 =I2

V2I1=0

I1

I1=0

Short-Ckt Open_Ckt

A =V1

V2I2=0

V1=A*V2 +I1= C*V2 +

Definition of A,B

C =I1

V2I2=0

Open-Ckt



Network Functions

V1=z11*I1+z12*I2V2=z21*I1 + z22*I2

z11 =V1

I1I2=0

z12 =V1

I2I1=0

Definition of z parameters:

z21 =V2

I1I2=0

z22 =V2

I2I1=0

z21 =V2

I1I2=0

z22 =V2

I2I1=0

Find the open-circ

for the T- circuit

1

1

V1I1

2

z11= Ra+Rb=12

Ra

Rb

z12=I2*Rb

I2

= 1

z22 = Rb+Rc = 1

z21 =I1*Rb

I1=

Example:

Driving PointTransfer



How do we obtain the z parameters of any two port (active or pa

I1 = b11V1 + b12V2 + b13V3 +.........+ b1nVnI2 = b21V1 + b22V2 + b23V3 + ........+ b2nVn0 = b31V1 + b32V2 + b33V3 + ........+ b3nVn................................................................................................................................................0 = bn1V1 + bn2V2 + bn3V3 + ....... + bnnVn

V3,....,Vn : N: depen

I1,I2, .... : C

A set of Node Equations:

: independe

bjk : operatVector Equation:

Solution:

I= GV

V=G-1

I=ZI

G-1

=

11

21

n1

12

1n

2n

nn-----------------------------------

----------------------------

Z=

z12z11

For passive network:

12= 21

z12 = z21

22

z21

z22

V1,V2 : term



Example: Open-circuit (Z parameters) of

I2

1

1

2

2

V1 V2

I1 GA

GC

GB

I1=(GA+GCI2=-GCV1+

= GA*GB+

11= GB+G

21= GC

z11=G

GA*GB

z12=

z22=

z21=

Delta-star(wye) Transformation:

z12 = Rb = GC/

G

GA*GBG

GA*GB

z22 = Rb+Rc = (GA+GC)/

=> Rc = GA/

z11 = Ra+Rb = (GB+GC)/

=> Ra = GB/



How do we obtain the y parameters of any two port (active or pa

V1 = a11I1 + a12I2 + a13I3 +.........+ a1nInV2 = a21I1 + a22I2 + a23I3 + ........+ a2nIn0 = a31I1 + a32I2 + a33V3 + ........+ a3nIn................................................................................................................................................0 = an1I1 + an2I2 + In3I3 + ....... + annIn

I1,I2,I3,....,In: depen

V1,V2, .... :

A set of Loop Equations:

: independe

ajk : operatoVector Equation:

Solution:

V= RI

I=R-1

V=YI

R-1

=

11

21

n1

12

1n

2n

nn-----------------------------------

----------------------------Y=

y12

For passive network:

12= 21

y12 = y21

I1,I2 : Port in

22

y21

y11 y22



Network Functions

Find the short-circuit parameters for the bridgExample:

1

1 2

V1 V2I1 I2

2

10

7

2

I3

5

V1 = (10+2)I1 +10I2 - 2I3V2 = 10I1 +(10+7)I2 +7I3

0 = -2I1+7I2 +(5+7+2)I3

R =12 10 -210 17 7-2 7 14

11

= 12(17*14 - -(10)(10*14+(-2)(10*7

11=17*14 - 7

y11= =

15

21y12= =

-5

22y22= =

15

When y11=y22 or z11=z12,

is s mmetrical



V1=h11*I1+h12*V2I2= h21*I1 + h22*V2

Definition of h (hybrid) parameters:

h11 = V1I1

V2=0

h12 = V1V2

I1=0

h21 =I2

I1V2=0

h22 =I2

V2 I1=0

I1

I1=0

Short-Ckt Open_Ckt

The h parameters : comparison

y11 =I1

V1 V2=0

=>

Extremely useful for describing bipolar junction trans

V1= Vbe = Base to em

I2 = Ic= Collector curI1= Ib = Base current

V2 = Vce = Collector

z22 =V2

I2I1=0

=> h

z12 =V1

I2 I1=0

=> h

=> h2y21 =I2

V1V2=0



Example: h parameters of the Pi Circuit

1

1

2

2

V1

V2I1

I2

GA

GC

GB

h11 =V1

I1V2=0

h12 =V1

V2I1=0

h21 =I2

I1V2=0

h22 =I2

V2I1=0

I1

I1=0

Short-Ckt Open_Ckt

h11 = 1/(G

h21 = -GC

h12 = GC/

h22 = GB + G

Note: h12=-h21 =



The A,B,C,D parameters : comparisonExtremely useful for describing transmission net

A =V1

V2I2=0

B =-V1

I2V2=0

V1=A*V2 + B*-I2I1= C*V2 + D*-I2

Definition of A,B,C,D parameters:

C = I1V2

I2=0

D = -I1I2

V2=0

Short-CktOpen-Ckt

V1, I1 are sendiand current variaV2, I2 are receivand current varia

A BC D

<= TransmissionMatrix

Transmission engineers

z11 =V1

I1I2=0

z21 =V2

I1 I2=0

z21 =V2

I1 I2=0

=

=



The A,B,C,D parameters : comparisonExtremely useful for describing transmission net

V1=A*V2 + B*-I2I1= C*V2 + D*-I2

A =V1

V2I2=0

B =-V1

I2V2=0

Definition of A,B,C,D parameters:

C = I1V2

I2=0

D = -I1I2

V2=0

Short-CktOpen-Ckt

V1, I1 are sendiand current variaV2, I2 are receivand current varia

A BC D

<= Transmission (T)Matrix

Transmission engineers

y11 =I1

V1V2=0

y21 =

I2

V1V2=0



Example: T parameters of the Pi Circuit

1

1

2

2

V1 V2

I1 I2

GA

GC

GB

A =V1

V2I2=0

B =-V1

I2V2=0

C =I1

V2I2=0

D =-I1

I2V2=0

Short-CktOpen-Ckt

A =1/GB

1/G

1

1

V1

I1

GA

G

Ix = I *1/

1/GA +

GA*GB +GGC

C=GA*GB+G

G

V2 = Ix*1/GB

=



Example: T parameters of the Pi Circuit

1

1

2

2

V1 V2

I1 I2

GA

GC

GB

A =V1

V2I2=0

B =-V1

I2V2=0

C =I1

V2I2=0

D =-I1

I2V2=0

Short-CktOpen-Ckt

1

1

V1

I1

GA

GC

I2 = - V1*

B =1

GC

I2 = - 1/G

1/GA+

D = GC+G

Note : AD-BC =Equivalent cond



Relationship Between Two Port Paramete

The z and y relationships can be obtained by using m

Z =z11 z12

z21 z22Y =

y11 y12

y21 y22

V = ZI = ZYV

; V = ZI ; I

Therefore, Y = Z 1

y11 = z22/ z y22 = z11/ z

y12 = -z12/ z y21 = -z21/ z

z = z11*z22-z12*z21

z11 = y22/ y

z12 = -y12/ y

y = y11*y2

h and ABCD parameters can be expressed in terms of



Interconnection of Two Port Parameters

Na NbV2a V2bV1

I1 I2a I2b

Cascade connection of Two Ports

Ab Bb

Cb Db

Aa Ba

Ca Da

Aa Ba

Ca Da

A B

C D

A=

The transmission matrix of overall two port networkproduct of the transmission matrices of the individua




Na V2aV1a

I1a I2a

Nb V2bV1b

I2bI1b

1:1

Parallel Connection of two ports

I1 I2 V1aV2aI1 =

I2 =

I =

I1

I2Ia

V =V1

V2

Va

I = Ia + Ib = YaVa + YbVb = (Ya+Yb)V

Y = (Ya+Yb)




Na V2aV1a

I1a I2a

Nb V2bV1b

I2bI1b

1:1

Series Connection of two ports

I1 I2 V1aV2aI1 =

I2 =

I =

I1

I2Ia

V =V1

V2

Va

V = Va + Vb = ZaIa + ZbIb = (Za+Zb)I

Z = (Za+Zb)

V1V2




1. When the two ports are connected in parallel, fiparameters first, and, from the y parameters deriv

two-port parameters

2. When two-ports are connected in series, it is us

easiest to find the z parameters

3. When two ports are connected in tandem, it is geasiest to find the transmission matrix.



Q5.

=> Ideal Transformer (2) I 2

b = I 1 / a

=> Circuit connection (2)V in = V 1 − bV 2

=> Ideal Transformer (2) I in = − I 1 / a

=> Definition Input Impedance (2) Z =V in I in

=V 1−bV 2

− I 1 / a

=> Definition Z parameters (2)−

V 1a

V 2=

z11 z12

z21 z22

I 1

I 2

=> Eq. of the 1st row (2)V 1 = −a( z11 +ba z12) I 1

=> Eq. of the 2nd. row (2)V 2 = ( z21 +

b

a z22) I 1

(4) Z =−a( z11+

ba z12) I 1−b( z21+

ba z22) I 1

− I 1 / a= a2 z11 + abz12 + abz21 + b2 z22

6

z z11 12

z z21 22

T1 T2

a:1 1:b1

1

2

2

Ideal

Transformer

Ideal

Transformer

Z

Fig. A

V1 -(V1/a) V2-(bV2)

VinIin

I1 I2

I1/a

I2/b



Q6.

=> Definition of Z parameters (1)V 1

V 2=

z11 z12

z21 z22

I 1

I 2

Let us set => 2-2’ open circuited I 2 = 0

=> (1) I 1 =2V 1

( Z a+ Z b) z11 =Z a+ Z b

2

and (1)V 21∏ =Z b

Z a+ Z bV 1 V 2∏1 ∏ =

Z a Z a+ Z b

V 1

Therefore (1)V 22∏ = V 21∏ − V 2 ∏1∏ =Z b− Z a

Z a+ Z bV 1 = V 2

(1) z21 =V 2 I 1

=

Z b− Z a

Z b+ Z aV 1

I 1=

Z b− Z a

Z b+ Z a

Z a+ Z b2 =

Z b− Z a

2

Since this is a passive network, (1) z12

= z21

Since this is a symmetric network, (1) z22 = z11

(1) z = z11 z22 − z12 z21 =( Z a+ Z b)2−( Z b− Z a)2

4 = Z a Z b

(2) y11 =12 (

1 Z b

+1 Z a

) =14 (1 − j) = y22

(2) y12 = −12 (

1 Z a

−1

Z b) = −

14 (1 + j) = y21

7

Za

Za

1

1

2

2

Zb

Zb

Fig.B

V1V2

I1I2



Q7.

(2)V 3 = I xr d

I x =V x−V 3r d + (−gmV 3) =

V xr d − I x − gmr d I x

(2)

(2) Rth =V x I x

= r d (2 + gmr d )

(2)V 3 = −r d gmV 1

(2)V th = −(−gmV 3)r d + V 3

(2)V th = (1 + gmr d )(−r d gm)V 1

8

V1 gmV1

-gmV3

rd

rd

RL RL

Rth

Vth

(i) (ii)

Fig. C

V3

gmV1

-gmV3

rd

rd

(i)

Fig. C

V3V1=0

=0

Vx

Ix

Determination of Rth

V1 gmV1

-gmV3

rd

rd

(i)

Fig. C

V3

Determination of Vth

Vth

-gmV3

gmV1



Q8.

Introduce two new variables , that is the current through and , that is the current through I 1 V 1 I k , so that KCL at every node (1,2,3,4,5,6 and k) can be written. (V 1 − V k )

KCL at node 1:

=> (since is known and is unknown)−V 1 R1

+V 2 R1

= I 1 − I 1 +V 2 R1

=V 1 R1

V 1 I 1

KCL at node 2:

=>V 1 R1

−V 2 R2

−V 2 R1

− C 1dV 2

dt = 0 −V 2 R2

−V 2 R1

− C 1dV 2

dt = −V 1 R1

KCL at node 3:

=> => (since is known and isV 4 R3

−V 3 R3

= I k −V k + (V 1−V k )

R3− I k = 0 −

(1− )V k R3

− I k =V 1

R3V 1 I k

unknown)

KCL at node 4:

=>V 3 R3

−V 4 R3

−V 4 R5

−V 4 R +

V 5 R +

V 6 R5

= 0(1− )V k

R3−

V 4 R3

−V 4 R5

−V 4 R +

V 5 R +

V 6 R5

= − V 1

R3

KCL at node 5:

V 4 R −

1 L ¶ V 5dt −

V 5 R = 0

KCL at node 6:

=>V 4 R5

−V 6 R6

−V 6 R5

+ i2 = 0V 4 R5

−V 6 R6

−V 6 R5

+ V 6 R6

= 0

KCL at node k:

−V k R4

+ I k = 0

9

V1

R1

C1 R2

R3 R5

R

L

R6

i2

R4

(V1-Vk)

k i2

Fig. D

Vk

12

3 4

5

6

I1

Ik



The unknown variables are . Therefore we can organize the KCLs as I 1,V 2, I k ,V 4,V 5,V 6,V k Bp = q

= p

I 1

V 2 I k

V 4

V 5

V 6

V k

= B −11

R10 0 0 0 0

0 (−1

R1−

1 R2

− C 1d

dt ) 0 0 0 0 0

0 0 − 11

R30 0 −

(1− )

R3

0 0 0 − (1

R3+

1 R5

+1 R )

1 R

1 R5

(1− )

R3

0 0 01 R − (

1 R +

1 L ¶ dt ) 0 0

0 0 01

R50 − (

1 R5

+1

R6−

R6

) 0

0 0 1 0 0 0 − 1 R4

=q

V 1 R1

−V 1 R1

V 1

R3

− V 1

R3

00

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