Electrical circuits wyklad 5

Electrical Circuits

Dr inż. Agnieszka Wardzińska

Room: 105 Polanka

agnieszka.wardzinska@put.poznan.pl

cygnus.et.put.poznan.pl/~award

Advisor hours: Monday: 9.30-10.15 Wednesday: 10.15-11.00

Nonlinear elements - example

Nonlinear resistor is given by:

Nonlinear circuits



_{Analysing nonlinear circuits is often difficult.}

Only a few simple circuits are adequately

described by equations that have a closed

form solution.



_{A trivial example is a circuit consisting of a}

current source and an exponential diode .



_{Voltages and currents in circuits containing}

only a few nonlinear circuit elements may be

found using graphical methods for solving

nonlinear equations that describe the

behaviour of the circuit.

Nonlinear elements - example

Nonlinear resistor is given by:

Nonlinear element - methods



_Analytical



_{Graphical using load line}



_{Graphical using graphical}

Analytical method



_{For the nodes A-B:}



_{For the voltage drops in left}

loop:



_{For the node A:}

For nonlinear element we have:

Then combining the above formulas (U1=I1R1) we can write:

We know that: then

We can combine the equation to calculate the current:

And for the given value of source E:

Substituting the values:

We take into account only positive value

The characteristic of the

nonlinear element was given for the I>0



_{The voltage and current of the nonlinear}

element are called the operating poit of the

circuit



_{Finding a circuit's dc operating points is an}

essential step in its design and involves

solving systems of nonlinear algebraic

equations.

Then the operating point for our exapmle:

U=2V

I=2A

U=2V

I=2A

Graphical method

(using load line)



_{First we find with Thevenin method the equivalent}

source and resistance of the circuit on the left

from AB-nodes (we remove the nonlinear element,

then the circuit that remain is linear and we can

use the Thevenin’s or Nortons’s Method)

Than we can redraw our circuit using the

calculated source and resistance:



_{We plot the characteristic of the nonlinear}

element and the summation characteristic of

the source with resistor.



U

_AB

=E

_T

—I*R

_T

The operating point is as

we can see the same

U=2V

I=2A

U=2V

I=2A

Graphical

using graphical summing for Kirchoffs laws

1. We draw the characteristic of the nonlinear element

2. We draw the characteristic of parallel connected resistor R2 3. We are adding this two

characteristics to obtain the relation of voltage U from the sum of I and I2; U=f(I+I2),

I+I2=I, then the red line is for U=f(I1)

Black line

Blue line

4.

We draw the characteristic of the

R1 resistor

5.

We adding it to the obtained

earlier sum of characteristic (

red

line

)

Brown line

4.

We draw the characteristic of the ideal source

5.

We mark the point the ideal source line and our sum of

characteristic crosses as P – this point gives us the value of

the current I1

4. We can also subtract the pink line from the brown one and see

when the substracted characteristic cross zero.

Then we can calculate all the curents and voltages in the circuit.

We do it in reverse order as we drew characteristics.

I1 – point P. I1 =4A

For I1 we can see the voltage drops of the elements though this

current flews U1 – point P1 – crosses of I1=4 with R1 characteristics

(brown dash-dot line) U1=8V

U – point P2 – crosses of I1=4 with function U=f(I1)=f(I+I2) (red

Then for given U (U=2V from last step) we can read the currents of the branches with U drop of voltage (the branches between the A and B node)

I2 – point P3 – crosses of U=2 with characteristic of resistor R1 (blue dotted), then I2=2A

I – point P4 – crosses of U=2V with characteristic of nonlinear resistor R_N

(black curve) I=2A.The voltage and current of the nonlinear element R_{operating point of the circuit.} N (point P4) is the

U=2V

I=2A

U=2V

I=2A

Comment:

For given set of parameters, the points: P3 and P4 are in the same

place. But it is only a coincidence.

Comments

1.

The methods can be combined, for example using

Thevenin’s method in analytical method or

calculating the fragment of the circuit with

Thevenin’s method and then summing the

characteristics.

2.

U-I characteristic of nonlinear element could be

given by a formula, in a table or on the graph

3.

There is possible, that the characteristic on the

graph is gives as U=f(I) (horizontal U and vertical

I) or I=f(U) (horizontal I and vertical U)

 _{In the first case the summation is done as in example, but in}

the second case summation of voltages will be done horizontally and summation of currents vertically

4.

The graphical method of summation is adecuate to

summing the two nonlinear elements.

Another example

An expamle taken from: AnantAgarwaland Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT OpenCourseWare(http://ocw.mit.edu/), Massachusetts Institute of Technology.

Incremental Analysis

An expamle taken from: AnantAgarwaland Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT OpenCourseWare(http://ocw.mit.edu/), Massachusetts Institute of Technology.

The incremental method (small signal

method)

Mathematical consideration



_{For the nonlinear:}



_{We assume:}



We can use Taylor’s expansion of f(v

_D

) near

v

_D

=V

_D

:



_{We neglect the higher order terms because}

Δv

_D

Static and dynamic resistance

D D I d I U di du R     0 D D s

I

U

R



Static resistance Dynamic resistance

Static Resistance is the normal ohmic resistance in accordance with Ohm's Law.  It is the ratio of voltage and current and is a

constant at a given temperature.

Dynamic Resistance is a concept of resistance used in nonlinear circuit.  Dynamic resistance refers to the change in current in response to a

change in voltage at a specific region of the VI curve. When the voltage is increased, the current may not increase proportionally.  In some cases, the current may actually decrease

D D U d

U

I

du

di

G









0 d d

R

G



1

Dynamic conductance

Negative resistance



_{Negative resistance is the phenomenon in which the}

current through a device decreases as the voltage

increases.  This is in contrast to conventional logic

in which current increases as the voltage increases.



_{While negative resistance exists, there is no such}

thing as a negative resistor.  Negative Resistance

exists along some are of the V-I curve in certain

electronic components such as the Gunn Diode used

in microwave electronics.  



_{In certain regions of the V-I curve, the current drops}

source: wikipedia

Example for small signals



_{For the given nonlinear resistor described by}



_{Flows the current:}



_{Using small singal method, calculate the voltage drop.}



Given a=2[MΩ/A

2

], b=40[kΩ/A], c=0.4[kΩ], I

0

=10mA,

I

_m

=0.1mA

ci

bi

ai

u



3



2



)

cos(

0

I

t

I

i





_m



1. DC analysis 0

I

i



SUPERPOSITION 0 2 0 3 0 0

aI

bI

cI

U

u









V

U

₀



2



10

6

(

10



10

3

)

3



40



10

3

(

10



10

3

)

2



0

.

4



10

3

(

10



10

3

)



10

Then operating point:

U0=10V

_I0=10m

A

U0=10V

I0=10m

A

1. DC analysis –dynamic resistance

c

bI

aI

c

bi

ai

di

du

R

I I d













0



2 0 2

₂

₃

₂

3

0 0





























₃

₂

₁₀

6

₍

₁₀

₁₀

3

₎

2

₂

₄₀

₁₀

3

₁₀

₁₀

3

₀

_.

₄

₁₀

3

₁

_.

₈

₁₀

3 d

R

2. AC analysis

The Ohms law !

)

cos(

)

(

'

t

I

t

i



_m



2 m I I 

V

I

R

U

m d

2

18

.

0

2

10

1

.

0

10

8

.

1

2

3 3



_







 3. Superposition

)

cos(

18

.

0

)

(

'

t

t

u





V

t

t

u

U

t

u

(

)



₀



'

(

)



10



0

.

18

cos(



)

Example



_{Tuneling diode with characteristic given at the graph is}

working in circuit system as shown below.Find the voltage

drop on resistor R.



J

₀

=4mA, J

_m

= 0.1mA, ω=10

6

rad/s, R=1kΩ, L=0.5mH,

C=2nF.

] [ ) 3 1 cos( 10 01 . 0 2 t arctg V u    cos( 0.4)[ ] 29 1 . 0 7 t arctg V u    Answer:

Istnieją dwa rozwiązania stabilne:

Trzecie rozwiązanie odpowiadające stałoprądowemu punktowi pracy (U_D=4V, I_D=4mA) jest niestabilne i należy je odrzucić.

)

cos(

0

J

t

J