Download Electrical circuits wyklad 5

Dr inż. Agnieszka Wardzińska
Room: 105 Polanka
[email protected]
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:
For I>0
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:
For I>0
Nonlinear element - methods
 Analytical
 Graphical using load line
 Graphical using graphical summing for
Kirchoffs laws
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:
For the current I=2 we calculate the voltage:
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
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)
1.
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.
 UAB=ET—I*RT
The operating point is as we
can see the same
U=2V
I=2A
Graphical using graphical summing for Kirchoffs laws
1. We draw the characteristic of the
nonlinear element
Black line
2. We draw the characteristic of parallel
connected resistor R2
Blue line
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)
Red line
4.
5.
We draw the characteristic of the R1
resistor
Brown line
We adding it to the obtained earlier sum
of characteristic (red line)
Pink line
4.
5.
We draw the characteristic of the ideal source Brown dashed line
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 dotted line) U=2V
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 RN (black curve)
I=2A.
The voltage and current of the nonlinear element RN (point P4) is the operating point
of the circuit.
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
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)
1.
 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.
For linear element:
~linear:
The incremental method (small
signal method)
Mathematical consideration
 For the nonlinear:
 We assume:
 We can use Taylor’s expansion of f(vD) near vD=VD:
=0
 We neglect the higher order terms because ΔvD
Mathematical consideration - 2
Static and dynamic resistance
Static 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.
Rs 
UD
ID
Dynamic resistance
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
du
U D
Rd 

di I 0 I D
Dynamic conductance
di
I D
Gd 

du U 0 U D
Gd 
1
Rd
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 as the
voltage rises.
Negative resistance examples
source: wikipedia
Example for small signals
 For the given nonlinear resistor described by
u  ai3  bi 2  ci
 Flows the current:
i  I 0  I m cos(t )
 Using small singal method, calculate the voltage drop.
 Given a=2[MΩ/A2], b=40[kΩ/A], c=0.4[kΩ], I0=10mA, Im=0.1mA
SUPERPOSITION
1. DC analysis
i  I0
u  U 0  aI 0  bI 0  cI 0
3
2
U 0  2 106 (10 10 3 )3  40 103 (10 10 3 ) 2  0.4 103 (10 10 3 )  10V
Then operating point:
U0=10V
I0=10mA
1. DC analysis –dynamic resistance
Rd 
du
di
 3ai 2  2bi  c
I0
 3aI 0  2bI 0  c
2
I0
Rd  3  2 106 (10 10 3 ) 2  2  40 103 10 10 3  0.4 103  1.8 103 
2. AC analysis
i' (t )  I m cos(t )
I
Im
2
The Ohms law !
3
Im
0.18
3 0.110
U  Rd
 1.8 10

V
2
2
2
u ' (t )  0.18 cos(t )
3. Superposition
u(t )  U 0  u' (t )  10  0.18 cos(t )V
Example
 Tuneling diode with characteristic given at the graph is working in
circuit system as shown below.Find the voltage drop on resistor R.
 J0=4mA, Jm = 0.1mA, ω=106rad/s, R=1kΩ, L=0.5mH, C=2nF.
j  J 0  J m cos(t )
Answer:
Istnieją dwa rozwiązania stabilne:
1
u  2  0.01 10 cos(t  arctg ) [V ]
3
0.1
cos(t  arctg 0.4) [V ]
29
Trzecie rozwiązanie odpowiadające stałoprądowemu punktowi pracy (UD=4V, ID=4mA) jest niestabilne i należy je odrzucić.
u 7