xOne Port Networks

8/2/2019 xOne Port Networks

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i-v Characteristics of

One-Port Networks

Topic 2


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The concept of a v-i characteristic need not be confined to anindividual element.

A group of elements connected together can be described by asingle equation (characteristic curve) that relates the voltage and

current variables at any one port.

Such a port can be considered as a single element with a specificv-i characteristic.

The composite characteristics of series- or parallel-connected

elements can be obtained by the method of graphical addition,except that we need to take into account the reference directionsof unilateral elements when drawing their characteristics.

Composite v-i characteristics


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Consider the circuit shown in Figure 1 where two two-terminal elementsare connected in series at node B. Nodes A and C are connected to themeasuring circuit comprising a variable voltage source, an ammeter, anda voltmeter.

Finding the Composite v-i Characteristics of Series-Connected Elements

A

C

v

i

Variable

voltage

source

Circuit

element 1

Circuit

element 2

B

v1

v2

#2

#1

Figure 1


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The one-port, consisting of elements 1 and 2, whose terminals arenodes A and C, is called the series connection of elements 1 and 2.

The two elements #1 and #2 are specified by their characteristics,which are assumed known.

A

C

Circuit

element 1

Circuit

element 2

B

v1

v2 #2

#1

1-port netwok formed by

series connection of

elements #1 and #2.

Figure 2


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To find the composite characteristic of the series connected elements,we note that the same current i flows through both elements. Theterminal voltage v is given by the sum of voltage drops v1 and v2. Thatis,

i = i1 = i2

andv = v1 + v2

A

C

v

Circuit

element 1

Circuit

element 2

B

v1

v2

Figure 3


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Thus, the composite characteristic can be obtained as follows:

Step 1. Select suitable values of current on the the i-v characteristiccurve of each element and draw horizontal lines passingthrough these selected current values to intersect the i-v

curve.

Step 2. Add the two curves on a point-by-point basis at eachselected values of current. The result is a combined curvefor the two elements.


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ExampleObtain the composite characteristics of the one-port network shown inFigure 4, which consist of a linear resistor connected in series with anideal diode (a unilateral circuit element).

1

Ideal

diode

v

i

Figure 4


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Step 1. Define the reference directions of the terminal voltage andterminal current of each circuit element.

For series-connected elements,

i. use the reference direction of the input current to define

the reference direction of the terminal current of eachelement.

ii. place the positive and negative signs of the terminalvoltage of each circuit element such that they are on thesame side as that of the input voltage.

Solution

i

1 v1

i

Ideal

diodev2

v

v1

i

Ideal

diodev2

1

Figure 5


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Step 2. Draw on the same graph the i-v characteristic curve of eachcircuit element based on their reference directions defined inStep 1.

Solution (continued)

v

i

1 A

1 V

0

1- resistor

Ideal diode

Figure 6


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Step 3. Add, at suitably selected values of terminal current, thecorresponding terminal voltage of each circuit element.

Step 4. Draw on a graph of i versus v, the composite i-v characteristiccurve of the one-port using the pairs of values of terminalcurrent and terminal voltage obtained in Step 3.

Figure 7 shows a plot of the composite characteristic curve forthe one-port circuit based on the above steps.


v

i

1 A

1 V

0

Composite

characteristic

curve

Figure 7


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ExampleObtain the composite characteristics of the one-port network shown inFigure 8, which consist of an resistor (a linear resistor) connected inseries with an ideal diode (a unilateral circuit element).

1

Ideal

diode

v

i

Figure 8


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Step 1. Define the reference directions for the terminal voltage andterminal current of each circuit element.





Solution

i

1 v1

i

Ideal

diodev2

v

v1

i

Ideal

diodev2

1

Figure 9


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v

i

1 A

1 V

0

1- resistor

Ideal diode

Figure 10


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Figure x shows a plot of the composite characteristic curve forthe one-port circuit based on the above steps.


v

i

1 A

1 V

0

Composite

characteristic

curve

-1 V

- 1 A

Figure 11


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Example 3Obtain the composite characteristics of the one-port network shown inFigure 12.

1

Ideal

diodev

i

2 V

Figure 12


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Step 1. Define the reference directions for the terminal voltage andterminal current of each circuit element.





Solution


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i

v1

i

Ideal

diodev2

1 i

1

v

v3 1 V

v1

v2

v3 1 V

i

Figure 13


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v

i

1 A

1 V battery

0

1- resistor

Ideal diode

Figure 14


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Figure 15 shows a plot of the composite characteristic curve forthe one-port circuit based on the above steps.


v

i

1 A

1 V

0

Composite

characteristiccurve

- 1 A

2 VFigure 15


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Example 4Obtain the composite characteristics of the one-port network shown inFigure 16.

Ideal zener diode

VZK = 5.6 V

v

i

ZD1

ZD2 Ideal zener diode

VZK = 9.4 V

Figure 16


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Step 1. Define the reference directions of the terminal voltage andterminal current of each circuit element.





Solution


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v

VZK = 9.4 V

VZK = 5.6 V

i

ZD1v1

ZD2v2

VZK = 5.6 V

i

ZD1v1

VZK = 9.4 V

i

ZD2v2

Figure 17


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v

i

0

ZD1

-5.6 V

Figure 18


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v

i

9.4 V

0

ZD2

Figure 18


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Step 3. Add, at suitably selected values of terminal current, thecorresponding terminal voltages of each circuit element.


Figure x shows a plot of the composite characteristic curve forthe one-port circuit based on the above steps.


v

i

-5.6 V

0

Compositecharacteristic

curve

9.4 V

Figure 19


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Finding the Composite characteristic curve of parallel-connected elements

Any two elements are called in parallel if and only if:

i. One terminal of each element is connected to a common node

ii. The other terminal of each element is connected to another commonnode.

When any two elements are connected in parallel, it follows that the voltageacross one is identical to the voltage across the other. Also, the totalcurrent of the combination is equal to the sum of the two individualelements.

i = i1 + i2

v = v1 = v2#1 #2

i

i1 i2

v

A

B

v1 v2

Figure 20


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Worked ExampleDraw the composite characteristics for the parallel-connected solarcells shown in Figure 21a. The solar cells are identical and their i-vcharacteristic is shown in Figure 21b. Determine its maximum outputvoltage and maximum output current.

i

v

(a)

v

i

1 V

-10 mA

0

i

v

(b)Figure 21


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For the parallel connected solar cells, the terminal current is the sumof the currents contributed by each cell; that is

i = i1 + i2 + i3 + i4

At v = 0 V,

i1(max) = i2(max) = i3(max) = i4(max) = 10 mA

Hence, at v = 0 V

i(max)

= i1(max)

+ i2(max)

+ i3(max)

+ i4(max)

= 10 mA + 10 mA + 10 mA + 10 mATherefore,

i(max) = 40 mA

Solution


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v

i

1 V

40 mA

0

i

v


The composite i-v characteristic is as shown in Figure 22b.

Figure 22

(a)

(b)


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Drill ExerciseDraw the composite characteristics for the circuit shown in Figure 23.

3 2 v

iA

B

Figure 23


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Composite characteristic of series-parallel elements

#1

A

B

#2

#3

The composite characteristic of a 1-port comprising of series- and parallel-connected elements can be found in a few steps by replacing eachparallel and series-connected elements with their circuit equivalents.

Figure 24

C


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#1

i

i1

A

v1 #2

i2

v2va

C

A

C

#A


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A

B

v

Circuit

element A

Circuit

element 3

C

va

v3

#1

A

B

#2

#3

C


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#1

i

i1

v

A

B

v1 #2

i2

v2

#3

i3

v3


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R1= 1

v

i

L1 L2

10

v (V)

i (A)

00

10 20

i

v 5

Example

Using the circuit of Figure xa and the incandescent lamp characteristicshown in Figure xb, obtain by graphical method the driving pointcharacteristic of the circuit. Assume the two incandescent lamps haveidentical i-v chacteristics.


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Solution

The incandescent lamps are bilateral devices and their voltage andcurrent reference directions are immaterial in this problem. Thus, wecan simply proceed to do graphical addition in a straight-forwardmanner.

i = i1 + i2

v = v1 = v2#1 #2

i

i1 i2

v

A

B

v1 v2

Step 1. Use graphical addition of parallel-connected elements tofind the equivalent i-v characteristic of the parallel-connectedlamps.


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v

VZK = 9.4 V

VZK = 5.6 V

i

ZD1v1

ZD2v2

VZK = 5.6 V

i

ZD1v1

VZK = 9.4 V

i

ZD2v2

Step 3. Use graphical addition of series-connected elements

to obtain the composite characteristic.


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1 k

RL= 1 kv

i

v

i

i

v- 5 V

ExampleConsider the one-port network shown in Figure x. The zener diode ismodeled by a three-segment piecewise-linear characteristic shownin Figure xb. Using the graphic method only, find the driving pointcharacteristic of the one-port network.


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Drill ExerciseAssuming an ideal zener diode characteristic with VZ = 5 V, find by using thegraphical method the driving point characteristic of the one-port N shown inFigure x.

1 k

E = 10 V

i

v


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Composite characteristic of linear One-Port networksFor linear one port networks, it is possible to obtain the i-v characteristicentirely by analytical method. The following examples illustrates the idea.

3

3 volts

v

iA

B

Worked Example

Figure x show a one-port network consisting of two linear elementsconnected in series. Draw the driving point characteristics for thecircuit shown in Figure x.


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ExampleFigure x show a one-port network consisting of two linear elementsconnected in parallel. Derive the equation describing the following one-port in terms of the port voltage and port current and plot it graphicallyin the i-v plane.

1 A2 v

iA

B


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2

4

6

8

10

i (ma)

v (V)0 1 2 3 4 5 6 7 8

6.67 V

i-v characteristic

of the 1-port

network


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2

4

6

8

10

i (ma)

v (V)0 1 2 3 4 5 6 7 8

6.67 V

i-v characteristic

of the 1-port

network


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For the circuits shown in Figure S2,i. derive the equation describing the one-port in terms of the port

voltage and the port current.ii. Plot the i-v characteristic in the i-v plane.

Example

10 V

A

1 k

2 k v

i

B


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10 V

A

1 k

2 k

i2

v

i

B

i1

Step 2. Derive the circuits current equation by applying KCL at node A.

21 iii

Solution

Step 1. Label all branch currents.

Step 3. Apply Ohms law to the 2 k and 1 k resistors.

1000

10

1

1

v

R

vEi

20002

2

v

R

vi


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Step 4. Substitute Eqs.() and () into Eq.().

20001000

1021

vviii

2000

20

2000

3

vi

Step 5. Simplify expression.

Upon simplifying Eq.(), we obtain the required i-v characteristicequation of the one-port as follows:


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Step 6. Plot the i-v characteristic equation in the i-v plane.

We note that since Eq.() represents that of a straight line, we caneasily plot the characteristic equation by simply finding its x- and y-intercepts and then connecting the two points on the graph with a

straight line.

Coordinates of the x-intercept is obtained by setting v = 0 in Eq.().Thus,

mA102000

20

0

vi

Similarly, coordinates of the y-intercept is obtained by setting i = 0in Eq.(). Thus,

V67.63

20

0

iv


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2

4

6

8

10

i (ma)

v (V)

0 1 2 3 4 5 6 7 8

6.67 V

i-v characteristic

of the 1-portnetwork

A plot of the i-v characteristic of the one-port is shown in Figure x.


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The one-port network shown in Figure x contains both ac and dcvoltage sources.

i. Derive the i-v characteristic of the network for the given voltageand current reference directions.

ii. Plot the i-v characteristic in the i-v plane.

Example

1 k

vs(t) = 10sin 10t

i

v


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Solution

Step 1. Apply KVL to the circuit to obtain an equation that relates

the terminal voltage and the terminal current to the circuit elements.

0510sin10 itv

or

tvi 10sin105

1

Step 2. Plot the i-v characteristic in the i-v plane.

We note that Eq() is that of a straight line with slope -5 and ay-intercept that is a function of time.


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2

4

6

8

10

i (A)

v (V)0 20 40 60

i-v characteristic of the 1-port

network at t = 3/20 s.

-10

- 8

- 6

- 4

- 2

-20-40

vi5

1(i) At t = 0,


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2

4

6

8

10

i (A)

v (V)0 20 40 60

i-v characteristic of the 1-portnetwork at t = /20 s.

-10

- 8

- 6

- 4

- 2

-20-40

105

1 vi(ii) At t = /20,


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2

4

6

8

10

i (A)

v (V)0 20 40 60

i-v characteristic of the 1-port

network at t = 3/20 s.

- 8

- 6

- 4

- 2

-20-40

105

1 vi(iii) At t = 3/20,

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