Passive components and circuits - CCP

Lecture 3

Introduction

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Index

Theorems for electric circuit analysis Kirchhoff theorems Superposition theorem Thevenin theorem Norton theorem

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Kirchhoff theoremshttp://www-groups.dcs.st-and.ac.uk/~history/Biographies/Kirchhoff.html

The theorems are applicable in circuit analysis for insulated circuits (the circuit is not exposed to external factors as electrical or magnetic fields).

Kirchhoff’s voltage law : The algebraic sum of the voltages at any instant around any loop in a circuit is zero.

Kirchhoff’s current lawThe algebraic sum of the currents at any instant at any node in a circuit is zero.

0:0: iTKIvTKV

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Applying of Applying of KirKircchhoffhhoff’s Theorem’s Theorem If a circuit has l branches and n nodes, then the complete

description of its operation is obtained by writing KVL for l-n+1 loops and KCL for n-1 nodes. The loops must form an

independent system.

Prior to the analysis of an electric circuit, the conventional directions of the currents in the circuit are not known. So, before writing the equations (Kirchhoff’s laws) for each loop, a positive arbitrary direction is selected for each branch of the circuit.

After performing the analysis of the circuit, if the value of the current is positive, the arbitrary and conventional directions of the current flow are identical. If the value of the current is negative, the conventional direction is opposite to the arbitrary selected direction.

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Applying of Applying of KirKircchhoffhhoff’s Theorem’s Theorem

Step I – choosing the voltages and currents arbitrary directions

Step II – choosing the loop’s cover direction Step III – writing the Kirkhhoff’s theorems

0

02

021

R3R2R1

R3R2

R2R1

III

VVV

VVVVR1 R2 R3

V1=5 V V2=9 V

VR1

VR2 VR3

IR1

IR3

IR2

A

B

330 150 1K

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Solving equation systemsSolving equation systems

In order to solve the equations, the Ohm’s Law is applied and the voltage across the resistors are substituted.

It is obtained a system with three equations and three variables, IR1, IR2 and IR3.

R3R3

R2R2

R1R1

3

2

1

IRV

IRV

IRV

0

0322

02211

321

32

21

RRR

RR

RR

III

IRIRV

VIRIRV

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The System SolutionsThe System Solutions

R1 R2 R3

V1=5 V V2=9 V

VR1

VR2

VR3

IR1

IR3

IR2

A

B

330 150 1K

The solutions are: IR1-6 mA IR2-13 mA IR37 mA

The voltages across the resistances: VR1-2 V VR2-2 V VR37 V

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Linear and nonlinear circuits

If the transmittances defined for a circuit are constant (are represented with linear segments in v-i, v-v or i-i planes), are called linear transmittances.

A circuit or a component with only linear transmittances is called linear circuit or linear component.

Important: generally, electronics devices and circuits made with them are nonlinear.

The method used to approximate a nonlinear circuit operation with a linear circuit operation is called linearization.

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The Superposition Theorem

The Superposition theorem states that the response in a linear circuit with multiple sources can be obtained by adding the individual responses caused by the separate independent sources acting alone.

The source passivation the sources are replaced by their internal resistance.

By passivation, the ideal voltage source is replaced with a short-circuit, and the ideal current source is replaced with an open-circuit.

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The Superposition Theorem

+

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Thevenin’s Theorem

Any two-terminal, linear network of sources and resistances can be replaced by a single voltage source in series with a resistance. The voltage source has a value equal to the open-circuit voltage appearing at the terminals of the network. The resistance value is the resistance that would be measured at the network’s terminals for passivated circuit.

The source passivation= the sources are replaced by their internal resistance

By passivation, the ideal voltage source is replaced with a short-circuit, and the ideal current source is replaced with an open-circuit.

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Thevenin’s Theorem

EQUIVALENTCIRCUITELECTRONIC CIRCUIT

R1 R2 R3

V1=5 V V2=9 V

VR3

IR3

A

B

330 150 1K

Ro R3

VR3

IR3

A

B

1K

VO=?

=?

Vo and Ro must be determined.

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Calculus of open-circuit voltage

In order to calculate the open-circuit voltage, the Kirchhoff’s theorems can be applied.

The superposition theorem will also be applied.

C IR C UIT E LE C T R O NIC

R 1 R 2

V 1=5 V V 2=9 V

A

B

330 150

Vgo l

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The superposition theorem for calculus of open-circuit voltage

V56,1121

2gol1

V

RR

RV

V19,6221

1gol2

V

RR

RV

V75,7gol2gol1golO VVVVS U B B -C IR C U IT1

R1

R2

V 1=5 V

A

B

330 150

Vgo l1

S U B B -C IR C U IT2

R1

R2

V 2=9 V

A

B

330 150

Vgo l2

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The equivalent resistance calculus The circuit is passivated.

A test voltage is applied (VTEST)

The current through the terminals is determinate (ITEST)

RO = VTEST / ITEST

21TEST

TEST RR

VI

10321

2121ECHO RR

RRRRRR

P A S S IV A TE D C IR C U IT

R 1 R 2

A

B

330 150 VT EST

IT EST

RECH

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Conclusion

CIRCUITECHIVALENT

Ro R3

VR3

IR3

A

B

1K

VO

=103

=7,75 V

From the R3 resistance point of view, the equivalent circuit will have the same effect:

V7KΩ1mA7

mA71103

75,7

3

R33R3

OR3

IRV

V

RR

VI O

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Norton’s Theorem

Any two-terminal, linear network of sources and resistances may be replaced by a single current source in parallel with a resistance.

The value of the current source is the current flowing between the terminals of the network when they are short-circuited.

The resistance value is the resistance that would be measured at the terminals of the network when all the sources have been replaced by their internal resistances.

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Norton’s Theorem

CIRCUITECHIVALENTCIRCUIT ELECTRONIC

R1 R2 R3

V1=5 V V2=9 V

VR3

IR3

A

B

330 150 1K RoR3

VR3

IR3

A

B

1K

IO

=?

=?

Io and Ro must be determined.

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Calculus of short-circuit current

In order to calculate the short-circuited current, the Kirchhoff’s theorems can be applied.

The Superposition theorem!

E LE C TR O N IC C IR C U IT

R1

R2

V 1=5 V V 2=9 V

A

B

330 150I sc

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The superposition theorem for calculus of the short-circuit current

S U B B -C IR C U IT1

R1

R2

V 1=5 V

A

B

330 150Isc1 mA15,15

1

1SC1

R

VI

mA602

2SC2

R

VI

mA15,75SC2SC1SCO IIII

S U B B -C IR C U IT2

R1

R2

V 2=9 V

A

B

330 150Isc2

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Calculus of equivalent resistance

10321

2121ECHO RR

RRRRRR

C IR C U IT P A S IV IZA T

R 1 R 2

A

B

330 150 VT E S T

IT E S T

RE C H

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Conclusion

EQUIVALENTCIRCUIT

RoR3

VR3

IR3

A

B

1K

IO

=103

=75,15 m A

From the R3 resistance point of view, the equivalent circuit will have the same effect:

mA73

V73

33

R3R3

SCO

OSCOR3

R

VI

IRR

RRIRRV

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Transfer from Thevenin to Norton equivalence Once having an equivalent circuit (Thevenin or Norton),

the other one is obtained using the relation:

O

OTheveninONorton R

VI

mA15,75103

V75,7

O

OTheveninONorton

R

VI

For previous example:

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Recommendation for individual study For the following circuit determine the current through R

resistor and the voltage across it, using: Kirchhoff’s theorem Thevenin and/or Norton equivalence (use the superposition theorem)

R3 R4

R1 R2

RV I9 V 1 mA

2 K 4 K

2 K7 K

1 K