Ch 06 electrical and electromechanical systems

of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien §1.Electrical Elements - Voltage and current are the primary variables used to describe acircuit’s behavior -Current: the

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6 Electrical and

Eletromechanical Systems

System Dynamics 6.01 Electrical and Electromechanical Systems

HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien

§1.Electrical Elements

- Voltage and current are the primary variables used to describe

acircuit’s behavior

-Current: the flow of electrons, the time rate of change of

electrons passing through a defined area

𝑖 =𝑑𝑄

𝑑𝑡, 𝑄 = 𝑖𝑑𝑡𝑖: current, 𝐴 𝑄: charge, 𝐶Electrons are negatively charged, the positive direction ofcurrent flow is opposite to that of the electron flow

-Voltage: energy is required to move a charge between two

points in a circuit The work per unit charge required to do this

§1.Electrical Elements1.Active and passive Elements

- Elements that provide energy aresources, and elements that

dissipate energy areloads

- Circuit elements may be classified asactiveorpassive

• Passive elements: resistors, capacitors, and inductors arenot sources of energy

• Active elements: energy sources that drive the system

- Several types of energy sources

• Chemical: batteries

• Mechanical: generators

• Thermal: thermocouples

• Optical: solar cells

- Active elements are modeled as eitherideal voltage sources

orideal current sources

Electrical quantities

-Power𝑃: work done per unit time, so the power generated by

an active element, or the power dissipated or stored by apassive element, can be calculated as

𝑝𝑜𝑤𝑒𝑟 =𝑤𝑜𝑟𝑘𝑡𝑖𝑚𝑒=

𝑤𝑜𝑟𝑘𝑢𝑛𝑖𝑡 𝑐ℎ𝑎𝑟𝑔𝑒=

𝑐ℎ𝑎𝑟𝑔𝑒𝑡𝑖𝑚𝑒 = 𝑣𝑜𝑙𝑡𝑎𝑔𝑒 × 𝑐𝑢𝑟𝑟𝑒𝑛𝑡

𝑃 = 𝑖𝑣 = 𝑖2𝑅 =𝑣

2𝑅

𝑃: power, 𝑊 ≡ 𝐽/𝑠 𝑖: current, 𝐴𝑅: resistance, Ω 𝑣: voltage, 𝑉System Dynamics 6.06 Electrical and Electromechanical Systems

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2.Modeling Circuits

- The dynamics of physical systems result from the transfer,

loss, and storage of mass or energy

- Basic laws used to model electrical systems

• Conservation of charge: Kirchhoff’s current law

• Conservation of energy: Kirchhoff’s voltage law

currents entering the node =

currents leaving the node

the sum of all the voltage drops around the loop is equal to zeroSystem Dynamics 6.07 Electrical and Electromechanical Systems

§1.Electrical Elements3.Resistances

1𝑅2

and

𝑖1= 𝑅2

𝑅1+ 𝑅2𝑖, 𝑖2= 𝑅1

𝑅1+ 𝑅2𝑖The current-divider rule

𝑖1

𝑖2=

𝑅2

𝑅1

𝑅1𝑅2

𝑑𝑖𝑑𝑣

𝑣=0

=0.16(0.12𝑒0.12𝑣)

𝑣=0= 0.0192Thus, for small voltages,𝑖 = 0.0192𝑣, and the resistance is

𝑅 = 𝑣/𝑖 = 1/0.0192 = 52ΩSystem Dynamics 6.10 Electrical and Electromechanical Systems

4.Capacitance

-Capacitance: the ability of a device to store charge for a

given voltage difference across the element

- A capacitor is designed to store charge Charge on a

𝑄0: the initial charge on the capacitor at time𝑡 = 0, 𝐶

𝑣 =1

𝐶 𝑖𝑑𝑡 𝑖 = 𝐶

𝑑𝑣𝑑𝑡

§1.Electrical Elements6.Power and Energy

- The power dissipated by or stored by an electrical element

𝑃 = 𝑖𝑣𝑃: power, 𝑊 ≡ 𝐽/𝑠 𝑖: current, 𝐴 𝑣: voltage, 𝑉

- Capacitors and inductors store electrical energy as storedcharge and in a magnetic field

• The energy 𝐸 stored in a capacitor

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Summarizes the voltage-current relations and the energy

expressions for resistance, capacitance, and inductance

elements

§2.Circuit Examples

- Ex.6.2.1 Current and Power in a Resistance Circuit

For the circuit shown in the figure, the appliedvoltage is𝑣𝑠= 6𝑉 and the resistance is 𝑅 =10Ω Determine the current and the powerthat must be produced by the power supplySolution

The current is found from

𝑖 =𝑣𝑠

𝑅=

6

10= 0.6𝐴The power is computed from

𝑃 =𝑣𝑠

𝑅 =

62

10= 3.6𝑊Note that we can also compute the power from𝑃 = 𝑖𝑣𝑠System Dynamics 6.14 Electrical and Electromechanical Systems

The figure shows a circuit for summingthe voltages𝑣1 and𝑣2 to produce 𝑣3.Derive the expression for 𝑣3 as afunction of𝑣1and𝑣2, for the case where

𝑅1= 𝑅2= 10Ω and 𝑅3= 20ΩSolution

The voltage-current relations

𝑖1=𝑣1− 𝑣3

𝑅1 , 𝑖2=𝑣2− 𝑣3

𝑅2 , 𝑖3=𝑣3

𝑅3Kirchoff’s current law gives 𝑖3= 𝑖1+ 𝑖2

- Ex.6.2.3 Application of the Voltage-Divider Rule

Consider the circuit shown in thefigure Obtain the voltage𝑣0as afunction of the applied voltage𝑣𝑠

by applying the voltage-dividerrule Use the values 𝑅1= 5Ω ,

𝑅2= 10Ω, 𝑅3= 6Ω, and 𝑅4= 2ΩSolution

Let𝑣𝐴be the voltage at the node as shown

The voltage-divider rule applied toresistors𝑅3and𝑅4gives

Resistors𝑅3and𝑅4are in series 𝑅𝑠= 2 + 6 = 8Ω

Resistors𝑅𝑠and𝑅2are in parallel

8

17𝑣𝑠=

2

17𝑣𝑠System Dynamics 6.17 Electrical and Electromechanical Systems

-Potentiometer: a resistance with a sliding electrical pick-off

• The resistance 𝑅1between the sliding contact and ground

is a function of the distance𝑥 of the contact from the end ofthe potentiometer

• Potentiometers, commonly calledpots, are used as linear

and angular position sensorsSystem Dynamics 6.18 Electrical and Electromechanical Systems

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𝐾: the gain of the pot

Assuming the potentiometer resistance𝑅1is proportional to𝑥

Derive the expression for the output voltage𝑣0as a function of𝑥Solution

The length of the pot is 𝐿 and its totalresistance is𝑅1+ 𝑅2

From the voltage-divider rule

𝑣𝑜= 𝑅1

𝑅1+ 𝑅2𝑉The resistance𝑅1proportional to𝑥: 𝑅1= 𝑅1+ 𝑅2𝑥/𝐿Substituting the above equation gives

- Ex.6.2.5 Maximum PowerTransferin a Speaker-AmplifierSystem

A common example of an electrical system is an amplifier and

a speaker

The load is the speaker, which requires current from the

amplifier in order to produce sound The resistance𝑅𝐿is that

of the load The source supplies a voltage𝑣𝑠and a current𝑖𝑆,

and has its own internal resistance𝑅𝑆 For optimum efficiency,

we want to maximize the power supplied to the speaker, for

given values of 𝑣𝑆 and 𝑅𝑆 Determine the value of 𝑅𝐿 to

maximize the power transfer to the load

SolutionFromKirchhoff’s voltage law

𝑣𝑠− 𝑖𝑠𝑅𝑠− 𝑖𝑠𝑅𝐿= 0From the voltage-divider rule

𝑣𝐿= 𝑅𝐿

𝑅𝑆+ 𝑅𝐿𝑣𝑆

The power consumed by the load

𝑃𝐿= 𝑖𝑆𝑅𝐿= 𝑣𝐿/𝑅𝐿Using the relation between𝑣𝐿 and𝑣𝑆 we can express𝑃𝐿interms of𝑣𝑆as

𝑃𝐿= 𝑅𝐿

𝑅𝑆+ 𝑅𝐿 𝑣𝑆System Dynamics 6.22 Electrical and Electromechanical Systems

To maximize𝑃𝐿for a fixed value of𝑣𝑆, we must choose𝑅𝐿to

maximize the ratio

This result for a resistance circuit is a special case of the more

general result known as impedance matching

Early in the twentieth century engineersstruggled to design vacuum-tube amplifierswhose gain remained constant at apredictable value The gain is the ratio ofthe output voltage to the input voltage Thevacuum-tube gain𝐺 can be made large but

is somewhat unpredictable and unreliabledue to heat effects and manufacturing variations A solution tothe problem is shown in the figure

Derive the input-output relation for 𝑣𝑜 as a function of 𝑣𝑖.Investigate the case where the gain𝐺 is very large

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§2.Circuit Examples1.Loop Currents

Sometimes the circuit equations can be simplified by using theconcept of aloop current, which is a current identified with aspecific loop in the circuit Use of loop currents usually reducesthe number of unknowns to be found, although when derivingthe circuit equations you must be careful to use the properalgebraic sum for each element

- Example 6.2.7 Analysis with Loop Currents

Given the values of the voltages and the resistances for the

circuit in the figure (a) Solve for thecurrents𝑖1, 𝑖2 and𝑖3 passing throughthe three resistors (b) Use the loop-current method to solve for thecurrents

Solution

a.ApplyingKirchhoff’s voltage law to the left-hand loop gives

𝑣1− 𝑅1𝑖1− 𝑅2𝑖2= 0For the right-hand loop

𝑣2− 𝑅2𝑖2+ 𝑅3𝑖3= 0From conservation of charge

𝑖1= 𝑖2+ 𝑖3Solving for the solution

b.Define the loop currents𝑖𝐴and𝑖𝐵positive clockwise

Note -voltage drop𝑅2𝑖𝐴due to𝑖𝐴

-voltage increase𝑅2𝑖𝐵due to𝑖𝐵

ApplyingKirchhoff’s voltage law to theleft-hand loop gives

𝑣1− 𝑅1𝑖𝐴− 𝑅2𝑖𝐴+ 𝑅2𝑖𝐵= 0

𝑣2+ 𝑅3𝑖𝐵+ 𝑅2𝑖𝐵− 𝑅2𝑖𝐴= 0The solution

2.Capacitance and Inductance in Circuits

The resistor and capacitor in the circuit shown

in the figure are said to be in series because

the same current flows through them Assume

that the supply voltage𝑣𝑆is known Obtain the

model of the capacitor voltage𝑣1

SolutionFrom Kirchhoff’s voltage law

𝑣𝑆− 𝑅𝑖 − 𝑣1= 0For the capacitor

𝑣1=1

𝐶 0

𝑡

𝑖𝑑𝑡 +𝑄0𝐶Differentiate this with respect to 𝑡 to obtain

𝑑𝑣1

𝑑𝑡 =

1

𝐶𝑖Then substitute for 𝑖 from the first equation

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- Ex.6.2.9 Pulse Response of a Series RC Circuit

A rectangular pulse input is a positivestep function that lasts a duration𝐷 Oneway of producing a step voltage input is

to use a switch like that shown in thefigure

The battery voltage𝑉 is constant and the switch is initially

closed at point𝐵 At 𝑡 = 0 the switch is suddenly moved from

point𝐵 to point 𝐴 Then at 𝑡 = 𝐷 the switch is suddenly moved

back to point𝐵 Obtain the expression for the capacitor voltage

𝑣1(𝑡) assuming that 𝑣10 = 0

response

𝑣1𝑡 = 𝑉(1 − 𝑒−𝑡/𝑅𝐶)When the switch is moved back to point𝐵 at time 𝑡 = 𝐷, thecircuit is equivalent to that shown in the figure, whose model isequation the same above with𝑉 = 0

The solution of this equation for 𝑡 ≥ 𝐷 is simply the free

response with the initial condition𝑣1(𝐷)

𝑣1𝑡 = 𝑣1 𝐷 𝑒−𝑡−𝐷𝑅𝐶 = 𝑉(1 − 𝑒−𝑅𝐶)𝑒−𝑡−𝐷𝑅𝐶

Pulse response of a series 𝑅𝐶 circuit

The resistor, inductor, and capacitor in thecircuit shown in the figure are in series becausethe same current flows through them Obtainthe model of the capacitor voltage𝑣1with thesupply voltage𝑣𝑆as the input

SolutionFromKirchhoff’s voltage law

𝑣𝑆− 𝑅𝑖 − 𝐿𝑑𝑖

𝑑𝑡− 𝑣1= 0For the capacitor

𝑅 and 𝐶 in the circuit shown in the figureare said to be in parallel because theyhave the same voltage𝑣1 Given supplycurrent𝑖𝑆, Obtain the model of the current

𝑖2passing through the inductorSolution

The currents𝑖1and𝑖2are defined in the figure Then

𝑣1= 𝐿𝑑𝑖2

𝑑𝑡 = 𝑅𝑖1From conservation of charge,𝑖1+ 𝑖2= 𝑖𝑆, we obtain

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- Ex.6.2.12 Analysis of a Telegraph Line

Figure shows a circuit representation of

a telegraph line The resistance𝑅 is theline resistance and𝐿 is the inductance

of the solenoid that activates thereceiver’s clicker The switch representstheoperator’s key Assume that when sending a “dot,” the key

is closed for0.1𝑠 Using the values 𝑅 = 20Ω,𝐿 = 4𝐻, obtain the

expression for the current𝑖(𝑡) passing through the solenoid

𝑣𝑖𝑡 = 1.2𝛿(𝑡)The Laplace transform of equation with𝑖(0) = 0 gives

4𝑠 + 20 𝐼 𝑠 = 1.2 ⟹ 𝐼 𝑠 = 1.2

4𝑠 + 20=

0.3

𝑠 + 5This gives the solution 𝑖 𝑡 = 0.3𝑒−5𝑡

Note that this solution gives𝑖(0+) = 0.3, whereas 𝑖(0) = 0

The difference is due to the impulsive inputSystem Dynamics 6.38 Electrical and Electromechanical Systems

- Ex.6.2.13 An𝑅𝐿𝐶 Circuit with Two Input Voltages

The𝑅𝐿𝐶 circuit shown in the figure hastwo input voltages Obtain the differentialequation model for the current𝑖3

SolutionApplyingKirchhoff’s voltage law to the left-hand loop gives

𝑣1− 𝑅𝑖1− 𝐿𝑑𝑖3

𝑑𝑡= 0For the right-hand loop

𝑣2−1

𝐶 𝑖2𝑑𝑡 − 𝐿

𝑑𝑖3

𝑑𝑡= 0Differentiate this equation with respect to𝑡

3.State Variable Models of Circuits

The presence of several current and voltage variables in a

circuit can sometimes lead to difficulty in identifying the

appropriate variables to use for expressing the circuit model

⟹ Use of state variables can often reduce this confusion

- Ex.6.2.14 State-Variable Model of a Series𝑅𝐿𝐶 Circuit

Consider the series 𝑅𝐿𝐶 circuit

Choose a suitable set of statevariables, and obtain the statevariable model of the circuit in matrixform The input is the voltage𝑣𝑠

SolutionThe energy stored in the capacitor is𝐶𝑣1/2 and the energystored in the inductor is𝐿𝑖2/2 ⟹ state variables: 𝑣1and𝑖FromKirchhoff’s voltage law

𝑑𝑡

𝑑𝑣1𝑑𝑡

=

−𝑅

1𝐿1

𝑖

𝑣1 +

1𝐿0

𝑣𝑠

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§3.Impedance and Amplifier

Determine the transfer function

𝑣𝑜(𝑠)/𝑣𝑠(𝑠) of the circuitSolution

The energy in this circuit is stored

in the two capacitorsThe energy stored in a capacitor is expressed by𝐶𝑣2/2

⟹ the state variables are the voltages 𝑣1and𝑣𝑜

The capacitance relations are

For the left-hand loop

𝑖1=𝑣𝑠− 𝑣𝑜𝑅From conservation of charge

1.Impedance

- A resistance resists or “impedes” the flow of current The

corresponding relation is 𝑣/𝑖 = 𝑅 Capacitance and

inductance elements also impede the flow of current

- In electrical systems an impedance is a generalization of the

resistance concept and is defined as the ratio of a voltage

transform𝑉(𝑠) to a current transform 𝐼(𝑠) and thus implies a

current source

- Standard symbol for impedance

𝑍(𝑠) ≡𝑉(𝑠)

𝐼(𝑠)

- The impedance of a resistor is its resistance

𝑍 𝑠 = 1𝐶𝑠

- For an inductor

𝑣 𝑡 = 𝐿𝑑𝑖

𝑑𝑡⟹ 𝑉 𝑠 = 𝐿𝐼 𝑠 𝑠The impedance of a inductor

𝑍 𝑠 = 𝐿𝑠

2.Series and Parallel Impedances

- The concept of impedance is useful because the impedances

of individual elements can be combined with series and

parallel laws to find the impedance at any point in the system

- The laws for combining series or parallel impedances are

extensions to the dynamic case of the laws governing series

and parallel resistance elements

- Series Impedances

• Two impedances are in series if they have the same

current If so, the total impedance is the sum of the

⟹𝑉(𝑠)𝐼(𝑠)≡ 𝑍 𝑠 =

𝑅𝐶𝑠 + 1𝐶𝑠and the differential equation model is

𝐶𝑑𝑣

𝑑𝑡= 𝑅𝐶

𝑑𝑖

𝑑𝑡+ 𝑖(𝑡)System Dynamics 6.48 Electrical and Electromechanical Systems

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- Parallel Impedances

• Two impedances are in parallel if they have the same

voltage difference across them Their impedances combine

by the reciprocal rule

11/𝐶𝑠+

1

𝑅⟹

𝑉(𝑠)𝐼(𝑠)≡ 𝑍 𝑠 =

𝑅𝑅𝐶𝑠 + 1and the differential equation model is

𝑅𝐶𝑑𝑣

𝑑𝑡+ 𝑣 = 𝑅𝑖(𝑡)System Dynamics 6.49 Electrical and Electromechanical Systems

- Ex.6.3.2 Circuit Analysis Using Impedance

For the circuit shown in the figure, determine the transferfunction between the input voltage𝑣𝑠and the output voltage𝑣𝑜

SolutionThe equivalent impedance𝑍(𝑠) for 𝑅 and 𝐶1

𝑍(𝑠)=

11/𝐶𝑠+

1

𝑅⟹ 𝑍 𝑠 =

𝑅𝑅𝐶𝑠 + 1

In this representation we may think of the impedance as asimple resistance, provided we express the relations inLaplace transform notation

Kirchhoff’s voltage law gives

𝑉𝑠𝑠 − 𝑅1𝐼 𝑠 − 𝑍 𝑠 𝐼 𝑠 = 0

The output voltage is related to the current by 𝑉𝑜𝑠 =

𝑍 𝑠 𝐼(𝑠) Eliminating 𝐼(𝑠) from these two relations gives

𝑉𝑠𝑠 = 𝑅1

𝑉𝑜(𝑠)𝑍(𝑠)− 𝑉𝑜𝑠 = 0which yields the desired transfer function

§3.Impedance and Amplifier3.Isolation Amplifier

- A voltage-isolation amplifier is designed to produce an output

voltage that is proportional to the input voltage

- Such an amplifier may be considered to be a voltage source

• if it does not affect the behavior of the source circuit that isattached to the amplifier input terminals, and

• if the amplifier is capable of providing the voltageindependently of the particular circuit (the“load”) attached

to amplifier output terminalsSystem Dynamics 6.52 Electrical and Electromechanical Systems

- Consider the system

𝑉𝑠𝑠 − 𝐼1𝑠 𝑍𝑠𝑠 − 𝑉𝑖𝑠 = 0, 𝐼1𝑠 = 𝑉𝑖(𝑠)/𝑍𝑖(𝑠)

• 𝑍𝑖(𝑠) is large ⟶ 𝑖1is small: the amplifier does not affect the

current𝑖1⟶ the amplifier does not affect the source circuit

• 𝑍𝑖(𝑠) is large

𝑉𝑖𝑠 = 𝑍𝑖𝑠

𝑍𝑖𝑠 + 𝑍𝑠𝑠 𝑉𝑠(𝑠) ≈ 𝑉𝑠(𝑠)

⟹ A voltage-isolation amplifier must have a high input impedance

- Denote the amplifier’s voltage gain as 𝐺 This means that the amplifier’s output voltage 𝑣𝑜is 𝑣𝑜= 𝐺𝑣𝑖 For the load circuit

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4.Operational Amplifier (Op Amp)

- Anoperational amplifieris a high-gain linear amplifier

- The op-amp approaches the ideal amplifier of the analog

designer’s dreams because it has such ideal characteristics

• Very high open-loop gain: 𝐺 = 100,000 +

• Very high input resistance: 𝑅𝑖𝑛> 1 𝑀Ω

• Low output resistance: 𝑅𝑜𝑢𝑡= 50 ÷ 75 Ω

- The output voltage

𝑉𝑜𝑢𝑡= 𝐺 𝑣2− 𝑣1

𝑣𝑜𝑢𝑡: output voltage, 𝑉 𝐺: open-loop gain

𝑣1: inverting input, 𝑉 𝑣2: non-inverting input, 𝑉

0.8 × (−𝑉𝑐𝑐) ≈ −𝑣𝑠𝑎𝑡≤ 𝑣𝑜𝑢𝑡≤ +𝑣𝑠𝑎𝑡≈ 0.8 × (+𝑉𝑐𝑐)

Determine the relation between the inputvoltage𝑣𝑖and the output voltage 𝑣𝑜ofthe op-amp circuit Assume that the opamp has the following properties

• very large gain 𝐺

From conservation of charge,𝑖1= 𝑖2+ 𝑖3≈ 𝑖2

Note that this circuit inverts the sign of the input voltage

§3.Impedance and Amplifier5.Generation Op-Amp Input-Output Relation

The impedance𝑍𝑖(𝑠) of the input elements is defined such that

𝑉𝑖𝑠 − 𝑉1𝑠 = 𝑍𝑖𝑠 𝐼1(𝑠)For the feedback elements

𝑉1 𝑠 − 𝑉𝑜𝑠 = 𝑍𝑓𝑠 𝐼2(𝑠)The high internal impedance: 𝑖3≈ 0 ⟹ 𝑖1≈ 𝑖2

The amplifier relation: 𝑣𝑜= −𝐺𝑣1⟹ 𝑉𝑜𝑠 = −𝐺𝑉1(𝑠)System Dynamics 6.58 Electrical and Electromechanical Systems

System Dynamics 6 Electrical and Electromechanical Systems

The gain of this multiplier is𝑅𝑓/𝑅𝑖, with a sign reversal

To eliminate the sign reversal, using an inverter, which is a

multiplier having equal resistancesSystem Dynamics 6.60 Electrical and Electromechanical Systems

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