1 Electronics AC Waveforms Copyright © Texas Education Agency, 2014. All rights reserved.

1

Electronics

AC Waveforms

Copyright © Texas Education Agency, 2014. All rights reserved.

2

Specific Objectives Understand how a sine wave of alternating

voltage is generated. Explain the three ways to express the amplitude

of a sinusoidal waveform and the relationship between them.

Calculate the RMS, average, and peak-to-peak values of a sine wave when the peak value is known.

Calculate the instantaneous value of a sine wave. Convert peak, peak-to-peak, average, and RMS

voltage and current values from one value to another.


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Specific Objectives (Continued) Explain the importance of the .707 constant and

how it is derived. Define frequency and period and list the units of

each. Calculate the period when the frequency is

known and frequency when period is known. Explain the sine, cosine, and tangent

trigonometric functions. Calculate the value of the sine of any angle

between 0° and 360°.


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Introduction AC and the characteristics of a sinusoidal

waveform Time and frequency measurement of a

waveform Trigonometric functions


What is AC?

Alternating Current (AC) is a useful form of voltage. AC is a flow of electric charge that periodically

changes direction. Direct Current (DC) is another useful form of

voltage. DC is the unidirectional flow of electric charge.

AC and DC are both useful power sources for heating, lights, and motors.


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AC vs DC

DC can power the same things AC can power. The devices may have to be constructed

differently, but they would work the same. AC cannot power some of the things that DC

can power. Example: electronic devices AC motors were not invented at the time Thomas

Edison built his first DC power station in 1882.


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Why Use AC Instead of DC?

AC is relatively easy to produce. It is created through rotational motion.

AC generation can produce large amounts of power economically.

AC voltage can be changed from one value to another relatively easily. DC voltage cannot be changed from one value to

another easily.


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Why Change Voltage?

AC voltage can be stepped up to a higher value using a transformer (up to 760 kV).

The basic concept is that “power in” equals “power out” of a transformer.

Power is voltage times current, so if voltage is made to increase, then current will decrease.

After the voltage is increased, AC is much more efficient to transmit over long distances.


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Why is AC More Efficient?

Electrical power (voltage and current) is sent over transmission wires from a generator to users.

Lower current sent through those transmission wires will produce less heat (Joule’s Law). Fewer moving electrons reduce the amount of friction. Voltage does not create heat.

Less heat means less power is lost during delivery.

We can use thinner wires for power transmission. Thin wires are more economical.


How Do You Make AC?

Most electricity is produced by induction. In a generator, induction occurs when a

conductor moves through a magnetic field. AC is the type of electricity generated by a

conductor moving in a circle through a static magnetic field. Circular motion is easy to produce. Like a wheel going around, it is a simple and efficient

process.Copyright © Texas Education Agency, 2014. All rights reserved.

Rotary Motion

The process of using a water wheel as a mechanical power source for milling and sawing has been used for thousands of years.

A water wheel produces circular motion. This is also called a water-driven turbine.

This process was applied to creating electricity in the late 1800s.

Two more things were necessary to make AC power viable: the alternator and the transformer.


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Mechanical Power Source An electrical power plant has a capacity, but

the actual amount of power produced is a function of user demand. Higher user demand creates a larger load on the

generator. The generator then needs to draw more

mechanical power from the prime mover. If the prime mover cannot provide the additional

mechanical power, the plant will shut down.


Hydroelectric Power The Ames Hydroelectric Generating Plant was

the world's first commercial system to produce and transmit AC electricity for industrial use.

In 1890, Westinghouse Electric supplied the station's generator and motor.

The AC was proven to be effective as it was transmitted two miles (3 km) at a loss of less than 5%. The maximum distance for DC was about a mile.


AC Generation

In an AC generator, one rotation of the rotor shaft creates one cycle of voltage.

This voltage is not steady over the cycle, it changes and reverses polarity depending on the direction of motion of the conductor through the magnetic field. The magnetic field goes in a line from the north pole to

the south pole. The rotor that contains the conductors move in a circle

through the linear magnetic field.Copyright © Texas Education Agency, 2014. All rights reserved.

15

AC Generation

Electricity is produced by induction. Induction occurs when a conductor cuts

through a magnetic field line. A conductor must move perpendicular to the

magnetic field line to cut through it. Conductor motion parallel to the magnetic

field does not cut through any magnetic field lines.


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Sine Wave Generation

During a rotation, the motion of a conductor changes from perpendicular to parallel. Perpendicular is a 90° angle. Parallel is a 0° (or 360°) angle.

Between these examples, the conductor motion is at some other angle relative to the direction of the magnetic field. Rotation goes through 0 to 360 degrees and then

repeats.Copyright © Texas Education Agency, 2014. All rights reserved.

17

Voltage Amplitude

The amount of voltage produced at any point is proportional to the sine of the angle of motion relative to the direction of the magnetic field lines.


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Voltage is a Function of Angle

There is a magnet in a generator. Field lines go from north to south.


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A conductor is placed in the magnetic field.

This is actually a single conductor that loops back through the magnetic field.


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Arrows show the direction of motion for the conductor in this position.

This motion is perpendicular to the direction of the field lines (at 90°).


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The sine of a 90° angle equals 1.

This represents the maximum or peak voltage out of the generator.


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We are here on the sine wave.

This represents the maximum or peak voltage out of the generator


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Here is the position of a conductor after a rotation of 45° (actual angle = 135°).

The motion is at an angle of 45° to the direction of the magnetic field lines.


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The sine of 45° (or 135°) is 0.707. We are here on the sine wave.

The amount of voltage produced at this angle is 0.707 of the peak voltage.


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This shows a conductor after another rotation of 45° from the previous example.

This motion is parallel to the magnetic field lines and represents an angle of 180°.


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The sine of 180° (or 0°) is zero. We are here on the sine wave.

At this instant the voltage produced by the generator is zero volts.


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The conductor rotates another 45°. Polarity starts to reverse.

This motion is now down through the magnetic field lines (the opposite direction).


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The Sine of 225° is 0.707. We are here on the sine wave.

Polarity is opposite because the direction of motion is going the opposite way.


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Here is the position of the conductor after another 45° rotation.

This more clearly shows that the direction of motion (red circle) is down.


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The sine of 270° is negative one. We are here on the sine wave.

This is the negative peak with equal but opposite amplitude of the positive peak.


AC Generation Example


To see animation, copy the link into your computer browser. http://commons.wikimedia.org/wiki/File:Dynamo.wechsel.wiki.v.1.00.gif

Waveform Plot At each point in the rotating cycle, the amount of voltage

produced is equal to the sine of the angle created by the direction of motion of the conductor relative to the direction of the magnetic field.

This produces the changing voltage used for electricity known as the sine wave.

As the angle between the conductor and the magnetic field changes, the voltage changes.

Peak

RMS


33Copyright © Texas Education Agency, 2014. All rights reserved.

To see animation, copy link into your computer browser.https://phet.colorado.edu/en/simulation/generator

https://phet.colorado.edu/en/simulation/generator

34

Common Terms A plot of the voltage vs time produces a wave

form in the shape of a sine wave. This wave form has many terms we need to learn. There are terms for the amplitude or size of the

wave: Peak, peak to peak, RMS, instantaneous, and average

voltage There are terms for how fast the wave changes:

Frequency, period


https://phet.colorado.edu/en/simulation/generator

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Waveform Values Graphically

The angle represents time. An angle is used because a time measurement would change with a change in frequency.

0.637a

VAVG


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Angular Measurement

We use an angle to measure instantaneous voltage values at an instant of time.

At a given angle, the relative amplitude at that point is the same for any sine wave regardless of the frequency.

A sine wave is created due to a conductor moving in a circle through a magnetic field. A circle always has 360 degrees. A sine wave always has 360 degrees for one cycle.


37

Radians

There is another way to measure the angle of a circle or sine wave.

It is called the radian measurement because of the relationship between the radius and the circumference of a circle.

There are 2π radians in 360 degrees.C = 2πr


Degrees to Radians


39

Time-based Waveform Terms Wave- a disturbance traveling through a

medium. For AC electricity, the movement of the electrons

back and forth in the wire Waveform- a graphic representation of a

wave. Waveform depends on both movement and time. Example: ripple on the surface of a pond A change in the vertical dimension of a signal is a

result of a change in the amount of voltage.


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Frequency (f)- the number of cycles of the waveform that occur in one second of time. Measured in hertz (Hz)

Period (P)- the time required to complete one cycle of a waveform. Measured in seconds, tenths

of seconds, milliseconds, or microseconds


Time-based Waveform Terms

Frequency and Period

There is an inverse relationship between frequency and period.

Example: A frequency of 100 Hz gives a period of 0.01 sec (10 ms)

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f=1P

P=1fand


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Waveform Terminology Amplitude- height of a wave

Expressed in one of the following methods Peak (PK, pk, or Pk) Peak-to-peak (P-P, PP, or pp) Root-mean-square (RMS) Average (AVG)


43

Waveform Amplitude Specification Peak- the maximum positive or negative

deviation of a waveform from its zero reference level. Sinusoidal waveforms

are symmetrical. The positive peak value

of sinusoidal will be equal to the value of the negative peak.

Measured at an instant of time


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Waveform Amplitude Specification (continued Peak-to-peak is the measurement from the

highest amplitude peak to the lowest peak. Sinusoidal waveform

If the positive peak value is 10 volts in magnitude, then the negative peak is also 10; therefore, peak-to- peak is 20 volts.

Non-sinusoidal waveform It is determined by adding

magnitude of positive and negative peaks.


45

Waveform Amplitude Specification (Cont’d.) Root-mean-square (RMS)

Measured over one (or more) cycles of the wave. Allows the comparison of AC and DC circuit values. RMS values of AC create the same heat as that

same numerical voltage value of DC.

RMS is most common method of specifying the value of sinusoidal waveforms.

Almost all voltmeter and ammeters are calibrated so that they measure AC values in terms of RMS amplitude.


VRMS = VDC = heat

46

RMS Heating Effect VRMS = 0.707 Vpeak

IRMS = 0.707 Ipeak

A sinusoidal voltage with peak amplitude of 1 volt has the same heating effect as a DC voltage of 0.707 volts. AC creates slightly more heat than DC. Comparing RMS value to average voltage.

Due to this, the RMS value of voltage is also referred to as the effective value.


47

Determining the 0.707 Constant

To determine the 0.707 constant, you must use the mathematical procedure suggested by the name, root-mean-square.

This has nothing to do with the sine function even though this is the same value as sin 45°.


VRMS =

48

Average Voltage

Average voltage is the DC equivalent voltage. The average voltage of AC over a full sine wave

equals zero. The positive half cycle is equal to the negative half

cycle. Because of this we only look at half the wave

(the positive half cycle). This means we are looking at an angle of π radians.


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Average Voltage

Average voltage over a half cycle:

Or, VAVG = .637 VPK VAVG is produced by rectifying then filtering the

AC in a voltage regulator. This is the process used in a DC power supply.

VAVG = =


50

Instantaneous Voltage

This is an important value when sampling a wave at several points. Example: analog to digital conversion

Different types of waves might have the same peak or average voltage but different instantaneous voltages Examples: square wave, triangle wave


51

Instantaneous Voltage

Instantaneous voltage is the voltage at a single point or instant of time.

To find the angle, take the inverse sine. Usually a 2nd function on a calculator

= sin θ

( )


52

Relationship Circles

P – P RMS

2 PK 0.707 PK

INST. AVG.

Sine° PK 0.637 PK


53

Relationship CalculationsEXAMPLES120 VAC = 170 Vpk

Formula: PK = RMS 0.707120 0.707 = 169.7 (round off to 170 Vpk)

18 V @ 72 = 19 VpkFormula: PK = Instantaneous Sine18 Sine(72°) = 18.9 (round off to 19)

30 Vpk = 21.2 VACFormula: RMS = 0.707 X PK0.707 X 30 = 21.2 350 V @ 23.5 = 30 Vpk

Formula PK = Instantaneous Sine350 Sine(23.5°) = 877.7(round off to 878)

50 Vpp = 17.7 VrmsStep 1: Need to find PKFormula: PK = P-P 250 2 = 25mStep 2: Find RMSFormula: RMS = 0.707 X PK0.707 X 25 = 17.675 (round off to 17.7 Vrms)

Find the angle with 454 V instantaneous and a PK of 908 V Formula: Sine (θ) = Instantaneous PK454 908 = 0.5 2nd Sine (.5) = 30

20 V Average = 22.2 Vrms= 31.4 Vpk = 62.8 Vp-pStep 1: Find PK, Formula: PK = Average 0.637 = 20 0.637 = 31.39 (round off to 31.4)Step 2: Find RMS, Formula: RMS = 0.707 X PK = 0.707 X 31.4 = 22.19 (round off to 22.2)Step 3: Find P-P, Formula: PK = 2 X PK= 2 X 31.4 = 62.8


54

Calculate peak voltage given RMS voltage

VRMS = 120 V Formula: PK = RMS 0.707 120 0.707 = 169.7

(round up to 170 Vpk) 120 VRMS = 170 Vpk

P – P RMS

2 PK 0.707 PK

INST. AVG.

Sine° PK 0.637 PK


55

Calculate RMS voltage given peak voltage

Vpk = 30 V Formula: RMS = 0.707 X PK 0.707 X 30 = 21.2 V 30 Vpk = 21.2 VRMS

P – P RMS

2 PK 0.707 PK

INST. AVG.

Sine° PK 0.637 PK


56

Calculate RMS given p-p

Vpp = 50 V Step 1: Find Vpk

Formula: PK = p-p 2 50 2 = 25 V

Step 2: Find RMS Formula: RMS = 0.707 X PK 0.707 X 25 = 17.675

(round off to 17.7 Vrms) 50 Vpp = 17.7 Vrms

P – P RMS

2 PK 0.707 PK

INST. AVG.

Sine° PK 0.637 PK


57

Calculate VRMS, Vpk, and Vpp given VAVG

VAVG = 20 V Step 1: Find Vpk,

Formula: PK = Average 0.637 = 20 0.637 = 31.39

(round up to 31.4) Step 2: Find RMS,

Formula: RMS = 0.707 X PK = 0.707 X 31.4 = 22.19 (round up to 22.2) Step 3: Find P-P,

Formula: P-P = 2 X PK = 2 X 31.4 = 62.8Copyright © Texas Education Agency, 2014. All rights reserved.

P – P RMS

2 PK 0.707 PK

INST. AVG.

Sine° PK 0.637 PK

58

Calculate peak voltage from instantaneous voltage

VINST = 18 V @ 72° Formula: PK = Instantaneous Sine

of the angle = 18 Sine(72°) = 18 .951 = 18.9 V

(round up to 19) 18 V @ 72 = 19 Vpk

P – P RMS

2 PK 0.707 PK

INST. AVG.

Sine° PK 0.637 PK


59

Calculate peak voltage from instantaneous voltage

VINST = 350 V @ 23.5° Formula: PK = Instantaneous

Sine(angle) = 350 Sine(23.5°) = 350 0.4 = 877.7

(round up to 878) 350 V @ 23.5 = 878 Vpk

P – P RMS

2 PK 0.707 PK

INST. AVG.

Sine° PK 0.637 PK


60

Calculate phase angle given instantaneous voltage and peak voltage

454 VINST at unknown angle with a Vpk of 908 V Formula: Sine θ =

Instantaneous PK 454 908 = 0.5 2nd Sine (.5) = 30


P – P RMS

2 PK 0.707 PK

INST. AVG.

Sine° PK 0.637 PK

61

Relationship Exercise

# rms peak pk-to-pk average instantaneous

200mV ________ ____ V @ 72º113V ________ V @ 90º

96.4 ________ V @ 235º1.5V @ 122º

689V ________ V @ 35º Go to Answers


62

Sine Wave and Sine Trigonometric Function The term sinusoidal has been used to describe a

waveform produced by an AC generator. The term sinusoidal comes from a trigonometric

function called the sine function. Sines, cosines, and tangents are numbers equal

to the ratio of the lengths of the sides. The sine function is used in AC because the

opposite side is the direction of motion of the conductor through a magnetic field relative to the angle.


63

Right-Triangle: Side and Angle Relationships A right triangle has a 90° angle. Each side is named with respect to the angle

you are using (called the angle theta, or θ). The side of the triangle across from the angle

theta is called the opposite side. The longest side of a right triangle is called the

hypotenuse. The remaining side is called the adjacent.

Each of these sides are commonly abbreviated to their initials, O, H, and A.


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Right Triangle

Hypotenuse

Opposite

Adjacent


θ

65

Basic Trigonometric Functions

In trigonometry, there are three common ratios used to study right triangles.

1. SineA. The sine of the angle theta is equal to the ratio formed by

the length of the opposite side divided by the length of the hypotenuse.

B. Sine θ =


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Basic Trigonometric Functions(Continued)

2. CosineA. The cosine of the angle theta is equal to the ration formed by

length of the adjacent side divided by the length of the hypotenuse.

B. Cosine θ = 3. Tangent

A. The tangent of the angle theta is equal to the ratio formed by length of the opposite side divided by the length of the adjacent side.

B. Tangent θ =


Simple Memory Aid

Remember the acronym SOH-CAH-TOA Pronounced “sock ah toa”

Sine equals opposite over hypotenuseCosine equals adjacent over hypotenuseTangent equals opposite over adjacent

67Copyright © Texas Education Agency, 2014. All rights reserved.

68

Basic Trigonometric Functions (Continued)

Opposite

Sine Hypotenuse

Adjacent

Cosine Hypotenuse

Opposite

Tangent Adjacent


69

Trigonometric ExerciseTriangle #1

Hypotenuse? 8’ Opposite

10’ AdjacentWhat is the Hypotenuse? ___________

Triangle #2

Hypotenuse? 5.3 Rods Opposite

6.8 Rods AdjacentWhat is the Hypotenuse? ___________

Triangle #3

125 Miles Hypotenuse Opposite?

85 miles AdjacentWhat is the Opposite? ____________

Triangle #4

56’ Hypotenuse 23.2’ Opposite

Adjacent?What is the Adjacent? _____________

Go to Answers

End of

Presentation


70

Answer Keys

Follow This Slide

End of

PresentationGo to Answers


71

Relationship Key

# rms peak pk-to-pk average instantaneous

1. 200mV 283mV 566mV 180mV 269mV @ 72º2. 80v 113V 226V 72V 113V @ 90º3. 96.4 136V 272V 87V 111V @ 235º4. 1.25V 1.77V 3.54V 1.13V 1.5V @ 122º5. 764µV 1.08mV 2.16mV 689V 619µV @ 35º

Go to the Trigonometric

Functions


72

Trigonometric Exercise KeyTriangle #1

Hypotenuse? 8’ Opposite

10’ AdjacentHypotenuse = 12.8’

Triangle #2

Hypotenuse? 5.3 Rods Opposite

6.8 Rods AdjacentHypotenuse = 8.62 Rods

Triangle #3

125 Miles Hypotenuse Opposite?

85 miles AdjacentOpposite = 91.7 miles

Triangle #4

56’ Hypotenuse 23.2’ Opposite

Adjacent?Adjacent = 51’

End of

Presentation

Go to the Calculations


73

Trigonometric Exercise Calculations

Triangle #1 Given Adjacent side = 10’ and Opposite side = 8’. What is the Hypotenuse?

Step 1 - Find the degree angleTangent = Opposite Adjacent

= 8_ = 0.8 10 = enter 2nd tangent (0.8) on your calculator = 38.65980825 round up to 38.66º = 38.66°

Step 2 - Change the degree angle to cosineHypotenuse = Adjacent (take 38.66°, enter cosine on your calculator, your answer is 0.780866719)

Cosine = 10’____________ .780866719 = 12.8’


74

Trigonometric Exercise Calculations (Continued)

Triangle #2Given Adjacent side = 6.8 rods and Opposite side = 5.3 rods. What is the Hypotenuse?

Step 1 - Find the degree angleTangent = Opposite

Adjacent = 5.3 rods

6.8 rod enter 2nd tangent 0.779411765 on your calculator

= 37.93°Step 2 - Change the degree angle to sineHypotenuse = Opposite Sine

= 5.3 rods sine 37.93 (enter 37.93, enter sine on the calculator ) = 5.3 rods

0.6146982793 = 8.63


= 0.779411765

75

Trigonometric Exercise Calculations (continued)

Triangle #3Given Hypotenuse side = 125 miles and Adjacent side = 8.5 miles What is the Opposite side?

Step 1 - Find the degree angleCosine = = 0.68

enter 2nd function button on calculator enter cosine 0.68 = 47.15635696 (round off to 47.16º)

Step 2 - Change the degree angle to sineOpposite = Sine x Hypotenuse

enter sine 47.16 on calculator = 0.7332553462 = 0.7332553462 x 125 miles = 91.65691828 (round off to 91.7) = 91.7 miles


=

76

Trigonometric Exercise Calculations (continued)

Triangle #4Given Hypotenuse side = 56’ and Opposite side = 23.2’ What is Adjacent side?

Step 1 - Find the degree angleSine = = 0.4142857143 = enter 2nd sin 0.4142857143 = 24.47ºStep 2 - Change the degree angle to cosineAdjacent = Cosine x Hypotenuse

enter cosine 24.47 on the calculator = 0.910178279 x 56’ = 50.96998362 (round off to 51’) = 51’


=

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