electronics fundamentals

circuits, devices, and applications

THOMAS L. FLOYD DAVID M. BUCHLA

Chapter 8 – AC Circuits 1 of 77

AC Circuits

Alternating Voltage is a voltage that: 1. Continuously varies in magnitude 2. Periodically reverses in polarity

2 of 77

AC Circuits

Symbol for a sinusoidal voltage source.

3 of 77

AC Circuits

Sine waves

The sinusoidal waveform (sine wave) is the fundamental alternating current (ac) and alternating voltage waveform.

Electrical sine waves are named from the mathematical function with the same shape.

5 of 77

AC Circuits Generation of a sine wave

Sinusoidal voltage sources

Sinusoidal voltages are produced by ac generators and electronic oscillators. When a conductor rotates in a constant magnetic field, a sinusoidal wave is generated. N S A B C D Motion of conductor Conduc tor induced.

6 of 77

AC Circuits Sine waves Sine waves are characterized by the amplitude and period. 1. The

amplitude

or current 2. The

period

is the maximum value of a voltage is the time interval for one complete cycle. 20 V The amplitude (

A

of this sine wave is 20 V ) The period is 50.0  s 15 V 10 V 0 V 0

A

25 37.5

50.0

t

( s) -10 V -15 V -20 V

T

7 of 77

AC Circuits Sine waves The period

(T)

of a sine wave can be measured between any two corresponding points on the waveform.

T T T T A T T A

By contrast, the amplitude of a sine wave is only measured from the center to the maximum point.

8 of 77

AC Circuits Frequency

Frequency ( f ) is the number of cycles that a sine wave completes in one second.

Frequency is measured in

hertz

(Hz).

If 3 cycles of a wave occur in one second, the frequency is 3.0 Hz 1.0 s 10 of 77

AC Circuits 11 of 77

AC Circuits Period and frequency The period and frequency are reciprocals of each other.

f



T

1 and

T

 1

f

If the period is 50  s, the frequency is 0.02 MHz = 20 kHz.

(The 1/

x

key on your calculator is handy for converting between

f

and

T

.) 12 of 77

AC Circuits Sine wave voltage and current values

• •

Instantaneous value (v): Voltage or current at any point on the curve.

Peak value (V

P

a sine wave.

for voltage): The amplitude of

20 V 15 V 10 V

v V

P The peak voltage of this waveform is 20 V.

0 V 0 25

v

37.5

50.0

t

( s) -10 V -15 V -20 V 13 of 77

AC Circuits Sine wave voltage and current values

Peak to peak value: Value from positive peak to negative peak. Equation =

20 V 15 V

V PP

 2

V P I PP

 2

I P

10 V 0 V -10 V 0 25 37.5

-15 V -20 V

RMS (root mean squared) value: Is the sinusoidal wave with the same heat value as a DC voltage source (known as the effective value )

V rm s

 0 .

707

V P V p

 1 .

414

V vms I I rm s p

  0 .

707 1 .

414

I I P vms

50.0

t

( s) 15 of 77

AC Circuits Sine wave voltage and current values

V

P

= 20 volts

20 V 15 V 10 V

The peak-to-peak voltage i

s

40 V.

The rms voltage is

14.1 V.

0 V 0 -10 V -15 V -20 V

V P

 1 .

414

V rms V

PP 25

V

rms 37.5

50.0

t

( s)

V PP

 2 .

828

V rms

This is magnitude V pp 16 of 77

AC Circuits

Phase of a Sine Wave

Phase: Angular measurement that specifies the position on the sine wave relative to a reference point.

23 of 77

AC Circuits

Phase shifts

Occurs when a sine wave is shifted right or left in relation to the base/reference sine wave.

25 of 77

AC Circuits

Phase shift – Lead/Lag

Occurs when a sine wave is shifted right or left in relation to the base/reference sine wave.

A - Leads B – Lags by 45 0 A - Lags B – Leads by 30 0 26 of 77

AC Circuits Phase shift Example of a wave that lags the reference (not on guided notes) …and the equation 40 Referenc e has a negative phase shift 30 20 Peak voltage

v

= 30 V sin ( q 45 o ) 10 0 0  45  90  135  180  225  270  315  360  405  -20 -30 - 40 Notice that a lagging sine wave is below the axis at 0 Angle (  ) o 27 of 77

AC Circuits Phase shift Example of a wave that leads the reference (not on guided notes) 40 Referenc e Notice that a leading sine wave is above the axis at 0 o 30 20

v

Peak voltage = 30 V sin ( q + 45 o ) 10 - 45  0 -10 -20 -30 -40 0  45  90  135  180  225  …and the equation has a positive phase shift Angle (  ) 270  315  360  28 of 77

AC Circuits

PolyPhase power

An important application of phase-shifted sine waves is in • electrical power systems. Electrical utilities generate ac with three phases that are separated by 120 o . • 3-phase power is delivered to the user with three hot lines plus neutral. The voltage of each phase, with respect to neutral is 120 V.

120 o 120 o 120 o 0 o 29 of 77

AC Circuits Sine wave equation Instantaneous values of a wave are shown as

v

or

i

. The equation for the instantaneous voltage (

v

) of a sine wave is

v



V p

sin q where

V p

= Peak voltage q (

theta

) = Angle in rad or degrees If the peak voltage is 25 V, the instantaneous voltage at 50 degrees is 19.2 V 30 of 77

AC Circuits Sine wave equation A certain sine wave has a positive-going zero crossing at 0 ° and an peak value of 40V. Calculate its instantaneous voltage for the degrees listed below for the sine wave below.

4 0

v



V p

sin q 3 0 2 0 1 0

V p

= Peak voltage q (

theta

) = -1 0 -2 0 Angle in rad or degrees -3 0 - 4 0 0 45 ° , 125 ° , 180 ° , 220 ° ,325 ° 31 of 77

AC Circuits Phasors Phasor (aka Phase Vector): Representation of a sine wave whose amplitude (

A

) and angular frequency (

ω -

omega ) are a constant rate. 90 180 0 0 90 180 360 33 of 77

AC Circuits 34 of 77

AC Circuits Power in resistive AC circuits

A sinusoidal voltage produces a sinusoidal current

35 of 77

AC Circuits Power in resistive AC circuits

Kirchhoff’s voltage law applies to AC circuits just like DC circuits

36 of 77

AC Circuits Power in resistive AC circuits

Power in AC circuits is calculated using

RMS

values for voltage and current.

The power formulas are:

P



V I

rms rms

P P

 2

V rms R

 2

I

rms

R

The dc and the ac sources produce the same power to the bulb: 120 V dc 0 V ac or dc source 170 V p = 120 V rms 0 V

WHY?

Bulb 37 of 77

AC Circuits Power in resistive AC circuits Assume a sine wave with a peak value of 40 V is applied to a 100 W resistive load. What power is dissipated?

4 0 3 0 2 0 1 0 0 -1 0 -2 0 -3 0 - 4 0

V

rms = 0.707 x

V

p = 0.707 x 40 V = 28.3 V

P

 2

V rms R

 28.3 V 2 100 W  8 W 38 of 77

AC Circuits AC generator (alternator) • • Generators convert rotational energy to electrical energy. The armature has an induced voltage, which is connected through slip rings and brushes to a load. • The armature loops are wound on a magnetic core (not shown for simplicity).

Small alternators may use a permanent magnet Others use field coils to produce the magnetic flux. 44 of 77

AC Circuits 45 of 77

AC Circuits AC generator (alternator) • • Increasing the number of poles increases the number of cycles per revolution. A four-pole generator will produce two complete cycles in each revolution.

46 of 77

AC Circuits Output Frequency of an AC Generator

f



Ns

120 1

f – frequency (Hz) N – number of poles s - speed in RPM

4 2 3 47 of 77

AC Circuits Alternators • • In vehicles, alternators generate ac, which is converted to dc for operating electrical devices and charging the battery.

AC is more efficient to produce and can be easily regulated, hence it is generated and converted to DC by diodes.

The output is taken from the Stator coils rotor through the slip rings.

Housing Rotor Diode plate Diodes Slip rings 48 of 77

AC Circuits AC Motors

There are two major classifications of ac motors. 1. Induction motor 2. Synchronous motor

.

3. Both types use a rotating field in the stator windings

.

Induction motors work because current is induced in the rotor by the changing current in the stator. This current creates a magnetic field that reacts with the moving field of the stator, which develops a torque and causes the rotor to turn. Synchronous motors have a magnet for the rotor. In small motors, this can be a permanent magnet, which keeps up with the rotating field of the stator. Large motors use an electromagnet in the rotor, with external dc supplied to generate the magnetic field.

49 of 77

AC Circuits

Rotating the stator produces a net magnetic field

50 of 77

AC Circuits

Rotating the stator produces a net magnetic field

51 of 77

AC Circuits

Induction vs. Stator Motors

Squirrel Cage Rotor 52 of 77

AC Circuits

Induction vs.

Stator

Motors

53 of 77

AC Circuits Pulse definitions Am plitude Ideal pulses Leading (rising) edge Trailing (falling) edge Leading (falling) edge Trailing (rising) edge Baseline Am plitude Pulse width (a) Positive-going pulse Baseline Pulse width (b) Negative-going pulse 55 of 77

AC Circuits

Repetitive pulse waveforms

• • •

Periodic waveforms repeat at fixed intervals.

Pulse repetition frequency: Rate at which the pulses repeat

.

Duty Cycle – Ratio of pulse width (t w ) to the period (T)

Percent Duty Cycle =   

tw T

   100 V avg = baseline = (duty cycle)( amplitude) 57 of 77

AC Circuits

Nonsinusoidal Wave Forms

•

Define terms on page 359

•

Rise time:

•

Fall time:

•

Pulse width:

58 of 77

AC Circuits

Voltage ramps

Ramp – Linear increase or decrease in voltage or current.

Slope =

Yaxis Xaxis

 

V t or



I t

59 of 77

AC Circuits Triangular and Sawtooth waves Triangular and sawtooth waveforms are formed by voltage or current ramps (linear increase/decrease) T T Triangular waveforms have positive-going and negative-going ramps of equal duration (same slope either increasing or decreasing).

T T The sawtooth waveform consists of two ramps, one of much longer duration than the other. (unequal slopes in either direction). 60 of 77

AC Circuits

Oscilloscope

A device that traces the graph of a measured electrical signal on its screen.

63 of 77

AC Circuits

Video on Osciloscope

•

http://www.cleanvideosearch.com/media/acti on/yt/watch?v=8VEg6L2QG5o

•

The name of video is: AC vs Dc explain how to use an oscilloscope

65 of 77

AC Circuits Reading an Oscilloscope 66 of 77

AC Circuits Calculate for each wave: Period, Peak, Peak to Peak, RMS T= 3.0 ms Vp = 1250 mv Vpp = 2500 mv RMS = 833.8 mV T= 20 ms Vp = 1.5 v Vpp = 3.0 v RMS = 1.06 V T= 6000 µs; 6 ms Vp = 20.4 v Vpp = 40.8 v RMS = 14.4 V 500 mv 0.5 ms T= 30 µs Vp = 24 v Vpp = 48 v RMS = 16.97 V 6 v 300 µs 12 v 15 µs 67 of 77

AC Circuits

Selected Key Terms

Sine wave

A type of waveform that follows a cyclic sinusoidal pattern defined by the formula

y

=

A

sin q .

Alternating current

Current that reverses direction in response to a change in source voltage polarity.

Period (T)

The time interval for one complete cycle of a periodic waveform.

Frequency (f)

A measure of the rate of change of a periodic function; the number of cycles completed in 1 s.

Hertz

The unit of frequency. One hertz equals one cycle per second.

68 of 77

AC Circuits

Selected Key Terms

Instantaneous value

The voltage or current value of a waveform at a given instant in time.

Peak value

The voltage or current value of a waveform at its maximum positive or negative points.

Peak-to-peak value

The voltage or current value of a waveform measured from its minimum to its maximum points.

rms value

The value of a sinusoidal voltage that indicates its heating effect, also known as effective value. It is equal to 0.707 times the peak value.

rms

stands for root mean square.

69 of 77

AC Circuits

Quiz

1. In North America, the frequency of ac utility voltage is 60 Hz. The period is a. 8.3 ms b. 16.7 ms c. 60 ms d. 60 s 70 of 77

AC Circuits

Quiz

2. The amplitude of a sine wave is measured a. at the maximum point b. between the minimum and maximum points c. at the midpoint d. anywhere on the wave 71 of 77

AC Circuits

Quiz

5. The time base of an oscilloscope is determined by the setting of the a. vertical controls b. horizontal controls c. trigger controls d. none of the above 74 of 77

AC Circuits

Quiz

6. A sawtooth waveform has a. equal positive and negative going ramps b. two ramps one much longer than the other c. two equal pulses d. two unequal pulses 75 of 77

AC Circuits

Quiz

8. For the waveform shown, the same power would be delivered to a load with a dc voltage of 60 V a. 21.2 V 45 V b. 37.8 V 30 V c. 42.4 V d. 60.0 V 0 V 0 25 37.5

50.0

t

( s) -30 V -45 V -60 V 77 of 77

AC Circuits

Quiz

10. A control on the oscilloscope that is used to set the desired number of cycles of a wave on the display is a. volts per division control b. time per division control c. trigger level control d. horizontal position control 79 of 77