circuitanimations - ibiblio this worksheet and ... the purpose of this animation is to let students...

Circuit animations

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Resources and methods for learning about these subjects (list a few here, in preparation for yourresearch):

1

Question 1

Animation: simple circuit with switch

This question consists of a series of images (one per page) that form an animation. Flip the pages withyour fingers to view this animation (or click on the ”next” button on your viewer) frame-by-frame.

The following animation shows a simple circuit with one battery, two switches, and a resistor. Watchwhat happens when both switches are closed, and when either switch opens. Here are some things to lookfor:

• How do we define what an ”open” switch is?• What is opposite of ”open” for an electric switch? Hint: it isn’t ”shut”• Where does a voltage ”drop” appear?• Which direction do the electrons move?• Compare the effects of the two switches: does one switch have any more effect on the circuit’s current

than the other? If so, which one?

2

Direction of electron motion

3


4


5


6


7


8


9


10


switch open!

11


switch open!

12


switch open!

13


switch open!

14


15


16


17


18


19


20


21


22


switch open!

23


switch open!

24


switch open!

25


switch open!

file 03231

26

Answer 1

Note how the opening of either switch is sufficient to halt current throughout the entire circuit. Notealso how a voltage drop appears across the greatest circuit resistance (and the battery terminals, of course).

Notes 1

The purpose of this animation is to let students study the behavior of this switch circuit and reachtheir own conclusions. Similar to experimentation in the lab, except that here all the data collection is donevisually rather than through the use of test equipment, and the students are able to ”see” things that areinvisible in real life.

27

Question 2

Animation: Soldering a wire to a lug


The following animation shows the procedure for properly soldering a wire to a lug.

28

Soldering iron

Solder

LugWire

29

Soldering iron

Solder

LugWire

30

Soldering iron

Solder

LugWire

31

Soldering iron

Solder

LugWire

32

Soldering iron

Solder

LugWire

33

Soldering iron

Solder

LugWire

34

Soldering iron

Solder

LugWire

35

Soldering iron

Solder

LugWire

36

Soldering iron

Solder

LugWire

37

Soldering iron

Solder

LugWire

38

file 03450

68

Answer 2

Nothing to note here.

Notes 2

The purpose of this animation is to let students see proper soldering technique: applying heat to the”work” and not the solder directly is the most important element of the technique.

69

Question 3

Animation: Applying Thevenin’s theorem


The following animation shows the steps involved in ”Thevenizing” a circuit.

70

RloadR1

R2

R3

18 V

10 kΩ

12 kΩ

14 kΩ

71

RloadR1

R2

R3

18 V

10 kΩ

12 kΩ

14 kΩ

This is our original circuit:

72

RloadR1

R2

R3

18 V

10 kΩ

12 kΩ

14 kΩ

We may use Thevenin’s theoremto simplify this portion of the circuit . . .

73

Rload


R1

R2

R3

18 V

10 kΩ

12 kΩ

14 kΩ

74

Rload


R1

R2

R3

18 V

10 kΩ

12 kΩ

14 kΩ

75

Rload


R1

R2

R3

18 V

10 kΩ

12 kΩ

14 kΩ

76

RloadR1

R2

R3

18 V

10 kΩ

12 kΩ

14 kΩ

77

RloadR1

R2

R3

18 V

10 kΩ

12 kΩ

14 kΩ

To this Thevenin equivalent circuit . . .

78

RloadR1

R2

R3

18 V

10 kΩ

12 kΩ

14 kΩ

To this Thevenin equivalent circuit . . .

VTH

RTH

79

RloadR1

R2

R3

18 V

10 kΩ

12 kΩ

14 kΩ

VTH

RTH

. . . to which we may attachthe same load and analyze.

80

RloadR1

R2

R3

18 V

10 kΩ

12 kΩ

14 kΩ

VTH

RTH


Rload

81

RloadR1

R2

R3

18 V

10 kΩ

12 kΩ

14 kΩ

VTH

RTH


Rload

82

RloadR1

R2

R3

18 V

10 kΩ

12 kΩ

14 kΩ

VTH

RTH


Rload

83

RloadR1

R2

R3

18 V

10 kΩ

12 kΩ

14 kΩ

VTH

RTH


Rload

84

RloadR1

R2

R3

18 V

10 kΩ

12 kΩ

14 kΩ

VTH

RTH

Rload

85

RloadR1

R2

R3

18 V

10 kΩ

12 kΩ

14 kΩ

VTH

RTH

Rload

the load resistor.First we disconnect

86

RloadR1

R2

R3

18 V

10 kΩ

12 kΩ

14 kΩ

VTH

RTH


87

Rload

R1

R2

R3

18 V

10 kΩ

12 kΩ

14 kΩ

VTH

RTH


88

Rload

R1

R2

R3

18 V

10 kΩ

12 kΩ

14 kΩ

VTH

RTH


89

Rload

R1

R2

R3

18 V

10 kΩ

12 kΩ

14 kΩ

VTH

RTH

90

Rload

R1

R2

R3

18 V

10 kΩ

12 kΩ

14 kΩ

VTH

RTH

Then we calculate howmuch voltage appearsacross the open load

terminals.

91

Rload

R1

R2

R3

18 V

10 kΩ

12 kΩ

14 kΩ

VTH

RTH


terminals.+

V-

92

Rload

R1

R2

R3

18 V

10 kΩ

12 kΩ

14 kΩ

VTH

RTH


terminals.+

V-

93

Rload

R1

R2

R3

18 V

10 kΩ

12 kΩ

14 kΩ

VTH

RTH


terminals.+

V- (18 volts)

10 kΩ

(14 kΩ + 12 kΩ + 10 kΩ)

94

Rload

R1

R2

R3

18 V

10 kΩ

12 kΩ

14 kΩ

VTH

RTH


terminals.+

V- (18 volts)

10 kΩ

(14 kΩ + 12 kΩ + 10 kΩ)

= 5 volts

95

Rload

R1

R2

R3

18 V

10 kΩ

12 kΩ

14 kΩ

VTH

RTH

+V

-5 V

96

Rload

R1

R2

R3

18 V

10 kΩ

12 kΩ

14 kΩ

VTH

RTH

+V

-5 V

This voltage becomesour Thevenin sourcevoltage . . .

97

Rload

R1

R2

R3

18 V

10 kΩ

12 kΩ

14 kΩ

RTH

+V

-5 V


5 V

98

Rload

R1

R2

R3

18 V

10 kΩ

12 kΩ

14 kΩ

RTH

+V

-5 V


5 V. . . in the Thevenin equivalent circuit.

99

Rload

R1

R2

R3

18 V

10 kΩ

12 kΩ

14 kΩ

RTH

5 V

100

Rload

R1

R2

R3

18 V

10 kΩ

12 kΩ

14 kΩ

RTH

5 V

Now we replace each sourcein the original circuit with itsown internal resistance.

101

Rload

R1

R2

R3

18 V

10 kΩ

12 kΩ

14 kΩ

RTH

5 V


For voltage sources, thismeans a short-circuit.

102

Rload

R1

R2

R3

18 V

10 kΩ

12 kΩ

14 kΩ

RTH

5 V



103

Rload

R1

R2

R3

18 V

10 kΩ

12 kΩ

14 kΩ

RTH

5 V



104

Rload

R1

R2

R3 10 kΩ

12 kΩ

14 kΩ

RTH

5 V



105

Rload

R1

R2

R3 10 kΩ

12 kΩ

14 kΩ

RTH

5 V

106

Rload

R1

R2

R3 10 kΩ

12 kΩ

14 kΩ

RTH

5 V

resistance across the openload terminals.

. . . and we calculate

107

Rload

R1

R2

R3 10 kΩ

12 kΩ

14 kΩ

RTH

5 V

resistance across the openload terminals.Ω


108

Rload

R1

R2

R3 10 kΩ

12 kΩ

14 kΩ

RTH

5 V



109

Rload

R1

R2

R3 10 kΩ

12 kΩ

14 kΩ

RTH

5 V



(14 kΩ + 12 kΩ) // 10 kΩ

110

Rload

R1

R2

R3 10 kΩ

12 kΩ

14 kΩ

RTH

5 V



(14 kΩ + 12 kΩ) // 10 kΩ= 7.22 kΩ

111

Rload

R1

R2

R3 10 kΩ

12 kΩ

14 kΩ

RTH

5 V

Ω 7.22 kΩ

112

Rload

R1

R2

R3 10 kΩ

12 kΩ

14 kΩ

RTH

5 V

Ω 7.22 kΩ

This resistance becomesour Thevenin source

resistance . . .

113

Rload

R1

R2

R3 10 kΩ

12 kΩ

14 kΩ

5 V

Ω 7.22 kΩ


resistance . . .

7.22 kΩ

114

Rload

R1

R2

R3 10 kΩ

12 kΩ

14 kΩ

5 V

Ω 7.22 kΩ


resistance . . .

7.22 kΩ

. . . in the Thevenin equivalent circuit.

115

Rload

R1

R2

R3 10 kΩ

12 kΩ

14 kΩ

5 V

7.22 kΩ

18 V

116

Rload

R1

R2

R3 10 kΩ

12 kΩ

14 kΩ

5 V

7.22 kΩ

Now that we have an equivalent circuit to

work with, we may insertthe load there to see what

happens!

18 V

117

Rload

R1

R2

R3 10 kΩ

12 kΩ

14 kΩ

5 V

7.22 kΩ



happens!

18 V

118

Rload

R1

R2

R3 10 kΩ

12 kΩ

14 kΩ

5 V

7.22 kΩ



happens!

18 V

119

Rload

R1

R2

R3 10 kΩ

12 kΩ

14 kΩ

5 V

7.22 kΩ



happens!

18 V

120

Rload

R1

R2

R3 10 kΩ

12 kΩ

14 kΩ

5 V

7.22 kΩ



happens!

18 V

121

Rload

R1

R2

R3 10 kΩ

12 kΩ

14 kΩ

5 V

7.22 kΩ



happens!

Calculate:

18 V

122

Rload

R1

R2

R3 10 kΩ

12 kΩ

14 kΩ

5 V

7.22 kΩ



happens!

Calculate:

Vload

18 V

123

Rload

R1

R2

R3 10 kΩ

12 kΩ

14 kΩ

5 V

7.22 kΩ



happens!

Calculate:

Vload

Iload

18 V

124

Rload

R1

R2

R3 10 kΩ

12 kΩ

14 kΩ

5 V

7.22 kΩ



happens!

Calculate:

Vload

Iload

Pload

18 V

125

Rload

R1

R2

R3 10 kΩ

12 kΩ

14 kΩ

5 V

7.22 kΩ

Calculate:

Vload

Iload

Pload

18 V

126

Rload

R1

R2

R3 10 kΩ

12 kΩ

14 kΩ

5 V

7.22 kΩ

Calculate:

Vload

Iload

Pload

18 VThese load calculations willreflect what happens in theoriginal circuit!

127

Rload

R1

R2

R3 10 kΩ

12 kΩ

14 kΩ

5 V

7.22 kΩ

Calculate:

Vload

Iload

Pload


Rload

128

Rload

R1

R2

R3 10 kΩ

12 kΩ

14 kΩ

5 V

7.22 kΩ

Calculate:

Vload

Iload

Pload


RloadVload

(same)

129

Rload

R1

R2

R3 10 kΩ

12 kΩ

14 kΩ

5 V

7.22 kΩ

Calculate:

Vload

Iload

Pload


RloadVload

(same)

Iload

130

Rload

R1

R2

R3 10 kΩ

12 kΩ

14 kΩ

5 V

7.22 kΩ

Calculate:

Vload

Iload

Pload


RloadVload

(same)

Iload

Pload

131

Rload

R1

R2

R3 10 kΩ

12 kΩ

14 kΩ

5 V

7.22 kΩ

Calculate:

Vload

Iload

Pload


RloadVload

Iload

Pload

132

Rload

R1

R2

R3 10 kΩ

12 kΩ

14 kΩ

5 V

7.22 kΩ Vload

Iload

Pload

18 V

Rload

Vload

Iload

Pload

133

Rload

R1

R2

R3 10 kΩ

12 kΩ

14 kΩ

5 V

7.22 kΩ Vload

Iload

Pload

18 V

Rload

Vload

Iload

Pload

134

Rload

R1

R2

R3 10 kΩ

12 kΩ

14 kΩ

5 V

7.22 kΩ Vload

Iload

Pload

18 V

Rload

Vload

Iload

Pload

The load cannot ‘‘tell’’any difference between the

Thevenin equivalent circuit.original circuit and the

file 03261

135

Answer 3


Notes 3

The purpose of this animation is to let students see how Thevenin’s theorem may be applied to thesimplification of a resistor network.

136

Question 4

Animation: Lissajous figures on an oscilloscope with 0 degrees phase shift


The following animation shows how two sinusoidal voltages with 0 degrees phase shift between themcan draw a Lissajous figure on the screen of an oscilloscope.

137

A B Alt Chop Add

Volts/Div A

Volts/Div B

DC Gnd AC

DC Gnd AC

Invert Intensity Focus

Position

Position

Position

Off

Beam find

LineExt.

AB

ACDC

NormAutoSingle

Slope

Level

Reset

X-Y

Holdoff

LF RejHF Rej

Triggering

Alt

Ext. input

Cal 1 V Gnd Trace rot.

Sec/Div0.5 0.2 0.1

1

10

5

2

20

50 m

20 m

10 m

5 m

2 m

0.5 0.2 0.11

10

5

2

20

50 m

20 m

10 m

5 m

2 m

1 m5 m

25 m

100 m

500 m

2.51

250 µ50 µ

10 µ

2.5 µ

0.5 µ

0.1 µ0.025 µ

off

Vvertical Vhorizontal

0o phase shift between Vvertical and Vhorizontal

138

A B Alt Chop Add

Volts/Div A

Volts/Div B

DC Gnd AC

DC Gnd AC


Position

Position

Position

Off

Beam find

LineExt.

AB

ACDC

NormAutoSingle

Slope

Level

Reset

X-Y

Holdoff

LF RejHF Rej

Triggering

Alt

Ext. input


Sec/Div0.5 0.2 0.1

1

10

5

2

20

50 m

20 m

10 m

5 m

2 m

0.5 0.2 0.11

10

5

2

20

50 m

20 m

10 m

5 m

2 m

1 m5 m

25 m

100 m

500 m

2.51

250 µ50 µ

10 µ

2.5 µ

0.5 µ

0.1 µ0.025 µ

off



139

A B Alt Chop Add

Volts/Div A

Volts/Div B

DC Gnd AC

DC Gnd AC


Position

Position

Position

Off

Beam find

LineExt.

AB

ACDC

NormAutoSingle

Slope

Level

Reset

X-Y

Holdoff

LF RejHF Rej

Triggering

Alt

Ext. input


Sec/Div0.5 0.2 0.1

1

10

5

2

20

50 m

20 m

10 m

5 m

2 m

0.5 0.2 0.11

10

5

2

20

50 m

20 m

10 m

5 m

2 m

1 m5 m

25 m

100 m

500 m

2.51

250 µ50 µ

10 µ

2.5 µ

0.5 µ

0.1 µ0.025 µ

off



140

A B Alt Chop Add

Volts/Div A

Volts/Div B

DC Gnd AC

DC Gnd AC


Position

Position

Position

Off

Beam find

LineExt.

AB

ACDC

NormAutoSingle

Slope

Level

Reset

X-Y

Holdoff

LF RejHF Rej

Triggering

Alt

Ext. input


Sec/Div0.5 0.2 0.1

1

10

5

2

20

50 m

20 m

10 m

5 m

2 m

0.5 0.2 0.11

10

5

2

20

50 m

20 m

10 m

5 m

2 m

1 m5 m

25 m

100 m

500 m

2.51

250 µ50 µ

10 µ

2.5 µ

0.5 µ

0.1 µ0.025 µ

off



141

A B Alt Chop Add

Volts/Div A

Volts/Div B

DC Gnd AC

DC Gnd AC


Position

Position

Position

Off

Beam find

LineExt.

AB

ACDC

NormAutoSingle

Slope

Level

Reset

X-Y

Holdoff

LF RejHF Rej

Triggering

Alt

Ext. input


Sec/Div0.5 0.2 0.1

1

10

5

2

20

50 m

20 m

10 m

5 m

2 m

0.5 0.2 0.11

10

5

2

20

50 m

20 m

10 m

5 m

2 m

1 m5 m

25 m

100 m

500 m

2.51

250 µ50 µ

10 µ

2.5 µ

0.5 µ

0.1 µ0.025 µ

off



142

A B Alt Chop Add

Volts/Div A

Volts/Div B

DC Gnd AC

DC Gnd AC


Position

Position

Position

Off

Beam find

LineExt.

AB

ACDC

NormAutoSingle

Slope

Level

Reset

X-Y

Holdoff

LF RejHF Rej

Triggering

Alt

Ext. input


Sec/Div0.5 0.2 0.1

1

10

5

2

20

50 m

20 m

10 m

5 m

2 m

0.5 0.2 0.11

10

5

2

20

50 m

20 m

10 m

5 m

2 m

1 m5 m

25 m

100 m

500 m

2.51

250 µ50 µ

10 µ

2.5 µ

0.5 µ

0.1 µ0.025 µ

off



143

A B Alt Chop Add

Volts/Div A

Volts/Div B

DC Gnd AC

DC Gnd AC


Position

Position

Position

Off

Beam find

LineExt.

AB

ACDC

NormAutoSingle

Slope

Level

Reset

X-Y

Holdoff

LF RejHF Rej

Triggering

Alt

Ext. input


Sec/Div0.5 0.2 0.1

1

10

5

2

20

50 m

20 m

10 m

5 m

2 m

0.5 0.2 0.11

10

5

2

20

50 m

20 m

10 m

5 m

2 m

1 m5 m

25 m

100 m

500 m

2.51

250 µ50 µ

10 µ

2.5 µ

0.5 µ

0.1 µ0.025 µ

off



144

A B Alt Chop Add

Volts/Div A

Volts/Div B

DC Gnd AC

DC Gnd AC


Position

Position

Position

Off

Beam find

LineExt.

AB

ACDC

NormAutoSingle

Slope

Level

Reset

X-Y

Holdoff

LF RejHF Rej

Triggering

Alt

Ext. input


Sec/Div0.5 0.2 0.1

1

10

5

2

20

50 m

20 m

10 m

5 m

2 m

0.5 0.2 0.11

10

5

2

20

50 m

20 m

10 m

5 m

2 m

1 m5 m

25 m

100 m

500 m

2.51

250 µ50 µ

10 µ

2.5 µ

0.5 µ

0.1 µ0.025 µ

off



145

A B Alt Chop Add

Volts/Div A

Volts/Div B

DC Gnd AC

DC Gnd AC


Position

Position

Position

Off

Beam find

LineExt.

AB

ACDC

NormAutoSingle

Slope

Level

Reset

X-Y

Holdoff

LF RejHF Rej

Triggering

Alt

Ext. input


Sec/Div0.5 0.2 0.1

1

10

5

2

20

50 m

20 m

10 m

5 m

2 m

0.5 0.2 0.11

10

5

2

20

50 m

20 m

10 m

5 m

2 m

1 m5 m

25 m

100 m

500 m

2.51

250 µ50 µ

10 µ

2.5 µ

0.5 µ

0.1 µ0.025 µ

off



file 03263

146

Answer 4


Notes 4

The purpose of this animation is to let students study the evolution of Lissajous figures and see how theyare created from the interrelationship between two sinusoidal waveforms. Similar to experimentation in thelab, except that here all the data collection is done visually rather than through the use of test equipment,and the students are able to ”see” things that are invisible in real life.

147

Question 5




148

A B Alt Chop Add

Volts/Div A

Volts/Div B

DC Gnd AC

DC Gnd AC


Position

Position

Position

Off

Beam find

LineExt.

AB

ACDC

NormAutoSingle

Slope

Level

Reset

X-Y

Holdoff

LF RejHF Rej

Triggering

Alt

Ext. input


Sec/Div0.5 0.2 0.1

1

10

5

2

20

50 m

20 m

10 m

5 m

2 m

0.5 0.2 0.11

10

5

2

20

50 m

20 m

10 m

5 m

2 m

1 m5 m

25 m

100 m

500 m

2.51

250 µ50 µ

10 µ

2.5 µ

0.5 µ

0.1 µ0.025 µ

off



149

A B Alt Chop Add

Volts/Div A

Volts/Div B

DC Gnd AC

DC Gnd AC


Position

Position

Position

Off

Beam find

LineExt.

AB

ACDC

NormAutoSingle

Slope

Level

Reset

X-Y

Holdoff

LF RejHF Rej

Triggering

Alt

Ext. input


Sec/Div0.5 0.2 0.1

1

10

5

2

20

50 m

20 m

10 m

5 m

2 m

0.5 0.2 0.11

10

5

2

20

50 m

20 m

10 m

5 m

2 m

1 m5 m

25 m

100 m

500 m

2.51

250 µ50 µ

10 µ

2.5 µ

0.5 µ

0.1 µ0.025 µ

off



150

A B Alt Chop Add

Volts/Div A

Volts/Div B

DC Gnd AC

DC Gnd AC


Position

Position

Position

Off

Beam find

LineExt.

AB

ACDC

NormAutoSingle

Slope

Level

Reset

X-Y

Holdoff

LF RejHF Rej

Triggering

Alt

Ext. input


Sec/Div0.5 0.2 0.1

1

10

5

2

20

50 m

20 m

10 m

5 m

2 m

0.5 0.2 0.11

10

5

2

20

50 m

20 m

10 m

5 m

2 m

1 m5 m

25 m

100 m

500 m

2.51

250 µ50 µ

10 µ

2.5 µ

0.5 µ

0.1 µ0.025 µ

off



151

A B Alt Chop Add

Volts/Div A

Volts/Div B

DC Gnd AC

DC Gnd AC


Position

Position

Position

Off

Beam find

LineExt.

AB

ACDC

NormAutoSingle

Slope

Level

Reset

X-Y

Holdoff

LF RejHF Rej

Triggering

Alt

Ext. input


Sec/Div0.5 0.2 0.1

1

10

5

2

20

50 m

20 m

10 m

5 m

2 m

0.5 0.2 0.11

10

5

2

20

50 m

20 m

10 m

5 m

2 m

1 m5 m

25 m

100 m

500 m

2.51

250 µ50 µ

10 µ

2.5 µ

0.5 µ

0.1 µ0.025 µ

off



152

A B Alt Chop Add

Volts/Div A

Volts/Div B

DC Gnd AC

DC Gnd AC


Position

Position

Position

Off

Beam find

LineExt.

AB

ACDC

NormAutoSingle

Slope

Level

Reset

X-Y

Holdoff

LF RejHF Rej

Triggering

Alt

Ext. input


Sec/Div0.5 0.2 0.1

1

10

5

2

20

50 m

20 m

10 m

5 m

2 m

0.5 0.2 0.11

10

5

2

20

50 m

20 m

10 m

5 m

2 m

1 m5 m

25 m

100 m

500 m

2.51

250 µ50 µ

10 µ

2.5 µ

0.5 µ

0.1 µ0.025 µ

off



153

A B Alt Chop Add

Volts/Div A

Volts/Div B

DC Gnd AC

DC Gnd AC


Position

Position

Position

Off

Beam find

LineExt.

AB

ACDC

NormAutoSingle

Slope

Level

Reset

X-Y

Holdoff

LF RejHF Rej

Triggering

Alt

Ext. input


Sec/Div0.5 0.2 0.1

1

10

5

2

20

50 m

20 m

10 m

5 m

2 m

0.5 0.2 0.11

10

5

2

20

50 m

20 m

10 m

5 m

2 m

1 m5 m

25 m

100 m

500 m

2.51

250 µ50 µ

10 µ

2.5 µ

0.5 µ

0.1 µ0.025 µ

off



154

A B Alt Chop Add

Volts/Div A

Volts/Div B

DC Gnd AC

DC Gnd AC


Position

Position

Position

Off

Beam find

LineExt.

AB

ACDC

NormAutoSingle

Slope

Level

Reset

X-Y

Holdoff

LF RejHF Rej

Triggering

Alt

Ext. input


Sec/Div0.5 0.2 0.1

1

10

5

2

20

50 m

20 m

10 m

5 m

2 m

0.5 0.2 0.11

10

5

2

20

50 m

20 m

10 m

5 m

2 m

1 m5 m

25 m

100 m

500 m

2.51

250 µ50 µ

10 µ

2.5 µ

0.5 µ

0.1 µ0.025 µ

off



155

A B Alt Chop Add

Volts/Div A

Volts/Div B

DC Gnd AC

DC Gnd AC


Position

Position

Position

Off

Beam find

LineExt.

AB

ACDC

NormAutoSingle

Slope

Level

Reset

X-Y

Holdoff

LF RejHF Rej

Triggering

Alt

Ext. input


Sec/Div0.5 0.2 0.1

1

10

5

2

20

50 m

20 m

10 m

5 m

2 m

0.5 0.2 0.11

10

5

2

20

50 m

20 m

10 m

5 m

2 m

1 m5 m

25 m

100 m

500 m

2.51

250 µ50 µ

10 µ

2.5 µ

0.5 µ

0.1 µ0.025 µ

off



156

A B Alt Chop Add

Volts/Div A

Volts/Div B

DC Gnd AC

DC Gnd AC


Position

Position

Position

Off

Beam find

LineExt.

AB

ACDC

NormAutoSingle

Slope

Level

Reset

X-Y

Holdoff

LF RejHF Rej

Triggering

Alt

Ext. input


Sec/Div0.5 0.2 0.1

1

10

5

2

20

50 m

20 m

10 m

5 m

2 m

0.5 0.2 0.11

10

5

2

20

50 m

20 m

10 m

5 m

2 m

1 m5 m

25 m

100 m

500 m

2.51

250 µ50 µ

10 µ

2.5 µ

0.5 µ

0.1 µ0.025 µ

off



file 03264

157

Answer 5


Notes 5


158

Question 6




159

A B Alt Chop Add

Volts/Div A

Volts/Div B

DC Gnd AC

DC Gnd AC


Position

Position

Position

Off

Beam find

LineExt.

AB

ACDC

NormAutoSingle

Slope

Level

Reset

X-Y

Holdoff

LF RejHF Rej

Triggering

Alt

Ext. input


Sec/Div0.5 0.2 0.1

1

10

5

2

20

50 m

20 m

10 m

5 m

2 m

0.5 0.2 0.11

10

5

2

20

50 m

20 m

10 m

5 m

2 m

1 m5 m

25 m

100 m

500 m

2.51

250 µ50 µ

10 µ

2.5 µ

0.5 µ

0.1 µ0.025 µ

off



160

A B Alt Chop Add

Volts/Div A

Volts/Div B

DC Gnd AC

DC Gnd AC


Position

Position

Position

Off

Beam find

LineExt.

AB

ACDC

NormAutoSingle

Slope

Level

Reset

X-Y

Holdoff

LF RejHF Rej

Triggering

Alt

Ext. input


Sec/Div0.5 0.2 0.1

1

10

5

2

20

50 m

20 m

10 m

5 m

2 m

0.5 0.2 0.11

10

5

2

20

50 m

20 m

10 m

5 m

2 m

1 m5 m

25 m

100 m

500 m

2.51

250 µ50 µ

10 µ

2.5 µ

0.5 µ

0.1 µ0.025 µ

off



161

A B Alt Chop Add

Volts/Div A

Volts/Div B

DC Gnd AC

DC Gnd AC


Position

Position

Position

Off

Beam find

LineExt.

AB

ACDC

NormAutoSingle

Slope

Level

Reset

X-Y

Holdoff

LF RejHF Rej

Triggering

Alt

Ext. input


Sec/Div0.5 0.2 0.1

1

10

5

2

20

50 m

20 m

10 m

5 m

2 m

0.5 0.2 0.11

10

5

2

20

50 m

20 m

10 m

5 m

2 m

1 m5 m

25 m

100 m

500 m

2.51

250 µ50 µ

10 µ

2.5 µ

0.5 µ

0.1 µ0.025 µ

off



162

A B Alt Chop Add

Volts/Div A

Volts/Div B

DC Gnd AC

DC Gnd AC


Position

Position

Position

Off

Beam find

LineExt.

AB

ACDC

NormAutoSingle

Slope

Level

Reset

X-Y

Holdoff

LF RejHF Rej

Triggering

Alt

Ext. input


Sec/Div0.5 0.2 0.1

1

10

5

2

20

50 m

20 m

10 m

5 m

2 m

0.5 0.2 0.11

10

5

2

20

50 m

20 m

10 m

5 m

2 m

1 m5 m

25 m

100 m

500 m

2.51

250 µ50 µ

10 µ

2.5 µ

0.5 µ

0.1 µ0.025 µ

off



163

A B Alt Chop Add

Volts/Div A

Volts/Div B

DC Gnd AC

DC Gnd AC


Position

Position

Position

Off

Beam find

LineExt.

AB

ACDC

NormAutoSingle

Slope

Level

Reset

X-Y

Holdoff

LF RejHF Rej

Triggering

Alt

Ext. input


Sec/Div0.5 0.2 0.1

1

10

5

2

20

50 m

20 m

10 m

5 m

2 m

0.5 0.2 0.11

10

5

2

20

50 m

20 m

10 m

5 m

2 m

1 m5 m

25 m

100 m

500 m

2.51

250 µ50 µ

10 µ

2.5 µ

0.5 µ

0.1 µ0.025 µ

off



164

A B Alt Chop Add

Volts/Div A

Volts/Div B

DC Gnd AC

DC Gnd AC


Position

Position

Position

Off

Beam find

LineExt.

AB

ACDC

NormAutoSingle

Slope

Level

Reset

X-Y

Holdoff

LF RejHF Rej

Triggering

Alt

Ext. input


Sec/Div0.5 0.2 0.1

1

10

5

2

20

50 m

20 m

10 m

5 m

2 m

0.5 0.2 0.11

10

5

2

20

50 m

20 m

10 m

5 m

2 m

1 m5 m

25 m

100 m

500 m

2.51

250 µ50 µ

10 µ

2.5 µ

0.5 µ

0.1 µ0.025 µ

off



165

A B Alt Chop Add

Volts/Div A

Volts/Div B

DC Gnd AC

DC Gnd AC


Position

Position

Position

Off

Beam find

LineExt.

AB

ACDC

NormAutoSingle

Slope

Level

Reset

X-Y

Holdoff

LF RejHF Rej

Triggering

Alt

Ext. input


Sec/Div0.5 0.2 0.1

1

10

5

2

20

50 m

20 m

10 m

5 m

2 m

0.5 0.2 0.11

10

5

2

20

50 m

20 m

10 m

5 m

2 m

1 m5 m

25 m

100 m

500 m

2.51

250 µ50 µ

10 µ

2.5 µ

0.5 µ

0.1 µ0.025 µ

off



166

A B Alt Chop Add

Volts/Div A

Volts/Div B

DC Gnd AC

DC Gnd AC


Position

Position

Position

Off

Beam find

LineExt.

AB

ACDC

NormAutoSingle

Slope

Level

Reset

X-Y

Holdoff

LF RejHF Rej

Triggering

Alt

Ext. input


Sec/Div0.5 0.2 0.1

1

10

5

2

20

50 m

20 m

10 m

5 m

2 m

0.5 0.2 0.11

10

5

2

20

50 m

20 m

10 m

5 m

2 m

1 m5 m

25 m

100 m

500 m

2.51

250 µ50 µ

10 µ

2.5 µ

0.5 µ

0.1 µ0.025 µ

off



167

A B Alt Chop Add

Volts/Div A

Volts/Div B

DC Gnd AC

DC Gnd AC


Position

Position

Position

Off

Beam find

LineExt.

AB

ACDC

NormAutoSingle

Slope

Level

Reset

X-Y

Holdoff

LF RejHF Rej

Triggering

Alt

Ext. input


Sec/Div0.5 0.2 0.1

1

10

5

2

20

50 m

20 m

10 m

5 m

2 m

0.5 0.2 0.11

10

5

2

20

50 m

20 m

10 m

5 m

2 m

1 m5 m

25 m

100 m

500 m

2.51

250 µ50 µ

10 µ

2.5 µ

0.5 µ

0.1 µ0.025 µ

off



file 03265

168

Answer 6


Notes 6


169

Question 7

Animation: Lissajous figures on an oscilloscope with 2:1 frequency ratio


The following animation shows how two sinusoidal voltages with a frequency ratio of 2:1 draw a Lissajousfigure on the screen of an oscilloscope.

170

A B Alt Chop Add

Volts/Div A

Volts/Div B

DC Gnd AC

DC Gnd AC


Position

Position

Position

Off

Beam find

LineExt.

AB

ACDC

NormAutoSingle

Slope

Level

Reset

X-Y

Holdoff

LF RejHF Rej

Triggering

Alt

Ext. input


Sec/Div0.5 0.2 0.1

1

10

5

2

20

50 m

20 m

10 m

5 m

2 m

0.5 0.2 0.11

10

5

2

20

50 m

20 m

10 m

5 m

2 m

1 m5 m

25 m

100 m

500 m

2.51

250 µ50 µ

10 µ

2.5 µ

0.5 µ

0.1 µ0.025 µ

off


2:1 frequency ratio between Vvertical and Vhorizontal

171

A B Alt Chop Add

Volts/Div A

Volts/Div B

DC Gnd AC

DC Gnd AC


Position

Position

Position

Off

Beam find

LineExt.

AB

ACDC

NormAutoSingle

Slope

Level

Reset

X-Y

Holdoff

LF RejHF Rej

Triggering

Alt

Ext. input


Sec/Div0.5 0.2 0.1

1

10

5

2

20

50 m

20 m

10 m

5 m

2 m

0.5 0.2 0.11

10

5

2

20

50 m

20 m

10 m

5 m

2 m

1 m5 m

25 m

100 m

500 m

2.51

250 µ50 µ

10 µ

2.5 µ

0.5 µ

0.1 µ0.025 µ

off



172

A B Alt Chop Add

Volts/Div A

Volts/Div B

DC Gnd AC

DC Gnd AC


Position

Position

Position

Off

Beam find

LineExt.

AB

ACDC

NormAutoSingle

Slope

Level

Reset

X-Y

Holdoff

LF RejHF Rej

Triggering

Alt

Ext. input


Sec/Div0.5 0.2 0.1

1

10

5

2

20

50 m

20 m

10 m

5 m

2 m

0.5 0.2 0.11

10

5

2

20

50 m

20 m

10 m

5 m

2 m

1 m5 m

25 m

100 m

500 m

2.51

250 µ50 µ

10 µ

2.5 µ

0.5 µ

0.1 µ0.025 µ

off



173

A B Alt Chop Add

Volts/Div A

Volts/Div B

DC Gnd AC

DC Gnd AC


Position

Position

Position

Off

Beam find

LineExt.

AB

ACDC

NormAutoSingle

Slope

Level

Reset

X-Y

Holdoff

LF RejHF Rej

Triggering

Alt

Ext. input


Sec/Div0.5 0.2 0.1

1

10

5

2

20

50 m

20 m

10 m

5 m

2 m

0.5 0.2 0.11

10

5

2

20

50 m

20 m

10 m

5 m

2 m

1 m5 m

25 m

100 m

500 m

2.51

250 µ50 µ

10 µ

2.5 µ

0.5 µ

0.1 µ0.025 µ

off



174

A B Alt Chop Add

Volts/Div A

Volts/Div B

DC Gnd AC

DC Gnd AC


Position

Position

Position

Off

Beam find

LineExt.

AB

ACDC

NormAutoSingle

Slope

Level

Reset

X-Y

Holdoff

LF RejHF Rej

Triggering

Alt

Ext. input


Sec/Div0.5 0.2 0.1

1

10

5

2

20

50 m

20 m

10 m

5 m

2 m

0.5 0.2 0.11

10

5

2

20

50 m

20 m

10 m

5 m

2 m

1 m5 m

25 m

100 m

500 m

2.51

250 µ50 µ

10 µ

2.5 µ

0.5 µ

0.1 µ0.025 µ

off



175

A B Alt Chop Add

Volts/Div A

Volts/Div B

DC Gnd AC

DC Gnd AC


Position

Position

Position

Off

Beam find

LineExt.

AB

ACDC

NormAutoSingle

Slope

Level

Reset

X-Y

Holdoff

LF RejHF Rej

Triggering

Alt

Ext. input


Sec/Div0.5 0.2 0.1

1

10

5

2

20

50 m

20 m

10 m

5 m

2 m

0.5 0.2 0.11

10

5

2

20

50 m

20 m

10 m

5 m

2 m

1 m5 m

25 m

100 m

500 m

2.51

250 µ50 µ

10 µ

2.5 µ

0.5 µ

0.1 µ0.025 µ

off



176

A B Alt Chop Add

Volts/Div A

Volts/Div B

DC Gnd AC

DC Gnd AC


Position

Position

Position

Off

Beam find

LineExt.

AB

ACDC

NormAutoSingle

Slope

Level

Reset

X-Y

Holdoff

LF RejHF Rej

Triggering

Alt

Ext. input


Sec/Div0.5 0.2 0.1

1

10

5

2

20

50 m

20 m

10 m

5 m

2 m

0.5 0.2 0.11

10

5

2

20

50 m

20 m

10 m

5 m

2 m

1 m5 m

25 m

100 m

500 m

2.51

250 µ50 µ

10 µ

2.5 µ

0.5 µ

0.1 µ0.025 µ

off



177

A B Alt Chop Add

Volts/Div A

Volts/Div B

DC Gnd AC

DC Gnd AC


Position

Position

Position

Off

Beam find

LineExt.

AB

ACDC

NormAutoSingle

Slope

Level

Reset

X-Y

Holdoff

LF RejHF Rej

Triggering

Alt

Ext. input


Sec/Div0.5 0.2 0.1

1

10

5

2

20

50 m

20 m

10 m

5 m

2 m

0.5 0.2 0.11

10

5

2

20

50 m

20 m

10 m

5 m

2 m

1 m5 m

25 m

100 m

500 m

2.51

250 µ50 µ

10 µ

2.5 µ

0.5 µ

0.1 µ0.025 µ

off



178

A B Alt Chop Add

Volts/Div A

Volts/Div B

DC Gnd AC

DC Gnd AC


Position

Position

Position

Off

Beam find

LineExt.

AB

ACDC

NormAutoSingle

Slope

Level

Reset

X-Y

Holdoff

LF RejHF Rej

Triggering

Alt

Ext. input


Sec/Div0.5 0.2 0.1

1

10

5

2

20

50 m

20 m

10 m

5 m

2 m

0.5 0.2 0.11

10

5

2

20

50 m

20 m

10 m

5 m

2 m

1 m5 m

25 m

100 m

500 m

2.51

250 µ50 µ

10 µ

2.5 µ

0.5 µ

0.1 µ0.025 µ

off



file 03266

179

Answer 7


Notes 7


180

Question 8

Animation: three-phase electric motor


The following animation shows how three sets of electromagnet poles create a rotating magnetic fieldwhen energized by three-phase AC power. Relative strengths of the magnetic fields produced by each pairof poles are indicated by the size of the ”N” (North) and ”S” (South) letters. Here are some things to lookfor in this animation:

• What determines the direction of the vector arrow?• Note when each pole pair (A and A’, etc.) reaches its peak magnetic field strength.• Is there any time where more than one pair of electromagnet poles is at its maximum field strength?

181

A

A

B

B C

C

N

SN

S

182

A

A

B

B C

C

N

S

N

SN

S

183

A

A

B

B C

C

N

S

N

S

184

A

A

B

B C

C

N

S

N

S

N

S

185

A

A

B

B C

C

N

S

N

S

186

A

A

B

B C

C

N

S

N

S N

S

187

A

A

B

B C

C

N

S N

S

188

A

A

B

B C

C

S

N

N

S N

S

189

A

A

B

B C

C

S

N

N

S

190

A

A

B

B C

C

S

NN

S

N

S

191

A

A

B

B C

C

S

N

N

S

192

A

A

B

B C

C

S

N

N

SN

S

file 03232

193

Answer 8

Note how each pole pair (A and A’, B and B’, C and C’) develops its peak magnetic field at differenttimes.

Notes 8

The purpose of this animation is to let students study the evolution of the rotating magnetic field andreach their own conclusions. Similar to experimentation in the lab, except that here all the data collection isdone visually rather than through the use of test equipment, and the students are able to ”see” things thatare invisible in real life.

194

Question 9

Animation: sketching characteristic curves for a transistor


The following animation shows how a family of characteristic curves are sketched for a bipolar junctiontransistor. Pay close attention to what parameters are varied as each curve is sketched.

195

IB = 10 µA

0 5 10 15 20 25 30 35

VCE (volts)

40

IC (mA)

0

1

2

3

4

5

6

7

IC

VCE

IB

196

IB = 10 µA

0 5 10 15 20 25 30 35

VCE (volts)

40

IC (mA)

0

1

2

3

4

5

6

7

IC

VCE

IB

197

IB = 10 µA

0 5 10 15 20 25 30 35

VCE (volts)

40

IC (mA)

0

1

2

3

4

5

6

7

IC

VCE

IB

198

IB = 10 µA

0 5 10 15 20 25 30 35

VCE (volts)

40

IC (mA)

0

1

2

3

4

5

6

7

IC

VCE

IB

199

IB = 10 µA

0 5 10 15 20 25 30 35

VCE (volts)

40

IC (mA)

0

1

2

3

4

5

6

7

IC

VCE

IB

200

IB = 10 µA

0 5 10 15 20 25 30 35

VCE (volts)

40

IC (mA)

0

1

2

3

4

5

6

7

IC

VCE

IB

201

IB = 10 µA

0 5 10 15 20 25 30 35

VCE (volts)

40

IC (mA)

0

1

2

3

4

5

6

7

IC

VCE

IB

202

IB = 10 µA

0 5 10 15 20 25 30 35

VCE (volts)

40

IC (mA)

0

1

2

3

4

5

6

7

IC

VCE

IB

203

IB = 10 µA

0 5 10 15 20 25 30 35

VCE (volts)

40

IC (mA)

0

1

2

3

4

5

6

7

IC

VCE

IB

204

IB = 20 µA

IB = 10 µA

0 5 10 15 20 25 30 35

VCE (volts)

40

IC (mA)

0

1

2

3

4

5

6

7

IC

VCE

IB

205

IB = 20 µA

IB = 10 µA

0 5 10 15 20 25 30 35

VCE (volts)

40

IC (mA)

0

1

2

3

4

5

6

7

IC

VCE

IB

206

IB = 20 µA

IB = 10 µA

0 5 10 15 20 25 30 35

VCE (volts)

40

IC (mA)

0

1

2

3

4

5

6

7

IC

VCE

IB

207

IB = 20 µA

IB = 10 µA

0 5 10 15 20 25 30 35

VCE (volts)

40

IC (mA)

0

1

2

3

4

5

6

7

IC

VCE

IB

208

IB = 20 µA

IB = 10 µA

0 5 10 15 20 25 30 35

VCE (volts)

40

IC (mA)

0

1

2

3

4

5

6

7

IC

VCE

IB

209

IB = 20 µA

IB = 10 µA

0 5 10 15 20 25 30 35

VCE (volts)

40

IC (mA)

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IB = 20 µA

IB = 10 µA

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IC (mA)

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VCE

IB

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IB = 20 µA

IB = 10 µA

0 5 10 15 20 25 30 35

VCE (volts)

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IC (mA)

0

1

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3

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6

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IC

VCE

IB

212

IB = 30 µA

IB = 20 µA

IB = 10 µA

0 5 10 15 20 25 30 35

VCE (volts)

40

IC (mA)

0

1

2

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4

5

6

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IC

VCE

IB

213

IB = 30 µA

IB = 20 µA

IB = 10 µA

0 5 10 15 20 25 30 35

VCE (volts)

40

IC (mA)

0

1

2

3

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5

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VCE

IB

214

IB = 30 µA

IB = 20 µA

IB = 10 µA

0 5 10 15 20 25 30 35

VCE (volts)

40

IC (mA)

0

1

2

3

4

5

6

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IC

VCE

IB

215

IB = 30 µA

IB = 20 µA

IB = 10 µA

0 5 10 15 20 25 30 35

VCE (volts)

40

IC (mA)

0

1

2

3

4

5

6

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VCE

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216

IB = 30 µA

IB = 20 µA

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VCE (volts)

40

IC (mA)

0

1

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VCE

IB

217

IB = 30 µA

IB = 20 µA

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VCE (volts)

40

IC (mA)

0

1

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5

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IC

VCE

IB

218

IB = 30 µA

IB = 20 µA

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VCE (volts)

40

IC (mA)

0

1

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3

4

5

6

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VCE

IB

219

IB = 30 µA

IB = 20 µA

IB = 10 µA

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VCE (volts)

40

IC (mA)

0

1

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3

4

5

6

7

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VCE

IB

220

IB = 40 µA

IB = 30 µA

IB = 20 µA

IB = 10 µA

0 5 10 15 20 25 30 35

VCE (volts)

40

IC (mA)

0

1

2

3

4

5

6

7

IC

VCE

IB

221

IB = 40 µA

IB = 30 µA

IB = 20 µA

IB = 10 µA

0 5 10 15 20 25 30 35

VCE (volts)

40

IC (mA)

0

1

2

3

4

5

6

7

IC

VCE

IB

222

IB = 40 µA

IB = 30 µA

IB = 20 µA

IB = 10 µA

0 5 10 15 20 25 30 35

VCE (volts)

40

IC (mA)

0

1

2

3

4

5

6

7

IC

VCE

IB

223

IB = 40 µA

IB = 30 µA

IB = 20 µA

IB = 10 µA

0 5 10 15 20 25 30 35

VCE (volts)

40

IC (mA)

0

1

2

3

4

5

6

7

IC

VCE

IB

224

IB = 40 µA

IB = 30 µA

IB = 20 µA

IB = 10 µA

0 5 10 15 20 25 30 35

VCE (volts)

40

IC (mA)

0

1

2

3

4

5

6

7

IC

VCE

IB

225

IB = 40 µA

IB = 30 µA

IB = 20 µA

IB = 10 µA

0 5 10 15 20 25 30 35

VCE (volts)

40

IC (mA)

0

1

2

3

4

5

6

7

IC

VCE

IB

226

IB = 40 µA

IB = 30 µA

IB = 20 µA

IB = 10 µA

0 5 10 15 20 25 30 35

VCE (volts)

40

IC (mA)

0

1

2

3

4

5

6

7

IC

VCE

IB

227

IB = 40 µA

IB = 30 µA

IB = 20 µA

IB = 10 µA

0 5 10 15 20 25 30 35

VCE (volts)

40

IC (mA)

0

1

2

3

4

5

6

7

IC

VCE

IB

228

IB = 40 µA

IB = 30 µA

IB = 20 µA

IB = 10 µA

0 5 10 15 20 25 30 35

VCE (volts)

40

IC (mA)

0

1

2

3

4

5

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7

IC

VCE

IB

file 03237

229

Answer 9


Notes 9

The purpose of this animation is to let students study the generation of characteristic curves and reachtheir own conclusions. Similar to experimentation in the lab, except that here all the data collection is donevisually rather than through the use of test equipment, and the students are able to ”see” things that areinvisible in real life.

230

Question 10

Animation: crossover distortion in a push-pull transistor amplifier


The following animation shows a simple push-pull, emitter-follower amplifier circuit exhibiting crossoverdistortion. Watch what happens as the input voltage goes through a whole cycle, noting when each transistorbegins to conduct, and when each transistor ceases conduction. Here are some things to look for:

• Which transistor handles which portion of the input waveform (positive versus negative)?• Why is there a ”flat” spot in the output waveform?• What would have to be done to this circuit to allow it to reproduce the waveform in full?

231

Vin Rload

-V

+V

Input Output

232

Vin Rload

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Input Output

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Vin Rload

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+V

Input Output

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Vin Rload

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Input Output

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Vin Rload

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Input Output

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Vin Rload

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+V

Input Output

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Vin Rload

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Input Output

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Vin Rload

-V

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Input Output

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Vin Rload

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Input Output

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Vin Rload

-V

+V

Input Output

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Vin Rload

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Input Output

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Vin Rload

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Input Output

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Vin Rload

-V

+V

Input Output

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Vin Rload

-V

+V

Input Output

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Vin Rload

-V

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Input Output

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Vin Rload

-V

+V

Input Output

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Vin Rload

-V

+V

Input Output

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Vin Rload

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+V

Input Output

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Vin Rload

-V

+V

Input Output

250

Vin Rload

-V

+V

Input Output

251

Vin Rload

-V

+V

Input Output

file 03233

252

Answer 10

For silicon transistors, the crossover distortion amounts to approximately 1.4 volts (from +0.7 to -0.7volts) in the input waveform.

Follow-up question: in terms of percentage, do you think crossover distortion increases as the inputsignal increases in peak-to-peak magnitude, or decreases? Explain your reasoning.

Notes 10

The purpose of this animation is to let students study the behavior of this amplifier circuit and reachtheir own conclusions. Similar to experimentation in the lab, except that here all the data collection is donevisually rather than through the use of test equipment, and the students are able to ”see” things that areinvisible in real life.

253

Question 11

Animation: PWM comparator circuit


The following animation shows a comparator used to generate a PWM pulse signal from a triangle waveand a DC reference voltage. Watch what happens as the input voltage goes through a whole cycle, notingwhen the comparator switches output states. Here are some things to look for:

• What input conditions are necessary for the comparator to output a ”high” (+V) state?• What input conditions are necessary for the comparator to output a ”low” (-V) state?• Which direction would you move the potentiometer wiper to increase the duty cycle of the PWM output?

254

−

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file 03235

296

Answer 11


Notes 11

The purpose of this animation is to let students study the behavior of this comparator circuit and reachtheir own conclusions. Similar to experimentation in the lab, except that here all the data collection is donevisually rather than through the use of test equipment, and the students are able to ”see” things that areinvisible in real life.

297

Question 12

Animation: telephony multiplexer


The following animation shows how multiple telephone conversations may be ”multiplexed” across asingle communication channel. A high switching speed is necessary to make the conversations seamless,which may be simulated by watching the animation at a high frame rate and seeing that the respectiveconversation outputs appear to be constant. Some questions to ponder:

• Why do you suppose anyone would do this? Why not just have five separate lines, one for eachconversation?

• How fast do you suppose the mux/demux pair would have to switch in order for the received conversationsto appear seamless?

298

Conversation A

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Mux DemuxConversation A

299

Conversation A

Conversation B

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Conversation B

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Conversation A

Conversation B

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Mux Demux

Conversation C

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Conversation A

Conversation B

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Mux Demux

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Conversation A

Conversation B

Conversation C

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Mux Demux

Conversation E

file 03236

303

Answer 12

This animation must be played with a very fast frame rate to do the principle justice. If this is notpossible, imagine the mux/demux pair moving at a blinding speed – so fast that your eyes could not followthe motions of the selector switches. What do you suppose the words next to the output lines (ConversationA, Conversation B, etc.) would look like?

Notes 12

The purpose of this animation is to let students study the behavior of this multiplexer circuit and reachtheir own conclusions. Similar to experimentation in the lab, except that here all the data collection is donevisually rather than through the use of test equipment, and the students are able to ”see” things that areinvisible in real life.

304

Question 13

Animation: Johnson ring counter


The following animation shows a 4-bit Johnson ring counter circuit. Watch what happens as the clocksignal oscillates. Here are some things to look for:

• Note when the logic state at each flip-flop input gets sent to the Q output.• Why do you think this is called a ”ring” counter circuit?

305

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Gnd

Q0

Q1

VDD

Gnd

VDD

Gnd

VDD

Gnd

VDD

Gnd

Q2

Q3

1 1 1

0

0 0 0 0 0

335

D

C

Q

Q D

C

Q

Q D

C

Q

Q D

C

Q

Q

Clk

2 310

ClockVDD

Gnd

Q0

Q1

VDD

Gnd

VDD

Gnd

VDD

Gnd

VDD

Gnd

Q2

Q3

1 1 1

0

0 0 0 0 0

336

D

C

Q

Q D

C

Q

Q D

C

Q

Q D

C

Q

Q

Clk

2 310

ClockVDD

Gnd

Q0

Q1

VDD

Gnd

VDD

Gnd

VDD

Gnd

VDD

Gnd

Q2

Q3

1 1 1

0

0 0 0 0 0

337

D

C

Q

Q D

C

Q

Q D

C

Q

Q D

C

Q

Q

Clk

2 310

ClockVDD

Gnd

Q0

Q1

VDD

Gnd

VDD

Gnd

VDD

Gnd

VDD

Gnd

Q2

Q3

1 1 1

0

0 0 0 0 0

338

D

C

Q

Q D

C

Q

Q D

C

Q

Q D

C

Q

Q

Clk

2 310

ClockVDD

Gnd

Q0

Q1

VDD

Gnd

VDD

Gnd

VDD

Gnd

VDD

Gnd

Q2

Q3

1 1 1

0

0 0 0 0 0 0 0

339

D

C

Q

Q D

C

Q

Q D

C

Q

Q D

C

Q

Q

Clk

2 310

ClockVDD

Gnd

Q0

Q1

VDD

Gnd

VDD

Gnd

VDD

Gnd

VDD

Gnd

Q2

Q3

1

0

0 0 0 0 0 0 0

340

D

C

Q

Q D

C

Q

Q D

C

Q

Q D

C

Q

Q

Clk

2 310

ClockVDD

Gnd

Q0

Q1

VDD

Gnd

VDD

Gnd

VDD

Gnd

VDD

Gnd

Q2

Q3

1

0

0 0 0 0 0 0 0

341

D

C

Q

Q D

C

Q

Q D

C

Q

Q D

C

Q

Q

Clk

2 310

ClockVDD

Gnd

Q0

Q1

VDD

Gnd

VDD

Gnd

VDD

Gnd

VDD

Gnd

Q2

Q3

1

0

0 0 0 0 0 0 0

342

D

C

Q

Q D

C

Q

Q D

C

Q

Q D

C

Q

Q

Clk

2 310

ClockVDD

Gnd

Q0

Q1

VDD

Gnd

VDD

Gnd

VDD

Gnd

VDD

Gnd

Q2

Q3

1

0

0 0 0 0 0 0 0

343

D

C

Q

Q D

C

Q

Q D

C

Q

Q D

C

Q

Q

Clk

2 310

ClockVDD

Gnd

Q0

Q1

VDD

Gnd

VDD

Gnd

VDD

Gnd

VDD

Gnd

Q2

Q3

1

0

0 0 0 0 0 0 0 0

1

1

344

D

C

Q

Q D

C

Q

Q D

C

Q

Q D

C

Q

Q

Clk

2 310

ClockVDD

Gnd

Q0

Q1

VDD

Gnd

VDD

Gnd

VDD

Gnd

VDD

Gnd

Q2

Q3

0 0 0 0 0 0 0

1

1

file 03234

345

Answer 13

Note that each rising edge of the clock pulse has its own frame in the animation sequence, to bettershow you what happens at those crucial times.

Notes 13

The purpose of this animation is to let students study the behavior of this counter circuit and reachtheir own conclusions. Similar to experimentation in the lab, except that here all the data collection is donevisually rather than through the use of test equipment, and the students are able to ”see” things that areinvisible in real life.

In this animation, I show each rising edge of the clock signal in its own frame, whereas the falling edgeof the clock shares a frame with the first half of the ”low” state. I do this because these are positive edge-triggered flip-flops, and so the rising edge of the clock pulse is most important. I could have slowed thingsdown on the falling edge of the clock as well, but since there is little ”action” happening then, I decided tosave a frame and make it a shorter animation.

346

Question 14

Animation: addressing 16 × 8 bit ROM memory


The following animation shows how setting the input (address) switches in particular combinationsselects individual memory ”cells” inside the ROM, resulting in data stored within those cells to appear atthe output (data) lines. Questions to ponder:

• What is the organization of this particular ROM chip? (e.g. 256 × 4, 1k × 1, etc.)• What is the relationship between the hexadecimal numbers stored inside each cell and the alphabetical

characters shown below the chip? What code is being used to represent these characters?• What character is represented by the hexadecimal code 20? (Hint: this code is used twice in the sequence

shown.)

347

VDD ROM memory IC

VDD

A0

A1

A2

A3

D0

D1

D2

D3

D4

D5

D6

D7

0

1

0

0

0

0 1 2 3

4 5 6 7

8 9 10 11

12 13 14 15

4D 65 6D 6F

72 79 20 49

73 55 73

65

20

66 75 6C

00

0

11

10

M

348

VDD ROM memory IC

VDD

A0

A1

A2

A3

D0

D1

D2

D3

D4

D5

D6

D7

1

0

0

0

0 1 2 3

4 5 6 7

8 9 10 11

12 13 14 15

4D 65 6D 6F

72 79 20 49

73 55 73

65

20

66 75 6C

0

0

1

101

Me

1

0

349

VDD ROM memory IC

VDD

A0

A1

A2

A3

D0

D1

D2

D3

D4

D5

D6

D7

1

0

0

0

0 1 2 3

4 5 6 7

8 9 10 11

12 13 14 15

4D 65 6D 6F

72 79 20 49

73 55 73

65

20

66 75 6C

0

0

1

10

1

1

1

Mem

350

VDD ROM memory IC

VDD

A0

A1

A2

A3

D0

D1

D2

D3

D4

D5

D6

D7

1

0

0

0 1 2 3

4 5 6 7

8 9 10 11

12 13 14 15

4D 65 6D 6F

72 79 20 49

73 55 73

65

20

66 75 6C

0

0

1

1

1

1

1

1 1

Memo

351

VDD ROM memory IC

VDD

A0

A1

A2

A3

D0

D1

D2

D3

D4

D5

D6

D7

1

0

0 1 2 3

4 5 6 7

8 9 10 11

12 13 14 15

4D 65 6D 6F

72 79 20 49

73 55 73

65

20

66 75 6C 0

1

1

1

0

0

100

0

Memor

352

VDD ROM memory IC

VDD

A0

A1

A2

A3

D0

D1

D2

D3

D4

D5

D6

D7

1

0

0 1 2 3

4 5 6 7

8 9 10 11

12 13 14 15

4D 65 6D 6F

72 79 20 49

73 55 73

65

20

66 75 6C 0

1

1

1

0

1

001

1

Memory

353

VDD ROM memory IC

VDD

A0

A1

A2

A3

D0

D1

D2

D3

D4

D5

D6

D7

0

0 1 2 3

4 5 6 7

8 9 10 11

12 13 14 15

4D 65 6D 6F

72 79 20 49

73 55 73

65

20

66 75 6C 0

1

1

0

00

1

Memory

0

0

0

0

354

VDD ROM memory IC

VDD

A0

A1

A2

A3

D0

D1

D2

D3

D4

D5

D6

D7

0

0 1 2 3

4 5 6 7

8 9 10 11

12 13 14 15

4D 65 6D 6F

72 79 20 49

73 55 73

65

20

66 75 6C 01

1

00

1

00

1

1

1

Memory I

355

VDD ROM memory IC

VDD

A0

A1

A2

A3

D0

D1

D2

D3

D4

D5

D6

D7

0 1 2 3

4 5 6 7

8 9 10 11

12 13 14 15

4D 65 6D 6F

72 79 20 49

73 55 73

65

20

66 75 6C 01

1

00

11

0

0

0

11

Memory Is

356

VDD ROM memory IC

VDD

A0

A1

A2

A3

D0

D1

D2

D3

D4

D5

D6

D7

0 1 2 3

4 5 6 7

8 9 10 11

12 13 14 15

4D 65 6D 6F

72 79 20 49

73 55 73

65

20

66 75 6C 0

1

000

0

1

Memory Is

1

0

0

00

357

VDD ROM memory IC

VDD

A0

A1

A2

A3

D0

D1

D2

D3

D4

D5

D6

D7

0 1 2 3

4 5 6 7

8 9 10 11

12 13 14 15

4D 65 6D 6F

72 79 20 49

73 55 73

65

20

66 75 6C 0

1

00

0

1

1

0

0

1

1

1

Memory Is U

358

VDD ROM memory IC

VDD

A0

A1

A2

A3

D0

D1

D2

D3

D4

D5

D6

D7

0 1 2 3

4 5 6 7

8 9 10 11

12 13 14 15

4D 65 6D 6F

72 79 20 49

73 55 73

65

20

66 75 6C 0

1

00

1

1 0

1

111

1

Memory Is Us

359

VDD ROM memory IC

VDD

A0

A1

A2

A3

D0

D1

D2

D3

D4

D5

D6

D7

0 1 2 3

4 5 6 7

8 9 10 11

12 13 14 15

4D 65 6D 6F

72 79 20 49

73 55 73

65

20

66 75 6C 0

1

0

1

1

01

1

1

0

0

0

Memory Is Use

360

VDD ROM memory IC

VDD

A0

A1

A2

A3

D0

D1

D2

D3

D4

D5

D6

D7

0 1 2 3

4 5 6 7

8 9 10 11

12 13 14 15

4D 65 6D 6F

72 79 20 49

73 55 73

65

20

66 75 6C 0

1

0

1

1

0

11

1

0

0

1

Memory Is Usef

361

VDD ROM memory IC

VDD

A0

A1

A2

A3

D0

D1

D2

D3

D4

D5

D6

D7

0 1 2 3

4 5 6 7

8 9 10 11

12 13 14 15

4D 65 6D 6F

72 79 20 49

73 55 73

65

20

66 75 6C 0

1

0

1

1

01

1

1

0

1

1

Memory Is Usefu

362

VDD ROM memory IC

VDD

A0

A1

A2

A3

D0

D1

D2

D3

D4

D5

D6

D7

0 1 2 3

4 5 6 7

8 9 10 11

12 13 14 15

4D 65 6D 6F

72 79 20 49

73 55 73

65

20

66 75 6C 0

1

0

1

1

011

1

1

1

0

Memory Is Useful

file 03244

363

Answer 14

Nothing to note here!

Notes 14

The purpose of this animation is to let students study the behavior of this switch circuit and reachtheir own conclusions. Similar to experimentation in the lab, except that here all the data collection is donevisually rather than through the use of test equipment, and the students are able to ”see” things that areinvisible in real life.

364

circuitanimations - ibiblio this worksheet and ... the purpose of this animation is to let students...

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