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Standard Grade Physics. Unit 4 Electronics. Exercise. Label the following signals as analogue or digital. ( a ) ( b ) ( c ). analogue. analogue. digital. analogue. digital. analogue.

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standard grade physics

Standard Grade Physics

Unit 4 Electronics

slide4

Exercise

Label the following signals as analogue or digital.

(a) (b) (c)

analogue

analogue

digital

slide5

analogue

digital

analogue

(d) (e) (f)

slide6

Label following devices as analogue or digital.

(a) (b)

analogue

digital

(c) (d)

digital

analogue

slide7

sound

digital

sound

analogue

kinetic(rotation)

analogue

light

digital (analogue with variable R)

light

digital

light

digital

kinetic

digital

kinetic (in straight line)

digital

the solenoid set up the circuit as shown below
The solenoid.Set up the circuit as shown below:

Solenoid Unit

5 V

Touch lead here

“flying” lead

0 V

slide11

Negative sign on its side.

LED only allows current to flow and light up when “negative connected to negative”.

slide13

on

off

on

For a current to flow there has to be a difference in voltage.

on

off

slide15

10

VS– VLED

0∙015

12 – 2 = 10 V

667 Ω

slide16

VR = I R

VS– VR

 VR = 20 × 10-3 × 140

5 – 2∙8 = 2∙2 V

 VR = 2∙8 V

slide18

0 0 0 0

0 0 0 1

0 0 1 0

0 0 1 1

0 1 0 0

0 1 0 1

0 1 1 0

0 1 1 1

1 0 0 0

1 0 0 1

slide19

T.U.R.D.

temperature up resistance down

L.U.R.D.

light up resistance down

slide20

sound to electrical

light to electrical

heat to electrical

capacitor investigation
Capacitor Investigation

V

Potential divider

5 V

resistor

component investigations

0 V

capacitor

switch

With power off, discharge capacitor using switch so that V = 0 V. When power is switched on, start timer.

slide24

Conclusions:

The time to charge a capacitor depends on the values of the capacitance and the series resistor.

If the value of R and/or C are increased, the time taken to reach the final voltage also increases.

Comparison with waterCopy diagrams from P.T.A. page 107

slide25

Discharging a capacitor

The quickest and easiest way to discharge a capacitor is to place a wire across both ends of it.

5V

0V

charged

discharged

set up the apparatus as shown below
Set up the apparatus as shown below:

Potential divider

5 V

V1

resistor R1

V2

0 V

resistor R2

switch

Complete the table by measuring the voltage V1 and V2 for each pair of resistors.

slide32

V/R = 12/36 (leave as fraction to avoid rounding off)

12/36 x 24 = 8 V

12/36 x 12 = 4 V

8 244 = 12 = 2 (check!)

slide33

Step 1. RT = 1k + 5k = 6 kΩ

Step 2. I = V/R = 4∙5/6000

Step 3. V = I R

 V = 4∙5/6000x 5000

 V = 3∙75 V

slide34

Voltage across R= 6 – 2 = 4 V

V1 R1V2= R2

4 R2 = 4

2 x R = 4 x 4

 R = 16/2 = 8 Ω

slide38

Adjust the knob on the potentiometer until Vbe = required value. Now measure the corresponding Vout.

5 V

Potential divider

Transistor

Vout

0 V

4∙7 kpotentiometer

Vbe

now draw a graph of your results
Now draw a graph of your results.

Vout (V)

ON

OFF

Vbe (V)

0∙7 V

slide40
A current cannot flow through the collector unless a current flows through the base. For a current to flow through the base, Vbe≥ 0∙7 V.
slide42

RT = 1800 + 200 = 2 kΩ

I = V/R = 5/2000

V1 = Vbe = I R

 Vbe = 5/2000x 200= 0∙5 V

 Off

5 – 0∙5 = 4∙5 V

If temperature increases, resistance of thermistor decreases

 V across thermistor decreases and V across 200Ω increases

 V2 decreases and V1increases (V2 + V1 = 5 V)

slide43

0 V

5 V

2∙5 V

slide44

Remember:When the voltage divides, the resistor with the biggest value will take the biggest share of the voltage.

high

high

low

low

low

high

a temperature sensor set up the circuit as shown below
A temperature sensor.Set up the circuit as shown below:

Potential divider

Transistor

5 V

thermistor

0 V

4∙7 k pot.

Adjust potentiometer until LED is just off. Now warm thermistor by rubbing with your finger.

slide46

decreases

V across thermistor decreases

V across variable R increases

Vbe increases

Vbe≥ 0∙7 V, transistorswitches ON

LED is ON

reducing

slide47

high

high

low

low

low

high

dark

a light sensor set up the circuit as shown below
A light sensor.Set up the circuit as shown below:

Potential divider

Transistor

5 V

4∙7 k pot.

0 V

LDR

Adjust potentiometer until LED is just off. Now cover the LDR with your finger.

slide49

increases

voltage across LDR increases

Vbe increases

Vbe≥ 0∙7 V transistor switches on

LED is on

swap the

positions of the LDR and variable resistor.

the moisture unit set up the circuit as shown below
The moisture unit.Set up the circuit as shown below:

200 k Ωsetting

Rain sensing unit

Ω

Observe what happens to the reading on the ohmmeter when water is added to the moisture unit.

slide51

high

high

low

low

low

high

dry

a moisture sensor set up the circuit shown below
A moisture sensor. Set up the circuit shown below:

Potential divider

Transistor

5 V

Rain sensing unit

22k pot.

0 V

Turn the knob on the potentiometer fully clockwise. LED should be off. Now add water to the moisture unit.

slide53

increases

voltage across probes increases

Vbe increases

Vbe≥ 0∙7 V transistor switches on

LED is on

wet

slide54

0 V

5 V

5 V

0 V

a time controlled circuit set up the circuit shown below
A time-controlled circuit. Set up the circuit shown below:

Potential divider

Transistor

5 V

Capacitor

1000 μF

component investigations

4∙7 k pot.

switch

0 V

Press the switch to put the LED on. Release switch, LED will go off after a time delay.

slide56

discharges

voltage across C falls to 0 V immediately

voltage across R and Vbe rise to 5 V immediately

transistor switches on, current flows in relay, switch closes to complete circuit, motor and heater turn on.

charges

voltage across C rises to 5 V slowly

voltage across R and Vbe fall to 0 V slowly

Vbe  0∙7 V, transistor switches off, no current in relay, switch opens to break circuit, so motor and heater turn off.

Increase the value of R and/or C.

slide57

high

high

low

low

low

high

a switch controlled circuit set up the circuit shown below
A switch controlled circuit. Set up the circuit shown below:

Potential divider

Transistor

5 V

switch

0 V

4.7 k pot.

slide59

high

V across switch = high

V across R = low

Vbe = low i.e.  0∙7 V, so transistor is off

low

V across switch = low

V across R = high

Vbe≥ 0∙7 V, transistor is on, current flows in relay, so relay switch closes to complete circuit

slide60

1

0

INVERTED (NOT the same as input)

slide63

0

1

1

1

A or B (or both) = 1

slide64

switch

lamp

switch

slide66

Notes page 25

a n D

0

0

0

1

A and B = 1

slide67

master switch

motor

drivers’ switch

slide68

A

C

LDR

D

0

1

1

1

light

1

0

0

0

0

0

1

0

1

1

0

0

B

engine switch

A NOT gate is needed because the output from the LDR is ‘0’ in the dark.

slide69

A

D

LDR

E

F

light

B

IR detector

C

master switch

slide70

1

0

0

0

0

0

1

1

1

1

0

1

0

0

0

1

1

1

0

0

1

0

0

1

slide72

A clock pulse generator. Set up the circuit shown below:

5 V

Signal potentiometer

Inverter

1 kresistor

0 V

1000 Fcapacitor

slide73

0 V

0

1

5 V

off

rise

1

0

on

fall

0

1

off

slide74

decrease

computers, timers, clocks

slide76

Now replace the switch with a clock pulse generator:

5 V

Signal potentiometer

Counter

Inverter

1 kresistor

0 V

1000 Fcapacitor

slide79

Output from light sensor in light = 0.X is an AND gate i.e. output will only be 1 when both inputs are logic 1.

5

0∙01s× 5= 0∙05 s

Length of the car.

slide80

TVs, radios, telephones, hi-fis, public announcement, intercom etc

Vo

Vg

Vin

Vo

Vg Vin

Vg = ?Vin = 5 mV

Vo = 0∙45 V

 Vg = Vo/Vin = 0∙45/(5 × 10–3)

 Vg = 90 (no unit)

slide81

Vg = 50Vin = 0∙1 V

Vo = ?

Vo

Vg Vin

 Vo = Vg× Vin

 Vo = 50 × 0∙1

 Vo = 5 V

100 Hz (no change in frequency)

slide83

Vpeak = 500 mV × 2

Vpeak = 2V × 3

 Vpeak = 1000 mV = 1V

Vpeak = 6V

 Vg = Vo/Vin = 6/1 =6

slide84

Po

Pg

Pin

Pg = 400Pin = 0∙01 W

Po = ?

Po

Pg Pin

 Po = Pg× Pin

 Po = 400 × 0∙01

 Po = 4 W

slide85

Pin = I V = 0∙005 × 0∙2 = 0∙001 W

Pout = I V = 0∙04 × 2 = 0∙08 W

Po 0∙08 Pg = Pin = 0∙001 = 80 (no unit needed)

slide86

V2P = R

V2

P

R

Vout2 R

Vin2 R

slide87

V2

Vin2 Pin = R

Po

 Po = Pg× Pin

P

R

Pg

Pin

(12 × 10-3)2 Pin = 10

 Po = 500 × 1∙44 × 10-5

 Pin = 1∙44 × 10-5 W

 Po = 7∙2 × 10-3 W