Standard Grade Physics

Unit 4 Electronics

Exercise 

Label the following signals as analogue or digital.

(a) (b) (c)analogue analogue digital

(d) (e) (f)analogue digital analogue

 

Label following devices as analogue or digital.

(a) (b)

(c) (d)

analogue digital

digital analogue

sound digital

sound analogue

kinetic(rotation) analogue

light digital (analogue with variable R)

light digital

lightdigital

kinetic digital

kinetic (in straight line)

digital

The solenoid.Set up the circuit as shown below:

Solenoid Unit5 V

0 V

“flying” lead

Touch lead here

Negative sign on its side.LED only allows current to flow and light up when “negative connected to negative”.

on off on

on off

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

VS – VLED

12 – 2 = 10 V

10 0∙015

667 Ω

VR = I R

VR = 20 × 10-3 × 140

VR = 2∙8 V

VS – VR

5 – 2∙8 = 2∙2 V

0 0 0 0 0 0 0 1

1 0 0 1

0 0 1 1 0 1 0 0 0 1 0 1 0 1 1 0 0 1 1 1 1 0 0 0

0 0 1 0

T.U.R.D.

temperature up resistance down

L.U.R.D.

light up resistance down

sound to electrical

light to electrical

heat to electrical

Capacitor Investigation

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

Potential divider

resistor

capacitor

5 V

0 V

component investigations

switch

V

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

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

Conclusions:

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

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

5 V 0 V

charged discharged

Discharging a capacitor

Copy the following table into the back of your jotter.

R1 (Ω) 1 k 10 k 1 k 100 10 k 100

R2 (Ω) 10 k 1 k 100 1 k 100 10 k

V1 (V)

V2 (V)

R1

R2

V1

V2

R1 (Ω) 1 k 10 k 1 k 100 10 k 100

R2 (Ω) 10 k 1 k 100 1 k 100 10 k

V1 (V)

V2 (V)

R1

R2

0∙1 10 10 0∙1 100 0∙01

V1

V2

R1 (Ω) 1 k 10 k 1 k 100 10 k 100

R2 (Ω) 10 k 1 k 100 1 k 100 10 k

V1 (V) 0.45 4.55 4.5 0.5 4.95 0.05

V2 (V) 4.55 0.45 0.5 4.5 0.05 4.95

R1

R2

0∙1 10 10 0∙1 100 0∙01

V1

V2

R1 (Ω) 1 k 10 k 1 k 100 10 k 100

R2 (Ω) 10 k 1 k 100 1 k 100 10 k

V1 (V) 0.45 4.55 4.5 0.5 4.95 0.05

V2 (V) 4.55 0.45 0.5 4.5 0.05 4.95

R1

R2

0∙1 10 10 0∙1 100 0∙01

V1

V2

0.1 10 10 0.1 100 0.01

Set up the apparatus as shown below:

Potential divider

resistor R1

resistor R2

5 V

0 Vswitch

V1

V2

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

24 + 12 = 36 Ω

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!)

Step 1. RT = 1k + 5k = 6 kΩ

Step 2. I = V/R = 4∙5/6000

Step 3. V = I R

V = 4∙5/6000 x 5000

V = 3∙75 V

Voltage across R = 6 – 2 = 4 V

V1 R1

V2 = R24 R2 = 4

2 x R = 4 x 4

R = 16/2 = 8 Ω

Potential divider

4∙7 kpotentiometer

5 V

0 V

Transistor

Vbe

Vout

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

Now draw a graph of your results.Vout (V)

Vbe (V) 0∙7 V

OFF ON

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.

RT = 1800 + 200 = 2 kΩ

I = V/R = 5/2000

V1 = Vbe = I R

Vbe = 5/2000 x 200 = 0∙5 V

Off5 – 0∙5 = 4∙5 V

If temperature increases, resistance of thermistor decreases V across thermistor decreases and V across 200 Ω increases

V2 decreases and V1 increases (V2 + V1 = 5 V)

0 V 5 V

2∙5 V

high high low

low low high

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

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

Potential divider

4∙7 k pot.

5 V

0 V

Transistor

thermistor

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

decreases

V across thermistor decreases

V across variable R increases

Vbe increases

Vbe ≥ 0∙7 V, transistor switches ON

LED is ON

reducing

high high low

low low high

dark

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

Potential divider5 V

0 V

Transistor

LDR

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

4∙7 k pot.

increases

voltage across LDR increases

Vbe increases

Vbe ≥ 0∙7 V transistor switches on

LED is on

swap thepositions of the LDR and variable resistor.

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

Rain sensing unit

Ω

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

200 k Ω setting

high high low

low low high

dry

A moisture sensor. Set up the circuit shown below:

Potential divider5 V

0 V

Transistor

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

22 k pot.

Rain sensing unit

increases

voltage across probes increases

Vbe increases

Vbe ≥ 0∙7 V transistor switches on

LED is on

wet

0 V 5 V

5 V 0 V

Potential divider5 V

0 V

Transistor

4∙7 k pot.

Capacitor

1000 μF

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

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

component investigations

switch

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 slowlyVbe 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.

high high low

low low high

Potential divider5 V

0 V

Transistor

4.7 k pot.

switch

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

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

1

0

INVERTED (NOT the same as input)

Debounced Switch5 V

0 V

INVERTER

A NOT (Inverter) gate. Set up the circuit shown below:

OR gate

An OR gate. Set up the circuit shown below:5 V

0 V

0

1

1

1

A or B (or both) = 1

switch

switch

lamp

AND gate

An AND gate. Set up the circuit shown below:5 V

0 V

0

0 0

1

A and B = 1

a n D

Notes page 25

master switch

drivers’ switch

motor

0LDR

engine switch

light A

B

C D 1

0 0

1

1 0

1

1 0 1 0

1 0 0 0

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

LDR

IR detector

A

B

C

D

master switch

light

E

F

1 0 1 1 0 1 0 0

0 0 1 0 0 1 0 0

0 0 1 1 0 1 1 1

Signal potentiometer

1000 Fcapacitor

5 V

0 V

Inverter

1 kresistor

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

0 V

0 1 5 V off

rise

1 0 on

fall

0 1 off

decrease

computers, timers, clocks

A binary counter. Set up the circuit shown below:

5 V

0 V

Now replace the switch with a clock pulse generator:

Signal potentiometer

1000 Fcapacitor

5 V

0 V

Inverter

1 kresistor

Counter

An electronic counter with display and decoder. Set up the circuit shown below:

5 V

0 V

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∙01 s × 5= 0∙05 s

Length of the car.

Vo Vg Vin

Vg = ?

Vin = 5 mV

Vo = 0∙45 V

Vo

Vg Vin

Vg = Vo/Vin = 0∙45/(5 × 10–3)

Vg = 90 (no unit)

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

Vg = 50

Vin = 0∙1 V

Vo = ?

Vo

Vg Vin

Vo = Vg × Vin

Vo = 50 × 0∙1

Vo = 5 V

100 Hz (no change in frequency)

Vpeak = 500 mV × 2

Vpeak = 1000 mV = 1 V

Vpeak = 2 V × 3

Vpeak = 6 V

Vg = Vo/Vin = 6/1 = 6

Po Pg Pin

Pg = 400

Pin = 0∙01 W

Po = ?

Po

Pg Pin

Po = Pg × Pin

Po = 400 × 0∙01

Po = 4 W

Pin = I V = 0∙005 × 0∙2 = 0∙001 WPout = I V = 0∙04 × 2 = 0∙08 W

Po 0∙08

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

V2

P = R V2

P R

Vin2

R Vout

2

R

V2

P R Vin

2

Pin = R

(12 × 10-3)2

Pin = 10

Pin = 1∙44 × 10-5 W

Po

Pg Pin Po = Pg × Pin

Po = 500 × 1∙44 × 10-5

Po = 7∙2 × 10-3 W