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# Standard Grade Physics - PowerPoint PPT Presentation

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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Presentation Transcript

Unit 4 Electronics

Label the following signals as analogue or digital.

(a) (b) (c)

analogue

analogue

digital

digital

analogue

(d) (e) (f)

(a) (b)

analogue

digital

(c) (d)

digital

analogue

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:

Solenoid Unit

5 V

0 V

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

off

on

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

on

off

VS– VLED

0∙015

12 – 2 = 10 V

667 Ω

VR = I R

VS– VR

 VR = 20 × 10-3 × 140

5 – 2∙8 = 2∙2 V

 VR = 2∙8 V

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

temperature up resistance down

L.U.R.D.

light up resistance down

light to electrical

heat to electrical

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.

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

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

5V

0V

charged

discharged

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.

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∙5/6000x 5000

 V = 3∙75 V

Voltage across R= 6 – 2 = 4 V

V1 R1V2= R2

4 R2 = 4

2 x R = 4 x 4

 R = 16/2 = 8 Ω

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

Vout (V)

ON

OFF

Vbe (V)

0∙7 V

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.

R flows through the base. For a current to flow through the base, VT = 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)

0 V flows through the base. For a current to flow through the base, V

5 V

2∙5 V

Remember: flows through the base. For a current to flow through the base, VWhen 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. flows through the base. For a current to flow through the base, VSet 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.

decreases flows through the base. For a current to flow through the base, V

V across thermistor decreases

V across variable R increases

Vbe increases

Vbe≥ 0∙7 V, transistorswitches ON

LED is ON

reducing

high flows through the base. For a current to flow through the base, V

high

low

low

low

high

dark

A light sensor. flows through the base. For a current to flow through the base, VSet 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.

increases flows through the base. For a current to flow through the base, V

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. flows through the base. For a current to flow through the base, VSet 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.

high flows through the base. For a current to flow through the base, V

high

low

low

low

high

dry

A moisture sensor. Set up the circuit shown below: flows through the base. For a current to flow through the base, V

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.

increases flows through the base. For a current to flow through the base, V

voltage across probes increases

Vbe increases

Vbe≥ 0∙7 V transistor switches on

LED is on

wet

0 V flows through the base. For a current to flow through the base, V

5 V

5 V

0 V

A time-controlled circuit. flows through the base. For a current to flow through the base, VSet 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.

discharges flows through the base. For a current to flow through the base, V

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.

high flows through the base. For a current to flow through the base, V

high

low

low

low

high

A switch controlled circuit. flows through the base. For a current to flow through the base, VSet up the circuit shown below:

Potential divider

Transistor

5 V

switch

0 V

4.7 k pot.

high flows through the base. For a current to flow through the base, V

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 flows through the base. For a current to flow through the base, V

0

INVERTED (NOT the same as input)

A NOT (Inverter) gate. flows through the base. For a current to flow through the base, VSet up the circuit shown below:

Debounced Switch

INVERTER

5 V

0 V

An OR gate. flows through the base. For a current to flow through the base, VSet up the circuit shown below:

5 V

OR gate

0 V

0 flows through the base. For a current to flow through the base, V

1

1

1

A or B (or both) = 1

switch flows through the base. For a current to flow through the base, V

lamp

switch

An AND gate. flows through the base. For a current to flow through the base, VSet up the circuit shown below:

5 V

AND gate

0 V

Notes page 25 flows through the base. For a current to flow through the base, V

a n D

0

0

0

1

A and B = 1

master switch flows through the base. For a current to flow through the base, V

motor

drivers’ switch

A flows through the base. For a current to flow through the base, V

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.

A flows through the base. For a current to flow through the base, V

D

LDR

E

F

light

B

IR detector

C

master switch

1 flows through the base. For a current to flow through the base, V

0

0

0

0

0

1

1

1

1

0

1

0

0

0

1

1

1

0

0

1

0

0

1

A clock pulse generator. flows through the base. For a current to flow through the base, VSet up the circuit shown below:

5 V

Signal potentiometer

Inverter

1 kresistor

0 V

1000 Fcapacitor

0 V flows through the base. For a current to flow through the base, V

0

1

5 V

off

rise

1

0

on

fall

0

1

off

decrease flows through the base. For a current to flow through the base, V

computers, timers, clocks

A binary counter. flows through the base. For a current to flow through the base, VSet up the circuit shown below:

5 V

0 V

Now replace the switch with a clock pulse generator: flows through the base. For a current to flow through the base, V

5 V

Signal potentiometer

Counter

Inverter

1 kresistor

0 V

1000 Fcapacitor

An electronic counter with display and decoder. flows through the base. For a current to flow through the base, VSet up the circuit shown below:

5 V

0 V

Output from light sensor in light = 0. flows through the base. For a current to flow through the base, VX 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.

TVs, radios, telephones, hi-fis, flows through the base. For a current to flow through the base, Vpublic 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)

V flows through the base. For a current to flow through the base, 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)

V flows through the base. For a current to flow through the base, Vpeak = 500 mV × 2

Vpeak = 2V × 3

 Vpeak = 1000 mV = 1V

Vpeak = 6V

 Vg = Vo/Vin = 6/1 =6

P flows through the base. For a current to flow through the base, Vo

Pg

Pin

Pg = 400Pin = 0∙01 W

Po = ?

Po

Pg Pin

 Po = Pg× Pin

 Po = 400 × 0∙01

 Po = 4 W

P flows through the base. For a current to flow through the base, Vin = 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)

V flows through the base. For a current to flow through the base, V2P = R

V2

P

R

Vout2 R

Vin2 R

V flows through the base. For a current to flow through the base, 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