standard grade physics
DESCRIPTION
Standard Grade Physics. Unit 4 Electronics. Exercise. Label the following signals as analogue or digital. ( a ) ( b ) ( c ). analogue. analogue. digital. analogue. digital. analogue. - PowerPoint PPT PresentationTRANSCRIPT
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Standard Grade Physics
Unit 4 Electronics
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Exercise
Label the following signals as analogue or digital.
(a) (b) (c)analogue analogue digital
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(d) (e) (f)analogue digital analogue
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Label following devices as analogue or digital.
(a) (b)
(c) (d)
analogue digital
digital analogue
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sound digital
sound analogue
kinetic(rotation) analogue
light digital (analogue with variable R)
light digital
lightdigital
kinetic digital
kinetic (in straight line)
digital
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The solenoid.Set up the circuit as shown below:
Solenoid Unit5 V
0 V
“flying” lead
Touch lead here
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Negative sign on its side.LED only allows current to flow and light up when “negative connected to negative”.
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on off on
on off
For a current to flow there has to be a difference in voltage.
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VS – VLED
12 – 2 = 10 V
10 0∙015
667 Ω
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VR = I R
VR = 20 × 10-3 × 140
VR = 2∙8 V
VS – VR
5 – 2∙8 = 2∙2 V
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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
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T.U.R.D.
temperature up resistance down
L.U.R.D.
light up resistance down
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sound to electrical
light to electrical
heat to electrical
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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
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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.
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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
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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
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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
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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
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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
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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.
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24 + 12 = 36 Ω
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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!)
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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
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Voltage across R = 6 – 2 = 4 V
V1 R1
V2 = R24 R2 = 4
2 x R = 4 x 4
R = 16/2 = 8 Ω
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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.
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Now draw a graph of your results.Vout (V)
Vbe (V) 0∙7 V
OFF ON
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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.
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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)
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0 V 5 V
2∙5 V
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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.
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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.
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decreases
V across thermistor decreases
V across variable R increases
Vbe increases
Vbe ≥ 0∙7 V, transistor switches ON
LED is ON
reducing
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high high low
low low high
dark
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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.
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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.
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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
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high high low
low low high
dry
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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
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increases
voltage across probes increases
Vbe increases
Vbe ≥ 0∙7 V transistor switches on
LED is on
wet
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0 V 5 V
5 V 0 V
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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
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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.
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high high low
low low high
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Potential divider5 V
0 V
Transistor
4.7 k pot.
switch
A switch controlled circuit. Set up the circuit shown below:
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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
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1
0
INVERTED (NOT the same as input)
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Debounced Switch5 V
0 V
INVERTER
A NOT (Inverter) gate. Set up the circuit shown below:
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OR gate
An OR gate. Set up the circuit shown below:5 V
0 V
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0
1
1
1
A or B (or both) = 1
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switch
switch
lamp
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AND gate
An AND gate. Set up the circuit shown below:5 V
0 V
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0
0 0
1
A and B = 1
a n D
Notes page 25
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master switch
drivers’ switch
motor
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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.
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LDR
IR detector
A
B
C
D
master switch
light
E
F
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1 0 1 1 0 1 0 0
0 0 1 0 0 1 0 0
0 0 1 1 0 1 1 1
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Signal potentiometer
1000 Fcapacitor
5 V
0 V
Inverter
1 kresistor
A clock pulse generator. Set up the circuit shown below:
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0 V
0 1 5 V off
rise
1 0 on
fall
0 1 off
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decrease
computers, timers, clocks
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A binary counter. Set up the circuit shown below:
5 V
0 V
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Now replace the switch with a clock pulse generator:
Signal potentiometer
1000 Fcapacitor
5 V
0 V
Inverter
1 kresistor
Counter
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An electronic counter with display and decoder. Set up the circuit shown below:
5 V
0 V
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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.
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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
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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)
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Vpeak = 500 mV × 2
Vpeak = 1000 mV = 1 V
Vpeak = 2 V × 3
Vpeak = 6 V
Vg = Vo/Vin = 6/1 = 6
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Po Pg Pin
Pg = 400
Pin = 0∙01 W
Po = ?
Po
Pg Pin
Po = Pg × Pin
Po = 400 × 0∙01
Po = 4 W
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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)
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V2
P = R V2
P R
Vin2
R Vout
2
R
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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