standard grade physics
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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
(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
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