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## VCE Physics

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### Title: VCE Physics Author: ol Last modified by: ollie Created Date: 2/6/2006 4:36:02 AM Document presentation format: On-screen Show Company: Department of Education ... – PowerPoint PPT presentation

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Title: VCE Physics

1
VCE Physics
• Unit 3
• Electronics Photonics
• Revision Questions

2
Electronics Photonics RevisionQuestion Type
Ohms Law
Figure 1 shows a resistor, a linear circuit
component, with resistance R 100 O. A DC
current, I 40 mA, passes through this resistor
in the direction shown by the arrows.
Q What is the voltage drop across this resistor?
A V IR (40 x 10-3)(100)
4.0 V
3
Electronics Photonics RevisionQuestion Type
Potential Difference
Q Which one of the following statements (A to D)
concerning the voltage across the resistor in
Figure 1 is true? A. The potential at point A is
higher than at point B. B. The potential at point
A is the same as at point B. C. The potential at
point A is lower than at point B. D. The
potential at point A varies in sign with time
compared to that at point B.
A Alternative A
4
Electronics Photonics RevisionQuestion Type
Electrical Energy
Q Determine the electrical energy dissipated in
the 100 O resistor of Figure 1 in 1 second. In
A Electrical energy W VQ VIt
(4.0)(40 x
10-3)(1) 0.16 Joule
5
Electronics Photonics RevisionQuestion Type
Voltage Divider Network
Q What is the RMS output voltage, VOUT?
A Step 1 Determine the equivalent resistance for
A and B and D and E
For A and B 1/RE 1/RA 1/RB
1/100 1/100
2/100 So RE 100/2
50 O
Replace A and B with one 50 O resistor, same with
D and E.
In Figure 2, five identical 100 O resistors are
used to construct a voltage divider. The voltage
source across this voltage divider is an AC
supply with an RMS voltage of 20 V. The resistors
are labelled by the letters A to E as shown.
You can now redraw the circuit.
6
Electronics Photonics RevisionQuestion Type
Voltage Divider Network
The series circuit in Figure 3 can be further
simplified as shown in Figure 4
The original question (value of VOUT) can now be
answered, using the Voltage Divider formula
15 V
7
Electronics Photonics RevisionQuestion Type
Current Flows
Q Which one of the following statements (A to D)
concerning the RMS currents in the circuit of
Figure 2 is true? A. The current in resistor A is
identical to the current in resistor C. B. The
current in resistor D is twice the current in
resistor C. C. The current in resistor B is twice
the current in resistor E. D. The current in
resistor A is identical to the current in
resistor D.
A Alternative D
8
Electronics Photonics RevisionQuestion Type
Ohms Law
Figure 3 shows a typical n-p-n transistor voltage
amplifier circuit. You may assume that the
transistor is correctly biased and the circuit is
operating in the linear-amplification region. The
DC collector current is 20 mA.
Q Show that the voltage between point C and
Earth is 20 V.
A A Collector current of 20 mA means RC has 20
mA flowing through it. V across Rc IR
(20 x 10-3)(500)
10 V
So, VC Vcc VRc 30 10
20 V
9
Electronics Photonics RevisionQuestion Type
Transistor Characteristics
Q Calculate the change in collector voltage, and
determine the small-signal voltage gain of the
The AC current gain of the transistor is 200. A
small time-varying AC voltage, VIN, is applied at
the input. As VIN rises from 0 to 10 mV this
causes the base current to increase from 0 to 5
µA.
A The rise in base current switches the
transistor on causing a current to flow in the
collector circuit. The base and collector
currents are related through the current gain.
IC ßIB (200)(5 x 10-6)
0.001 A
Thus the Voltage drop across RC is VC IRC
(0.001)(500) 0.5 V
50
10
Electronics Photonics RevisionQuestion Type
Transducer Properties
Q Describe the basic purpose of each of the
following electronic transducers. i.
Light-Emitting Diode (LED) ii. Photodiode
A (i) Emits visible light when a current flows
through it. Converts electrical energy
to light. (ii) Switches on (allows a
current to flow through it) when exposed to
light. Converts light to electrical
energy.
11
Electronics Photonics RevisionQuestion Type
The information on an audio CD is represented by
a series of pits (small depressions) in the
surface that are scanned by laser light. When
there is no pit the reflected light gives a
maximum light intensity, I1, detected by
a photodiode circuit. When the laser light
strikes a pit, the light intensity is reduced to
I0. A plot of a typical light intensity incident
on the photodiode is shown in Figure 4.
12
Electronics Photonics RevisionQuestion Type
Ohms Law
The variation in current as a function of light
intensity for the photodiode is shown in Figure
5a, together with the circuit used to determine
this, which is shown in Figure 5b.
Q With no light incident upon the photodiode,
the current in the photodiode circuit, the dark
current, is 5 µA. What is the output voltage,
VOUT, across the 100 O resistor in the circuit of
Figure 5b?
A The photodiode and resistor are in series. The
same current flows through each. VOUT
V100O IR (5 x
10-6)(100) 5 x 10-4 V
13
Electronics Photonics RevisionQuestion Type
Linear and Non Linear Devices
A resistor is a linear device. An example of a
non-linear device is a light-emitting diode (LED).
Q On the axes provided, sketch a typical
current-voltage characteristic curve for each of
the devices mentioned. In both cases label the
axes and indicate appropriate units.
I(mA)
I(Amp)
V(volts)
V(volts)
a non linear device an LED
a linear device a resistor
14
Electronics Photonics RevisionQuestion Type
Voltage Divider Network
An essential component in some of the practical
circuits covered in this exam paper is the
voltage divider. A DC voltage divider circuit is
shown in Figure 1.
For the circuit of Figure 1, VIN 30 V, R1 5
kO and the output voltage VOUT 6 V. Q What is
the value of the resistance R2? Show your working.
R2 20,000 O 20kO
15
Electronics Photonics RevisionQuestion Type
Parallel Resistors
You wire up the circuit shown in Figure 1 but
only have 10 kO resistors to work with. Q
Explain how you would construct the R1 5 kO
resistor using only 10 kO resistors. Include a
sketch to show the connections between the
appropriate number of 10 kO resistors.
A Connect two 10k resistors together in parallel
16
Electronics Photonics RevisionQuestion Type
Ohms Law
In Figure 1 the 30 V DC input to the voltage
divider is replaced by a 100 mV (peak-to-peak)
sinusoidal AC input voltage. The resistance
values are now R1 5 kO and R2 15 kO. Q What
is the current through resistor R2? Show your
peak-to-peak current in µA.
A Series circuit add resistances so RT 20kO
Use Ohms Law to find current
V IR so I
5µA
17
Electronics Photonics RevisionQuestion Type
Capacitor Properties
Figure 1 is modified so that R1 has a capacitor,
C, placed in parallel with it. The new circuit is
shown in Figure 2, and can now have an AC or DC
input voltage, VIN.
Q It is observed that if VIN is a DC voltage,
then in the R1C parallel combination, a DC
current passes through R1, but not through C.
However, if VIN is a high frequency AC voltage,
then an AC current passes mainly through C with
very little through R1. Explain this observation.
A At low frequency the Capacitor acts like an
open circuit (infinite resistance) forcing all
current to flow through the resistor At high
frequency the capacitor acts like a short circuit
(zero resistance) allowing most current to flow
through it and very little through the resistor.
18
Electronics Photonics RevisionQuestion Type
These ideas are incorporated into the design of
an n-p-n transistor amplifier circuit, shown in
Figure 3, to amplify the small, high frequency AC
voltage from a microphone. You may assume that
the transistor is correctly biased. The collector
and base currents in Figure 3 are denoted
respectively using lower case iC and iB to
emphasise their AC time variation and small
magnitude.
In this circuit the AC collector current, iC, and
the AC base current, iB, are related by iC 100
iB. The 10 mV (peak-to-peak) input voltage from
the microphone gives rise to a time-varying base
current of iB 10 µA (peak-to-peak).
19
Electronics Photonics RevisionQuestion Type
Transistor Characteristics
Q Calculate the AC voltage across the collector
resistor, RC, and show that the magnitude of the
voltage gain of this transistor amplifier circuit
is 200.
A iC 100iB and iB 10 µA So iC
(100)(10 x 10-6) 0.001 A
VRc iCRC (0.001)(2000) 2.0
V
Gain VOUT/VIN 2.0/(10 x10-3) 200
20
Electronics Photonics RevisionQuestion Type
Sketch Graphs
Q Which one of the following cathode ray
oscilloscope (CRO) traces of the output voltage
(A to D) correctly identifies the distortion
arising from transistor cut-off for the circuit
of Figure 3? The input to this transistor
amplifier circuit is a sinusoidal voltage.
A Alternative B
21
Electronics Photonics RevisionQuestion Type
Ohms Law
You are asked to investigate the properties of an
optical coupler, sometimes called an
opto-isolator. This comprises a light-emitting
diode (LED) that converts an electrical signal
into light output, and a phototransistor (PT)
that converts incident light into an electrical
output. Before using an opto-isolator chip you
consider typical LED and PT circuits
separately. A simple LED circuit is shown in
Figure 4 along with the LED current-voltage
characteristics. The light output increases as
the forward current, IF , through the LED
increases.
Q Using the information in Figure 4, what is the
value of the resistance, RD, in series with the
LED that will ensure the forward current through
the LED is IF 10 mA?
A For 10 mA to flow through LED requires a
voltage of 1.5 V (read from graph) Because LED
and R are in series VRD 10 1.5 8.5 V
VRD IRD so RD VRD/I 8.5/(10 x 10-3)
850 O
22
Electronics Photonics RevisionQuestion Type
Ohms Law
Q Will the light output of the LED increase or
decrease if the value of RD is a little lower
than the value you have calculated in the last
A Increased output
Justification Reducing the value of RD will not
affect the voltage drop across it. The Voltage
across RD is controlled by the LED which will
remain at 1.5 V thus VRD will still equal 8.5 V.
So if V remains the same and R goes down I must
go up. So a larger current flows through the LED
meaning an increased light output
23
Electronics Photonics RevisionQuestion Type
The LED in Figure 4 is an electro-optical
converter. Which one of the following statements
(A to D) regarding energy conversion for the LED
is correct?
All the electrical energy supplied from the DC
power supply is converted A. only to heat energy
in both the resistor, RD, and the LED. B. partly
to heat energy in the resistor, RD, the remainder
to light-energy output from the LED. C. partly to
heat energy in both the resistor, RD, and the
LED, with the remainder to light-energy output
from the LED. D. to heat energy in the LED, with
the remainder to light-energy output from the
LED.
A Alternative C
24
Electronics Photonics RevisionQuestion Type
Ohms Law
You now consider the phototransistor (PT) circuit
of Figure 5 with RC 2.2 kO. The light is
incident upon the base region of the PT and
produces a collector current, IC.
Q As the light intensity incident on the PT
increases, which one of the following statements
concerning the PT-circuit of Figure 5 is
correct? A. The collector current remains
constant, but VOUT increases. B. The collector
current remains constant, but VOUT decreases. C.
The collector current increases, but VOUT
decreases. D. The collector current decreases and
VOUT decreases.
A Alternative C
25
Electronics Photonics RevisionQuestion Type
Rather than use a separate LED and PT, you choose
an opto-isolator chip (the region within the
dotted lines) shown in the circuit of Figure 6.
The opto-isolator is a linear device, that means
that the PT collector current, IC, and the LED
forward current, IF , are related by IC const
IF.
In addition you are also told that the
opto-isolator has a switching time of 8 µs. This
represents the time it takes for the PT collector
current, IC, to respond to a sudden change in
light output from the LED.
26
Electronics Photonics RevisionQuestion Type
Sketch Graphs
Q Figure 7a shows the LED forward current, IF ,
as a function of time. On Figure 7b sketch the
opto-isolator output collector current, IC, as a
function of time, given the initial condition (at
t 0) IC 4 mA when IF 20 mA.
A As shown the slope on the fall and rise
represents the 8 µs delay in the response of the
phototransistor