Download Physics – Further Electronics

Physics – Further Electronics
“the essentials” lectures, 2014
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The principal author of this booklet is:
Dr. Greg Wilmoth, B. Sc (Hons)., Ph.D., Dip. Ed., Grad. Dip. Computing
(Senior VCE Teacher – Haileybury College).
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REGULATED DC POWER SUPPLY
Electricity comes to our home as 240 volt 50 Hz alternating current (AC).
AC has many advantages over DC:
•
More efficiently transmitted.
•
Easily transformed.
•
AC generators are simpler and cheaper.
•
More efficient safety devices work on AC.
(The generation and distribution of AC electricity will be covered in Unit 4.)
Most of our electronic devices require a low voltage direct current (DC). This unit of work
looks at the design of a regulated DC power supply that is able to convert an AC supply to
DC.
Generally there are four main stages involved in the establishment of a regulated DC supply
from an AC input.
Transformer:
Converts the AC supply to a suitable value for conversion.
Rectification:
Converts alternating current into current flowing in a single direction.
Filtering:
Smoothing of the pulsating voltage.
Regulation:
Maintaining a constant voltage over a range of input and load conditions.
Block diagram for a regulated
DC power supply
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The Essentials – Physics – Further Electronics
Page 1
WHY IS A VOLTAGE DIVIDER AN
UNREGULATED POWER SUPPLY?
1.
Effect of varying load on a voltage divider.
Consider the following voltage divider circuit.
What is the voltage output when
the voltage divider is unloaded
(no load resistance placed across
VOUT)? Enter your answer into the
table below.
Now consider a single 100 Ω load placed across VOUT. This load resistance is
effectively arranged in parallel to R2. Determine the combined resistance and apply
voltage divider principles to find VOUT? Enter your answer in the table below.
Add another, then another 100 Ω load across VOUT. Again determine combined
resistance with R2, as well as the output voltage.
Number of 100
ohm loads
Combined R2 / load
Resistance (Ω)
0
100
Output
voltage
1
2
3
What happens to VOUT when the voltage divider is loaded?
2.
Effect of varying input voltage.
Consider what would happen if the input voltage reduces to 10 V from the original
12 V. Voltage divider calculations will reveal that VOUT decreases to 6.7 V. Dropping the
input even further to 9 V gives an output voltage of only 6.0 V. Clearly there is a direct
relationship between VIN and VOUT.
In an unregulated power supply, as the load current increases, the voltage across the
load decreases. This is unsuitable for many applications. Voltage regulation involves
the supply of a constant DC voltage over a range of input voltages and varying load
conditions.
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ALTERNATING CURRENT
The current in an AC circuit changes
direction many times each second.
The free electrons in the conducting
wires move first in one direction and
then the other, creating an oscillating
voltage ‘wave’. In an AC circuit the
electrons simply oscillate back and
forth. At 50 Hz, the electrons change
direction 100 times per second.
In a DC circuit, such as a
battery, the electrons
move in only one
direction.
Alternating
supply
voltage
The peak value of the domestic AC voltage supply oscillates between +340 V and –340 V.
Sometimes the supply is described by its peak to peak variation, in this case 680 V.
i.e. V peak-to-peak or V p-p = 680 V
Both the Vp and the V p-p values are of limited use because they are not representative of the
effective voltage delivered. These values are only maintained for an instant and a lesser
amount is supplied at all other times.
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An arithmetic average would yield zero voltage and current because of the alternating
nature.
The most useful measurement of an AC voltage is what is called its root mean square
(RMS) value. This term refers to the equivalent DC voltage that would be needed to produce
the same power.
Unless otherwise specified, it is usual to assume RMS values as this is the most effective
and useful value.
RMS voltage
V rms =
V peak
2
RMS current
I rms =
I peak
2
Note that the relationship for powerRMS is somewhat different:
PRMS = VRMS × I RMS =
VPEAK I PEAK
V
×I
P
×
= PEAK PEAK = PEAK
2
2
2
2
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QUESTION 1
A cathode ray oscilloscope was used to measure the maximum voltage of an AC power
supply at 18 V. What is the RMS voltage of the power supply?
A
9.0 V
B
12.7 V
C
25.5 V
D
3.0 V
QUESTION 2
What is the peak to peak voltage of a 25 V RMS AC supply?
A
17.7 V
B
35.4 V
C
50.0 V
D
70.7 V
The following information refers to Questions 3 to 5.
An AC current with an RMS voltage of 12 V is applied to a 50 ohm load.
QUESTION 3
What is the RMS current?
A
0.24 A
B
0.03 A
C
0.01 A
D
0.48 A
QUESTION 4
What is the peak current?
A
0.04 A
B
0.34 A
C
0.10 A
D
0.68 A
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QUESTION 5
What is the peak power dissipated in the load?
A
4.08 W
B
2.89 W
C
5.77 W
D
8.15 W
QUESTION 6
What is the RMS equivalent of 0.50 A peak current?
A
0.250 A
B
0.353 A
C
0.707 A
D
0.354 A
QUESTION 7
If the voltage at point A is 200 V, what is the RMS voltage for this supply?
A
141 V
B
283 V
C
71 V
D
566 V
QUESTION 8
The frequency for this supply is 50 Hz. What is the time represented by point D on the
graph? Give your answer in milliseconds (ms).
A
20 ms
B
60 ms
C
30 ms
D
40 ms
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Page 6
TRANSFORMER PRINCIPLES
A transformer is a device that can change the size of AC voltage.
A transformer is a very simple device that consists of two sets of insulated coils that are
wound onto a soft iron core. The first coil is called the primary or input, and the other is the
secondary coil or output.
Load
Primary Winding
Secondary Winding
AC Power Supply
Iron Core
The operating principles of a transformer will be explored more thoroughly in Unit 4. In this
unit it is only a requirement to know what the transformer does, not how it does it.
For the point of this exercise, it is worth noting that the number of coils in the primary coil
compared to the secondary coil will affect the ratio of voltages and currents according to the
following relationship:
Transformer equation:
N P VP I S
=
=
N S VS
IP
You may also come across the term ‘an ideal transformer’. This simply means that it is a
transformer that is 100% efficient, i.e. power in = power out.
QUESTION 9
A transformer has 1000 primary coils and an input of 240 V rms. How many secondary coils
would be needed in order to have an output of 6 volts?
A
18 coils
B
25 coils
C
17 coils
D
40 coils
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QUESTION 10
The diagram below shows a simple transformer, which may be regarded as 100% efficient.
The input terminals P and Q are connected to a 50 V AC power supply. The output terminals
R and S are connected to an AC voltmeter. Based on this diagram, what is the approximate
voltmeter reading?
A
13 V
B
1.0 V
C
10 V
D
14 V
QUESTION 11
The power input into an ideal transformer is 50 W. If the output voltage is 10 V, what current
would be flowing in the secondary circuit?
A
0.5 A
B
2.0 A
C
0.2 A
D
5.0 A
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MEASURING DEVICES
VOLTMETER
Voltmeters are placed in parallel across the component being measured. Their purpose is to
measure the difference in electrical potential (energy level) between the two points of
connection.
It is important that any measuring device being used has minimal effect on the actual circuit
being used. Voltmeters achieve this by having very high resistance. This way they draw very
little current and place negligible additional load on the circuit being tested. However, under
circumstances of high circuit resistance, inclusion of a voltmeter can affect the amount of
current drawn.
AMMETER
Ammeters have very low resistance and are designed to be placed in series within the circuit
being tested.
The low resistance of the ammeter minimises any effect it may have on the resistive load of
the circuit. If the circuit has low resistance however, some effect may be noted.
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OHMMETER
Ohmmeters are designed to measure the resistance of components or sections of circuits.
Ohmmeters can be used with individual components or on open circuits such as the one
shown below. The ohmmeter applies a potential difference (hence it needs its own power
source) to the component and measures the current that flows. The amount of current is
related to the resistance of the circuit.
An ohmmeter can be used to measure continuity, i.e. whether a circuit is continuous or open.
An example of this is the testing of a faulty lead. If the lead has an internal break, the
ohmmeter will record this as infinite resistance and go off scale. If the lead is continuous, the
ohmmeter will record a near zero resistance.
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MULTIMETER
(From Jacaranda Physics 2 – 2nd Ed)
A multimeter is a device that can perform a range of functions. Most can measure DC and
AC voltages, current and resistance; however, others can also test diodes and transistors,
as well as measure temperature and frequency.
Multimeters also have a range of scales. If the approximate value being measured is
unknown, start with the maximum range, and then progressively switch to an appropriate
narrower range.
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CATHODE RAY OSCILLOSCOPE
An oscilloscope is a laboratory instrument commonly used to display and analyse the
waveform of electronic signals. In effect, the device draws a graph of the instantaneous
signal voltage as a function of time.
The CRO produces a beam of electrons that makes a spot on the screen. The beam can be
made to sweep across the screen using the time base control. The vertical deflection results
from the applied input.
(Heinemann Physics 12 – 2nd Ed)
Since a CRO measures voltage, it is connected in parallel with the component across which
the voltage variation is to be measured.
If connected to an AC source it would produce an image such as this.
The grid allows voltage and time to be measured and calculated in conjunction with the
horizontal and vertical scale settings.
A cathode ray oscilloscope is a particularly useful measuring device in this unit because it
not only reveals the AC nature of the input voltage, but it visualises rectification, the extent of
filtering, and the effectiveness of the final regulated output.
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An oscilloscope is connected to an AC signal generator. The vertical scale is set on 5 V/cm,
and the horizontal scale on 20 ms/cm.
QUESTION 12
What is the peak-to-peak voltage of the signal generator?
A
35 V
B
17.5 V
C
7.5 V
D
15 V
QUESTION 13
What is the voltage RMS of the signal generator?
A
5.30 V
B
24.7 V
C
10.6 V
D
12.4 V
QUESTION 14
What is the frequency of the signal generator?
A
20 Hz
B
12.5 Hz
C
50 Hz
D
80 Hz
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RECTIFICATION
Power diode – has a very high breakdown voltage to prevent reverse flow of current.
Note the different scales for reverse and forward voltages, and for forward and
reverse current.
When the voltage across the diode is positive, the diode is said to be forward biased. Before
a silicon diode starts to conduct, or behave like a low resistance conductor, it must have a
voltage of about 0.7 V across it.
When a negative voltage is placed across the diode, it is said to be reverse biased; a small
leakage current of only a few microamperes will flow at best. If the negative voltage is large
enough, the semiconductor will break down and a large current will flow.
Note that when a diode is reverse biased, as shown below, very little current flows through
the diode and the resistor. The voltage drop across the resistor will be negligible and all of
the voltage drop will be across the diode.
In forward bias the diode conducts well and the Vdiode is around 0.7 V.
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A rectifier is a circuit element that is connected to the output of a transformer and converts
the AC voltage into a DC voltage. Rectification can be half-wave or full-wave. Half-wave
rectification simply blocks the negative part of the AC cycle, full-wave rectification converts
the negative cycle into a positive one.
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HALF-WAVE RECTIFICATION
A half-wave rectifier uses a single power diode to pass only one direction of the AC voltage
from the transformer onto another part of the device that is sometimes called the load. The
remaining part of the voltage signal is across the diode, since the diode has a very large
resistance when reverse biased. A half-wave rectifier circuit is shown below. Since the load
has an effective resistance, it is represented as a resistor.
If the input voltage from the transformer has a peak voltage of +9.0 V, the maximum voltage
drop across the load resistance will be +8.3 V, because silicon-based diodes will have a
voltage drop of 0.7 V when they are forward biased. When the input voltage is negative, the
diode effectively prevents current from flowing through the load resistance and the voltage
drop across the load will be zero.
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FULL-WAVE RECTIFICATION
A bridge rectifier uses four diodes to produce full-wave rectification.
Trace the current through the ‘four diode bridge’ for both halves of the AC cycle to determine
the direction of current flow across the load.
There will be an approximate voltage drop of 0.7 V across each diode that the current
passes through. Each half of the AC cycle passes through two diodes in a bridge rectifier.
This means that the peak voltage from a bridge rectifier will be approximately 1.4 V less than
the peak voltage of the AC signal.
If the frequency of the AC supply is 50 Hz, what is the frequency of the rectified output?
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A centre-tapped transformer uses a different method to achieve full-wave rectification.
The centre-tap full-wave rectifier uses only two diodes.
The diodes are connected to either end of the secondary coil of a transformer. The centre of
the coil is connected (tapped) to the earth, hence it is maintained at 0 V. In this case there
will also be an approximate voltage drop of 0.7 V across each diode that the current passes
through.
Here each half of the AC cycle passes through only one diode in the centre-tap rectifier. This
means that the peak voltage from a centre-tap rectifier will be approximately 0.7 V less than
the peak voltage of the AC signal. The output of a centre-tap rectifier can also be smoothed
using a capacitor, as shown in the following section.
QUESTION 15
Which one or more of the above circuits will produce a full wave DC output?
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QUESTION 16
The bridge rectifier consists of four diodes mounted side by side on a heat sink, and
connected by wires. Which one of the following circuits shows the diodes correctly
connected?
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EXAMPLE 1
What is the peak voltage across the load?
Don’t forget to allow for the voltage drop across the rectifying diodes when assigning a value
to VPEAK beyond the rectifier.
Solution
This problem requires three steps to be completed.
Step 1: Determine the RMS voltage output of the transformer.
N P VP I S
=
=
N S VS I P
20 249
=
1
VS
Vs = 12 V RMS
Step 2: Determine the peak voltage output of the transformer.
VPEAK = VRMS × 2 = 12 × 2
= 17.0V
Step 3: Allow for voltage drop across the two diodes.
At any one time current passes through two diodes. Therefore, two potential
decreases of 0.7 V occur.
The peak voltage across the load will be 17.0 – 2 x 0.7 = 15.6 V.
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CAPACITORS
The process of converting the varying DC voltage from the rectifying diodes into a steady DC
voltage is called filtering or smoothing. They can be achieved by placing a capacitor into
the circuit.
In general, circuits containing a resistor and a capacitor are known as RC circuits. This is
an example of a simple RC circuit.
When the switch is closed, current flows through the circuit. Electrons are pulled off one
plate, flow through the battery, and are deposited onto the other plate of the capacitor. This
is called charging the capacitor and continues until the voltage across the plates is equal to
the emf of the cell. The closer the voltage across the capacitor gets to the supply voltage,
the smaller the current becomes.
When the voltages of the cell and the capacitor are equal, the current ceases. When the
battery is disconnected the capacitor retains its charge.
If the capacitor is now placed into a circuit, because it has potential difference across its
plates, it can release this stored up charge. This is known as discharging the capacitor.
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The magnitude of the current is affected by the resistance of the resistor. The bigger the
resistance, the smaller the current will be and the slower the capacitor will charge (and
discharge). If this resistance is decreased, the larger current will allow faster charging.
The time constant, τ, (the Greek letter tau) for an RC circuit is the time in seconds it takes
for the capacitor to reach 63% (or approximately two-thirds) of its final voltage, E, when
charging. The time constant can be calculated by finding the product of the resistance in
ohms and capacitance in farads for the RC circuit.
τ = RC
In a discharging cycle, the time constant is the time it takes for 63% of the charge to
discharge from the plates.
As a rule of thumb, 5 time constants is the time to either fully charge or discharge the
capacitor, but note however, that different time constants may apply to the charging and
discharging cycles as these may involve different resistances.
The charge (Q) stored on a capacitor is proportional to the voltage drop (V) across the
plates.
Q=CxV
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The relationship between the potential difference across the capacitor and the current
flowing to or from the capacitor is interesting:
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Capacitors in series behave like one capacitor with a plate separation greater than that of
the individual capacitors. Because capacitance is inversely related to plate separation, the
capacitance of capacitors in series is less than that of the individual capacitor.
1
C SERIES
=
1
1
+
C1 C 2
Capacitors in parallel behave like one capacitor with a greater plate area than the
individual capacitors. The capacitance of parallel capacitors is the arithmetic sum of the
individual capacitors i.e.
CPARALLEL = C1 + C2
In this unit the most likely arrangement is in parallel. For example, if there is insufficient
capacitance in a filtering circuit it could be handled by placing an additional capacitor in
parallel (or replacing with a larger capacitor).
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The following information refers to Questions 17 to 19.
When switch S1 is closed, capacitor C charges as depicted in the graph below.
QUESTION 17
What is the value of capacitor C? Give your answer in mF.
A
10 mF
B
25 mF
C
15 mF
D
20 mF
QUESTION 18
What is the current passing through the 200 ohm resistor 5.0 seconds after switch S1
closes?
A
0.082 A
B
0.041 A
C
0.091 A
D
0.009 A
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QUESTION 19
After one minute of charging switch S1 is opened and switch S2 closes. What is the voltage
across the capacitor 30 seconds after the closing of S2?
A
10 V
B
3.7 V
C
2.5 V
D
9.0 V
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FILTERING (OR SMOOTHING)
The varying voltage from the rectifier is unsuitable for most electronic circuitry. It is important
to be able to smooth out the signal and make it a constant DC value. This is where the
capacitor is used.
This is a smoothing circuit for a half-wave rectified load.
While the voltage is reaching its peak the capacitor is charging as shown with the half-wave
rectified signal shown below.
As the voltage from the input starts to drop from its peak value the potential difference of the
capacitor is now larger than from the source, so the capacitor starts to release its charge,
maintaining a current flow through the resistor and potential difference across the load.
The size of the time constant determines how quickly this charge is released. The larger the
time constant of the capacitor and resistor combination the smoother the current will be.
Full-wave rectifier output signals can be smoothed in the same way as half-wave rectifier
signals. The main difference is that for a given RC combination, the capacitor has less time
to discharge before it is recharged by the output signal.
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Using a capacitor with a time constant five times longer than the period of the supply
gives a ripple voltage about 20% of Vpeak. This is an acceptable value for many
devices.
QUESTION 20
How long will it take the capacitor to fully charge when switch S1 is closed?
A
0.01 s
B
0.21 s
C
0.05 s
D
0.50 s
QUESTION 21
Switch S1 is opened when the capacitor was fully charged, then switch S2 closes. How long
will it take for the capacitor to discharge?
A
1.0 s
B
0.2 s
C
0.5 s
D
1.4 s
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The following circuit was built to provide high voltage DC to a cathode-ray tube. The DC
voltage has to be approximately 10 kV and the effective resistance of the cathode-ray tube in
the circuit is 5 M Ω (5 x 106 Ω).
QUESTION 22
The primary windings of the transformer have 50 turns. Which of the following best gives the
number of windings required in the secondary windings?
A
5
B
30
C
1500
D
7200
The bridge rectifier consists of four diodes connected appropriately, and the smoothing
capacitor has a value of 10 µF (10 x 10-6 F).
QUESTION 23
Which of the following best gives the time constant for the smoothing circuit of the cathoderay tube?
A
5 x 10-6s
B
0.5s
C
5s
D
50s
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QUESTION 24
Which of the following graphs best shows the signal as observed across XY?
QUESTION 25
Which of the following graphs best shows the signal as observed across CD?
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QUESTION 26
Initially the circuit operates correctly providing the required smoothed output, however, a
fault then develops and the output across EF is shown below:
Which of the following is the most likely cause of this fault? A failure of:
A
B
C
D
The transformer.
The capacitor used for smoothing the signal.
All the diodes in the bridge rectifier are blown.
Only one of the diodes in the bridge rectifier is blown.
QUESTION 27
After fixing this fault another develops, causing the following signal across EF:
Which of the following is the most likely cause of this fault? A failure of:
A
B
C
D
The transformer.
The capacitor used for smoothing the signal.
All the diodes in the bridge rectifier are blowing.
Only one of the diodes in the bridge rectifier blowing.
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QUESTION 28
What would be the lowest power transformer that would be able to provide the cathode-ray
tube with the required power?
A
B
C
D
1W
5W
25 W
100 W
QUESTION 29
If the cathode-ray tube and its power supply, including the transformer, was connected to the
110 VRMS 60 Hz mains supply in the USA, some changes would need to be made? The best
reason(s) for this change is:
A
B
C
D
A different transformer is needed to allow for the different power required.
A different transformer is needed to allow for the different input voltage only.
A different transformer is needed for the different frequency only.
A different transformer is needed to allow for both the different input voltage and
different frequency.
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RIPPLE VOLTAGE
Ripple voltage is the periodic variation in a DC voltage that results from the
rectification/smoothing of an AC voltage.
The peak-to-peak value of the ripple voltage is:
Vripple = Vmax - Vmin
The average voltage will be approximately the mid value between the maximum and
minimum voltage.
The peak-to-peak value of the ripple voltage depends on the following:
•
The maximum value of the voltage (Vmax) supplied by the diode(s) of the rectifier.
•
The period (T) of the unsmoothed voltage supplied by the diode(s) of the rectifier.
•
The capacitance of the capacitor.
•
Load resistance.
And can be expressed as:
Since:
V
= I LOAD
R
Then:
Vr ( p − p ) ∝ I LOAD
Vr ( p − p ) =
V max × T
R×C
=
I LOAD × T
C
Note that with a 50 Hz supply the period is:
Half-wave rectified:
Full-wave rectified:
T = 20 ms
T = 10 ms
Note:
The study design clearly specifies that only a qualitative understanding of ripple voltage is
required. Therefore you won’t be asked to perform calculations such as the one above. The
formula has deliberately been included here because it provides a convenient means for
addressing some of the more likely qualitative questions.
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QUESTION 30
The following circuits have identical AC input, loads, capacitance and use similar diodes.
CIRCUIT A
CIRCUIT B
How would the ripple voltages across each compare? Justify your choice.
QUESTION 31
Indicate whether the following changes to CIRCUIT A from the previous question will alter
the magnitude of the ripple voltage measured across the LOAD.
(a)
(b)
Place a second capacitor in parallel to the original capacitor.
A
Increase
B
Decrease
C
No change
Place a second LOAD in parallel to the original LOAD.
A
Increase
B
Decrease
C
No change
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(c)
(d)
Place a second LOAD in series with the original LOAD.
A
Increase
B.
Decrease
C
No change
Double voltage across the LOAD and double the capacitance.
A
B
C
Increase
Decrease
No change
QUESTION 32
A rectifier supplies power to a 1 kΩ resistor. Which of the following capacitors connected
across the output would give the smoothest DC supply?
A
B
C
D
200 μF
2000 μF
20000 μF
200000 μF
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VOLTAGE REGULATION
ZENER DIODE SHUNT
Zener diodes have a relatively low reverse breakdown voltage, meaning that they can
conduct in reverse bias when a specified voltage is applied. The breakdown voltage of a
zener diode is significantly lower than a power diode.
In this example, the lamp
placed in parallel to the
zener has a potential
difference equivalent to
the zener breakdown
voltage.
There are limitations which will restrict the use of this circuit. The current flowing through the
dropping resistor, as well as through the zener diode, will result in power lost as wasted
heat. If too much power is lost this way either the dropping resistor or the zener diode may
burn out.
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EXAMPLE 2
A zener with a breakdown voltage of 6 V is placed in series with a voltage dropping resistor
connected to a 9 V DC supply.
This circuit delivers a constant 120 mA current. The load is an ohmic conductor with a
resistance of 60 ohms.
(a)
What value voltage dropping resistor should be used?
(b)
How much power is dissipated by the load?
(c)
How much power is dissipated across the zener diode?
(d)
How much energy is wasted as heat across the dropping resistor and the zener diode
during operation of this circuit in 60 seconds?
Solution
(a)
The voltage across the resistor should be 3.0 V and the current 0.12 A.
Applying Ohm’s law gives: R =
(b)
V
3
=
= 25 Ω
I
0.12
How much power is dissipated by the load?
I =
V
6
=
= 0 .1 A
R 60
PLOAD = VLOAD x ILOAD = 6 x 0.1 = 0.6 W
(c)
(or PLOAD =
V 2 62
=
= 0.6 W )
R 60
The current through the zener will equal the total current less what flows through the
load i.e. 0.02 A.
PZENER = VZENER x IZENER = 6 x 0.02 = 0.12 W
(d)
The total power consumption of the circuit is:
Ptotal = VI = 9 x 0.12 = 1.08 W
With 0.6 W dissipated by the load, the balance of the electrical power, which is 0.48 W, is
dissipated as heat through the rest of the circuit (i.e. within the zener and the dropping
resistor).
Energy wasted = power x time = 0.48 x 60 = 28.8 J
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The following information refers to Questions 33 to 35.
QUESTION 33
What is the voltage observed across R1?
A
3.2 V
B
6.0 V
C
8.0 V
D
0. 7 V
QUESTION 34
What is the voltage observed across R2?
A
7.3 V
B
1.0 V
C
4.8 V
D
2.0 V
QUESTION 35
Which one of the following best gives the current through the Zener diode?
A
B
C
D
1 mA
3 mA
8.5 mA
10 mA
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The following information refers to Questions 36 to 38.
QUESTION 36
A 12 V battery is used to run a 9 V mobile phone by using a voltage dropping resistor in
series with a zener diode. The zener has a breakdown voltage of 9.0 V. The telephone uses
180 mA of current and 20 mA flows through the zener diode. What should be the resistance
of the voltage dropping resistor?
A
0.15 Ω
B
2.40 Ω
C
0.54 Ω
D
15.0 Ω
QUESTION 37
How much power is the mobile phone using?
A
0.18 W
B
2.40 W
C
180 W
D
1.62 W
QUESTION 38
How much energy is being wasted in the circuit each second?
A
2.40 J
B
0.78 J
C
1.62 J
D
2.22 J
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The following information refers to Questions 39 to 43.
An oscilloscope is placed across the 500 ohm load and the output is shown in the graph
below:
QUESTION 39
Which one of the following is the best estimate of the average power dissipated in the 500 Ω
load resistor?
A
B
C
D
0.02 W
0.2 W
0.4 W
5000 W
QUESTION 40
Which one of the following is the best estimate of the peak-to-peak ripple voltage?
A
B
C
D
0.5 V
5.0 V
8.5 V
13.5 V
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Voltage regulation is attempted by adding R1 and a zener diode as shown below.
QUESTION 41
Which one of the following best shows the output she will observe?
QUESTION 42
What is the function of resistor R1?
QUESTION 43
Which one of the following changes would best reduce the ripple voltage?
A
B
C
D
Replace R1 with a 50 ohm resistor.
Replace the 6 V Zener diode with a 9 V one.
Replace the capacitor with a 500 uF capacitor.
Replace the transformer with one with a 9.0 VRMS output.
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IC REGULATION
A voltage regulator is an integrated circuit (IC) device that delivers a steady terminal voltage
despite variations in the input voltage or variations in the current drawn by the load.
The three terminals of an IC regulator are:
•
The input from the power supply.
•
The common or ground (0 V).
•
The output, the smoothed and regulated voltage ready for use.
There are two types of commonly used IC voltage regulators. A fixed regulator gives a set
output voltage, and an adjustable regulator can give a range of output voltages.
It is important that the input into the IC regulator is maintained a few volts above the
regulated output voltage. The minimum amount of difference required by the IC regulator is
called the drop-out voltage.
The ripple must not go below the output voltage of the IC regulator.
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Because the input to the regulator must be maintained higher than the regulated output, it
goes that there must be a potential drop across the regulator. The power dissipated in the IC
regulator equals:
PLOSS = Vdrop x ILOAD
But how can you determine Vdrop if the input is not constant?
VIN
≈
Vmax + Vmin
i.e. the input voltage is approximately the mid-ripple voltage
2
and Vdrop = V IN − VOUT .
EXAMPLE 3
A 9 volt regulator is connected to a full-wave rectified power supply with a 500 μF smoothing
capacitor. The load has resistance of 100 Ω and the diodes have a forward threshold of
0.7 V.
12V rms
LOAD – 100 ohms
(a)
What is the maximum voltage supplied to the regulator?
(b)
With a ripple voltage of 3.1 V, what is the power loss in the voltage regulator?
(c)
What is the power loss in the load?
Solution
(a)
V peak = Vrm × 2 = 16.97 less 2 x Vdiode = 15.6 V
(b)
The minimum voltage is the maximum voltage less the ripple voltage.
i.e. 15.6 – 3.1 = 12.5 V
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To find the power loss we need to determine the average potential drop across the
regulator, as well as the load current.
Vav =
Vmax + Vmin 15.6 + 12.5
=
= 14.05 V (a close approximation)
2
2
Vdrop = 14.05 – 9.0 = 5.05 V
I LOAD =
V LOAD
9
=
= 0.09 A
R LOAD 100
PLOSS = VDROP x ILOAD = 5.05 x 0.09 = 0.45 W
Note that the value ‘V’ used in Ohm’s Law and in power formulae does not represent
voltage, rather potential difference. It is the drop in electrical potential across the
element considered.
(c)
PLOAD = VLOAD x ILOAD = 9.0 x 0.09 = 0.81 W
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The following information applies to Questions 44 to 51.
The RMS output of the transformer is 12 V AC 50 Hz. The diodes are typical silicon power
diodes with a forward threshold of 0.7 V. The load has 200 Ω, and C = 1000 μF.
QUESTION 44
What is the peak voltage at point X?
A
10.6 V
B
16.3 V
C
24.0 V
D
15.6 V
QUESTION 45
With S1 closed and S2 open, what is the value of Vripple at X?
A
12.0 V
B
0.0 V
C
15.6 V
D
10.6 V
QUESTION 46
With S1 open and S2 closed, what is Vripple?
A
15.6 V
B
12.0 V
C
0.80 V
D
0.0 V
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With both S1 and S2 closed there is a ripple voltage of about 0.8 V.
QUESTION 47
If a second 1000 μF capacitor is now arranged in parallel to the original capacitor, what is
the closest value of the ripple voltage?
A
1.6 V
B
0.4 V
C
0.6 V
D
0.9 V
QUESTION 48
A second identical load is added in parallel to the original. What happens to the ripple
voltage?
A
1.6 V
B
0.4 V
C
0.6 V
D
0.9 V
QUESTION 49
Explain why is this circuit classified as unregulated?
It is decided to add a 7812 voltage regulator (12 V) to the circuit. The ripple voltage received
at the input pin to the regulator is 0.8 V and the regulator has a drop out voltage of 2.4 volts.
QUESTION 50
With a single 200 Ω load across the voltage regulator, what power loss is dissipated within
the regulator?
A
0.72 W
B
0.14 W
C
0.89 W
D
0.19 W
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QUESTION 51
With another load applied to the circuit it was found that a 2.5 V ripple was observed at the
input to the regulator. What does minimum voltage value at the input pin drop down to?
A
15.5 V
B
13.1 V
C
14.4 V
D
12.0 V
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HEAT DISSIPATION
Under certain circumstances, voltage regulators can get so hot that they can lead to circuit
failure.
Voltage regulators are often mounted on heat sinks that help to disperse the heat produced.
If for instance the supply averaged 3 V above the output for a 1 amp current, that would
equate to 3 W of heat being dissipated within the regulator (P = V x I). This is a considerable
amount that would cause it to heat up considerably.
Heat can move from one place to another by three processes: Conduction, radiation and
convection. Features of heat sinks that facilitate heat dissipation include:
•
They are usually made from aluminium because this material is an excellent conductor
of heat. Copper is sometimes used.
•
The component that needs to have thermal energy removed from it must make good
contact with the heat sink. Thermal grease smeared between the surfaces facilitates
heat transfer.
•
Heat sinks may be coloured black because this enables them to better radiate heat
energy.
•
Heat sinks are designed so that they have a large surface area in contact with the air.
This allows the thermal energy to be better conducted from the sink to the air. The hot
air expands, becomes less dense and rises, carrying the energy away with it. This
process is known as convection.
Electronic circuits that use heat sinks usually need to have ventilation holes in the appliance
casing so that cool air can enter the casing, come into contact with the heat sink, heat up
and carry heat energy away through the ventilation holes. Cooling fans can also be added in
cases of extreme heat accumulation.
If voltage regulators do overheat, they are designed to automatically shutdown if their
operating temperature exceeds a certain value. As they cool down they switch back on
again.
A drop in their output voltage may be experienced prior to thermal shutdown, as well as a
reduction in the amount of current that they can conduct.
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SOLUTIONS
FOR ERRORS AND UPDATES, PLEASE VISIT
WWW.TSFX.COM.AU/VCE-UPDATES
QUESTION 1
Answer is B
12.7 V
QUESTION 2
Answer is D
70.7 V
QUESTION 3
Answer is A
0.24 A
QUESTION 4
Answer is B
0.34 A
QUESTION 5
Answer is C
5.76 W
QUESTION 6
Answer is D
0.354 A
QUESTION 7
Answer is A
141 V
QUESTION 8
Answer is C
30 ms
QUESTION 9
Answer is B
25 coils
QUESTION 10
Answer is D
14 V
The reading on the voltmeter will be less than 50 V because the number of turns of wire is
less.
An approximate calculation is: Vs = Vp ×
Ns
3
= 50 × = 14 V (approximately).
Np
11
QUESTION 11
Answer is D
5.0 A
QUESTION 12
Answer is A
35 V
QUESTION 13
Answer is D
12.4 V
QUESTION 14
Answer is C
50 Hz
QUESTION 15
Answer is A and D
QUESTION 16
Answer is B
QUESTION 17
Answer is C
15mF
One time constant is 3.0 s = RC; C = 0.015 F = 15 mF
QUESTION 18
Answer is D
0.009 A
At 5.0 s VCAP = 8.2 V, VRESISTOR = 1.8 V; Current = 1.8 / 200 = 0.009 A
QUESTION 19
Answer is B
3.7 A
30 seconds = one discharge time constant. Therefore it has discharged 63%, giving
remaining voltage at 3.7 V.
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QUESTION 20
Answer is C
0.05 s or 50 ms
QUESTION 21
Answer is A
One second.
QUESTION 22
Answer is C
QUESTION 23
Answer is D
QUESTION 24
Answer is D
QUESTION 25
Answer is A
QUESTION 26
Answer is B
QUESTION 27
Answer is D
QUESTION 28
Answer is C
QUESTION 29
Answer is B
QUESTION 30
Circuit A has a frequency of 50 Hz, whereas Circuit B has a frequency of 100 Hz. Therefore,
the period of Circuit A (20 ms) is double that of Circuit B (10 ms).
Since VRIPPLE ∝ Period , Circuit A has approximately double the ripple of Circuit B.
QUESTION 31
(a)
Answer is B
Decrease
(b)
Answer is A
Increase
(c)
Answer is B
Decrease
(d)
Answer is C
No change
QUESTION 32
Answer is C
2000 µF
QUESTION 33
Answer is B
6.0 V
QUESTION 34
Answer is D
2.0 V
QUESTION 35
Answer is C
8.5 mA
QUESTION 36
Answer is D
15.0 Ω
Answer is D
1.62 W
R=
V
3
=
= 15 Ω
I 0.2
QUESTION 37
P = V × I = 9 × 0.18 = 1.62 W
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QUESTION 38
Answer is B
2.22 J
Ptotal = VI = 12 x 0.2 = 2.4 W
Power wasted is 2.4 – 1.62 = 0.78 W. Therefore 0.78 J is wasted each second.
QUESTION 39
Answer is B
0.2 W
QUESTION 40
Answer is C
8.5 V
QUESTION 41
Answer is A
QUESTION 42
R1 is a dropping resistor which takes the potential difference between the zener diode and
the rectifier output.
QUESTION 43
Answer is C
Replace the capacitor with a 500 µF capacitor
QUESTION 44
Answer is D
15.6 V
15.6 V (convert Vrms to Vpeak then subtract 2 x Vdiode)
QUESTION 45
Answer is B
0.0 V
Zero ripple because there is no load current, therefore the capacitor cannot discharge.
QUESTION 46
Answer is A
15.6 V (as there is no smoothing.)
QUESTION 47
Answer is B
0.4 V
With an inverse relationship between ripple voltage and capacitance, the ripple will
approximately halve.
QUESTION 48
Answer is A
1.6 V
The second load doubles the current (or halves the total resistance depending on how you
want to look at it). This doubles the ripple voltage to about 1.6 V.
QUESTION 49
It is classed as unregulated because the average voltage supplied to the load will decrease
as the current through the load increases.
With a high resistive load: Vaverage ≈ V peak = 15.6 V .
While the peak voltage will be maintained, the minimum voltage decreases as the current
increases due to increasing ripple. Consequently the mean voltage falls as the current
increases.
Also, if the primary supply voltage changes, so too will the peak voltage, and hence the
mean voltage to the load.
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QUESTION 50
V AVERAGE =
Answer is D
0.19 W
VMAX + V MIN
15.6 + 14.8
=
= 15.2
2
2
V regulator = 15.2 − 12.0 = 3.2 V
I LOAD =
VLOAD
12
=
= 0.06 A
RLOAD 200
PLOSS = Vregulator × I LOAD = 3.2 × 0.06 = 0.19 W
QUESTION 51
Answer is B
13.1 V
There will be a small voltage fluctuation from the regulator.
With a dropout voltage of 2.4 V, the output of the regulator will fall if its input drops below
14.4 V. With a ripple of 2.5 V, the minimum voltage at the input pin of the regulator drops to
13.1. This is below the level needed to maintain constant output.
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