Download Summary Notes for Further Electronics

Detailed study 3.3 - Further electronics

An understanding of electronic systems may be approached through the study of the functions of
the basic building blocks of such systems, both separately and combined.

The following construction project to be used for the development of understanding of
component principles and demonstration of practical skills is a low voltage AC to DC voltage
regulated power supply system. Knowledge and understanding of the role of the transformer in
the power supply system is required. Knowledge and understanding of internal working of the
transformer is not required.

The project will require the connection of components into a functional electronic system, as well
as the use of appropriate test and measuring equipment.

Students will use electronic devices, circuits, test and measuring equipment in the context of the
design and evaluation of a low voltage AC to DC voltage regulated power supply system. They
will use safe and responsible practices when completing experiments and/or investigations.

On completion of this unit the student should be able to design an AC to DC voltage regulated
power supply system; and describe and explain the operation of the system and its components,
and the effects of test equipment on the system.
To achieve this outcome the student should be able to:
 design and investigate an AC to DC voltage regulated power supply system, given a range of AC
input voltages (specified as root mean square, peak, and peak-to-peak), smoothing conditions and
resistive loads;
 describe the role of a transformer, including the analysis of voltage ratio,
N 1 / N 2 = V 1 / V 2 , but not including induction;
 describe effects on the DC power supply system of changes to the components used;
 interpret information from the display of an oscilloscope in terms of voltage as a function of time;
 analyse circuits, including fault diagnosis, following selection and use of appropriate measuring
devices, including analogue meters, multimeters, oscilloscope;
 evaluate the operation of a circuit in terms of its design brief by selecting measurements of potential
difference (voltage drop) and current (using analogue meters, multimeters and an oscilloscope) in the
DC power supply circuit;
 explain the function of diodes in half-wave and full-wave bridge rectification:
 explain the effect of capacitors in terms of
- potential difference (voltage drop) and current when charging and discharging
- time constant for charging and discharging, τ = RC
smoothing for DC power supplies;
 apply the current—voltage characteristics of voltage regulators, including Zener diodes and integrated
circuits, to circuit design;
 describe, qualitatively, the effect on the magnitude of the ripple voltage of changing the effective
load, the capacitance and the input supply voltage (magnitude and period);
 describe the use of heat sinks in electronic circuits;
 calculate power dissipation in circuit elements. P = VI, P =I2R, P = V2/R;
 identify and apply safe and responsible practices when undertaking investigations involving electrical
and electronic equipment.
Summary: FURTHER ELECTRONICS
A.C. power supplies. Distinction between A.C and D.C current

A D.C. current is one that flows in one direction at all times. The size of the current may vary. An
A.C. current is one that flows in either direction at different times. A.C. current is powered by an A.C.
power supply. The most common waveform used in an A.C. supply is a sinusoidal one but square
wave or sawtooth waveforms are used. In audio, the waveform is complex and mimics the sound
being recorded or sent to speakers.
Frequency, period, amplitude and RMS values for A.C. Voltage and A.C. Current

the frequency is the number of times per second that a periodic signal repeats itself. For example 50
times per second means a frequency of 50 hertz.

the period is the time taken to repeat one cycle of a periodic signal. It can be calculated by taking the
reciprocal of the frequency. For example a 50 Hz signal would have a period of 1 50 th of a second,
that is 0.020 s.

The amplitude is the magnitude of the signal from 0 to its peak value.

The RMS value is the D.C. equivalent voltage that would deliver the same power to a load. The peak
value and RMS value are related for a sinusoidal signal. Peak Value = 2  RMS value. This is valid
for both voltage signals and current. For example the RMS voltage of mains supply in Victoria is 240
V and thus its peak value is 2240  340 V
Household circuits : Active, Neutral and Earth lines
In Victoria, as elsewhere in the world electricity is available from commercial grids in A.C. form. In
Australia a 50 Hz frequency, 240 V RMS supply is used.
period
amplitude
t
1
period = frequency
Vrms
= Vamplitude
2

This is delivered to houses via high voltage transmission and step-down transformers, but more on
that later.

Mains voltage to a house is 240V A.C. It is delivered along the Active line, using the Neutral return
line and a third line called the Earth. The Earth is used as a safety net to protect against short circuits
to the casing of appliances, for example.

The fuse protects the household wiring as well as the appliance. Any fault in an appliance could cause
a surge of electricity. The fuse will burn out, rather than some expensive component, and cut off the
supply. This prevents excessively large currents from flowing through the household wiring and being
a potential fire risk. Fuses are always placed in the active line preceding the appliance, so that if it
blows it isolates the appliance from the active.

Remember the active line's potential varies from + 340 V to - 340 V 50 times per second.

The earth lead protects the user. Usually the case of an appliance is plastic, or, if metal, the case must
be earthed. A fault in the appliance could cause the case to be connected to the live wire - any current
will go to earth rather than through the user and the fuse would also blow.
Power plug

The live or active wire: is attached to the Active line.

The neutral wire: Power is extracted between the live and neutral wires.

The earth is attached to the casing of the appliance.
The Switch: Always in the active line again to isolate the Active from the circuit.
Fuses
Fuses should always be placed in the active line so that in the event that they blow they isolate the device
that they are protecting from the active line. Fuses may be simply pieces of metal that melt if a current
greater than a designated amount passes through it. They fuse may be one which triggers open if the
current on the active line differs from the current on the neutral line indicating that charge is flowing
where it ought not to.
The transformer( ideal case) and qualitative understanding of operation

The transformer is a device that changes the amplitude of an AC voltage. The input is an AC voltage
and the output is also an AC voltage. It consists of two coils called the primary and secondary coil and
a core of soft iron.

The device acts to either increase the amplitude of the AC signal ( a step-up transformer) or decrease
the amplitude of the AC signal (a step-down transformer). On the primary side, the input voltage Vp
is dropped across the primary coil with turns Np. The input current is Ip. On the output side referred
to as the secondary side, the output voltage VS is dropped across the secondary coil with turns NS.
The output current is IS. The AC output voltage arises as a consequence of magnetic effects that occur
in the soft iron core that you will learn about later in Unit 4 when magnetism is studied.

The transformer shown below is a step up transformer with a coil ratio of 2. That is a 12 V RMS
sinusoidal voltage across the primary will produce a 24 V RMS sinusoidal voltage across the secondary
output.
Np =4 turns
Ns = 8 turns
Governing equations for an ideal transformer are
Power in = Power out ( ideal transformer). This leads to the result
VS I P

VP I S
ideal transformer equation 1
For example if the secondary current is 4.0 A RMS with a secondary voltage 12 V RMS then the power
output of the transformer is 4 × 12 = 48 W and so the input power would also be 48 W. If the primary
voltage was 240 V RMS the input current would only need to be 0.20 A RMS. Thus the voltage ratio
12:240 ie 12/240 = 0.05 is precisely the inverse of the current ratio 4:0.20 ie 4/0.20 = 20.
Conservation of magnetic flux  around a closed loop with no leakage This leads to.
V
N
V P VS

or S  S
N P N S VP N P
ideal transformer equation 2
For example in the above transformer the ratio
NS
= 8/4 = 2. Thus an input voltage of 12 V RMS
NP
would yield an output voltage of 24 V RMS. In this case the transformer would be classified as a step-up
transformer.
Common transformers are step-down ones that take 240 V RMS inputs from mains supplies and output 12
V RMS or 6 V RMS for a myriad of applications.
Lets try a few simple problems.
1) Find Vs
2) Find Is and thus Ip
3) Find the power delivered to the
transformer.
Answers:
1) 792 V
2) 99A, 17.8A
3)78.4 kW.
Vp = 4400 V
Vs
factory
8.0 
transformer : Np = 1000, Ns = 180
For non-ideal transformers, heat is transferred to the core via eddy currents and hence the secondary
power output is less than the power input. Also the secondary voltage Vs would be less than indicated by
equation 2 above.
diodes
Diodes are semiconductors that have negligible resistance to the flow of current in one direction and very
large resistance in the other. In this way they can convert an AC current into a DC current.
A second useful way of looking at a diode is that it is a device that passes a positive potential difference,
that is negligible potential difference drops across the diode [about 0.70 V drop is sufficient for it to
conduct] when positively biased and when negatively biased causes all the available potential difference to
drop across the diode rather than elsewhere in the circuit. The use of one diode is often referred to as half
wave rectification and the use of a 4-diode circuit as full wave rectification. Half wave rectification of an
AC signal is examinable but the 4-dide bridge rectification circuit is not.
diode
load
50 Hz
half wave rectification using a single diode
full wave rectification with four diodes
voltage
voltage
time
time
Capacitors

Capacitors are devices that store charge when a voltage is placed across their terminals. They can
“charge up” and “discharge” by acquiring and losing charge respectively. The size of a capacitor is
determined by the amount of charge it can store per unit volt of potential difference across the
terminals.

The size of a capacitor is called the capacitance of the capacitor.

The capacitance C has the unit farad named after Michael Faraday. Typical capacitors used in circuits
range from nanofarad (nF) to millifarad (mF). The rate that a capacitor charges or discharges jointly
depends on the size of the capacitor and the size of the resistance of the charging or discharging
circuit.

Capacitors store charge by having insulated pairs of plates that store positive excess charge on one
plate and negative excess charge on the other. The size of a capacitor is called its capacitance C and is
measured by the amount of charge stored Q per volt V of potential difference applied across the
capacitor. Thus : C 
Q
C V -1 or F .
V
+


C = 10 mF = 10 mC per V
Capacitors take time to charge and discharge. In a circuit such as the one shown below the
characteristic charging time would be 2.0 second. The discharge time would be 1.0 second. This time
can be calculated by multiplying the size of the electrical resistance and the size of the capacitor.
o
In the charging circuit this is 2 k  1 mF = 2.0 s.
o
In the discharge circuit this is 1 k  1 mF = 1.0 s.
Note the unit ΩF, the ohm-farad has the same unit as time, second.
charging circuit
Discharging circuit
The charging and discharging curves are shown for the capacitor circuit above with the appropriate pairs
of switches opened and closed to facilitate charging and discharging.
Switch closed
to discharge
Switch
closed to
charge
The time it takes a capacitor to charge to 63% capacity or to discharge to 37% of capacity is called the time
constant .
The constant can be calculated from the capacitance C and resistance R of the circuit.
  RC
For the circuit shown above, the time constant for the charging circuit is 2  103  1  10-3 = 2 second
while the time constant for the discharge circuit is one second.
diodes and capacitors: converting AC into DC.
Used together in a circuit, a diode/capacitor network can convert a sinusoidal AC signal into a steady DC
voltage. The diode converts the AC to DC and the capacitor smooths the variable voltage to make it more
constant. This is shown below.
S1
Vrms = 10 V
50 Hz
heating
element
capacitor
Circuit 1
Circuit 2
However the DC voltage is not fully constant and a “ripple” appears. You can readily observe this with a
power supply in a school laboratory.
size of ripple
voltage is
approx 2 V
With a larger capacitor the size of the ripple can be decreased to give a more stable DC voltage.
voltage ripple
The output voltage in the diagram above results from an AC supply rectified with a bridge diode and
smoothed with a capacitor. The average voltage is about 7.0 V and the peak to peak ripple is about 2 V.
For many systems a steady DC voltage is required and hence the amount of ripple is required to be
minimised. This can generally be achieved by using a large smoothing capacitor. However large capacitors
require a large movement of charge before they can smooth and hence place demands on the power supply
initially.
1) Smother output voltages are obtained with the use of larger capacitors.
2) If the load resistor using the power supply decreases, this will cause the current to the load to increase
and hence the size of the ripple voltage will also increase as well as the mean value decreasing. You need
to refer specifically to the practical exercise done using a real power supply.
3) Comparing the time constant  = RC of the DC power supply to the period T of the AC supply voltage
gives a good indication of how smooth the DC voltage is. If

T
is much greater than 1 then good
smoothing is achieved. If it is less than one the smoothing is not great. Using full-wave rectifying bridges
also helps to improve the amount of smoothing.
voltage regulators: zener diodes
Zener diodes are tailor made diodes with a predetermined break-down voltage in reverse bias. This can be
controlled by the amount of doping of impurities into the semi-conducting crystal. Zener diodes are used
to provide steady DC voltages after an AC voltage has been rectified and smoothed. They are placed in
series with a second resistor again acting as a potential divider circuit and are oriented in reverse-bias.
If the break-down voltage of a particular zener diode is 4.5 volt and the input to the divider is V(t) = 5 +
0.2sin t, that is a 5 volt signal with a 0.2 ripple then the output of the zener will be 4.5 V and the remaining
voltage 0.5 + 0.2sin t will drop across the upper resistor. In this way a constant 4.5 V can be maintained at
the output.
active, V(t)
voltage across resistor will be
0.5 + 0.2 sint ie the remaining
voltage
Voutput = 4.5 V
zener diode
Integrated circuit voltage regulators
With the development of miniaturised semiconductor circuits, voltage regulators have been manufactured
as integrated circuits for millions of different applications. They typically are designed to have an
unregulated and unstable voltage input, but constant to within a predetermined value and a set constant
output over a wide range of output currents. A typical example is the Toshiba 7805 which gives a constant
output of 5.0 V for a variety of inputs. The output voltage is also, to a large extent, independent of the
resistance of the load and hence is constant for a wide range of applications where the output current may
vary. The device has 3 terminals, an input voltage and output voltage and a neutral and is shown below.
The typical operating characteristic curve for the 7805 voltage regulator is shown below with 3 different
output currents.
Note the output current becomes stable provided the input is big enough and the larger the current drawn
the higher the input voltage required for a 5 V stable output.
output voltage (V)
8
output current = 0 A
6
4
output current = 1.0 A
2
output current = 0.5 A
3
6
9 12 15
input voltage (V)
The diagram below gives the circuit for a typical integrated circuit of which the 7805 is an example
A good symbol for a voltage regulator is shown below. It is a three terminal device with an input (any DC
voltage above the specified output will do) an output which gives the specified DC voltage and a ground
terminal
in
out
gnd
heat sinks
Since a lot of integrated circuits are small a large amount of heat is generated in a small volume. For
example if a chip has several million transistors and there are microamp sized-currents in the myriads of
circuits then is plausible that several joules of heat energy may be produced each second. To deal with this
and remove heat from the chip heat sinks are physically attached to the chip. These are typically black with
fins to conduct heat from the chip and they may involve dc motors with fans to convect the heat away.
Semiconductor circuits, like most things only operate within a limited band of temperatures and so for the
correct operation they need to be cooled they generate too much heat. Heat generation can also be
minimised if the input voltage is as close as possible to the desired output voltage so that the excess voltage
is not large. Hence the excess power dissipated in the regulator will not be large either.