Добавил:
ivanov666
Опубликованный материал нарушает ваши авторские права? Сообщите нам.
Вуз:
Предмет:
Файл:Basics of electronics. Study aid
.pdf
One application of a step-down transformer is in a low-voltage power
L
L
L
supply, in which the supply mains voltage is reduced to a lower voltage
(and then converted to d.c.) for use in electronic circuits. Such transformers
also improve the safety of mains-driven equipment because the two windings
are electrically isolated from one another.
Note the graphical symbol for a transformer with a ferromagnetic core
shown in fig. 2.15(a). If the transformer does not have such a core, the
vertical line in the symbol is omitted.
i
1
v
S
Fig. 2.15. Transformer circuits: (a) a transformer with open-circuit secondary. This
primary current (i) is called the magnetizing current i. (b) A transformer with a load
N
1
e
1
L
1
(a) (b)
TV
N
2
v
e
2
resistance
i
1
S
Primary Circuit Secondary Circuit
e
N
1
1
TV
N
2
e
2
i
2
R
L
If the secondary coil of a transformer is not connected to a load, as in
fig. 2.15(a), and is left open-circuit, the secondary current must be zero.
Under these circumstances the secondary coil can be ignored and the primary
v L di dt
coil simply behaves like an inductor in which
primary inductance. The primary current
zero, is called the
magnetizing current. Transformers are usually designed
i, when the secondary current is
where L1 is the
1s
to have relatively large inductances so that
i is small compared with the
current that flows when the secondary is connected to its intended load. So
we can ignore it for the moment.
Figure 2.15(b) shows the transformer connected to the load resistor which
resistance is
the current in
That is
ieR
22
. Since the voltage appearing across the secondary winding is
R
, and in the secondary circuit as a whole, is
R
divided by RL.
e
2
.
,
e
2
Thus, although there is no direct connection between these two
windings, the primary current is affected. The law of conservation of energy,
demands that, to a first approximation, the output power of the transformer
41
must be equal to the input power. But the instantaneous output power is
p
and the instantaneous input power is
. So, if there is no power loss,
vi
1S
ei
22
or, substituting for
,
v
S
vi ei
122S
iN iN
12
.
s
These equations are applied to any waveform. If the input waveform is a
sinusoid, then it follows from these equations that all waveforms are
sinusoids. Both the input and the output voltages are either in phase with each
other or completely out of phase, depending on the way the secondary
winding is connected. Suppose that
e
2
and
are in phase.
e
1
If the load is a resistor, the output current must be in phase with the
output voltage, that is
e
2
and
are in phase. Then, all the currents and
i
2
voltages are in phase. The only out-of-phase current is the magnetizing current
which lags
inductor. The total input current is then the sum of two currents,
/2 radians behind the input voltage, as in the case with any
and
i
1
i
,
which are /2 radians out of phase with each other. You should know how to
calculate the sum of two sinusoidal currents that are not in phase.
The resistance ‘seen’ by the source is called the
input resistance. If
i
is negligible, the input resistance can be calculated as
eN N N
input resistance
R
ee
in
2
12
iNiNiN
122
pp p
ss s
So if the load is a resistor of the resistance
2
.
R
, then e2/i2 = R
L
and the
L
input resistance is
R
L
R
in
NN
sp
.
2
A step-up transformer, therefore, apparently reduces the resistance of a
load resistor by the square of the turns ratio.
Consequently, if a transformer has a capacitor as a load, the source
‘sees’ a capacitance rather than a resistance. That is, the phase of the current
i
leads the input voltage vS by /2 rad. But the capacitance it sees is
1
2
NN
sp
times the capacitance of the actual load. So a transformer can be
used to increase or decrease the effective capacitance of a capacitor. In
general, therefore, the phase difference between the input current and voltage
(assuming
i is negligible) is determined by the load, not by the transformer.
42
Self-Assessment Questions
1. Draw the sinusoidal signal. What is the amplitude if it is known that
V
equals 230 V? How can we calculate the average voltage?
rms
2.
What is the frequency of a signal if the period is equal to 1 ms?
3.
What does the term ‘phase difference’ mean? Explain it by using
waveforms.
4.
How much energy is dissipated by a 100 resistor in 10 seconds if
2 A of current are flowing?
5.
Ten volts is placed across the terminals of a 200 F capacitor. How
much charge can be stored by the capacitor? What is the highest possible
energy stored by this capacitor?
6.
What is the capacitance of the equivalent capacitor in the series
connection of two capacitors?
7.
Write the dependences between the current flow and voltage across
the inductor.
8.
A 3 mH, a 2.2 mH, and a 0.3 mH inductor are connected in series.
What is the total inductance?
What is the voltage across a coil when di/dt = 200 mA/s and
9.
L = 80 H?
10.
What is the electrical transformer? What types of transformers do
you now?
11.
Which materials can be used for transformer magnetic core?
12.
Give the expression for the turn ratio of a transformer. What is the
difference between step-up and step-down transformers?
43
CHAPTER 3. SIMPLE A.C. CIRCUITS
I
R
After studying this chapter you will be able to understand the operation
of elementary low-pass and high-pass networks containing a resistor and a
capacitor or a resistor and an inductor.
3.1. Low-Pass Networks
A.C. Response of CR
Network
Figures 3.1 and 3.2 show two single CR (capacitor-resistor) and LR
(inductor-resistor) networks driven by sinusoidal input signals. In each case,
the capacitor and the resistor (or inductor and resistor) form a potential
divider, so that the output signal is an attenuated version of the input signal.
R
Gain=V
1
C
v
S
v
out
0
Fig. 3.1. The low-pass CR circuit and its frequency response
out
/ V
S
Frequency
In fig. 3.1 the capacitor’s reactance is much greater than the resistor’s
resistance at low frequencies, so there is little attenuation at low frequencies.
At high frequencies, the capacitive reactance
and attenuates the output signal. So, this is a
/C falls with the frequency,
low-pass filter, which ‘passes’
low-frequency sinusoids but attenuates high frequencies.
The amplitude ratio is
VIX X
out C C
V
s
22 22
RX RX
CC
.
An alternative form of this expression can be obtained by using
1
XC
C
. First, divide top and button of the expression by XC, giving
V
out
V
s
11
2222
1
2
X
C
1
CR
.
This expression represents the graph of the amplitude ratio (or gain)
versus the frequency which is shown in fig. 3.1. This graph is called the
frequency response. Note the following features.
44
(a) At frequencies which are sufficiently low, such that
R
R
R
amplitude ratio .
(b)
At frequencies which are sufficiently high, such that
the amplitude ratio approaches , which is inversely proportional to the
1
CR
1
222
222
1CR
, the
1CR
frequency.
At the frequency at which
(c)
222
1CR
, that is when
1RC
, the
,
amplitude ratio is
12. This frequency is called the cut-off frequency of the
network.
1
The angular cut-off frequency
is determined by
c
C
C
. It
corresponds to the frequency at which the reactance of the capacitor becomes
equal to the circuit resistance. Thus, when
X
C
or
1
, we have
C
C
.
C
The product of the resistance and the capacitance has dimensions of
time. The product RC is known as the
time constant of the RC circuit. The
time constant, denoted by the symbol , affects the rate of change of the
output waveform. The greater , the slower the output rises in response to the
input step voltage.
A.C. Response of LR Network
The LR circuit of fig. 3.2 acts as a low-pass filter, with the inductive
reactance attenuating high frequencies.
L
1
Gain=V
out
/ V
S
v
S
R
v
out
0
Frequency
Fig. 3.2. The low-pass LR circuit and its frequency response
The amplitude ratio is
VIR R
out
V
s
22 22
IR X R X
LL
.
Another form of this expression can be obtained by substituting
XL
L
. First, divide top and bottom of the expression by R, so
V
out
V
s
11
2222
XLR
1
L
2
R
45
1
.
This expression represents the graph of the amplitude ratio versus the
R
R
L
R
R
frequency (frequency response) which is shown in fig. 3.2. Note the
following features.
(a)
At frequencies which are sufficiently low, such that
22 2
1LR
,
the amplitude ratio tends to the unity.
(b) At frequencies which are sufficiently high, such that
22 2
1LR
,
the amplitude ratio approaches , which is inversely proportional to the
LR
frequency.
At the frequency at which
(c)
amplitude ratio is
12.
The angular cut-off frequency
L
C
. It corresponds to the frequency at which the reactance of the
22 2
of this circuit is determined by
c
1LR
, that is when
inductor becomes equal to the circuit resistance. Thus, when
L
, the
, we have
C
the relationship
The ratio
X
or
L/R has dimensions of time and is called the time constant of the
L
.
C
LR circuit. The greater the value of the time constant, the longer it takes the
current to get to its final value. The greater the value of
R, the less time it takes.
3.2. High-Pass Networks
A.C. Response of CR
Network
In fig. 3.3, the positions of the capacitor and the resistor are swapped
(compare with fig. 3.1). Now the high reactance of the capacitor at low
frequencies attenuates the output, but at high frequencies its reactance is
negligible and there is little attenuation. So this is a
high-pass filter. Its
frequency response is also shown in fig. 3.3.
C
Gain=V
out
/ V
S
1
v
S
R
v
out
0
Frequency
Fig. 3.3. The high-pass CR circuit and its frequency response
46
A.C. Response of LR Networks
L
L
The inductor-resistor circuit of fig. 3.4 acts as a high-pass filter
to sinusoidal inputs. At low frequencies, where the inductive reactance
is much lower than R, the output voltage is much smaller than the input. But
at high frequencies, where
R , the output voltage is nearly equal to the
input. The frequency response of this circuit is shown in fig. 3.4.
R
Gain=V
out
/ V
S
1
v
S
L
v
out
0
Frequency
Fig. 3.4. The high-pass LR circuit and its frequency response
Self-Assessment Questions
1. Time constant: what does it mean? Find time constant for CR- and
LR-circuit.
Give the circuit of CR low-pass filter and its frequency response.
2.
Explain the curve.
3.
15 V
In a series RC-circuit, 12 V
is measured across the capacitor. Find the r.m.s source voltage.
(rms)
A 20 kHz sinusoidal voltage is applied to a series RC-circuit. What
4.
is measured across the resistor and
(rms)
is the frequency of the voltage across the resistor?
5.
The frequency of the voltage applied to a series RC-circuit is
decreased. How does the impedance of this circuit changed?
Draw the frequency response of a high-pass filter. Explain how to
6.
find the cut-off frequency.
7.
An RC low-pass filter consists of a 110 resistor and a 0.002 F
capacitor. The output is taken across the capacitor. How can we find the
circuit’s cut-off frequency?
8.
An RC high-pass filter consists of an 8.2 k resistor. What is the
value of
C so that impedance X
is ten times less than R at an input frequency
C
of 12 kHz?
9.
What is the difference between Pass Band and Stop Band?
What is the term for a ten-times change in frequency?
10.
47
CHAPTER 4. DIODES
This chapter concentrates on the simplest semiconductor device – the
diode – which has one basic property: it conducts the current much more easily
in one direction, if compared to the other. This depends above all on properties
of p-n junctions, which are the junctions between two different types of the
semiconductor formed within a single crystal. In this chapter, the properties of
p-n junction diodes will be described, as a first step towards the understanding
of the structure and operation of more complex integrated circuits.
Huge numbers of diodes are made as discrete components, and one of their
widest applications is the rectification, i. e. the conversion of a.c. into d.c.
4.1. Semiconductors. P-N Junction
The fundamental characteristic of a diode is that it passes the current
more easily in one direction than in the other. In the early days of electronics
this behaviour was achieved by means of an evacuated tube containing two
electrodes with the property that electrons could be emitted only from one
electrode to cross the gap between them, but not from the other one.
Nowadays diodes are ‘solid state’, usually consisting of a junction between
two differently doped pieces of a semiconductor such as silicon. This basic
property of passing current in one direction only is fundamental to the
operation of many analogue and digital circuits.
Electrons and Holes
In a metal, the electric current consists of a flow of electrons from a
more negative to a more positive end. However in silicon, as in other
semiconductors, the electric current appears to be carried not only by
electrons but also by means of something else, called
some new kind of sub-atomic particles: silicon still consists of electrons,
protons and neutrons and nothing else. Holes are created by missing electrons
in some covalent bonds which hold the crystal structure together. When a
potential is applied across the crystal, the movement of electrons between
these incomplete bonds
gives rise to the conduction occurring by means of
independent, positive charge carriers. So a hole behaves like a mobile
positive charge in the crystal.
The current will consist of a quantity of negatively charged electrons
moving from negative to positive, and a different quantity of positively
charged holes moving from positive to negative. The direction of the current
conventionally marked on circuits is, of course, the same as the direction of
the holes.
holes. Holes are not
48
Pure silicon has an equal number of free holes and electrons that can
contribute to the conduction. However the relative proportions of holes and
electrons can be changed by
atoms. Doping with the
doping the silicon crystal with certain impurity
acceptor atoms, such as boron, produces p-type
silicon with a huge number of free holes and very few free electrons. In this
case holes are the
Doping with
donor atoms, such as phosphorus, produces n-type silicon with
majority carrier and electrons are the minority carrier.
a great number of electrons (the majority carrier) and very few holes (the
minority carrier). Most diodes consist of a junction between n-type material
and p-type material formed with a continuous crystal structure.
The Structure of P-N Junctions
A p-n junction is a junction, within a single crystal, between p-type
silicon and n-type silicon. In other words, the doping changes from primarily
acceptors to primarily donors at a plane within the crystal.
A thin slice or wafer of silicon is first cut from an ingot of a silicon
crystal which has been produced in a special furnace. This starting wafer
is doped during the growth of the single crystal by adding either donors
or acceptors to the molten silicon so that, as it crystallizes, the silicon
becomes either of n-type or p-type. Then each wafer forms the substrate for
p-n junctions. These junctions are formed by a high temperature diffusion
process, in which acceptor atoms are diffused into an n-type substrate,
or donor atoms into a p-type substrate.
The diffusion process is used in the fabrication of silicon p-n junctions
in the following way. First a film of silicon dioxide is grown over the surface
of the silicon by heating silicon in the atmosphere of oxygen. Then a hole, or
‘window’, is etched in the oxide layer.
When the silicon is heated to over 1000 °C in the atmosphere of the
dopant, the donor or acceptor atoms diffuse into the silicon through this
‘window’ in the oxide. If the starting material is n-type and the diffused
material is an acceptor, such as boron, the surface is converted to p-type
silicon because the acceptor density there exceeds the original donor density.
But deeper down the donors still dominate, so a boundary is formed between
the p-region and n-region at which the dominant dopant changes from donors
to acceptors. This boundary is called the metallurgical junction. Even if the
change of doping from donors to acceptors is abrupt, the change of majority
carriers from electrons to holes is not so abrupt. The result is that the p-n
junction as a whole extends a little on either side of the metallurgical
junction, to form the so-called
transition region which properties are at the
heart of the p-n junction and transistor action.
49
With both forms of the p-n junction, if wires are attached to the two
regions, it turns out that it is possible to pass a much greater current through
the device in one direction than in the opposite direction for the same
magnitude of the applied voltage.
4.2. Forward and Reverse Bias. V-I Characteristics
A typical d.c. characteristic of a silicon p-n junction diode is shown in
fig. 4.1. Forward direction of the current is when the p-region is made more
positive than the n-region. This is called
of the voltage
VAK up to about 0.5 V a little current flows, while above this
forward bias. For a positive value
value the current increases with the voltage more steeply.
I
DC
I
R
A
I
FSM
I
FRM
I
F
Continuous
forward current
Forward
voltage
drop
Single pulse
forward current
Maximum repe titive
forward current
V
AK
A
Cathode Anode
breakdown
voltage
non-repetitive
V
RSM
V
RRM
V
BR
Peak
reverse
voltage
V
RDC
blocking
voltage
V
K
Reverse
leakage
current
Peak repetitive
reverse
voltage
F
Fig. 4.1. V-I Characteristic of a Diode
The conventional symbol for the diode is also included in fig. 4.1, where
the arrow-shaped part indicates the direction in which the current can flow
easily. The bar-shaped end of the symbol always corresponds to the n-type
material in a p-n junction.
When the voltage is reversed in polarity, called
small but measurable current flows. This is called the
current
(or reverse leakage current), IS, and is typically 10
reverse bias, a very
reverse saturation
–13
amper in small
silicon diodes.
50
Соседние файлы в предмете [НЕСОРТИРОВАННОЕ]