Industrial Electronics
i
N4
Industrial Electronics
Learner Book
Johann Kraft
© Future Managers 2012
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Industrial Electronics
N4 Industrial Electronics
iii
CONTENTS
Chapter 1 – Kirchhoff ’s Laws ............................................................................................1
Chapter 2 – Superposition Theorem...............................................................................21
Chapter 3 – Thevenin’s Theorem.....................................................................................29
Chapter 4 – Series RLC-networks...................................................................................41
Chapter 5 – Parallel RLC-networks................................................................................69
Chapter 6 – Q-factor, Bandwidth and Complex Notation...........................................89
Chapter 7 – Basic Atomic Theory.................................................................................105
Chapter 8 – PN–Junction Theory..................................................................................115
Chapter 9 – Semi-conductor Diodes............................................................................127
Chapter 10 – Diode Applications..................................................................................139
Chapter 11 – Special Diodes and Applications...........................................................183
Chapter 12 – Transistors.................................................................................................199
Chapter 13 – Amplification Classes, Coupling Methods and Feedback..................231
Chapter 14 – Hybrid-Parameters .................................................................................249
Chapter 15 – Uni-Junction- and Field Effect Transistors .........................................263
Chapter 16 – Power Control .........................................................................................271
Chapter 17 – Operational Amplifiers ..........................................................................289
Chapter 18 – Function Generator and Oscilloscope..................................................309
Chapter 19 – Transducers..............................................................................................331
iv
KIRCHHOFF’S LAWS
KIRCHHOFF’s laws
N4 Industrial Electronics
CHAPTER
1 Kirchhoff’s Laws
Learning Outcomes
On completion of this module you will be able to:
•
•
•
•
•
1.1
Describe using applicable network diagrams the concept of Kirchhoff ’s:
– Current Law; and
– Voltage Law.
Describe using suitable expressions and network diagrams the concept of:
– Voltage division; and
– Current division.
Solve current magnitudes in a given network containing one power source;
Solve current magnitudes in a given network containing two power sources; and
Solve voltage magnitudes, power magnitudes and resistive magnitudes in a given
network diagram using Ohm’s Law as well as Kirchhoff ’s voltage and current laws.
Introduction
The concept of Kirchhoff ’s Laws is not a new concept since you have studied the
module on Ohm’s Law and the basis was laid there. It would be true to say that these
laws (Kirchhoff ’s Laws) have as origin Ohm’s Law. You will find that as you proceed
with further network theorems that all the theorems have as origin Ohm’s Law.
1.2
Kirchhoff’s Laws
1.2.1 Kirchhoff’s Current Law (KCL)
Definition 1.1 Kirchhoff ’s Current Law (KCL)
Kirchhoff ’s Current Law states that the algebraic sum of currents entering a point
will be equal to the algebraic sum of the currents leaving that point.
1
2
KIRCHHOFF’S LAWS
Figure 1.1(a) and (b) will illustrate this principle.
i1
i3
i2
i4
i2
i3
i1
(a)
i5
(b)
Figure 1.1
According to figure 1.1 (a): IT = I1 + I2 + I3, and
According to figure 1.1 (b): I1 + I2 = I3 + I4 + I5.
Important to note is that this current law is only applicable to a parallel network since
current has to divide.
Example 1.1
Consider the network below and determine by calculation all the unknown
values. All calculations must be shown.
I1 = 0,6 A
I5
I9 = 0,4 A
I10
I3 = 0,3 A
I6 = 0,7 A
I11
I2
I4 = 0,4 A
I12 = 0,3 A
I7
I13 = 0,9 A
I8
Solution:
I2 = I3 + I4 + I6
I5 = I1 + I3
I7 = I12 + I13
= 0,3 + 0,4 + 0,7 = 0,6 + 0,3 = 0,3 + 0,9
= 1,4 ampere = 0,9 ampere = 1,2 ampere
I8 = I7 - I4
I10 = I5 - I9
I11 = I6 + I10 +I12
= 1,2 - 0,4 = 0,9 - 0,4 = 0,7 + 0,5 + 0,3
= 0,8 ampere = 0,5 ampere = 1,5 ampere
N4 Industrial Electronics
1.2.2 Kirchhoff’s Voltage Law (KVL)
Definition 1.2 Kirchhoff ’s Voltage Law (KVL)
Kirchhoff ’s Voltage Law states that the algebraic sum of the individual voltage drops
in a closed network is equal to the algebraic sum of the applied voltage.
Figure 1.2 will illustrate this principle.
R1
R2
R3
V1
V2
V3
VS
Figure 1.2
According to figure 1.2: VS = V1 + V2 + V3
Something that also need to be addressed here is that should we have a current in the
network which happen to be in the opposite direction to that which we assume, it will
still cause a voltage drop across that resistor but that it will be of the opposite polarity
and should we then define Kirchhoff ’s Voltage Law in an expression we will find that:
ΣVS - ΣIR = 0
This expression will be very useful when working on networks which have no supply
present. Important to note is that the voltage law is only applicable to a series network
since the voltage has to divide between the respective resistive values.
1.3Application of Kirchhoff’s Laws
We once again find that the easiest and most convenient way to apply Kirchhoff ’s Laws
will be by doing a couple of activities. In doing these activities the basic concepts will
be applied as well as expanded. It must however be noted that as with many concepts
that we will be dealing with that we need to commence with the most basic of those
concepts as will be the case with Kirchhoff ’s Laws.
3
4
KIRCHHOFF’S LAWS
1.3.1 Kirchhoff’s Laws with a single sourcer
Consider the network illustrated in figure 1.3. There is more than one way in which to
solve this activity and you are advised to seek more than one solution. The secret lies
therein to set up two mathematical equations and then to solve them mathematically. In
setting up the two equations you must understand the concepts that were explained in
the previous section since that theory will form the basis of setting up the two equations
you will require. Furthermore, the concepts of Ohm’s Law are equally applicable since
Kirchhoff ’s Laws has as origin Ohm’s Law. The two equations will be set up using what
is termed ‘loops’ and you will also notice that the network has been labelled from A
through to H.
R2
I1
C
D
3 ohm
R1
B
G
3 ohm
R3
E I2
I1 + I2
F
6 ohm
V1
A
V2
-
VS = 10 V
+
H
Figure 1.3
Example 1.2
Determine with the given information of the network the magnitude of the
current produced by the battery (I1 + I2).
Solution:
Loop ABCDGHA
Loop ABEFGHA
V1 + V2
= VS (KVL)
= R1 × (I1+I2)
But V1
= R2 × I1
V2
R1 × (I1+I2) + R2 × I2 = VS
V1 + V2 =
VS (KVL)
But V1 =
R1 × (I1+I2) and
V2 =
R3 × I2
R1× (I1 + I2) + R3 × I2 = VS
But R1 = 3 ohm; R2 = 3 ohm; R3 = 6 ohm and VS = 10 V
N4 Industrial Electronics
3 × (I1 + I2) + 3 × I1
3I1 + 3I2 + 3I1
6I1 + 3I2
= 10
= 10
= 10 (1)
3 × (I1 + I2) + 6 × I2
3I1 + 3I2 + 6I2
3I1 + 9I2
5
= 10
= 10
= 10 (2)
We now have two equations with two unknowns and are solved in the following
manner:
(1) × 1: 6I1 + 3I2
(2) × 2: 6I1 + 18I2
(1) - (3)
- 15I2
I2
Substitute I2 = 0,667
=
6I1 + 3I2
6I1 + 3 (0,667) =
6I1 + 3 (0,667) =
=
6I1 + 2,001
6I1 =
6I1 =
I1 =
= 10 (1)
= 20 (3)
= - 10
= 0,667 ampere
ampere in (1)
10
10
10
10
10 - 2,001
7,999
1,333 ampere
Current produced by the battery = I1 + I2
= 1,333 + 0,667
= 2 ampere
Having obtained the magnitudes of the currents indicated it is of utmost
importance that these answers be verified and this is done in the following manner.
Use any one of the equations you set up and substitute the obtained values therein
and you must then have the left hand side equal to the right hand side of the
equation.
= 10 (1)
6I1 + 3I2
Substitute I1=1,333 ampere and I2 = 0,667 ampere
6 (1,333) + 3 (0,667) = 10
7,998 + 2,001
= 10
9,999
= 10
10
= 10
It must be noted however that should you have set up your equations incorrectly
from the onset, that, when substituting the current magnitudes in any of the
equations will result in the left hand side being equal to the right hand side.
Therefore, be sure that the equations you set up are correct!
6
KIRCHHOFF’S LAWS
Example 1.3
R3
R1
A
1,2 k-ohm
B
I1
1 k-ohm
C
V1
I1 - i2
I1
V2
R2
VS = 30 V
F
4 k-ohm
E
D
Consider the illustrated network and determine by calculation the:
(a)
Magnitude of the currents in the parallel section of the network by
making use of Kirchhoff ’s Laws;
(b)
Magnitude of the voltage drop V2; and
(c)
Power dissipated by resistor R3.
Solution:
(a)
Loop ABEFA
Loop ABCDEFA
V1 + V2 = VS (KVL)
V1 + V2 = VS (KVL)
But V1 = R1 × I1 and V2 = R3 × I2
But V1 = R1 × I1 and V2 = R2 × (I1 - I2)
R1 × I1 + R2 × (I1 - I2) = VS
R1 × I1 + R2 × I2
= VS
3
3
3
3
1,2 × 10 I1 + 4 × 10 (I1 - I2) = 30
1, 2 × 10 I1 + 1 × 10 I2 = 30 (2)
1,2 × 103I1 + 4 × 103I1 -1 × 103I2 = 30
= 30 (1)
5,2 × 103I1 - 4 × 103I2
(1) × 1:
5,2 × 103I1 - 4 × 103I2
(2) × 4:
4,8 × 103I1 + 4 × 103I2
(1) + (3)
10 × 103I1
I1
=
=
=
=
30 (1)
120 (3)
150
15 m-ampere
Substitute I1 = 15 × 10-3 ampere in (1)
5, 2 × 103 (15 × 10-3) - 4 × 103I2
78 – 4 × 103I2
4 × 103I2 I2
=
=
=
=
30
30
- 48
12 m-ampere
N4 Industrial Electronics
7
Current through R3 = I2 = 12 m-ampere
Current through R2 = I1 - I2 = 15 × 10-3 - 12 × 10-3 = 3 m-ampere
(b) V1 = R1 × I1
= 1,2 × 103×15 × 10-3
= 18 volt
(c) PR1 = I22 × R3
= (12 × 10-3)2 × 1 × 103
= 144 m-watt
The question may now arise why we make use of Kirchhoff ’s Laws when Ohm’s Law
may be applied in obtaining the same result. This statement is confirmed when you
look at example done on the application of Kirchhoff ’s Laws. We do however find times
when the application of Ohm’s Law is impossible and we then have to find a method to
solve the current magnitudes and it is then that Kirchhoff ’s Laws must be applied. Up to
now we have only really had a look at Kirchhoff ’s Voltage Law except for example 1.1.
Our aim will now be to solve a network in which no source included.
1.3.2 Kirchhoff’s Laws without a source
Up to now we have only really had a look at Kirchhoff ’s Voltage Law except for example
1.1. Our aim will now be to solve a network in which no source included.
Example 1.4
Determine, with the aid of Kirchhoff ’s Laws the magnitude of the currents I1
and I2 and I1 - I2.
B
10 ohm
I1 - I2
I1
3A
A
Loop 1
20 ohm
Loop 2
20 ohm
3 - I1
30 ohm
24 ohm
D
I2
C
3 - I2
3A
8
KIRCHHOFF’S LAWS
Solution:
Before we commence with the solution to this network it is important to note that
there is no supply source present and one would assume that it will be very easy to
solve this type of network. Yes, you are correct that it is very easy to solve since VS
= 0, but there is a more important factor that need to be taken into account. You
will notice that the current direction is indicated by means of the arrow points on
the network. Now, should you select the loops at random and you follow the loop
you must indicate a minus when you move against the arrow indicated on the
network. Please follow the steps in the solution.
Loop 1 (ABDA)
Loop 2 (BCDB)
20I2 - 24 (3 - I2) - 20 (I1 - I2) = 0
10I1 + 20 (I1 - I2) - 30 (3 - I1) = 0
20I2 – 72 + 24I2 - 20I1 + 20I2 = 0
10I1 + 20I1 - 20I2 - 90 + 30I1 = 0
= 90 (1) 20I1 + 64I2
= 72 (2)
60I1 - 20I2
(1) × 1: 60I1 - 20I2
(2) × 3: 60I1 + 192I2
(1) + (3):
172I2
I2
=
=
=
=
90 (1)
216 (3)
306
1,779 ampere
Substitute I2 =1,779 ampere in (1)
= 90
60I1- 20I2
= 90
60I1- 20 (1,779)
= 90
60I1 - 35, 58
= 125, 58
60I1
= 2,093 ampere
I1
I1 - I2 = 2,093 - 1,779 = 0,314 ampere
It should now be obvious why we need to be able to apply Kirchhoff ’s Laws and you
are now advised to apply the same principles that you applied in activity 1.2 and see
if you can solve all the current magnitudes through each resistor in the network.
N4 Industrial Electronics
Example 1.5
Calculate with the aid of Kirchhoff ’s Laws the magnitude of the currents I1 and
I2.
0,2 A
40 ohm
B
A
D
C
I1 - 0,2 A
I1
10 ohm
I2
10 ohm
I1 - I2 - 0,2 A
80 ohm
VS = 4 V
F
E
Solution:
Loop ABCEFA
Loop BDCB
10 (I1 - 0, 2) + 80I2 = 4 40 × 0, 2 - 10 (I1 - I2 - 0, 2) - 10 (I1 - 0, 2) = 0
= 4 8 - 10I1 + 10I2 + 2 - 10I1 + 2
= 0
10I1- 2 + 80I2
- 20I1 + 10I2
=-8
10I1 + 80I2 = 6 (1)
= 8 (2)
20I1 - 10I2
(1) × 2:
(2) × 1:
(3) - (2)
20I1 + 160I2
20I1 - 10I2
170I2
I2
Substitute I2 = 23, 53 × 10-3
10I1 + 80I2
10I1 + 80 (23, 53 × 10-3)
10I1 + 1,882
10I1
I1
=
=
=
=
12 (3)
8 (2)
4
23,53 m-ampere
ampere in (1)
= 6
= 6
= 6
= 4, 12
= 0,412 ampere
9
10
KIRCHHOFF’S LAWS
1.3.3 Kirchhoff’s Laws with two sources
It was obvious that some networks cannot be solved by means of Ohm’s Law since there
is no supply present but yet we were able to set up equations and then solve for the
current magnitudes. The question may now arise of how to solve current magnitudes
as well as direction of current flow should two or more voltage sources are present.
Important to note here is that one very often has to assume certain conditions and it
may be possible that these assumptions be proved incorrect once an answer had been
obtained.
One assumption that we can assume from the onset is that current will flow from the
negative terminal of the voltage source toward the positive terminal of that voltage
source. There may be arguments that this is not a valid assumption since some theories
may indicate that current flow is from the positive terminal of the voltage source toward
the negative terminal of that source. The latter theory is therefore termed conventional
current flow and the first theory is termed current flow. The most productive method of
explanation will once again be to solve a given network and such a network is illustrated
in figure 1.4.
In this given network the assumptions that we spoke about will mainly be indicated by
the direction of the assumed current flow which of course may be incorrect but more
than enough examples will be given in order to rectify our assumptions made at the
onset. For purposes of clarity and understanding of the assumptions to be made more
than one solution will be given and this will assist you as learner to fully understand the
concepts. Important to note is that specific theoretical concepts are embedded and these
should be fully understood in order to gain success.
R1
A
R2
B
10 ohm
10 ohm
I1 + I2
I1
R3
V1 = 10 V
F
10 ohm
E
Figure 1.3
C
I2
V2 = 20 V
D
N4 Industrial Electronics
Example 1.6
Solve for the given network the magnitude of the current I1 + I2 through the
resistor R3.
Solution:
Loop ABEFA
R1 × I1 + R3 (I1 + I2)
10I1 + 20 (I1 + I2)
10I1 + 20I1 + 20I2
30I1 + 20I2
=
=
=
=
Loop CBEDA
V1R2 × I2 + R3 (I1 + I2)
10
15I2 + 20 (I1 + I2)
10
15I2 + 20I1 + 20I2
10 (1)
20I1 + 35I2
(1) × 20:
600I1 + 400I2
(2) × 30:
600I1 + 1050I2
(3) - (4)
- 650I2
I2
=
=
=
=
=
=
=
=
V2
20
20
20 (2)
200 (3)
600 (4)
- 400
0,615 ampere
Substitute I2 = 0,615 ampere in (1)
= 10
30I1 + 20I2
= 10
30I1 + 20 (0,615)
= 10
30I1 + 12,3
30I1 = - 2,3
I1 = - 0,0766 ampere
Current through resistor R3 = I1 + I2 = - 0,0766 + 0,615 = 0,5384 ampere.
This solution proves to us that assumptions we made were wrong! Initially we
assumed that the current I1 flows from A to B but the negative value obtained
indicates that the current in fact flows from B to A. Secondly we assumed that
the currents will be additive between B and E where in fact they subtract from
one another. So what does this prove? Source V2 is the stronger of the two
sources and therefore negates the current produced by V1.
The following solution to this network (figure 1.5) will illustrate that should we
have made the correct assumptions initially that all values obtained will have a
positive value. It must however be noted that the negative answer obtained is
not incorrect but that it is an indication that the current actually flows in the
opposite direction as was assumed. Also important is that when that negative
value is substituted into any expression that the negative sign must be used.
11
12
KIRCHHOFF’S LAWS
R1
A
R2
B
10 ohm
10 ohm
I2 - I1
I1
R3
V1 = 10 V
F
10 ohm
E
C
I2
V2 = 20 V
D
Loop CBEDC
Loop CBAFEDC
R2 × I2 + R3 (I2 - I1)
15I2 + 20 (I2 - I1)
15I2 + 20I2 - 20I1
35I2 - 20I1
=
=
=
=
V2
20
20
20 (1)
R2 × I2 + R1 × I1
15I2 + 10I1
15I2 + 10I1
(1) × 1: 35I2 - 20I1
(2) × 2: 30I2 + 20I1
(1) + (3) 65I2
I2
=
=
=
=
20 (1)
20 (3)
40
0,615 ampere
Substitute I2 = 0,615
35I2 - 20I1 = 20
35 (0,615) - 20I1
21,525 - 20I1
- 20I1
I1
= V2 - V1
= 20 - 10
= 10 (2)
ampere in (1)
=
=
=
=
20
20
- 1,525
0,07625 ampere
Current through resistor R3 = I2 - I1 = 0,615 - 0,07625 = 0,5388 ampere
N4 Industrial Electronics
Example 1.7
Make use of Kirchhoff ’s Laws and determine the magnitude of the current
through RL.
R2
A
B
15 ohm
I2 - I1
V1 = 10 V
C
I2
RL
20 ohm
V2 = 20 V
I1
10 ohm
F
E
D
R1
Solution:
Loop FEBAF
R1 × I1 - RL (I2 - I1)
10I1 - 20 (I2 - I1)
10I1 - 20I2 + 20I1
30I1 - 20I2
=
=
=
=
Loop CBEDC
V1
10
10
10 (1)
(1) × 20:
600I1 - 400I2
(2) × 30:
- 600I1 + 1050I2
(3) + (4)
650I2
I2
R2 × I2 + RL (I2 - I1)
15I2 + 20 (I2 - I1)
15I2 + 20I2 - 20I1
- 20I1 + 35I2
=
=
=
=
200 (3)
600 (4)
800
1,23 ampere
Substitute I2 = 1,23 ampere in (1)
= 10
30I1 - 20I2
30I1 - 20 (1, 23) = 10
= 10
30I1 - 24, 6
= 34,6
30I1
= 1,15 ampere
I1
Current through resistor RL = I2 - I1 = 1, 23 - 1,15 = 0,08 ampere
=
=
=
=
V2
20
20
20 (2)
13
14
KIRCHHOFF’S LAWS
Example 1.8
R1
A
R2
B
4 ohm
3 ohm
I1 + I2
I1
C
I2
R3
V1 = 15 V
F
2 ohm
E
V2 = 10 V
D
Calculate with the given information the:
(a)
Magnitude of the current through resistor R3 using Kirchhoff ’s Laws;
(b)
Voltage drop across resistor R1; and
(c)
Power consumed by resistor R2.
Solution:
(a) Loop ABEFA
R1 × I1 + R3 (I1 + I2)
4I1 + 2 (I1 + I2)
4I1 + 2I1 + 2I2
6I1 + 2I2
(1) × 1:
(2) × 3:
(1) – (3)
6I1 + 2I2
6I1 + 15I2
-13I2
I2
=
=
=
=
Loop CBEDC
V1
15
15
15 (1)
R2 × I2 + R3 (I1 + I2)
3I2 + 2 (I1 + I2)
3I2 + 2I1 + 2I2
2I1 + 5I2
=
=
=
=
=
=
=
=
V2
10
10
10 (2)
15 (1)
30 (3)
-15
1,154 ampere
Substitute I2 = 1,154 ampere in (1)
= 15
6I1 + 2I2
= 15
6I1 + 2 (1,154)
= 15
6I1 + 2,308
= 12,692
6I1
= 2,115 ampere
I1
Current through resistor R3 = I1 + I2 = 2,115 + 1,154 = 3,269 ampere
N4 Industrial Electronics
(b)
Although this section is concerned with Kirchhoff ’s Laws the quantities that need to be calculated here must be done using Ohm’s Law.
VR1
=
=
=
I1 × R1
1,154 × 4
4,616 volt
(c) PR2
=
=
=
I22 × R2
(2,115)2 × 3
13,419 watt
Example 1.9
Determine from the following network diagram with the assistance of
Kirchhoff ’s Laws:
I1 - I2
D
4 ohm
I2
12 ohm
I1
R5
1,1 A
R2
A
I2 - 1,1 A
VS = 20 V
10 ohm
R3
R1
4 ohm
R4
C
R6
(a)
(b)
(c)
12 ohm
4 ohm
B
I1 - 1,1 A
The magnitude of the currents I1 as well as I2 (use the loops DABCD
and DACD); and
The potential difference (voltage drop) across points A and B.
Which of the terminals in (b) above is positive?
15
16
KIRCHHOFF’S LAWS
Solution:
(a)
Loop DABCD
R2 × I2 + R3 (I2 - 1,1) + R4 (I1 - 1,1) + R1 × I1
12I2 + 10 (I2 - 1,1) + 12 (I1 - 1,1) + 4I1
12I2 + 10I2 - 11 + 12I1 - 13,2 + 4I1
16I1 + 22I2 - 24,2
16I1 + 22I2
=
=
=
=
=
VS
20
20
20
44,2 (1)
Loop DACD
R2 × I2 + R6 (1,1) + R1 × I1
12I2 + 4 (1,1) + 4I1
12I2 + 4,4 + 4I1
4I1 + 12I2 + 4,4
4I1 + 12I2 (1) × 1: 16I1 + 22I2
(2) × 4: 16I1 + 48I2
(1) - (3)
- 26I2
I2
Substitute I2 = 0, 7
16I1 + 22I2
=
16I1 + 22 (0,7) =
=
16I1 + 15,4
=
16I1
=
I1
=
=
=
=
=
=
=
=
=
VS
20
20
20
15,6 (2)
44,2 (1)
62,4 (3)
-18,2
0,7 ampere
ampere in (1)
44,2
44,2
44,2
28,8
1,8 ampere
(b)
VAB = R3 (I1 - 1,1)
= 10 (1, 8 - 1, 1)
= 7 volt
(c)
Observing the indicated direction point B will be positive.
N4 Industrial Electronics
Example 1.10
The following Wheatstone-bridge network is given.
R1
B
R3
20 ohm
I1 - I2
I1
0,4 A
I2
A
G
C
0,4 A
0,4 - I1
0,4 - I2
20 ohm
R2
D
RX
The variable resistor R3 is adjusted until the galvanometer reading indicates
zero. It is then determined by measurement that R3 has a value of 180 ohm.
Determine the value of the resistor RX by making use of Kirchhoff ’s Laws only.
NB: You may not use Ohm’s Law!
Solution:
Loop ABDA
R1 × I1 - R2 (0,4 - I1)
20I1 - 20 (0,4 - I1)
20I1 - 8 + 20I1
40I1 I1
I2
=
=
=
=
=
=
Loop BCDB
0
R3 × I2 - RX (0,4 - I2)
0
180 × 0,2 - RX (0,4 - 0,2)
0
90 - 0,4RX - 0,2RX
8
- 0,6RX
0,2 ampere RX
0,2 ampere
=
=
=
=
=
0
0
0
- 90
150 ohm
Exercise 1.1
1.
Describe Kirchhoff ’s:
1.1
Voltage Law; and
1.2
Current Law in your own words.
2.
Use applicable network diagrams and explain the concepts of:
2.1
Current division; and
2.2
Voltage division.
17
18
KIRCHHOFF’S LAWS
3.
Determine by calculation the magnitude of all the unknown current magnitudes in the following diagram.
I1
I5 = 0,9 A
I9
I10 = 0,5 A
I3
I6 = 0,7 A
I11 = 1,6 A
I2 = 1,4 A
I4 = 0,4 A
I12
I7 = 1,2 A
I13
I8
4.
Two resistors having values of 5 ohm and 15 ohm respectively are connected
in parallel and this combination is connected in series with a 12 ohm resistor across a direct current source of 12 volts.
4.1
Draw a neat labelled network diagram that will illustrate the above information.
4.2
Make use of Kirchhoff ’s Laws to determine the current magnitudes through each of the parallel resistors.
4.3
Make use of Ohm’s Law to determine the voltage drop across the parallel section of the network.
4.4
Calculate the power consumed by the 12 ohm resistor.
5.
B
4 ohm
I1 - I2
I1
2A
A
Loop 1
12 ohm
Loop 2
12 ohm
2 - I1
18 ohm
5.1
5.2
5.3
5.4
16 ohm
D
I2
C
2A
2 - I2
Use Kirchhoff ’s Laws and determine the magnitude of the currents I1 and I2.
Determine the current through each resistor of the network.
Determine the voltage drop across each resistor of the network.
Determine the power consumed by each resistor of the network.
N4 Industrial Electronics
6.
I1
A
6.1
6.2
6.3
6.4
B
I2
10 ohm
VS = 12 V
I1 - I2 - 0,128 A
I1 - I2
40 ohm
H
C
0,128 A
10 ohm
30 ohm
G
D
20 ohm
F
E
Use Kirchhoff ’s Laws and calculate the magnitude of the currents I1 and I2.
Determine the current through each resistor of the network. All calculations must be shown.
Determine the voltage drop across each resistor of the network. All calculations must be shown.
Determine the power consumed by each resistor of the network. All calculations must be shown.
7.
Consider the network below and determine by calculation the:
7.1 Current using Kirchhoff ’s Laws that is produced by the voltage sources V1 and V2;
7.2 Voltage drop across the 6 ohm resistor R3; and
7.3 Power consumed by R1 and R2.
R1
A
R2
B
10 ohm
I2 - I1
I1
V1 = 10 V
F
4 ohm
C
I2
R3
6 ohm
E
V2 = 20 V
D
19
20
KIRCHHOFF’S LAWS
8.
8.1
8.2
Consider the given network and determine the following quantities.
Use Kirchhoff ’s Laws and determine the magnitude of the current
I1 - I2.
It was found that through a change in temperature that the value
of the load resistor RL had been reduced to 1 ohm. Re-calculate the magnitude of the current I1 - I2.
R2
A
B
I1 - I2
C
I2
I1
RL
V2 = 20 V
F
3 ohm
4 ohm
R1
2 ohm
E
V1 = 10 V
D
SUPERPOSITION THEOREM
N4 Industrial Electronics
CHAPTER
2 Superposition Theorem
Learning Outcomes
On completion of this module you will be able to:
•
•
•
•
2.1
Define the Superposition Theorem;
Use and explain by applicable network diagrams and calculations the principle of:
– Voltage division; and
– Current division.
Solve current magnitudes and directions with the aid of the Superposition
Theorem; and
Solve voltage magnitudes, power magnitudes and resistive magnitudes in a given
network using Ohm’s law.
Introduction
Definition 2.1 Superposition Theorem
The Superposition Theorem states that all current magnitudes and directions may be
determined by considering each supply on its own.
Once we have solved the magnitudes and directions we superimpose the network
on top of one another in order to obtain the final solution. When considering each
supply on its own we will actually make use of Ohm’s Law from which this theorem
also originates. Before we can actually commence with this theorem there are two very
important aspects that need to be addressed namely voltage and current division. You
will recall that this concept was also discussed in the previous module but we will now
take this concept one step further.
21
22
Superposition Theorem
2.2
Current- and voltage division
2.2.1 Current division
In the definition of Kirchhoff ’s current law (KCL) we found that it may only be applied
to a parallel network. What is of further significance is that the way in which the
current will divide is based on the ratio to one another of the resistive elements in that
particular parallel network. We also know from Ohm’s Law that in order to determine
the current through the respective resistors in a parallel section that we need to know
the voltage drop across that parallel section. The current division rule bypasses this
requirement. Consider the resistive network illustrated in figure 2.1.
I1
IT
I2
R1
R2
Figure 2.1
We know that according to Kirchhoff ’s current law that the current will divide between
the two resistors and a point that should also need to be mentioned is that current will
always take the path of least resistance. The magnitudes of I1 and I2 are given by the
following mathematical expressions:
I1 =
R2
× ITI2 =
R1 + R2
R1
× IT
R1 + R2
Example 2.1
Determine from the given information of the network the magnitude of the
currents I1 and I2.
10 ohm
10 A
IT
I1
I2
R1
15 ohm
R2
N4 Industrial Electronics
23
Solution:
I1
=
R2
× IT
I2 =
R1
× IT
R1 + R2
R1 + R2
=
15 × 10
=
10
× 10
15 + 10 15 + 10
= 6 ampere
=
4 ampere
In the definition of Kirchhoff ’s voltage law (KVL) we found that it may only be
applied to a series network. What is of further significance is that the way in which the
voltage will divide is based on the ratio to one another of the resistive elements in that
particular series network. We also know from Ohm’s Law that in order to determine
the voltage drop across the respective resistors in a series section that we need to know
the current flowing in the network. The voltage division rule bypasses this requirement.
Consider the resistive network illustrated in figure 11.2.
We know that according to Kirchhoff ’s voltage law that the voltage will divide between
R1
R2
V1
V2
VS
Figure 2.2
the two resistors and a point that should also need to be mentioned is that the largest
resistor will also have the largest voltage drop. The magnitudes of V1 and V2 are given by
the following mathematical expressions:
V1 =
R1
× VS V2 =
R1 + R2 R2
R1 + R2
× VS
Example 2.2
Determine from the given information of the network the magnitude of the
voltages V1 and V2
R1
R2
18 ohm
12 ohm
V1
V2
VS = 50 V
24
Superposition Theorem
Solution:
R1
× VS
V2
R1 + R2
18
=
× 50
18 + 12
= 30 volt
V1 =
=
=
R2
× VS
R1 + R2
12
× 50
18 + 12
= 20 volt
We have once again reached the point where we need to look at the application of the
Superposition Theorem and we will once again do it by means of a number of activities.
It does not matter where you start but most important is that you must short circuit one
of the supplies and insert your assumed current directions. Important to note again is
that you may safely assume that the current will flow from the negative terminal of the
supply to the positive terminal of the supply. For the purpose of our explanation we will
short circuit supply V2 and what is most important is to redraw your network as follows.
Example 2.3
Apply the Superposition Theorem to find the direction and current
magnitudes through each resistor in the following network.
V1 = 10 V
R1
R2
10 ohm
15 ohm
R3
20 ohm
V2 = 20 V
Solution:
Short circuit V2 and indicate current directions and calculate values.
N4 Industrial Electronics
R2
R1
IA
IB
10 ohm
V1 = 10 V
IC
15 ohm
20 ohm
R3
R2 × R3
VS
RTA
=
R1 +
IA
=
R + R3
R
2
TA
15 × 20
=
10 +
=
10
15 + 20
22,857
=
22,857 ohm
=
0,438 ampere
R2
× IAIC
R2 + R3
15
=
× 0,438
15 + 20
= 0,188 ampere
IB
=
R3
× IA
R2 + R3
20
=
× 0,438
15 + 20
=
= 0,25 ampere
Step 2
Replace V2 and short circuit V1 and indicate current directions and calculate
values.
R2
R1
IE
10 ohm
ID
15 ohm
IF
R3
RTB
=
=
=
IE
20 ohm
R1 × R3
VS
ID
=
R2 + R3
R
TB
10 × 20
10
15 +
=
10 + 20
21,667
21,667 ohm
=
0,462 ampere
R2 +
R1
× IDIF
R1 + R3
10
=
× 0,462
10 + 20
= 0,154 ampere
=
V2 = 20 V
R3
× ID
R1 + R3
20
=
× 0,462
10 + 20
=
= 0, 308 ampere
25
26
Superposition Theorem
Having obtained the current directions and magnitudes by considering each supply
on its own we have now reached the stage where the two networks need to be
superimposed onto one another in order to find the final current directions as well
as the magnitudes thereof.
Step 3
R2
R1
IF
10 ohm
IB
IC
15 ohm
IE
IA
ID
V1 = 10 V
IR1
=
=
=
R3
V2 = 20 V
20 ohm
IA - IF
IR2 =
0,438 - 0,308
=
0, 13 ampere
=
ID - IC
IR3
0,462 - 0, 25
0,212 ampere
=
=
=
IB + IE
0,188 + 0,154
0,342 ampere
Example 2.4
Determine with the aid of the Superposition Theorem the magnitude of the
current through the load resistor RL.
R2
2 ohm
V1 = 10 V
RL
4 ohm
V2 = 20 V
3 ohm
R1
Solution:
R2
2 ohm
V1 = 10 V
RL
4 ohm
IA
IC
3 ohm
R1
IB
N4 Industrial Electronics
R2 × RL
VS
IA =
IB
=
R2 + RL
RTA
2×4
10
= 3 +
=
=
2+4
4,333
=
= 4,333 ohm
= 2,307 ampere
=
RTA = R1 +
R2
× IA
R2 + RL
2
× 2,307
2+4
0,769 ampere
R2
IF
2 ohm
RL
3 ohm
4 ohm
ID
V1 = 20 V
IE
R1
R1 × RL
VS
R1
ID =
IE
=
× ID
R1 + RL
R
R + RL
TB1
3×4
= 2 +
= 20
= 3 × 5,385
3+4
3,714 3 + 4
= 3,714 ohm
= 5,385 ampere
= 2,308 ampere
RTB = R2 +
Current through the load resistor RL
= IE - IB
= 2,308 - 0,769
= 1,539 ampere
Exercise 2.1
1.
Explain using suitable network diagrams and applicable mathematical
expressions the concept of:
1.1
Current division; and
1.2
Voltage division.
2.
Define the Superposition Theorem.
3.
Two resistors having values of 15 ohm and 35 ohm respectively are used in
a voltage divider network driven from a source of 100 volt. Determine
using the voltage divider rule the magnitude of the voltage across each
resistor.
Use any other method (Ohm’s Law) to prove that the magnitudes you
obtained are the correct values.
27
28
Superposition Theorem
4.A constant current source supplies 15 ampere to a resistive network consisting of two parallel resistors having values of 25 ohm and 40 ohm respectively. Determine using the current division rule the magnitude of the
current through each resistor.
Use any other method (Ohm’s Law) to prove that the magnitudes you obtained are the correct values.
5.
Give a step-by-step description to illustrate the application of the Superposition Theorem. Use a network and values for the resistors of your choice.
6.
Solve for each of the networks below the current magnitude and direction through each resistor by making use of the Superposition theorem.
V1 = 40 V
R1
R2
85 ohm
R3
47 ohm
3 ohm
100 ohm
R3
5 ohm
V1 = 80 V
R2
RL
2 ohm
V2 = 50 V
V2 = 60 V
2 ohm
2 ohm
R1
R2
V1 = 10 V
V2 = 6 V
R3
2 ohm
V3 = 4 V
THEVENIN’s THEOREM
N4 Industrial Electronics
29
CHAPTER
3 Thevenin’s Theorem
Learning Outcomes
On completion of this module you will be able to:
•
•
•
3.1
Define Thevenin’s theorem;
Solve current magnitudes with the aid of Thevenin’s theorem; and
Solve voltage magnitudes, power magnitudes and resistive magnitudes in a given
network using Ohm’s Law.
Introduction
Definition 3.1 Thevenin’s Theorem
Thevenin’s Theorem specifies that a complex network consisting of impedances and
voltage sources may be replaced by a constant voltage source with a series impedance.
Thevenin’s Theorem has the advantage that once the equivalent network of impedances
and constant voltage source had been obtained that the current through a load resistor
may be calculated by merely inserting the new load resistor.
3.2
Application of Thevenin’s Theorem
3.2.1 Thevenin’s Theorem with a single source
As was the case in the preceding modules, Thevenin’s Theorem will once again be
explained by means of a number of activities. It is however very important that you
must have a thorough background of Ohm’s Law since Thevenin’s Theorem has that law
as basis.
30
Thevenin’s Theorem
Example 3.1
Consider the network illustrated in figure 3.1 and determine with the aid of
Thevenin’s Theorem the magnitude of the current through resistor RL.
R1
RL
10 ohm
6 ohm
V1 = 20 V
R2
4 ohm
Figure 3.1
In order to solve for the current magnitude through resistor RL we have to
follow the step-by-step approach in that we must determine VTHEVENIN as well as
RTHEVENIN in order to obtain the Thevenin equivalent network.
Solution:
Remove resistor RL and label the points AB.
R1
A
10 ohm
V1 = 20 V
R2
4 ohm
B
Determine VTHEVENIN which is the voltage drop across resistor R2 by making use
of the voltage divider rule.
R2
=
VAB
=
× V1
VTHEVENIN
R + R2
1
4
=
× 20
10 + 4
=
5,714 volt
N4 Industrial Electronics
Step 2
Short circuit the supply V1 and determine RTHEVENIN by looking in at points AB.
R1
A
10 ohm
R2
4 ohm
RTHEVENIN
B
RTH
=
=
=
R1 × R2
R1 + R2
10 × 4
10 + 4
2,857 ohm
Step 3 Equivalent network
Having obtained the values for VTH and RTH we can now draw the equivalent
network and re-insert the load resistor RL and then use Ohm’s law to determine
the current through the load resistor RL.
RTH
A
2,857 ohm
VTH = 5,714 V
RL
B
IRL =
=
=
VTH
RTH + RL
5,714
2,857 + 6
0,645 ampere
6 ohm
31
32
Thevenin’s Theorem
Example 3.2
R1
R2
4 k-ohm
1 k-ohm
A
VS = 30 V
B
Consider the given network and:
(a)
Obtain the equivalent Thevenin’s network; and
(b)
Determine the magnitude of the current through the following load
resistors:
(i)
10 kilo-ohm;
(ii)
25 kilo-ohm; and
(iii)
100 ohm.
Solution:
(a)
VTH =
=
=
R2
R1 × R2
× VS
RTH
=
R1 + R2
R + R2
1
1 × 103
4 × 103 × 1 × 103
×
30
=
4 × 103 + 1 × 103
4 × 103 + 1 × 103
6 volt
Equivalent network
RTH
A
800 ohm
VTH = 6 V
RL
B
=
800 ohm
N4 Industrial Electronics
VTH
(b)(i) IRL
=
R
+ RL
TH
6
=
800 + 10 × 103
=
0,556 m-ampere
VTH
(ii) IRL =
RTH + RL
6
=
800 + 25 × 103
=
0,233 m-ampere
VTH
(iii) IRL =
R + RL
TH
6
=
800 + 100
=
6,667 m-ampere
Example 3.3
VS = 60 V
R1
R2
36 ohm
6 ohm
12 ohm
R3
5 ohm
RL
Consider the given network and determine with the aid of Thevenin’s Theorem
the:
(a)
Magnitude of the current through the 5 ohm resistor; and
(b)
Power dissipated by the 5 ohm resistor.
33
34
Thevenin’s Theorem
Solution:
(a)
R1
R2
36 ohm
6 ohm
A
VS = 60 V
R2
12 ohm
B
R3
VTH
=
× VS
R1 + R3
12
=
× 60
36 + 12
=
15 volt
R1
R2
36 ohm
6 ohm
A
RTHEVENIN
12 ohm
R2
B
RTH
R1 × R3
R1 + R3
36 × 12
6 +
36 + 12
R2 +
=
=
=
15 ohm
Equivalent network
RTH
A
15 ohm
VTH = 15 V
RL
B
5 ohm
N4 Industrial Electronics
VTH
=
IRL
RTH + RL
15
=
15 + 5
=
0, 75 ampere
(c)
PRL
=
=
=
IRL2 × RL
(0, 75)2 × 5
2,813 watt
3.2.2 Thevenin’s Theorem with two sources
Example 3.4
Obtain for the network the Thevenin’s equivalent network and then calculate
the magnitude of the current through the load resistor RL in figure 3.2.
V1 = 4 V
V2 = 6 V
RL
20 ohm
R1
4 ohm
3 ohm
R2
Figure 3.2
A point that should be noted with this type of problem is that we now have two
voltage sources which will both contribute to VTH and we have to consider each
supply on its own initially since both sources will contribute to the total of VTH.
Solution:
Step 1
Remove RL, mark terminals AB and short circuit any one of the voltage sources.
In this instance we will short circuit V2 and calculate VTH1.
35
36
Thevenin’s Theorem
V1 = 4 V
A
B
R1
4 ohm
3 ohm
R2
R2
VTH1
=
× V1
R1 + R2
4
=
× 4
3+4
=
2,286 volts
Step 2
Short circuit V1, replace V2 and calculate VTH2
V2 = 6 V
A
R1
3 ohm
VTH2
=
=
=
VTHT
=
=
=
R1
× V2
R1 + R2
3
× 6
3+4
2,571 volt
VTH1 + VTH2
2,286 + 2,571
4,857 volt
B
4 ohm
R2
N4 Industrial Electronics
37
VTH1 and VTH2 may be added together since they will be supplying current in the
same direction through RL. But, be aware that this will not always be the case and
this will be dealt with in a later activity.
Step 3
Short circuit both voltage supplies and calculate RTH by looking in at terminals AB.
A
RTHEVENIN
R1
4 ohm
B
3 ohm
R2
R1 × R2
RTH
=
R1 + R2
3×4
=
3+4
=
1,714 ohm
Step 4
Equivalent network
You can now draw the equivalent network, re-insert the load resistor RL and
determine the magnitude of the current.
RTH
A
1,714 ohm
VTH = 4,857 V
RL
B
IRL
=
=
=
VTHT
RTH + RL
4,857
1,714 + 20
0,224 ampere
20 ohm
38
Thevenin’s Theorem
Example 3.5
Make use of Thevenin’s Theorem and determine the magnitude of the current
through the load resistor RL.
R2
2 ohm
RL
V1 = 10 V
4 ohm
V2 = 20 V
3 ohm
R1
Solution:
R2
2 ohm
A
V1 = 10 V
B
3 ohm
R1
R2
VTH1
=
× V1
R1 + R2
2
=
× 10
2+3
=
4 volt
R2
2 ohm
A
B
3 ohm
R1
V2 = 20 V
N4 Industrial Electronics
R1
VTH2
=
× V2
R1 + R2
=
3
× 20
2+3
=
12 volt
VTHT
=
=
=
VTH2 - VTH1
12 - 4
8V
R2
2 ohm
A
RTHEVENIN
B
3 ohm
R1
R1 × R2
RTH
=
R1 + R2
3×2
=
3+2
=
1,2 ohm
Equivalent network
RTH
A
1,714 ohm
VTH = 4,857 V
RL
B
IRL
=
=
=
VTHT
RTH + RL
8
1,2 + 4
1,538 ampere
20 ohm
39
40
Thevenin’s Theorem
Exercise 3.1
1.
Give a suitable definition or description of Thevenin’s Theorem.
2.
Use a network containing:
2.1
One voltage source; and
2.2
Two voltage sources and give a step-by-step description of
the approach you would follow in order to apply Thevenin’s
Theorem.
3.
Consider each of the networks below and determine with the aid of
Thevenin’s Theorem the magnitude of the current through the load
resistor RL.
V1 = 40 V
R1
12 ohm
3 ohm
R2
80 ohm
RL
R1
R2
5 ohm
36 ohm
VS = 80 V
V2 = 60 V
R1
2 ohm
V2 = 4 V
4 ohm
V1 = 10 V
R2
3 ohm
RL
2 ohm
RL
SERIES RLC-NETWORKS
N4 Industrial Electronics
CHAPTER
4 Series RLC-networks
Learning Outcomes
On completion of this module you will be able to:
•
•
•
•
•
•
•
•
•
Illustrate using circuit diagrams, graphical representation and phasor diagrams the
effect that an alternating quantity has on a single:
– Resistor;
– Inductor; and
– Capacitor.
Illustrate with the aid of a circuit diagram and graphical representation a series
circuit containing a resistor and an inductor;
Determine by calculation the:
– Impedance;
– Current flow;
– Phase angle;
– Voltage drops; and
Supply voltage in a series circuit containing a resistor and an inductor.
Draw a phasor diagram for a series circuit containing a resistor and an inductor;
Illustrate with the aid of a circuit diagram and graphical representation a series
circuit containing a resistor and a capacitor;
Determine by calculation the:
– Impedance;
– Current flow;
– Phase angle;
– Voltage drops; and
– Supply voltage in a series circuit containing a resistor and a capacitor.
Draw a phasor diagram for a series circuit containing a resistor and a capacitor;
Illustrate with the aid of a circuit diagram and graphical representation a series
circuit containing a resistor, an inductor and a capacitor;
Determine by calculation the:
– Impedance;
– Current flow;
– Phase angle;
41
42
Series RLC-networks
– Power factor;
– True power;
– Apparent power;
– Voltage drops; and
– Supply voltage in a series circuit containing a resistor, an inductor and a capacitor.
Draw a phasor diagram for a series circuit containing a resistor, an inductor and a
capacitor;
Explain by means of a graphical representation the concept of resonance in a series
circuit containing a resistor, an inductor and a capacitor;
Explain the conditions for resonance in a series circuit containing a resistor, an
inductor and a capacitor; and
Calculate the resonant frequency in a series circuit containing a resistor, an
inductor and a capacitor.
•
•
•
•
4.1
The effect of an alternating quantity on a resistor
Figure 4.1 illustrates a resistor connected across an alternating current supply (a) and
(b) illustrates a graphical representation of the phase relationship between the current
and the supply voltage whereas (c) illustrates a phasor diagram.
It is of utmost importance that you grasp and understand, not only for the resistor
but also for the inductor and capacitor, these graphical representations and phasor
diagrams.
VS
R
IR
+
IR
t
IR
VS
VS
(a)
(b)
(c)
Figure 4.1
Obvious is that the voltage and current are in phase and therefore makes it possible
that Ohm’s Law as used in dc networks can also be used in this instance. These
expressions will be given again just to refresh your memory.
VS
where I
I
=
R
R
VS
=
=
=
current in ampere
resistance in ohm
supply voltage in volt
N4 Industrial Electronics
43
VS
Also: R
=
and
VS
=
I×R
I
Before we go any further it is important to remember that we are working with
alternating current and that all values that will be given will be as RMS-values unless
otherwise stated! Also of importance is that the frequency of the alternating supply has
no effect on the resistor.
Example 4.1
A resistor having a value of 100 ohm is connected across a 380V/100Hz
alternating current supply. Determine the current that will flow through this
resistor.
Solution:
I
VS
R
380
100
=
=
=
3, 8 ampere
4.2
The effect of an alternating quantity on an inductor
Figure 4.2 illustrates an inductor connected across an alternating current supply (a) and
(b) illustrates a graphical representation of the phase relationship between the current
and the supply voltage whereas (c) illustrates a phasor diagram.
VS
R
+
VS
IL
IL
t
IL
VS
(a)
(b)
(c)
Figure 4.2
Upon careful inspection you will notice that there is a phase displacement between the
alternating supply and the current through the inductor and is at a maximum of 90º
with the voltage leading the current. In order to remember when voltage is leading or
lagging we will look at the word CIVIL where I is used to identify current, C is used to
44
Series RLC-networks
identify a capacitor, V is used to identify voltage, and L is used to identify an inductor.
Since we are presently working with an inductor and we underline part of the word
CIVIL as indicated and we read it from left to right and we say:
The voltage (V) will lead the current (I) in an inductor (L). This phase angle between
the alternating supply voltage and the current will never be greater that 90º.
Should we assume that we are working with a pure inductor we will find that it has no
resistance (dc) but that it does offer an opposition to the flow of current which is termed
the ‘inductive reactance’ and is given mathematically as:
XL =
2×π×f×L
where
XL
π
f
L
=
=
=
=
inductive reactance of inductor in ohm
3,142
frequency of supply in hertz
value of inductor in Henry
Using Ohm’s Law as our basis we can also formulate the following expressions:
VS
VS
IL =
and
VS = IL × XL
and
XL
=
XL
I
L
Example 4.2
A 250 mH inductor is connected across a 250V/100 Hz alternating current
supply. Determine the:
(a)
Inductive reactance;
(b)
Current that will flow through the inductor.
Solution:
(a)
XL
=
=
=
(b)
IL
=
=
=
2×π×f×L
2 × 3,142 × 100 × 250 × 10-3
157, 1 ohm
VS
XL
250
157, 1
1,591 ampere
N4 Industrial Electronics
Example 4.3
Re-calculate the value of the inductive reactance and current for example 4.2
above if it is given that the frequency of the alternating current supply changes
to 50 Hz.
Solution:
VS
=
2 × π × f × L IL
=
X
=
2 × 3,142 × 50 × 250 × 10-3 L
250
=
78, 55 ohm
=
78, 55
=
3,182 ampere
XL
Therefore, as our calculations indicate, the opposition to the flow of current is purely
dependant upon the frequency of the supply. Also note that the inductive reactance of
the inductor is directly proportional to the frequency of the supply.
4.3
The effect of an alternating quantity on a capacitor
Figure 4.3 illustrates a capacitor connected across an alternating current supply (a) and
(b) illustrates a graphical representation of the phase relationship between the current
and the supply voltage whereas (c) illustrates a phasor diagram.
IC
VS
C
IC
+
IC
t
VS
(a)
(b)
VS
(c)
Figure 4.3
Upon careful inspection you will again notice that there is a phase displacement
between the alternating supply and the current through the inductor and is at a
maximum of 90º with the current leading the voltage. In order to remember when
voltage is leading or lagging we will look at the word CIVIL where I is used to identify
current, C is used to identify a capacitor, V is used to identify voltage, and L is used to
identify an inductor.
45
46
Series RLC-networks
Since we are presently working with a capacitor and we underline part of the word
CIVIL as indicated and we read it from left to right and we say:
In a capacitor (C) the current (I) will lead the voltage (V). This phase angle between
the alternating supply voltage and the current will never be greater that 90º.
Should we assume that we are working with a pure capacitor we will find that it has no
resistance (dc) but that it does offer an opposition to the flow of current and is termed
‘capacitive reactance’ and is given mathematically as:
1
where
2×π×f×C
XC =
XC
π
f
C
=
=
=
=
capacitive reactance of capacitor in ohm
3,142
frequency of supply in hertz
value of capacitor in Farad
Using Ohm’s Law as our basis we can also formulate the following expressions:
VS
VS
IC =
and
VS = IC × XC
and
XC
=
XL
I
C
Example 4.4
A 250 µF capacitor is connected across a 250V/100 Hz alternating current
supply. Determine the:
(a)
Capacitive reactance;
(b)
Current that will flow through the capacitor.
Solution:
1
(a)
XC
=
2×π×f×C
1
=
2 × 3,142 × 100 × 250 × 10-6
=
6,365 ohm
VS
(b) IC
=
X
C
=
250
6,365
= 39,277 ampere
N4 Industrial Electronics
Example 4.5
Re-calculate the value of the capacitive reactance and current for example 4.4
above if it is given that the frequency of the alternating current supply changes
to 50 Hz.
Solution:
1
XC
=
IC
=
2×π×f×C
=
1
=
2 × 3,142 × 50 × 250 × 10-6
=
=
127,307 ohm
VS
XC
250
127,307
1,963 ampere
Therefore, as our calculations indicate, the opposition to the flow of current is purely
dependant upon the frequency of the supply. Also note that the capacitive reactance of
the inductor is inversely proportional to the frequency of the supply.
Another important factor that needs to be mentioned here is the opposition being
offered to the flow of current when supplied from an alternating quantity is termed
‘reactance’ and not resistance as was the case with a direct current supply.
4.4
The series RL-network
A network consisting of an inductor and resistor connected in series is illustrated in
figure 4.4. Also indicated are all the variables that may be found in such a network.
R
L
VL
VR
I
VS
Figure 4.4
The graphic representation and phasor diagram for a series RL-network is given in
figure 4.5 (a) and (b).
47
48
Series RLC-networks
VL
VR
I
VS
VL
+
θ
t
I
(a)
VR
(b)
Figure 4.5
It is very important that when we draw phasor diagrams of series networks, irrespective
of the combination of components in the network, that we use our current in the
network as our reference. This may be done since the current in a series network is the
same throughout the network. In order to derive our mathematical expressions for a
series RL-network we also need to look at the reactance/impedance phasor diagram and
that is illustrated in figure 4.6.
XL
Z
θ
R
I
Figure 4.6
For the purpose of obtaining mathematical expressions for series RL-networks we will
refer to figure 4.5 (b) and figure 4.6.
A new concept namely impedance now comes to the fore. This impedance of the
network may be defined in this instance (RL-series networks) as the opposition offered
to the flow of an alternating current and is the vector sum of the reactance of the
inductor and the resistance of the resistor and is mathematically given by:
where
Z
Z = (R2 + XL2) ½
R
XL
=
=
=
impedance of the network in ohm
resistance of resistor in ohm
reactance of inductor in ohm
Observe figure 4.6 and you will notice that the above expression is derived by applying
Pythagoras to the triangle. Similarly the supply voltage expression can be derived by
N4 Industrial Electronics
49
applying Pythagoras using figure 4.5 (b) and is mathematically given by:
VS = (VR2 + VL2) ½
where VS
VR
VL
=
=
=
supply voltage in volt
voltage drop across resistor in volt
voltage drop across inductor in volt
where
=
I×R
VL
=
I × XL
and
VR
The current in the network is mathematically given by:
VS
where
Z
I
=
VS
Z
I
=
=
=
supply voltage in volt
impedance of the network in ohm
current flow in network in ampere
You will further notice that there is a phase angle θ between the current flowing in the
network and the supply voltage. This is termed the phase angle and is mathematically
given by:
θ = tan-1XL or θ = tan-1VL where
R
VR
XL
VL
VR
R
=
=
=
=
reactance of inductor in ohm
voltage drop across inductor in volt
voltage drop across resistor in volt
resistance of resistor in ohm
Both these expressions are derived with trigonometry using figure 4.5 (b) and figure 4.6.
Example 4.6
Consider the network diagram given.
I
R
L
10 ohm
25 mH
VR
VL
VS = 250 V/ 50 Hz
50
Series RLC-networks
Determine from the given information on the network diagram the:
(a)
Inductive reactance of the inductor;
(b)
Impedance of the network;
(c)
Current flow in the network;
(d)
Voltage drops across the:
(i)
Resistor; and
(ii)
Inductor.
(e)
Supply voltage
(f)
Phase angle between the supply voltage and the current.
(g)
Draw a voltage phasor that will represent the quantities you have calculated
above and insert all relevant values.
Solution:
(a)
XL
=
=
=
(b) Z
=
=
=
2×π×f×L
2 × 3,142 × 50 × 25 × 10-3
7,855 ohm
(R2 + XL2)½
(102 + 7, 8552)½
12,716 ohm
VS
(c)
I
=
Z
250
=
12,716
= 19, 66 ampere
(d)(i) VR
=
=
=
I × R
19, 66 × 10
196, 6 volt
(d)(ii) VL
=
=
=
I × XL
19, 66 × 7,855
154, 43 volt
(e)
VS
=
=
=
(VR2 + VL2)½
(196, 62 + 154,432)½
250 volt
The calculation done in (e) proves that the previous answers obtained are indeed
correct since the vector sum of the individual voltage across the resistor and
inductor is equal to the given supply voltage.
N4 Industrial Electronics
(f)
θ
=
=
=
tan-1 XL
or
θ
= tan-1 VL
R
VR
tan-1 7,855
= tan-1 154, 43
10
96, 6
38,15º
= 38,15º
What do these two answers mean? Since this is a series RL-network it will behave
inductively and the supply voltage will lead the current by 38,15º, or alternatively
the current will lag the supply voltage by 38,15º. This will become clearer when
we draw the voltage phasor in 5.
(g)
VL = 154,43 V
VS = 250 V
38,15º
I = 19,66 A
4.5
VR = 196,6 V
The series RC-network
A network consisting a capacitor and resistor connected in series is illustrated in figure
4.7. Also indicated are all the variables that may be found in such a network.
C
R
VC
VR
I
VS
Figure 4.7
The graphic representation and phasor diagram for a series RC-network is given in
figure 4.8 (a) and (b).
51
52
Series RLC-networks
VR
I
+
VC
I
t
θ
VR
(a)
VC
VS
(b)
Figure 4.8
It is very important that when we draw phasor diagrams of series networks, irrespective
of the combination of components in the network, that we use our current in the
network as our reference. This may be done since the current in a series network is the
same throughout the network. In order to derive our mathematical expressions for a
series RC-network we also need to look at the reactance/impedance phasor diagram
and that is illustrated in figure 4.9.
I
θ
XC
R
Z
Figure 4.9
For the purpose of obtaining mathematical expressions for series RC-networks we will
refer to figure 4.8 (b) and figure 4.9.
A new concept namely impedance now comes to the fore. This impedance of the
network may be defined in this instance (RC-series networks) as the opposition offered
to the flow of an alternating current and is the vector sum of the reactance of the
capacitor and the resistance of the resistor and is mathematically given by:
where
Z = impedance of the network in ohm
Z = (R2 + XC2)½
R = resistance of resistor in ohm
XC = reactance of capacitor in ohm
N4 Industrial Electronics
53
Observe figure 12.9 and you will notice that the above expression is derived by applying
Pythagoras to the triangle. Similarly the supply voltage expression can be derived by
applying Pythagoras using figure 4.8 (b) and is mathematically given by:
VS = (VR2 + VC2)½ where VS
VR
VC
=
=
=
supply voltage in volt
voltage drop across resistor in volt
voltage drop across capacitor in volt
where VL =
=
I×R
I × XC
and
VR
The current in the network is mathematically given by:
VS
where
Z
I
=
VS
Z
I
=
=
=
supply voltage in volt
impedance of the network in ohm
current flow in network in ampere
You will further notice that there is a phase angle θ between the current flowing in the
network and the supply voltage. This is termed the phase angle and is mathematically
given by:
θ = tan-1 XC or θ = tan-1 VC
R VR
where
XC
VL
VR
R
=
=
=
=
reactance of capacitor in ohm
voltage drop across capacitor in volt
voltage drop across resistor in volt
resistance of resistor in ohm
Both these expressions are derived with trigonometry using figure 4.8 (b) and figure 4.9.
Example 4.7
Consider the network diagram given.
R
C
10 ohm
130 µF
VC
VR
I
VS = 250 V/50 Hz
54
Series RLC-networks
Determine from the given information on the network diagram the:
(a)
Capacitive reactance of the capacitor;
(b)
Impedance of the network;
(c)
Current flow in the network;
(d)
Voltage drops across the:
(i)
Resistor; and
(ii)
Capacitor.
(e)
Supply voltage;
(f)
Phase angle between the supply voltage and the current.
(g)
Draw a voltage phasor that will represent the quantities you have calculated
above and insert all relevant values.
Solution:
1
(a)
XC
=
2×π×f×C
1
=
2 × 3,142 × 50 × 130 × 10-6
=
24, 48 ohm
(b)
Z
=
=
=
(c)
I
=
=
=
(R2 + XC2)½
(102 + 24, 482)½
26,443 ohm
VS
Z
250
26,443
9,454 ampere
(d)(i) VR
(d)(ii) VL
=
=
=
I × R
9, 454 × 10
94, 54 volt
=
=
=
I × XC
9,454 × 24, 48
231,434 volt
(e)
VS
=
=
=
(VR2 + VC2)½
(94, 54 + 231, 434)½
249,999 volt
N4 Industrial Electronics
55
The calculation done in (e) proves that the previous answers obtained are indeed
correct since the vector sum of the individual voltage across the resistor and
inductor is almost equal to the given supply voltage.
or
θ
=
tan-1 XC
R
tan-1 24, 48
=
10
67,62º=
(f)
θ
=
=
=
tan-1 VC
VR
-1
tan 231,434
94, 54
67,78º
What do these two answers mean? Since this is a series RC-network it will behave
capacitively and the supply voltage will lead the current by 67,62º, or alternatively
the current will lag the supply voltage by 67,62º. This will become clearer when we
draw the voltage phasor in (g).
(g)
I = 9,454 A
VR = 94,54 V
67,62º
VS = 250 V
VC = 231,434 V
4.6
The series RLC-network
A network consisting a resistor, capacitor and inductor connected in series is illustrated
in figure 4.10. Also indicated are all the variables that may be found in such a network.
R
I
L
VR
VL
VS
Figure 4.10
C
VC
56
Series RLC-networks
With all three components connected in the circuit it will either behave like an RLnetwork or an RC-network. This will depend on the frequency of the alternating supply
in that either the inductive reactance (XL) or the capacitive reactance (XC) will be the
greater. Should the inductive reactance be the greater then the network will behave
inductively as if there was only an inductor and resistor in the network. On the other
hand, should the capacitive reactance be the greater then the network will behave
capacitively as if there was only a capacitor and resistor in the network. These two
conditions will now be discussed in turn and you are referred to figure 4.10 and 4.11 (a)
and (b).
4.6.1 Inductive behaviour
The conditions for inductive behaviour include:
•
•
XL > XC; and
VL > VC
The phasor diagram for inductive behaviour is illustrated in figure 4.11 (a) and (b).
XL
VL
Z
XL - XC
θ
I
XC
VS
VL - VC
θ
R
I
VC
(a)
VR
(b)
Figure 4.11
If you refer back to figure 4.5 (b) and figure 4.6 you will notice that they are identical.
The only difference being is that VL and VC have been subtracted vectorially to yield VL VC. The reason for this is that they are 180º apart and cannot be subtracted algebraically.
4.6.2 Capacitive behaviour
The conditions for capacitive behaviour include:
•
•
XC > XL; and
VC > VL
The phasor diagram for capacitive behaviour is illustrated in figure 4.12 (a) and (b).
N4 Industrial Electronics
XL
VL
I
I
R
θ
XC - XL
θ
Z
XC
57
VC - VL
VS
VC
(a)
VR
(b)
Figure 4.12
If you refer back to figure 4.8 (b) and figure 4.9 you will notice that they are identical.
The only difference being is that VC and VL have been subtracted vectorially to yield VC VL. The reason for this is that they are 180º apart and cannot be subtracted algebraically.
Referring to figure 4.11 and 4.12 we can once again use Pythagoras as basis and derive
the following mathematical expressions:
Z = [R2 + (XC ~ XL)2]½ where Z
R
XC
XL
~
=
=
=
=
=
impedance of the network in ohm
resistance of resistor in ohm
reactance of capacitor in ohm
reactance of inductor in ohm
difference between
VS = [VR2 + (VC ~ VL)2]½ where VS
VR
VL
VC
~
=
=
=
=
=
supply voltage in volt
voltage drop across resistor in volt
voltage drop across inductor in volt
voltage drop across capacitor in volt
difference between
where VC = I × XC VR =
I × R
VL
=
I × XL
The current in the network is mathematically given by:
VS
where
Z
I
=
VS
Z
I
=
=
=
supply voltage in volt
impedance of the network in ohm
current flow in network in ampere
You will further notice that there is a phase angle θ between the current flowing in the
network and the supply voltage. This is termed the phase angle and is mathematically
given by:
θ
= cos-1 R
Z
58
Series RLC-networks
The power factor is mathematically given by:
Cos θ = R
Z
The true power in a circuit containing an inductor and a capacitor is the power
consumed taking into account the phase angle and is mathematically given by:
True power
=
V × I Cos θ watt
The true power also termed apparent power in a circuit is the power in a pure resistive
circuit and there is no phase angle and is mathematically given by:
Apparent power = V × I watt
A very important aspect concerning the behaviour of a series RLC-network behaving
either inductive or capacitive needs to be noted. In a series RLC-network it is very easy
to see whether XL or XC is the greater and that will determine whether the network is
inductive or capacitive. It is further obvious that should XL be the greater that VL will
also be the greater and vice versa. However, when we do parallel networks and we
work in terms of current this direct proportion does not hold well. We will find there
that should XL be the greater then IL will be smaller and the circuit actually behaves
capacitive. Therefore, to determine whether a network is inductive in a series RLCnetwork we look at VL and VC and for a parallel RLC-network we look at IL and IC.
The following two activities will serve to capture what we have discussed thus far. It
should be noted that there is a lot of mathematical, as well as theoretical knowledge
that need to be mastered. It is further noteworthy that a lot of ‘other’ work needs to be
done before we can draw a phasor diagram. In order to illustrate this logical process the
following two activities will be handled in this manner.
Example 4.8
A network consists of the following components connected in series across a
200V/200Hz alternating current supply:
• R
=
30 ohm
• L
=
5 mH
• C
=
10 µF
Use this data and draw a fully labelled phasor diagram that will represent the
conditions in this network. All relevant information must be indicated on the
phasor diagram.
N4 Industrial Electronics
Solution:
1
2xπxfxC
XL = 2 x π x f x L
XC =
-3
= 2 × 3,142 × 200 × 50 × 10 = 62,84 ohm
=
=
Z
=
=
=
79,56 ohm
[R2 + (XC - XL)2]½
[302 + (79, 56 - 62, 84)2]½
34, 34 ohm
VS
I
=
θ
=
Z
200
=
=
34,34
= 5,82 ampere
=
VC =
=
=
1
2 × 3,142 × 200 × 10 ×10-6
I × XC
VR =
5,82 × 79,56
=
463 volt
=
cos-1
cos-1
R
Z
30
34,34
29,11º
I × R
VL
5,82 × 30
174,6 olt
=
=
=
VL = 365,73 V
I = 5,82 A
29,11º
VC - VL = 97,27 V
VC = 463 V
VR = 174,6 V
VS = 200 V
I × XL
5,82 × 62,84
365,73 volt
59
60
Series RLC-networks
Example 4.9
Consider the following network diagram.
I
R
L
10 ohm
14 ohm
150 V
180 V
C
8 ohm
90 V
100 Hz
Use the information in the above network diagram and determine the:
(a)
Supply voltage;
(b)
Impedance;
(c)
Line current;
(d)
Value of the inductor in mH;
(e)
Value of the capacitor in µF; and
(f)
Phase angle between the line current and supply voltage.
(g)
You are further required to answer the following questions.
(i)
Does the network behave inductive or capacitive? Motivate
your answer!
(ii)
Is the line current leading or lagging the supply voltage? Motivate your answer!
Solution:
(a) VS
=
=
=
[VR2 + (VC2 ~ VL2)]½
[1502 + (180 - 90)2]½
174,9 volt
(b) Z
[R2 + (XC ~ XL)2]½
[102 + (14 - 8)2]½
11,66 ohm
=
=
=
VS
(c) I
=
Z
174, 9
=
11, 66
=
15 ampere
N4 Industrial Electronics
(d)
XL
=
61
2×π×f×L
XL
L
=
2×π×f
4
=
2 × 3,142 × 100
=
22,27 mH
(e)
XC =
=
=
=
(f)
θ
=
=
=
(g)(i)
1
2×π×f×C
1
2 × π × f × XC
1
2 × 3,142 × 100 × 8
198,9 µF
cos-1
R
Z
cos-1
10
11, 16
30,95º
The circuit is behaving inductive since VL > VC.
(g)(ii)
The line current is lagging the supply voltage since the circuit behaves
inductive.
Resonance is a condition that will occur in a series- or parallel RLC-network when the
inductive reactance equals the capacitive reactance. In order to analyse the phenomena
of resonance we need to look at the mathematical expressions we use to determine
inductive- and capacitive reactance.
It has already been mentioned that the inductive reactance is directly proportional to
the supply frequency and we will find that as frequency increases so does the inductive
reactance increase. The capacitive reactance however is indirectly proportional to the
supply frequency and will have a decrease in capacitive reactance as the supply frequency
increases. This concept is illustrated in figure 4.13. The graphic representation of
resonance illustrated in figure 4.13 shows that the graph for inductive reactance (XL) is a
straight line whereas the graph for the capacitive reactance (XC) is an exponential curve.
62
Series RLC-networks
There is the point where these two graphs intersect one another and is termed the
resonant frequency and it is also this point where XL = XC. We will now use this
information to derive a formula to determine this frequency.
R
e
a
c
t
a
n
c
e
XL
XC
f
fr
XL
Figure 4.13
= XC ............... (1)
But XL = 2 x π x f x L and XC
=
Substitute in (1)
1
2xπxfxC
2xπxfxL =
1
2xπxfxC
=
1
f2
(2 x π)2 x L x C
fr
=
1
2 x π x (L x C)½
The expression derived above clearly indicates that only the inductor and capacitor
plays a role in the resonance of a series RLC-network and that resistance plays no role at
all.
4.7
Conditions for resonance
At this point we will only consider the conditions for resonance in series RLC-networks.
The following conditions will exist for resonance in a series RLC-network.
•
•
•
•
XL
VL
VS
Z
=
=
=
=
XC
VC and is maximum
VR
R and is minimum
N4 Industrial Electronics
• θ =
0º
• I is maximum
• fr = 1
2 x π x (L x C)½
The explanation that will follow now we will prove mathematically as well as by means
of graphic representation some of the conditions mentioned above.
Z = [R2 + (XC ~ XL)2]½
but XL = XC
θ =
= [R2 + (0)2]½
= R =
=
=
VS = [VR2 + (VC2 ~ VL2)½
= [VR2 + (0)2]½
= VR
cos-1 R
Z
-1
cos R
R
-1
cos 1
0º
but Z = R
but VL = VC
The following two graphs illustrated in figure 4.14 (a) and (b) illustrates two of the
above conditions namely frequency versus impedance and frequency versus current.
Also illustrated in figure 4.14 (c) is the phasor diagram for a series resonant network.
R
I
m
p
e
d
a
n
c
e
R
XL > XC
XC > XL
RC
XC > XL
C
u
r
r
e
n
t
RL
VL
XL > XC
RC
RL
VR
I
f
f
fr
fr
(a)
(b)
VC
(c)
Figure 4.14
Example 4.10
A series RLC-network consists of the following components connected across
a 300V alternating current supply:
• R =
10 ohm
• L =
20 mH
• C =
75 µF
63
64
Series RLC-networks
Use this information to determine the:
(a)
Resonant frequency of the network;
(b)
Impedance of the network (prove your answer);
(c)
That the supply voltage VS = 300 V (prove your answer); and
(d)
That the phase angle θ = 0º (prove your answer).
Solution:
1
(a) fr =
2 x π x (L x C)½
1
=
2 × 3,142 (20 × 10-3 × 75 × 10-6)½
=
(b) Z =
=
=
129, 93 Hz
[R2 + (XC ~ XL)2]½
[102 + (0)2]½
10 ohm
but XL = XC = 0
(c) XL = 2 × π × f × L
= 2 × 3,142 ×129,93 × 20 × 10-3
= 16,33 ohm
1
XC =
2×π×f×C
=
1
2 × 3,142 ×129, 93 × 75 × 10-6
=
I =
=
=
VS =
=
=
16,33 ohm
VS
VC
Z
300
10
= I × XC
VR = I × R
VL = I × XL
= 30 × 16,33
= 30 × 10
= 30 × 16,33
= 489,9 volt
= 300 volt
= 489,9 volt
30 ampere
[VR2 + (VC2 ~ VL2 )]½
[3002 + (0)2]½
300 volt
but VL = VC = 0
N4 Industrial Electronics
(d) θ
=
=
=
=
R
Z
R
cos-1
R
10
cos-1
10
cos-1
but
Z
=
R
=
10
0º
Exercise 4.1
1.
An alternating quantity will behave in a prescribed manner when
applied to different components. Use applicable diagrams to illustrate
this behaviour and draw the graphical- as well as phasor diagrams when
an alternating quantity is applied to a (n):
1.1
Resistor;
1.2
Capacitor; and
1.3
Inductor.
2.
Use a graph and give a detailed explanation of series resonance. You are
further required to derive an expression that can be used to determine the
resonant frequency of a series RLC-network.
3.
Explain the concept of inductive- and capacitive behaviour of a series
RLC-network by means of two fully labelled phasor diagrams. Use your
own imaginary values.
4.
A series network consists of a 25 mH inductor and a 15 ohm resistor
connected across a 100V/60Hz alternating supply. Calculate the:
4.1
Impedance of the network;
4.2
Current that will flow in the network;
4.3
Voltage drop across the:
4.3.1
Resistor; and
4.3.2
Inductor.
4.4
Phase angle between the supply voltage and the current.
4.5
Draw a phasor diagram that will represent the quantities for the
above network and insert all relevant values.
5.
A series network consists of a 25 µF capacitor and a 10 ohm resistor
connected across a 150V/60Hz alternating supply. Calculate the:
5.1
Impedance of the network;
5.2
Current that will flow in the network;
65
66
Series RLC-networks
5.3
Voltage drop across the:
5.3.1
Resistor; and
5.3.2
Capacitor.
5.3.3
Phase angle between the supply voltage and the current.
5.3.4
Draw a phasor diagram that will represent the quantities for the above network and insert all relevant values.
6.
Mention seven conditions that will exist in as series resonant RLC-network.
7.
Consider the following phasor diagram.
VL = 55 V
VS
VL - VC
I-3A
VR = 30 V
VC = 20 V
Determine using the available date the:
7.1 Magnitude of the supply voltage;
7.2 Capacitive value of the capacitor in µF; and
7.3 Inductive value of the inductor in mH.
Note: Use a supply frequency of 50 Hz.
8.
A series RLC-network consists of the following components:
• Resistor =
2 ohm
• Capacitor =
100 µF
• Inductor =
20 mH
This combination is connected across a 100mV alternating supply at a frequency of 500 radians/second. Calculate the:
8.1 Impedance of the network;
8.2 Current that will flow in the network;
8.3 Voltage drop across the:
8.3.1 Resistor;
8.3.2 Capacitor; and
8.3.3 Inductor.
8.4 Phase angle between the line current and the supply voltage.
8.5 Draw a phasor diagram that will represent the quantities for the above
network and insert all relevant values.
N4 Industrial Electronics
9.
Draw neat labelled graphs that will illustrate:
9.1 Current versus frequency; and
9.2 Frequency versus impedance for a series RLC-resonant circuit.
10. A series RLC-network draws a current of 1,6 mA from a 1,6 V alternating current supply when the network is at resonance at a frequency of 1 kHz. Should 16 volt be measured across the inductor at this frequency determine
the:
10.1 Value of the inductor;
10.2 Value of the capacitor; and
10.3 Value of the resistor.
67
68
Series RLC-networks
PARALLEL RLC-NETWORKS
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69
CHAPTER
5 Parallel RLC-networks
Learning Outcomes
On completion of this module you will be able to:
•
•
•
•
•
•
•
•
•
Illustrate with the aid of a circuit diagram and graphical representation a parallel
circuit containing a resistor and an inductor;
Determine by calculation the:
– Impedance;
– Current flow;
– Phase angle;
– Voltage drops; and
– Supply voltage in a parallel circuit containing a resistor and an inductor.
Draw a phasor diagram for a parallel circuit containing a resistor and an inductor;
Illustrate with the aid of a circuit diagram and graphical representation a parallel
circuit containing a resistor and a capacitor;
Determine by calculation the:
– Impedance;
– Current flow;
– Phase angle;
– Voltage drops; and
– Supply voltage in a parallel circuit containing a resistor and a capacitor.
Draw a phasor diagram for a parallel circuit containing a resistor and a capacitor;
Illustrate with the aid of a circuit diagram and graphical representation a parallel
circuit containing a resistor, an inductor and a capacitor;
Determine by calculation the:
– Impedance;
– Current flow;
– Phase angle;
– Voltage drops; and
– Supply voltage in a parallel circuit containing a resistor, an inductor and
a capacitor.
Draw a phasor diagram for a parallel circuit containing a resistor, an inductor and a
capacitor;
70
parallel RLC-networks
•
•
•
5.1
Explain by means of a graphical representation the concept of resonance in a
parallel circuit containing a resistor, an inductor and a capacitor;
Explain the conditions for resonance in a parallel circuit containing a resistor, an
inductor and a capacitor; and
Calculate the resonant frequency in a parallel circuit containing a resistor, an
inductor and a capacitor.
The parallel RL-network
A parallel network consisting of an inductor and resistor connected in parallel is
illustrated in figure 5.1. Also indicated are all the variables that may be found in such a
network.
IR
R
IL
L
IT
VS
Figure 5.1
The graphic representation and phasor diagram for a parallel RL-network is given in
figure 5.2 (a) and (b).
VS
IL
+
IR
VS
t
IR
θ
IL
IT
(a)
(b)
Figure 5.2
It is very important that when we draw phasor diagrams of parallel networks,
irrespective of the combination of components in the network, that we use our supply
voltage to the network as our reference. This may be done since the voltage across a
parallel network is the same across the network. In order to derive our mathematical
N4 Industrial Electronics
expressions for a parallel RL-network we also need to look at the reactance/impedance
phasor diagram and that is illustrated in figure 5.3.
I
R
θ
Z
XL
Figure 5.3
For the purpose of obtaining mathematical expressions for parallel RL-networks we
will refer to figure 5.2 (b) and figure 5.3. A new concept namely impedance now comes
to the fore. This impedance of the network may be defined in this instance (RL-parallel
networks) as the opposition offered to the flow of an alternating current and is the
vector sum of the reactance of the inductor and the resistance of the resistor and is
mathematically given by:
where
R × XL
(R2 + XL2)½
Z
=
Z = impedance of the network in ohm
R = resistance of resistor in ohm
XL = reactance of inductor in ohm
VS
Z =
where Vs = supply voltage in volt
I
TIT = total current in ampere
The total current expression can be derived by applying Pythagoras using figure 5.2 (b)
and is mathematically given by:
IT = (IR2 + IL2)½
where
IT
IR
IL
= total current in ampere
= current through resistor in ampere
= current through inductor in ampere
VS
VS
where IL =
and
IR =
XL
R
The total current in the network is also mathematically given by:
VS
IT =
where
Z
VS
Z
IT
= supply voltage in volt
= impedance of the network in ohm
= current flow in circuit in ampere
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72
parallel RLC-networks
You will further notice that there is a phase angle θ between the current flowing in the
network and the supply voltage. This is termed the phase angle and is mathematically
given by:
I
I
where IL = current through inductor in ampere
θ = tan-1 L or θ = cos-1 R
IR
IT
IR = current through resistor in ampere
IT = total current in circuit in ampere
Example 5.1
Consider the network and determine from the given information on the
network diagram the:
R
IR
10 ohm
L
IL
25 mH
IT
100 V/60 Hz
(a)
(b)
(c)
(d)
(e)
(f)
Inductive reactance of the inductor;
Impedance of the network;
Current flow through the:
(i)
Resistor; and
(ii)
Inductor.
Total current
Phase angle between the supply voltage and the current.
Draw a current phasor that will represent the quantities you have
calculated above and insert all relevant values.
Solution:
(a)
XL
=
=
=
2xπxfxL
2 × 3,142 × 60 × 50 × 10-3
18,852 ohm
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73
(b) Z
=
R × XL
Z
= VS
2
2 ½
(R + XL )
IT
=
10 × 18,852
= 100
11, 32
(102 + 18,8522)½
= 8,834 ohm
= 8,834 ohm
VS
(c)(i) IL
=
X
L
100
=
18,852
= 5,3 ampere
VS
(c)(ii) IR
=
R
100
=
10
= 10 ampere
VS
(d) IT
= (IR2 + IL2)½
IT
=
Z
= (102 + 5,32)½
100
= 11,32 ampere
=
8,834
= 11, 32 ampere
The calculation done in (d) proves that the previous answers obtained are indeed correct since the vector sum of the individual voltage across the resistor and inductor is equal to the total current flow.
(e) θ =
=
=
IL
or
θ =
IR
5,3
tan-1 =
10
27,92º =
tan-1
cos-1
IR
IT
cos-1
10
11,32
27,95º
What do these two answers mean? Since this is a parallel RL-network it will behave
inductively and the supply voltage will lead the current by 27,92º, or alternatively
the current will lag the supply voltage by 27,92 º. This will become clearer when we
draw the voltage phasor in (f).
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parallel RLC-networks
(f)
VS = 100 V
IR = 10 A
27, 92º
IT = 11,32 A
IL = 5,3 A
5.2
The parallel RC-network
A network consisting a capacitor and resistor connected in parallel is illustrated in
figure 5.4. Also indicated are all the variables that may be found in such a network.
The graphic representation and phasor diagram for a parallel RC-network is given in
figure 5.5 (a) and (b).
IR
R
IC
C
IT
VS
Figure 5.4
IC
VS
+
IR
IT
IC
t
θ
VS
-
(a)
(b)
Figure 5.5
IR
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75
It is very important that when we draw phasor diagrams of parallel networks,
irrespective of the combination of components in the network, that we use our supply
voltage to the network as our reference. This may be done since the supply voltage in
a parallel network is the same across the network. In order to derive our mathematical
expressions for a parallel RC-networks we also need to look at the reactance/impedance
phasor diagram and that is illustrated in figure 5.6.
XC
Z
θ
R
I
Figure 5.6
For the purpose of obtaining mathematical expressions for parallel RC-networks we
will refer to figure 5.5 (b) and figure 5.6. A new concept namely impedance now comes
to the fore. This impedance of the network may be defined in this instance (parallel
RC-networks) as the opposition offered to the flow of an alternating current and is the
vector sum of the reactance of the capacitor and the resistance of the resistor and is
mathematically given by:
where Z =
R × Xc
(R2 + XC2)½ R =
XC =
impedance of the network in ohm
resistance of resistor in ohm
reactance of inductor in ohm
Z = VS
where Vs =
ITIT =
supply voltage in volt
total current in ampere
Z
=
The total current expression can be derived by applying Pythagoras using figure 13.2 (b)
and is mathematically given by:
IT = (IR2 + IC2)½
where IT
IR
IC
where
IC
=
=
=
=
total current in ampere
current through resistor in ampere
current through inductor in ampere
VS
and IR = VS
XC R
The total current in the network is also mathematically given by:
76
parallel RLC-networks
IT = VS where
VS
Z Z
IT
=
=
=
supply voltage in volt
impedance of the network in ohm
current flow in circuit in ampere
You will further notice that there is a phase angle θ between the current flowing in the
network and the supply voltage. This is termed the phase angle and is mathematically
given by:
θ = tan-1 IC or θ = cos-1 IR
where IC = current through inductor in ampere
IT
IR = current through resistor in ampere
IR
IT = total current in circuit in ampere
Example 5.2
Consider the network and determine from the given information on the
network diagram the:
R
IR
100 ohm
IC
IT
C
10 µF
120 V/100 Hz
(a)
(b)
(c)
(d)
(e)
(f)
Capacitive reactance of the capacitor;
Impedance of the network;
Current flow through the:
(i)
Capacitor; and
(ii)
Resistor.
Current flow in the network;
Phase angle between the supply voltage and the current.
Draw a current phasor that will represent the quantities you have
calculated above and insert all relevant values.
N4 Industrial Electronics
Solution:
(a)
XC
= ½xπxfxC
= 1/2 × 3,142 × 60 × 10 × 10-6
= 159,134 ohm
VS
Z
=
(b) Z
=
R × XC IT
(R2 + XC2)½ 120
100 × 159,134
= 1,417
=
(102 + 159,134)½
= 84,67 ohm
= 84,67 ohm
VS
VS
(c)(i) IC
=
(ii) IR =
XC
R
120
120
=
=
159,134
10
= 0,754 ampere
= 1,2 ampere
VS
(d) IT
= (IR2 + IC2)½
IT
=
Z
= (1,22 + 0,754)½
100
= 1,417 ampere
=
84,68
= 1,417 ampere
The calculation done in (d) proves that the previous answers obtained are indeed
correct since the vector sum of the individual voltage across the resistor and
inductor is equal to the total current flow.
(e)
θ
=
=
=
IC
or
θ
=
IR
0,754
tan-1
=
1, 2
32,14º
=
tan-1
IR
IT
1,2
cos-1
1,417
cos-1
32,13º
What do these two answers mean? Since this is a parallel RC-network it will
behave capacitively and the current will lead the supply voltage by 32,14º, or
alternatively the supply voltage will lag the current by 32,14 º. This will
become clearer when we draw the voltage phasor in (f)
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78
parallel RLC-networks
(f)
IC = 0,754 A
IT = 1,417 A
32, 13º
IR = 1,2 A
VS = 120 V
5.3
The parallel RLC-network
A network consisting a resistor, capacitor and inductor connected in series is illustrated
in figure 5.7. Also indicated are all the variables that may be found in such a network.
R
IR
C
IC
L
IL
IT
VS
Figure 5.7
With all three components connected in the network it will either behave like an RLnetwork or an RC-network. This will depend on the frequency of the alternating supply
in that either the inductive reactance (XL) or the capacitive reactance (XC) will be the
greater. Should the inductive reactance be the greater then the network will behave
capacitively as if there was only a capacitor and resistor in the network. On the other
hand, should the capacitive reactance be the greater then the network will behave
inductively as if there was only an inductor and resistor in the network. These two
conditions will now be discussed in turn and you are referred to figure 5.8 (a) and (b)
and 5.9 (a) and (b).
5.3.1 Capacitive behaviour
The conditions for capacitive behaviour include:
N4 Industrial Electronics
•
•
79
XL > XC; and
IC > IL
The phasor diagram for capacitive behaviour is illustrated in figure 5.8 (a) and (b).
XL
IC
Z
XL - XC
θ
θ
R
I
XC
IT
IC - IL
IL
(a)
IR
VS
(b)
Figure 5.8
If you refer back to figure 5.5 (b) and figure 5.6 you will notice that they are identical.
The only difference being is that IC and IL have been subtracted vectorially to yield IC IL. The reason for this is that they are 180º apart and cannot be subtracted algebraically.
5.3.2 Inductive behaviour
The conditions for inductive behaviour include:
•
•
XC > XL; and
IL > IC
The phasor diagram for inductive behaviour is illustrated in figure 5.9 (a) and (b).
the inductor is directly proportional to the frequency of the supply.
XL
IC
I
θ
XC - XL
Z
XC
VS
R
θ
IL - IC
IT
IL
(a)
IR
(b)
Figure 5.9
If you refer back to figure 5.2 (b) and figure 5.3 you will notice that they are identical.
The only difference being is that IL and IC have been subtracted vectorially to yield IL IC. The reason for this is that they are 180º apart and cannot be subtracted algebraically.
80
parallel RLC-networks
A new concept namely impedance now comes to the fore. This impedance of the
network may be defined in this instance (parallel RLC-networks) as the opposition
offered to the flow of an alternating current and is the vector sum of the reactance of the
capacitor and the resistance of the resistor and is mathematically given by:
Z =
1
where Z
2 ½
[1 + (1 ~ 1 ) ] R
[R (XC XL ) ] XC
XL
=
=
=
=
VS
Z =
where Vs =
I
TIT =
impedance of the network in ohm
resistance of resistor in ohm
reactance of capacitor in ohm
reactance of inductor in ohm
supply voltage in volt
total current in ampere
The total current expression can be derived by applying Pythagoras using figure 5.5 (b)
and is mathematically given by:
where IT
IT = [IR2 + (IC ~ IL)2]½
orIR
IT = [IR2 + IX2]½ IC
IL
IX
where IC
=
VS
XC
=
=
=
=
=
and IR =
total current in ampere
current through resistor in ampere
current through capacitor in ampere
current through inductor in ampere
difference between IC and IL
VS
VS
and
IL
=
R
X
L
The current in the network is mathematically given by:
VS
IT =
where VS =
Z
Z =
IT =
supply voltage in volt
impedance of the network in ohm
current flow in circuit in ampere
You will further notice that there is a phase angle º between the current flowing in the
network and the supply voltage. This is termed the phase angle and is mathematically
given by:
θ = tan-1 IX or θ = cos-1 IR where
IT
IR
IC
IR
IT
IX
= current through capacitor in ampere
= current through resistor in ampere
= total current in network in ampere
= difference between IC and IL
N4 Industrial Electronics
Example 5.3
A parallel RLC-network has the following components connected across a
150V/150 Hz alternating current supply:
• Resistor =
12 ohm
• Capacitor =
70 µF
• Inductor =
15 mH
Determine the:
(a)
Inductive reactance of the inductor;
(b)
Capacitive reactance of the capacitor;
(c)
Current flowing through the:
(i)
Inductor;
(ii)
Resistor; and
(iii)
Capacitor.
(d)
Total current flowing in the network;
(e)
The phase angle between the supply voltage and the line current (also
indicate whether the network is capacitive or inductive and motivate
your answer); and
(f)
Impedance of the network.
(g)
Draw a current phasor that will represent the quantities you have
calculated above and insert all relevant values.
Solution:
(a)
XL
=
=
=
2×π×f×L
2 × 3,142 × 150 × 15 × 10-3
14,139 ohm
1
(b) XC
=
2×π×f×C
1
=
2 × 3,142 × 150 × 70 × 10-6
=
15,155 ohm
VS
(c)(i) IL =
(ii)IR =
X
L
150
=
=
14,139
= 10, 61 ampere
=
VS
VS
(iii)IC =
R
XC
150
150
=
12
15,155
12, 5 ampere
= 9,897 ampere
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82
parallel RLC-networks
(d)
IT
=
=
=
[IR2 + (IL2 ~ IC2)]½
12,52 + (10,61 - 9,897)2]½
12,52 ampere
(e)
θ
=
=
=
or
θ
=
tan-1 IX
IR
tan-1 0,713=
12,5
3,2º=
cos-1 IR
IT
cos-1 12,5
12,52
3, 2º
The phase angle will indicate that the network is inductive and the line current is
lagging behind the supply voltage.
(f)
Z
=
1
Z
=
VS
[ 1 + ( 1 ~ 1 )2]­IT
[R2 (XL
XC ) ]
=
150
=
1
12, 52
=
11,98 ohm
[ 1 + ( 1
- 1 ) 2]½
[122 (14,139 15,155) ]
= 11, 98 ohm
(g)
IC = 9,897 A
VS = 150 V
3,2º
IL - IC
IL = 10,61 A
IR =12,5 A
IT = 12,52 A
N4 Industrial Electronics
5.4
83
Resonance
There that there is no difference between series- and parallel resonance and we can
therefore use similar expressions and graphic representations as was illustrated for
series resonance.
5.5
Conditions for resonance
The following conditions will exist for resonance in a parallel RLC-network.
• XL =
• θ =
• IL =
• IT =
• Z =
• I is minimum
• fr =
XC
0º
IC and is maximum
IR
is minimum
1
2 × π × (L × C)½
The following three graphs in figure 5.10 (a), (b) and (c) shows two of the above
conditions namely impedance versus frequency (a) and frequency versus current (b)
whereas (c) illustrates the phasor diagram for a parallel resonant network.
R
I
m
p
e
d
a
n
c
e
XC > XL
R
XL > XC
RC
C
u
r
r
e
n
t
RL
IC
XL > XC
XC > XL
VR
IT and IR
RC
RL
f
f
fr
fr
(a)
(b)
IL
(c)
Figure 5.10
5.6
The tuned network (tank network)
A more practical parallel network is illustrated in figure 5.11 and is also termed a ‘tuned
network’ or a ‘tank circuit’.
84
parallel RLC-networks
R
L
IR
IC
IT
C
VS
Figure 5.11
It is generally accepted that a capacitor has no ‘dc resistance’ and is considered as a pure
component. The inductor on the other hand contains ‘dc resistance’ since the inductor
is wound with a conductor in the shape of a coil and this conductor has a resistance.
However, should the inductive reactance of the inductor be ten times greater than the
resistance of the inductor the resistance can be ignored. This will however depend on
the frequency at which the network is operated. It could be generally accepted that
when the inductor is indicated as is illustrated in figure 5.11 with a dotted block, that
the resistance of the inductor can be ignored. Important to note is that although we
say that the resistance of the inductor plays no role it will only be valid for the resonant
frequency and it will play a role in the dynamic impedance and hence the total current
that will be produced in the network. The following mathematical expressions are
applicable for parallel resonant RLC-networks:
1
where
½
2 × π × (L × C) =
=
=
=
frequency in Hertz
3,142
value of the inductor in Henry
value of the capacitor in Farad
where IL
IC
VS
XL
XC
=
=
=
=
=
current through the inductor in ampere
current through the capacitor in ampere
supply voltage in volt
inductive reactance of the inductor in ohm
capacitive reactance of the capacitor in ohm
L
ZD =
where ZD
R×C
R
C
L
=
=
=
=
dynamic impedance of the network in ohm
value of the resistance in ohm
value of the capacitor in Farad
value of the inductor in Henry
fr
=
f
π
L
C
VS
VS
and IC =
IL =
X
X
LC
N4 Industrial Electronics
VS
IT =
where IT = total current in the network in ampere
ZD
VS = supply voltage in volt
ZD = dynamic impedance of the network in ohm
Example 5.4
A parallel resonant tuned network consists of an inductor of 25 mH and a
negligible resistance of 3 ohm and a capacitor of 100 µF connected across a
120V alternating supply. Use the given data to determine the:
(a)
Resonant frequency of the network;
(b)
Inductive reactance of the inductor;
(c)
Capacitive reactance of the capacitor;
(d)
Current through the inductor;
(e)
Current through the capacitor;
(f)
Dynamic impedance of the network; and
(g)
Total current produced by the network.
Solution:
(a)
=
fr
=
=
(b)
XL
=
=
=
=
(c)
XC
=
=
1
2 × π × (L × C)½
1
2 × 3,142 × (25 × 10-3 × 100 × 10-6)½
100,645 Hz
2×π×f×L
2 × 3,142 × 100,645 × 25 ×10-3
15,811 ohm
1
2×π×f×C
1
2 × 3,142 × 100 × 100 × 10-6
15,811 ohm
The question may now arise why these two values are the same? The reason for
that is that the network is at resonance and at the resonant frequency XL = XC!
85
86
parallel RLC-networks
VS
(d) IL =
X
L
120
=
15,811
= 7,589 ampere
VS
(e) IC =
X
C
120
=
15,811
= 7,589 ampere
Note that the two current are also the same for the same motivation as given above!
L
(f) ZD =
R×C
25 × 10-3
=
3 × 100 × 10-6
=
83,333 ohm
VS
(g) IT =
Z
D
120
=
83,333
= 1, 44 ampere
Exercise 5.1
1. A parallel RL-network has the following components connected across
a 200V/120 Hz alternating supply:
• inductor = 60 mH
• resistor= 15 ohm
Use the given values and calculate the:
1.1 Impedance of the network;
1.2 Total current flowing in the network; and
1.3 Phase angle between the supply voltage and line current.
1.4 Draw a phasor diagram that will represent the quantities you calculated above and insert all relevant values.
N4 Industrial Electronics
2.
A parallel RC-network has the following components connected across a 150V/100 Hz alternating supply:
• Capacitor =
60 µF
• Resistor
=
10 ohm
Use the given values and calculate the:
2.1
Impedance of the network;
2.2
Total current flowing in the network; and
2.3
Phase angle between the supply voltage and line current.
2.4
Draw a phasor diagram that will represent the quantities you calculated above and insert all relevant values.
3.
Name seven conditions that may exist for RLC-parallel resonance.
4.
Draw neat labelled graphs that will illustrate:
4.1
Current versus frequency; and
4.2
Frequency versus impedance for a parallel RLC resonant network.
5.
Consider the following network diagram and incomplete phasor diagram.
R
4A
C
4A
L
6A
IT
VS
Complete the phasor diagram using the available data and then determine the total current flowing in the network.
6.
Complete the following expressions with reference to a series resonant RLC-network.
=
(i)
XL
(ii)
θ
=
=
(iii)
VL
(iv)
Z
=
(v)
I
=
7.
A parallel resonant tuned network consists of an inductor of 75 mH and a negligible resistance of 10 ohms and a capacitor of 150 µF connected across a 200V alternating supply. Use the given data to determine the:
87
88
parallel RLC-networks
7.1
7.2
7.3
7.4
7.5
7.6
7.7
8.
Resonant frequency of the network;
Inductive reactance of the inductor;
Capacitive reactance of the capacitor;
Current through the inductor;
Current through the capacitor;
Dynamic impedance of the network; and
Total current produced by the network.
Consider the following network.
R
L
10 ohm
80 mH
C
8.1
8.2
8.3
VS
Find the value of the variable capacitor should the dynamic impedance be equal to 800 ohm.
Determine the value of the capacitor should the network resonate at 10 kHz.
What will the phase angle between the supply voltage and line current
be at 10 kHz?
89
Q-factor, Bandwidth and Complex Notation
N4 Industrial Electronics
CHAPTER
6 Q-factor, Bandwidth and Complex Notation
Learning Outcomes
On completion of this module you will be able to:
•
•
•
•
•
•
•
6.1
Define and do relevant calculations for Q-factor;
Explain graphically the concept of bandwidth;
Do relevant calculations to determine bandwidth;
Explain the concept of complex notation and reasons why it is used;
Perform by means of complex notation:
– Addition;
– Multiplication; and
– Division.
Solve for series RLC-networks using complex notation;
Solve for parallel RLC-networks using complex notation;
Introduction
In the previous two modules we had a look at series- and parallel networks as well as
resonance for these two types of networks. There is however a number of other concepts
regarding alternating current networks that need to be studied.
6.2
The Q-factor
The Q-factor of a network is also termed the ‘magnification factor’ and is applicable to
either a series- or parallel resonant network. This factor is mathematically expressed by:
XL
1
Q =
or
Q =
(L/C) ½
R
R
where
XL
= inductive reactance of inductor in ohm
R
= resistive value of resistor in ohm
L
= value of inductor in Henry
C
= value of capacitor in Farad
90
Q-factor, Bandwidth and Complex Notation
Since XL = XC at resonance it is obvious that:
Q =
XC
or
R
XL
R
It will be found that since this phenomenon is termed the magnification factor that the
following conditions will apply for series or parallel resonant networks:
Series networks:
The voltage across the inductor and capacitor is mathematically given by:
VL or VC
where VS
=
=
Parallel networks:
The current through the inductor and capacitor is mathematically given by:
IL or IC
where IT
6.3
VS × Q
value of supply voltage in volt
=
=
IT × Q
value of total line current in ampere
Bandwidth
The graphic representation in figure 6.1 illustrates the concept of bandwidth.
V or I
0 dB
- 3 dB
0,707 VM or IM
f
f1
fr
Bandwidth
Figure 6.1
f2
N4 Industrial Electronics
91
Definition 6.1 Bandwidth
The bandwidth may be defined as that range of frequencies between f1 and f2 where
the power has fallen or dropped to half its value.
Assuming absolutely pure components (L and C), this bandwidth will lie symmetrically
around the resonant frequency fr. Also of importance is that the bandwidth range of
frequencies will have a current or voltage value which will be equal to 0,707 × VM or IM.
In order to prove that this point where the bandwidth is taken is also termed the halfpower points or –3 dB points we need to substantiate this by means of the following
explanation. Assume an amplifier has an output of 1 watt and we reduce the power by
half to 0,5 watt. We know that:
P
dB =
10 log O
Pi
=
10 log 0, 5
1
=
- 3,010
The bandwidth of a network is mathematically given by:
BW =
R
fr
or
BW =
2×π×L
Q
A number of activities will now be given in order to explain these concepts. There are a
number of approaches that may be used and you are requested to experiment with all
approaches in solving the problems.
Example 6.1
A series RLC-network has the following component values and is connected
across an alternating current supply.
=
100 ohm
R
L
=
0,5 H
C
=
40 µF
Use the above data to determine the lower- and upper half-power frequencies.
92
Q-factor, Bandwidth and Complex Notation
Solution:
1
R
fr
=
BW
=
2 × π × (L × C)½
2×π×L
100
=
1
2 × 3,142 × 0,5
=
2 × 3,142 × (0, 5 × 40 × 10-6)½
=
35,584 Hz
=
31,826 Hz
Since we assume that the curve is symmetrical around the resonant frequency fr
we can now calculate the lower- and upper half-power frequencies in the following
manner:
BW
BW
Lower frequency = fr Upper frequency
= fr +
2
2
31,826
31,826
=
35,584 = 35,584 +
2
2
=
19,667 Hz
= 51,493 Hz
Example 6.2
A series RLC-network has the following component values and is connected
across an alternating current supply.
R = 3 ohm;
L = 0, 2 mH;
C = 0,323 µF
Use the above data to determine the bandwidth of the network if it is given
that the network resonates at a frequency of 19, 802 kHz
Solution:
R
BW =
2×π×L
3
=
2 × 3,142 × 0, 2 ×10-3
=
2,387 kHz
N4 Industrial Electronics
A second approach, somewhat more complex, will yield the same result.
BW = fr
Q
But Q
= XL
R
= 2×π×f×L
But XL
Q
= 2×π×f×L
R
BW =
fr
2×π×f×L
R
=
fr × R
2×π×f×L
=
19,803 × 103 × 3
2 × 3,142 × 19,802 × 103 × 0,2 × 10-3
= 2,387 kHz
Example 6.3
A series network consists of a resistance having a value of 30 ohm, an inductor
having a value of 10 mH and a capacitor that has a capacitive reactance of 300
ohm. This combination is connected across an alternating supply of 30V and
resonates at a frequency of 4,774 kHz. Determine the:
(a)
Q-factor of the network;
(b)
Bandwidth of the network; and
(c)
The voltage drop across the inductor and the capacitor.
Solution:
XL
(a)
Q
=
R
300
=
30
= 10
93
94
Q-factor, Bandwidth and Complex Notation
fr
(b) BW =
Q
4,774 × 103
=
10
=
477, 4 Hz
(c)
VL
=
=
=
VS × Q
VC
30 × 10
300 volt
=
=
=
VS × Q
30 × 10
300 volt
Example 6.4
A parallel resonant network resonates at an unknown frequency and consists
of a 10 ohm resistor, a 30 mH inductor and a 100 µF capacitor. Determine the
Q-factor of this network.
Solution:
Q
=
1 (L/C) ½
R
=
1 (30 × 10-3 / 100 × 10-6)½
10
=
1,732
Example 6.5
Consider the following RLC-network connected across a 600V alternating
supply resonating at 2,387 kHz.
6A
R
600 V
2,387 kHz
100 ohms
L
10 mH
C
150 ohms
N4 Industrial Electronics
95
Use the above information and calculate the:
Q-factor of the network;
(a)
(b)
Lower- and upper half-power frequencies;
(c)
Bandwidth;
(d)
Current through the inductor and the capacitor; and;
(e)
Value of the capacitor.
Solution:
XL
Q
=
R
150
=
100
= 1,5
(a)
fr
(b) BW =
Q
= 2,387 × 103
1,5
= 1,591 kHz
=
(c)
Lf
=
=
fr - BW
Uf = fr + BW
2
2
= 2,387 × 103 + 1,591 × 103
2,387 × 103 - 1,591 × 103 2
2
1,592 kHz
= 3,183 kHz
(d) IL
=
=
=
IT × Q
IC
6 × 1,5
9 ampere
(e)
=
=
=
=
½ ×π×f×C
½ × π × f × XC
½ × 3,142 × 2,387 × 103 × 150
444,45 nF
XC
C
=
=
=
IT × Q
6 × 1,5
9 ampere
6.4Complex notation
Complex notation is a method used to calculate different quantities in alternating
current networks in modulus and angle form which gives us a much easier method of
calculation. Complex notation is very often also referred to as the ‘j’ notation.
96
Q-factor, Bandwidth and Complex Notation
The most common forms to be used include: α ± jβ referred to as the rectangular form
and r/θº which is termed the polar form. Consider the following two networks as
illustrated in figure 6.2 (a) and (b).
-j
R
R
j
Z
Z
(a)
(b)
Figure 6.2
According to figure 6.2 (a):
Z
=
=
According to figure 6.2 (b):
α + jβ (rectangular)
Z
r/θº (polar)
=
=
α – jβ (rectangular)
r/θº (polar)
Rectangular to polar
Polar to rectangular
r2 =
θ =
α
jβ
α2 + β2
tan-1 jβ
α
(Moduli)
(Arguments)
=
=
r cos θ (Real value)
r sin θ (Imaginary value)
Although the conversion can be done this way most modern pocket calculator makes
provision for these conversions. All addition or subtractions of quantities are done
using the rectangular form whereas division and multiplication are done using the polar
form.
•
Addition (Z1 + Z2)
Z1 + Z2
=
=
(α1 + jβ1) + (α2 + jβ2)
(α1 + α2) + (jβ1 + jβ2)
All real values are added or subtracted whatever the case may be and all imaginary
values are added or subtracted whatever the case may be.
α =
jβ =
real values
imaginary values
The sign of the ‘j’ term must always be taken into account.
N4 Industrial Electronics
•
97
Multiplication (Z1 × Z2)
Z1 × Z2
=
=
r1/θ1º × r2/θ2º
r1 × r2/ θ1º + θ2º
All moduli values are multiplied and all argument values are added.
•
Division (Z1 ÷ Z2)
Z1 ÷ Z2
=
=
r1/θ1º ÷ r2/θ2º
r1 ÷ r2/ θ1º - θ2º
All moduli values are divided and all argument values are subtracted.
6.5RLC-networks
6.5.1 Series RLC-networks
Consider the network in figure 6.3
R
L
α
j
VR
VL
C
-j
VC
I
VS
Figure 6.3
The impedance of the network is mathematically given by:
Z
=
=
α ± jβ (rectangular form)
r/θº (polar form)
where
α
=
R
jβ =
reactance where
XL
XC
=
=
j(2 × π × f × L)
- j(1/2 × π × f × C)
VS
I =
where
Z
VS
Z
=
=
r/θº
r/θº
You will notice that r/θº indicates the angle of the current in relation to the supply
voltage and we can therefore draw a phasor diagram with much greater ease as was the
case earlier.
98
Q-factor, Bandwidth and Complex Notation
Example 6.6
A series RLC-network has the following component values connected across a
250/θºV/50 Hz alternating quantity.
R
=
20 ohm; L
=
150 mH; C
=
100 µF
Use this information to:
(a)
Calculate the impedance of the network;
(b)
Determine the current that will flow in the network; and
(c)
Determine the voltage drop across each component in the network.
(d)
Draw a phasor diagram indication all the relevant values calculated.
Solution:
XL =
=
=
=
j(2 × π × f × L)
XC
j2 × 3,142 × 50 × 150 × 10-3
j47, 13 ohm
47, 13/90º ohm
=
=
=
=
- j(1/2 × π × f ×C)
- j(1/2 × 3,142 × 50 × 100 ×10-6)
- j31, 83 ohm
31, 83/-90º ohm
(a) Z = α ± jβ
= 20 + j47,13 - j31,83
= 20 + j15,3 ohm
=25,18/37,42º ohm
V
(b) I = S
Z
250/θº
=
25,18/37,42º
=9,929/-37,42º ampere
(c)(i) VR =
=
=
I × R
9,929/-37,42º × 20/0º
198,58/-37,42º volt
(ii) VL =
=
=
I × XL
9,929/-37,42º × 47,13/90º
467,954/52,58º volt
(iii) VC =
=
=
I × XC
9,929/-37,42º × 31,83/-90º
316,04/-127,42º volt
N4 Industrial Electronics
(d)
VL = 467,954/52, 58º V
VL - VC
VS = 250/θº V
I =9,929 /-37, 42º A
VC = 316, 04/-127, 42º V
VR =198, 58/- 37, 42º
6.5.2Parallel RLC-networks
Example 6.7
The following parallel RLC-network is given.
R
L
10 ohms
J 8 ohms
C
- j 2 ohms
250 V
Determine with reference to the network above the:
(a)
Impedance of the network; and
(b)
Current that will flow in the network.
(c)
Use the information obtained in above and draw a phasor diagram
indicating all relevant values.
Solution:
(a)
Z1
=
=
=
α + jβ
Z2
10 + j8 ohm
12,8/38,66º ohm
=
=
=
α - jβ
0 - j2 ohm
2/90º ohm
99
100
Q-factor, Bandwidth and Complex Notation
Z1 × Z2
ZT
=
Z + Z2
1
12,8/38,66º × 2/-90º
=
10 + j8 + 0 - j2
25,6/-51,34º
=
10 + j6
25,6/-51,34º
=
11,66/ 0,96º
= 2,196/-82, 3º ohm
VS
(b) I
=
Z
250/θº
=
2,196/-82,3º
= 113,843/82,3º ampere
(c)
I = 113,843/82, 3º A
VS = 250/θº V
Example 6.8
The following RLC-network is given.
C
R
L
- j 2 ohms
5 ohms
j 20 ohms
L
j 10 ohms
A
ZOUT
B
Determine the impedance ZOUT that will be measured across terminals AB and
leave the answer you obtain in rectangular form.
N4 Industrial Electronics
Solution:
Z1
=
=
=
α + jβ
Z2
5 + j20 ohm
20,615/75,96º ohm
ZOUT =
=
=
=
=
=
=
=
=
=
α - jβ
Z3
0 - j2 ohm
2/-90º ohm
=
=
=
α + jβ
0 + j10 ohm
10/90º ohm
Z2 × Z3
Z2 + Z3
5 + j 20 + 2/-90º × 10/90º
0 - j 2 + 0 + j 10
5 + j 20 + 20/0º
0 + j8
5 + j 20 + 20/0º
8/90º
Z1 +
5 + j 20 + 2, 5/-90º
5 + j 20 + 0 - j 2, 5
5 + j17, 5 ohm
Example 6.9
An RLC-network is constructed in the following manner. Z1 = 5 - j2 ohm;
Z2 = 10 + j10 ohm; Z3 = 2 - j8 ohm. Z1 and Z2 are connected in parallel and
this parallel combination is connected in series with Z3 across an alternating
quantity of 100V.
(a)
Make a neat labelled network diagram of this combination and indicate
all relevant values.
(b)
Calculate the:
(i)
Impedance of the network;
(ii)
Current that will flow in the network; and
(iii)
The voltage drop across the parallel section of the network.
(c)
Draw a neat labelled phasor diagram that will illustrate the relationship
between the applied voltage and the current in the network.
101
102
Q-factor, Bandwidth and Complex Notation
Solution:
(a)
Z1
R
C
5 ohms
- j 2 ohms
R
R
L
2 ohms
10 ohms
j 10 ohms
- j 8 ohms
Z3
Z2
100 V
Z1 × Z2 + Z3
(b)(i)
ZT
=
Z + Z2
1
5 - j2 × 10 + j 10
=
+ 2 - j8
5 - j2 + 10 + j 10
5,385/-21, 8º × 14,142/45º
=
+2-j8
15 + j 8
76,155/23, 2º
=
+2-j8
17/28º
= 4,479/-4,8º + 2 - j 8
= 4,463 - j0, 375 + 2 - j8
= 6,463 - j 8,375
= 10,578/-52,34º ohms
VS
VZ1Z2
(c) I =
Z
100/θº
=
10,578/-52,34º
= 9,455/52,34º ampere
(d)
= I × Z Z1Z2
= 9,455/52,34º × 4,497/-4,8 º
= 42,35/47,54º volt
I = 9,455 / 52,34º A
VS = 100 / θº V
N4 Industrial Electronics
Exercise 6.1
1.
Define and then graphically represent the concept of bandwidth.
2.
A series RLC-network has the following component values and is connected across an alternating current supply.
R = 200 ohm;
L = 0, 15 H;
C
=
60µF
Use the above data to determine the lower- and upper half-power frequencies.
3.
A series network consists of a resistance having a value of 20 ohm, an inductor having a value of 15 mH and a capacitor that has a
capacitive reactance of 400 ohm. This combination is connected across
an alternating supply of 30V and resonates at a frequency of 4,774 kHz.
Determine the:
3.1 Q-factor of the network;
3.2 Bandwidth of the network; and
3.3 The voltage drop across the inductor and the capacitor.
4.
Two impedances are given as (α1 + jβ1) and (α2 + jβ2). Explain how these
values are converted to the polar form and then give the rules for:
4.1 Division;
4.2 Addition; and
4.3 Multiplication.
5.
A series RLC-network has the following component values connected
across a 250/θº V/50 Hz alternating quantity.
R = 20 ohm; L = 150 mH;
C
=
100 µF
Use this information to:
5.1 Calculate the impedance of the network;
5.2 Determine the current that will flow in the network; and
5.3 Determine the voltage drop across each component in the network.
5.4 Draw a phasor diagram indication all the relevant values calculated.
6.
An RLC-network has the following impedance values:
Z1 = 6 - j2 ohm; Z2 = 8 + j4 ohm; Z3 = 2 + j6 ohm. Z1 and Z2 are
connected in parallel and this parallel combination is connected in series
with Z3 across an alternating quantity of 100V.
6.1 Make a neat labelled network diagram of this combination and
indicate all relevant values.
103
104
Q-factor, Bandwidth and Complex Notation
6.2 Calculate the:
6.2.1
Impedance of the network;
6.2.2
Current that will flow in the network; and
6.2.3
The voltage drop across the parallel section of the network.
6.3 Draw a neat labelled phasor diagram that will illustrate the relationship between the applied voltage and the current in the network.
7.
An alternating current network consists of the following components:
=
100 – j 26,25 ohm
• Z1
=
120 + j 29,35 ohm
• Z2
=
100 + j 35,85 ohm
• Z3
Z1 and Z2 are connected in parallel and then in series with Z3 across a 220V/50 Hz alternating current supply. Determine from this information:
7.1 The impedance of the parallel section;
7.2 The total impedance of the network;
7.3 The total current flow in the network;
7.4 The voltage that will be measured across the parallel section; and
7.5 The current that will flow in each branch of the parallel section.
BASIC ATOMIC THEORY
N4 Industrial Electronics
CHAPTER
7 Basic Atomic Theory
Learning Outcomes
On completion of this module you will be able to:
•
•
•
•
•
•
•
•
•
•
7.1
Describe and identify the nature of matter by naming forms and examples;
Ascertain atomic weight and number of elements by using the Periodic Table of
Elements;
Define:
– Compounds;
– A molecule; and
– An atom.
Draw a two dimensional layout of any atom indicating all the particles as well as
major- and sub-shells;
Determine the number of particles in the major- and sub-shells;
Discuss or explain the term valency and valency shell;
Explain a covalent bond by means of a suitable sketch;
Discuss the difference between electron flow and hole flow by means of an
applicable sketch;
Understand and be able to explain by means of simple sketches the energy bands
which exist in atoms and the placing of these energy bands in:
– A conductor;
– An insulator; and
– A semi-conductor.
Define the electron volt.
The structure of matter
Definition 7.1 Matter
Matter may be defined as anything that has mass and that occupies space and can be
composed of elementary substances that are found in nature.
105
106
BASIC ATOMIC THEORY
These compositions may be in pure form or it may be a compound or mixture of
elements found in nature. Matter cannot be generated or destroyed but it may change
from one form into another. Matter can be divided into the following groups:
•
•
•
•
Solids;
Liquids;
Gasses; and
Plasma.
Examples of such elements found are:
•
•
•
•
•
Iron (Fe)
Silver (Ag)
Carbon (C)
Gold (Au)
Aluminum (Al)
Classification of all elements in nature, are found in the Periodic Table of Elements
and are classified by its atomic number and atomic weight. The atomic number of an
element is given by the number of electrons orbiting the atom whereas the atomic
weight is given by the number of protons and neutrons contained in the nucleus of
an atom. All elements consist only of one type of atom as is indicated by the Periodic
Table of Elements. Combining of different elements may form compounds by chemical
reaction with one another and the most common one would be water. Water is a
compound of Oxygen and Hydrogen.
Definition 7.2 Molecule
A molecule can be defined as the smallest part of an atom that will have the same
characteristics of the original element without breaking up into atoms again.
7.2
Atoms
Definition 7.3 Atoms
An atom may be defined as the smallest part of an element that can participate in a
normal chemical reaction.
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All atoms consist of minute particles of electrical charges arranged in a set pattern and
consist of:
•
•
•
Electrons, moving around the nucleus of an atom and that has a negative electrical
charge of 1,602 × 10-19 coulomb and are kept in its orbit by electrostatic forces of
attraction;
Protons in the nucleus of an atom that have a positive electrical charge; and
Neutrons in the nucleus of an atom that have no electrical charge.
It will be found that all atoms have an equal number of protons and electrons, which
indicates its atomic number, which makes all atoms electrically neutral. All electrons,
protons and neutrons are identical and it is possible that they be replaced in other atoms
that are not of the same element. A point of interest is that it is very easy for an atom
to loose or gain outer orbit or valence electrons. Therefore, should an atom loose an
electron it will loose some of its negative electrical charge and in the process become
positively charged! This is then known as a positive ion or cation. The opposite however
is also true in that when an atom gains an electron it will increase its negative electrical
charge and will then be known as a negative ion or anion. The only difference between
atoms is a numerical difference. Figure 7.1 illustrates a two dimensional layout of an
atom.
Negatively charged
orbiting electrons
Protons (+)
and neutrons
(no charge)
K - shell
L - shell
Nucleus
Figure 7.1
7.3
Energy Shells
All electrons are arranged around the nucleus of an atom in a pre-determined manner
as mentioned earlier but we should however have a closer look at these arrangements.
Together with figure 7.1 we should also refer to figure 7.2 to find out how the
arrangement of the electrons, protons and neutrons in an atom is accomplished.
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BASIC ATOMIC THEORY
Letter designation of
major shells
M
L
K
S
SP
SPD
Letter designation of
sub - shells
2
8
18
Number of orbiting
electrons per shell
Figure 7.2
In the representation of figure 7.2 the centre circle will represent the nucleus consisting
of the protons and neutrons. The outer circle or circles will indicate the shells for the
orbiting electrons. Although these representations are two dimensional, the shape of
a tennis ball can almost represent an atom in that the major shells of an atom are also
sub-divided into sub-shells. This concept is illustrated in figure 7.2.
The major shells of an atom are designated the K, L M and N shells and the sub-shells
are designated the S, P, D and F shells. Each shell, major or sub-shell, may only have a
pre-determined number of electrons per shell and the number of electrons in the major
shells is determined by the expression 2.n2 where n will represent the orbit number
counting from the inner circle toward the outer circle. The number of electrons per
major shell can thus be calculated in the following manner.
K shell: 2.n2
L shell: 2.n2
M shell: 2.n2
N shell: 2.n2
=
=
=
=
2 ×12
2 × 22
2 × 32
2 × 42
=
=
=
=
2
8
18
32
Thus, the K shell may only contain 2 electrons.
Thus, the L shell may only contain 8 electrons.
Thus, the M shell may only contain 18 electrons.
Thus, the N shell may only contain 32 electrons.
Since the number of electrons per major shell can be determined and predicted
it follows that the number of electrons per sub-shell can also be determined and
predicted. The number of electrons per sub-shell is given as follows:
S shell
P shell
D shell
F shell
=
=
=
=
2 electrons
6 electrons
10 electrons
14 electrons
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109
Electrons can be made to jump to other orbits by means of heat, light energy or an
electric field at or near an atom. Whatever the form of energy used, it is referred to as
ionization potential. This arrangement of electrons orbiting an atom will now allow one
to determine the conducting properties of the element and especially the outer orbit or
valence shell of an atom.
7.4
Valency
The number of electrons in the outer shell of an atom, called the valence shell, will
determine the valency of that element and may be defined as an indication of the ability
of an atom to gain or loose electrons and will determine the electrical properties of that
element. Should an atom not have the full complement of electrons in the valence shell,
that element is deemed to be a conductor since that particular atom will always be on
the lookout for electrons to fill the vacancies, which exist. Should an atom however have
the full compliment of electrons in the valence shell it will not need to look for electrons
to fill up vacancies since no vacancies exist and is then deemed to be an insulator.
The above statements are not always true. A conductor may also be an insulator under
certain specific conditions. Two or more atoms may combine to form a covalent bond.
The best manner in which to explain this concept is to look at two hydrogen atoms as
illustrated in figure 7.3. This is not a compound that will be formed but is the sharing of
electrons by two or more nuclei of atoms. In this instance the only shell of electrons will
contain two electrons, which make that particular shell complete. This phenomenon
is particularly important in the semi-conductor field and will be explained in more
detail when semi-conductors are discussed. Therefore, we will find that there are no free
electrons available since each atom shares its electrons with one another and that they
are firmly attached in a crystal lattice structure by this covalent bond.
Orbiting electrons
Hydrogen atom 1
Hydrogen atom 2
Figure 7.3
7.5
Conduction
A very important phenomena concerning conduction should however also be
explained. In the incomplete valence shell of an element we refer to the spaces, which
exist where, electrons should be, as holes. Strange as it may sound we will find that these
holes also appear to be moving. This concept is illustrated in figure 7.4.
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BASIC ATOMIC THEORY
Movement of electrons or conduction can and will take place in any given conducting
material, in a desired direction, should a source of power be applied across such
material. This conduction process can be by one or both of the following processes
namely hole flow or transfer or by electron motion.
Free electrons in the conduction band will, under the influence of an applied potential,
move around. The reason for this action is that since the electron has a negative charge
it will be repelled by the negative charge of the applied potential but at the same time be
attracted by the positive charge of the applied potential. In this manner we will find that
electrons will always move from a negative charge toward a positive charge. This action
is termed electron flow. Should we now observe figure 7.4 in more detail, it will be seen
that as the electron moves from the negative charge of the applied potential toward the
positive charge of the applied potential, that the hole moves from the positive charge of
the applied potential toward the negative charge of the applied potential. This action is
termed hole flow and both these concepts are of great importance and we will deal with
these principles again when we discuss semi-conductors.
Electron motion
Hole motion
-
+
-
+
-
+
Applied potential
Figure 7.4
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7.6
111
Energy bands
In an isolated atom the electrons will be acted upon from within. Should we now however
bring a number of atoms together, as in a solid material like a conductor; the electrons of
the various atoms will come under the influence of forces from the other atoms within
that solid. In this manner energy levels are created and electrons will merge into energy
bands of distinct energy levels. This concept is illustrated in figure 7.5.
Conduction
band
Forbidden gap
Energy level
Valence band
Figure 7.5
In any given material, conducting or insulating, there are two distinct energy bands
in which electrons may exist, namely the conduction band and the valence band but
they will be separated by the forbidden gap. The electrons in the conduction band are
drifting around in the material since they have virtually become disconnected from
the atom and they are easily moved around by the application of a very small amount
of energy. The electrons in the valence band however, are tightly bound around the
nucleus of the atom and a large amount of energy will be required to move these
electrons.
Definition 7.4 Electron Volt
The energy required to move any electron, whether it is in the conduction band
or in the valence band, is defined by the electron volt (eV) and is the amount of
energy required to move one electron through a potential difference of one volt.
Between the conduction band and valence band we find an area termed the forbidden
gap. This forbidden gap will vary in size for different materials, conductors or insulators,
and may be small, large or non-existent as is indicated in figure 7.6 (a), (b) and (c).
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BASIC ATOMIC THEORY
Conduction
band
Conduction
band
Conduction
band
Forbidden gap
Forbidden gap
Valence band
Valence band
Valence band
( a ) Insulator
( b ) Semi-conductor
( c ) Conductor
Figure 7.6
In the conductor we find that the forbidden gap is non-existent and that the conduction
band and the valence band actually overlap one another creating a large number of
electrons available for conduction. Very little energy is required for conduction to take
place and will even conduct at the slightest variation in temperature.
In the insulator we find a fairly large forbidden gap between the conduction band and
the valence band and no electrons are found in the conduction band. Since all the
electrons are found in the valence band only, which is tightly bound to the nucleus, a
large amount of energy, in the region of 6 eV, is needed to cause an electron to move
from the valence band to the conduction band.
The third material shown is that of a semi-conductor and has a fairly narrow forbidden
gap and requires less energy for conduction to take place. Here we have to differentiate
between the two types of semi-conductor materials commonly used in semi-conductor
manufacture namely, Silicon (Si) and Germanium (Ge). Typical energy levels required
to cause conduction is 0,7 eV for Germanium and 1,19 eV for Silicon. Since this is
termed a semi-conductor it is neither a conductor nor an insulator and has a typical
resistance of 10 ohm/cm3.
The forbidden gap will therefore determine whether a material can be termed as a
conductor, insulator or semi-conductor.
N4 Industrial Electronics
Exercise 7.1
1.
Name the four groups into which matter can be divided.
2.
Define the following terms:
2.1 Molecules; and
2.2 An atom.
3.
Indicate the charges that are found on the particles of an atom.
4.
An atom contains 54 electrons. Calculate the number of electrons per major orbit and then draw a two dimensional layout indicating all relevant information.
5.
The term valency will determine the behaviour of an element in that it may be a conductor or it may be an insulator. Express your opinion on this statement.
6.
Explain the concept of a covalent bond and illustrate your explanation with
a suitable labeled sketch.
7.
Use suitable sketches to illustrate the principle of conduction and differentiate between:
7.1 Electron motion; and
7.2 Hole motion.
8.
The conduction band and the valence band are separated by the forbidden gap in any given material and it makes no difference whether the material
is a conductor or an insulator. Draw three neat labeled sketches that will illustrate this concept for:
8.1 An insulator;
8.2 A semi-conductor; and
8.3 A conductor.
113
114
BASIC ATOMIC THEORY
PN–Junction Theory
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115
CHAPTER
8 PN–Junction Theory
Learning Outcomes
On completion of this module you will be able to:
•
•
•
•
•
•
8.1
Represent an atom as a crystal lattice structure;
Use crystal lattice structures to explain the concept of:
– Donor doping; and
– Acceptor doping.
Clarify the concept of the Fermi-levels;
Explain using suitable sketches the formation of a PN-junction;
Explain the principle of diffusion current and the formation of the depletion layer;
and
Explain using suitable sketches the following concepts of a PN-junction:
– No bias;
– Reverse bias; and
– Forward bias.
Introduction
In the field of electronics there are two main elements that are used in the manufacture
of semi-conductor devices or components namely Silicon and Germanium. Although
these two elements do not have the same atomic weight or number of orbiting electrons
they do however have one aspect in common: there are four valence electrons in its
valence shell although the orbit designation is different.
In order to understand how the conduction will take place in semi-conductor devices
you are advised to sharpen up on you’re already gained knowledge of Atomic Theory.
As the name Semi-conductor suggests, it is not a very good conductor and something
needs to be done in order to improve on its conducting capabilities. This process of
improvement in conducting capabilities is termed doping but can be best explained by
considering the elements (germanium and silicon) as a crystal lattice structure.
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PN–junction theory
8.2Crystal lattice structures
We will only deal with the two most common semi-conductor manufacturing elements
namely Germanium and Silicon. Since all elements are electrically neutral we will only
consider the valence electrons of each element since they have the same number of
valence electrons.
A simplified illustration of the crystal lattice structure of Germanium and Silicon
is depicted in figure 8.1. This illustration is interpreted as follows: the circle in the
centre represents the nucleus of the atom as well as the electrons contained in the
electrons orbits or shells that have their full compliment of electrons for those shells.
The squares indicate the valence electrons in the valence shell. Although only four
atoms are indicated here you must realise and accept that there are many more atoms
in a very small piece of Germanium or Silicon. Covalent bonding will now take place
among adjoining atoms at random. This type of crystal lattice structure is found in
all crystalline elements. Germanium and Silicon in its pure form is also known as
an intrinsic material and as already mentioned does not have very good conducting
properties. In order to improve the conducting properties the Germanium or Silicon is
mixed with other elements in a manufacturing process termed doping.
Orbiting electrons
of four nuclei
(same for silicon
and germanium)
Figure 8.1
8.3
Donor doping
Donor doping is a mixing process that will generate a free (extra) electron in the
conduction band of the atom as well as crystal lattice structure. The number of such free
electrons will depend on the extent to which this doping is performed. The concept of
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117
donor doping is illustrated in figure 8.2. Plainly said, how much impurities are added
to the Germanium or Silicon. The impurity added to the Germanium or Silicon has five
valence electrons in its valence shell and is therefore referred to as Penta-valent atoms.
Penta- is the Greek word for five. Typical elements used for donor doping include:
• Antimony
• Arsenic
• Phosphorous
Penta-valent
impurity atom
Orbiting electrons
of three nuclei
(same for silicon
and germanium)
Extra free
electron
Figure 8.2
Since covalent bonds are formed in the crystalline structure the impurities form part
of this covalent bond and create an extra electron for conduction purposes for each
impurity atom added and will enter the conduction band of the atom as a free electron.
Donor doping creates what is termed an N-type material and this material is negatively
charged since the electron number has increased by the introduction of the impurity
atoms. This material is therefore no longer an intrinsic material but is referred to as an
extrinsic material. What this process has now brought about is an extra or free electron
which will contribute to the better conducting properties of the material.
8.4
Acceptor doping
Acceptor doping is a mixing process that will generate a hole in the conduction band
of the atom as well as crystal lattice structure. The number of such holes will depend
on the extent to which this doping is performed. Plainly said, how much impurities are
added to the Germanium or Silicon.
118
PN–junction theory
The impurity added to the Germanium or Silicon has three valence electrons in its
valence shell and is therefore referred to as tri-valent atoms. Tri- is the Greek word for
three. Typical elements used for acceptor doping include:
• Gallium
• Boron
• Aluminium
The concept of acceptor doping is illustrated in figure 8.3.
Tri-valent
impurity atom
Orbiting electrons
of three nuclei
(same for silicon
and germanium)
Hole
Figure 8.3
Since covalent bonds are formed in the crystalline structure the impurities form part
of this covalent bond and create a hole for conduction purposes for each impurity
atom added and will enter the conduction band of the atom as a hole. Acceptor doping
creates what is termed a P-type material and this material is positively charged since the
electron number has decreased by the introduction of the impurity atoms. This material
is therefore no longer an intrinsic material but is referred to as an extrinsic material.
What this process has now brought about is a hole which will contribute to the better
conducting properties of the material.
Important to know is that although the pure Silicon or Germanium (intrinsic material)
has been doped in order to obtain an N-type material or P-type material (extrinsic
material) that the doped material is still electrically neutral in that the number of
electrons will still equal the number of protons. Of further importance is the fact that
in an N-type material conduction is by means of electrons as majority carrier and in a
P-type material conduction is by means of holes as majority carrier. Should a suitable
supply now be connected across this PN-junction a current flow will be measured.
N4 Industrial Electronics
8.5
119
Fermi-levels
Once the process of doping have been accomplished another interesting phenomena
namely the Fermi-level comes to light which should also be investigated. It has been
mentioned that when these materials are manufactured that the free electrons as well
as the holes introduced move into the conduction band. This conduction band also has
a band termed the valence band and is of great importance. But, the question may now
arise of what this Fermi-level is?
Definition 8.1 Fermi-levels
The Fermi-level may be defined as the amount of energy the free electrons as well
as the holes possess within the material.
8.5.1The Fermi-level in an intrinsic material
Since an extrinsic material is electrically neutral we find that the Fermi-level lies exactly
in the middle between the conduction band and the valence band. This is brought
about by the fact that the number of free electrons in the conduction band will be equal
to the number of holes and the valence band and that the energy levels will decrease
exponentially toward the centre between the conduction band and the valence band.
This concept is illustrated in figure 8.4.
Conduction band
-
-
-
-
-
-
Electron energy
Fermi-level
+
+
+
+
+
+
Hole energy
Valence band
Figure 8.4
8.5.2The Fermi-level in an extrinsic material
The position of this Fermi-level will however be influenced by the type of doping and
material being obtained. This concept is illustrated in figure 8.5 (a) and (b).
120
PN–junction theory
Conduction band
-
-
-
-
-
-
Conduction band
Electron energy
-
-
-
-
-
-
Electron energy
Fermi-level
Fermi-level
+
+
+
+
+
+
+
Hole energy
+
+
+
+
Valence band
Valence band
(a)
(b)
+
Hole energy
Figure 8.5
We find that in an N-type extrinsic material that more electrons have been added which
makes the material negatively charged that there is no longer a balance between the
conduction band and the valence band. Therefore the Fermi-level will move closer to
the conduction band. This concept is illustrated in figure 8.5 (a). On the other hand
we find the same phenomena with an extrinsic P-type material with the difference
that more holes have been created which now makes the material positively charged.
However, in this instance the Fermi-level will move closer to the valence band. This
concept is illustrated in figure 8.5 (b).
8.5.3The Fermi-level in a PN-junction
Should a PN-junction be created by joining the two types of extrinsic material then the
Fermi-level will be placed in this junction as illustrated in figure 8.6.
Conduction band
-
-
-
-
-
-
-
-
-
-
-
Electron energy
Fermi-level
Valence band
+
+
+
+
+
+
N-type
Figure 8.6
+
+
+
+
+
P-type
+
Hole energy
+
N4 Industrial Electronics
8.6The PN-junction
8.6.1The formation of a PN-junction
A PN-junction is formed when a P-type material and an N-type material is joined
together. This joining together is not an electrical junction but is a junction which
is achieved through a manufacturing process. In this manner a single junction is
created between the N-type and the P-type material. Electrons and holes are uniformly
distributed in the two types of material provided they have been doped to the same
extent. This concept is illustrated in figure 8.7.
P-type
Holes
+
+
+
+
+
+
+
+
+
+
N-type
Electrons
+ + + -
-
-
-
-
Junction
Figure 8.7
During the manufacturing process this evenly distribution of electrons and holes are
disturbed since some of these holes and electrons are very close to one another in the
immediate vicinity of the junction. Electrons from the N-type material now cross the
junction into the holes in the P-type material and holes from the P-type material also
cross the junction toward the N-type material. This natural process is termed diffusion
current and may be defined as a process whereby charge carriers in a region of high
density will move to a region of low density until they are evenly balanced. However,
this will only take place in the immediate vicinity of the junction. This results in the
creation of negative ions on the P-side as well as positive ions on the N-side in the
immediate vicinity of the junction. This concept is illustrated in figure 8.8.
121
122
PN–junction theory
P-type
Holes
+
+ +
+
+ +
+
+
+
+ +
+
N-type
Electrons
+
+ +
+ +
+ -
Holes moved across
Electrons moved across
Junction
Figure 8.8
This movement of charge carriers across the junction leaves on either side of the
junction a region which is completely depleted or void of charge carriers and is termed
the depletion layer. It is generally accepted that this depletion layer is of equal thickness
around the junction provided that both extrinsic materials have been doped to the same
extent. Therefore, the depletion layer on the N-side will consist of donor impurities
which have lost electrons and have a positive charge and the depletion layer on the
P-side will consist of donor impurities which have lost holes and have a negative charge.
This concept is illustrated in figure 8.9.
P-type
Holes
+
+
+
+
+
+
+
+
+
+
N-type
Electrons
-
+
+
+
+
+
-
-
-
-
Positive charge
Donor impurities
Negative charge
Junction
Depletion layer
Figure 8.9
The explanations discussed above constitute the behaviour of the PN-junction under no
bias conditions. Bias is a method applied to semi-conductor components that will allow
for it to either conduct or not to conduct.
N4 Industrial Electronics
123
8.6.2Reverse-biased junction
A reverse-biased junction condition will not allow for any current to flow except for
a leakage current which for all practical purposes may be ignored for our discussion.
A voltage source is connected to the PN-junction in such a manner that the negative
terminal of the source is connected to the P-type material of the junction and the
positive terminal of the source is connected to the N-type material of the source. This
concept is illustrated in figure 8.10.
Junction
P-type
Holes
+
Negative potential
+
+
+
+
+
+
+
N-type
Electrons
-
+
+
+
+
+
-
-
-
Positive potential
-
Depletion layer
Depletion layer
widened
Figure 8.10
Under these reverse-bias conditions no current will be flowing for the following
reasons. Electrons from the N-type material will be attracted to the positive terminal of
the source and holes from the P-type material will be attracted to the negative terminal
of the source. The result is that the depletion layer will widen as is illustrated. Although
it was mentioned earlier that under reverse-bias we have a leakage current which may
be ignored it is of utmost importance that this reverse-bias not exceed a given value
since the junction will then be destroyed. An interesting fact that need to be mentioned
is that under reverse-bias conditions a PN-junction will act as a variable capacitor since
the two doped materials are actually two conductors and since the junction is not an
electrical junction the junction will act as a capacitor and the distance between the two
conducting materials will be governed by the magnitude of the reverse bias.
8.6.3 Forward-biased junction
A forward-biased junction condition will allow for current to flow. A voltage source is
connected to the PN-junction in such a manner that the negative terminal of the source
is connected to the N-type material of the junction and the positive terminal of the
source is connected to the P-type material of the source. This concept is illustrated in
figure 8.11.
124
PN–junction theory
Junction
P-type
Holes
+
Positive potential
+
+
+
+
+
+
+
N-type
Electrons
-
+
+
+
+
+
-
-
-
Negative potential
-
Depletion layer
Depletion layer
narrowed
Figure 8.11
At a given point a voltage magnitude is reached when these charge carriers are able to
cross the junction and the electrons originating from the N-type material are attracted
by the positive terminal of the voltage source on the P-type material and at the same
time holes originating from the P-type material are attracted toward the negative
terminal of the voltage source on the N-type material. As long as this required supply
voltage is maintained a current will flow for as long as required.
The given points at which conduction will commence are defined as follows:
•
•
Germanium ± 0,3 volt
Silicon ± 0,7 volt
These values are the approximate values of the supply voltage that need to be reached
before any conduction will take place.
N4 Industrial Electronics
Exercise 8.1
1.
Germanium and Silicon may be represented in a crystal lattice structure. Make use of a neat labelled sketch to illustrate this statement.
2.
Make use of two neat labelled crystal lattice structure sketches and explain the concept of:
2.1 Donor doping; and
2.2 Acceptor doping.
3.
Differentiate between an intrinsic- and extrinsic material.
4.
Define the Fermi-level and then indicate by means of neat labelled sketches
the placement of the Fermi-level in:
4.1 An intrinsic material;
4.2 An extrinsic N-type material;
4.3 An extrinsic P-type material; and
4.4 A PN-junction.
5.
During the formation of a PN-junction a depletion layer is formed. Use suitable sketches and explain how this depletion layer is formed.
6.
A PN-junction may either be forward- or reverse biased. Use applicable sketches to illustrate this biasing concept.
125
126
PN–junction theory
Semi-conductor Diodes
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127
CHAPTER
9 Semi-conductor Diodes
Learning Outcomes
On completion of this module you will be able to:
•
•
•
•
•
•
•
•
•
•
9.1
Give the IEC circuit symbol and a suitable definition of a diode;
Discuss the operation of a diode with the aid of a characteristic curve;
Mention the basic rating factors for diodes;
Explain the electrical characteristics of a diode with the aid of a specification sheet;
Give the mathematical expression for a diode and explain the meaning of each
symbol;
Do calculations and derive conclusions using the standard expression as well as
derivatives thereof;
Give the mathematical expression for the forward resistance a diode and explain the
meaning of each symbol;
Do calculations and derive conclusions using the standard expression as well as
derivatives thereof;
Explain and draw a graphical representation of the diode loadline; and
Explain the concept of a series limiting resistor when using diodes.
Introduction
Definition 9.1 Diode
A diode may be defined a single PN-junction two terminal device which will offer a
low resistance when forward-biased and a high resistance when reverse-biased.
Therefore, should a diode be forward-biased it will act as a conductor and when it is
reverse-biased it will act as an insulator (non-conductor).
128
Semi-conductor Diodes
9.2The diode and characteristic curve
The IEC circuit symbol and characteristic curve for a diode is illustrated in figure 9.1 (a)
and (b).
Our first reference will be to figure 9.1 (a). The anode is always manufactured from a
P-type material and the cathode from an N-type material. The electron flow through
the diode is always against the arrow head indicated by the circuit symbol and for such
electron flow (current flow) to take place the anode must always be positive with respect
to the cathode. This does not necessarily mean that the anode must have a positive
potential supplied to it. Remember that minus 10 volt is more positive than minus 30
volt. The characteristic curve in figure 9.1 (b) is typical of a diode and will describe its
operation. The curve has two distinct areas in which it operates and these two areas
have another two points which are very significant. Area A on the curve constitutes the
forward bias area of the device and it is in this area that a diode is normally operated.
Should a diode now be correctly forward-biased then current will flow freely for as
long as this forward-bias is maintained. Point B which is also in the forward-bias
area is termed the forward break over point and it is at this point that the diode starts
conducting freely. This point differs between Germanium and Silicon diodes and is
illustrated in figure 9.2 and has the following approximate values and only the forward
characteristics are indicated in order to differentiate between the two types of diodes:
• Germanium ± 0,3 volt
• Silicon ± 0,6 volt
This value also indicates the voltage drop across the diode during conduction.
I
A
Forward bias area
Anode (P-type)
Cathode (N-type)
IF
VR
Electron motion
(a)
-V
V
VF
C
IR
D
Reverse bias area
-I
(b)
Figure 9.1
B
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129
Point C (figure 9.1 (b) is probably the most important point on the characteristic curve
and also falls in the reverse-bias area of the diode and is termed the reverse break down
point or the peak-inverse-voltage point and this point is specified by the manufacturer
and will differ between diodes. Should this point be reached the diode will be destroyed.
Area D (figure 9.1 (b) on the curve constitutes the reverse-bias area of the device and
the diode is normally not operated in this area. Should a diode be reverse-biased no
current will flow except a small reverse current which may be ignored.
I (mA)
20
Germanium
10
Silicon
Q-point
Q-point
V
1
2
Figure 9.2
9.3
Basic rating factors of a diode
Rating factors will assist during the design phase of circuits so that reliable as well as
satisfactory operation can be assured.
RangeRating
Low current
Medium current
High current
Up to 49 ampere
50 ampere to 199 ampere
200 ampere and higher
All these rating are supplied in the manufacturers’ specification sheets and are based on
the absolute maximum rating which must never be exceeded.
9.4Electrical characteristics of a diode
The electrical characteristics of a diode are based on the absolute maximum rating system
and provide information pertaining to the maximum values that may not be exceeded
for a given diode. These specifications are always contained in the manufacturers’
specification sheets and only the most important specifications are listed below.
130
Semi-conductor Diodes
•
•
•
•
•
•
•
•
Peak reverse voltage (VRM)
Also termed the peak-inverse-voltage (PIV) and is the maximum permissible
reverse voltage that may be applied across the diode after which it will be destroyed.
Reverse breakdown voltage (VBR)
This is the minimum voltage that may be applied across the diode before it will
break down.
Steady-state forward current (F or IO)
This specifies the maximum current that may be passed through the diode under
normal forward bias conditions.
Peak surge current (IFM (SURGE))
This is the maximum specified current which may pass through the diode for a
specified time and is higher than the specified steady state forward current and will
occur at switch-on or short circuit conditions.
Static reverse current (IR)
This is the reverse saturation current for a specified reverse bias voltage.
Static forward voltage drop (VF)
This is the voltage drop across the diode for a given steady-state forward current.
Continuous power dissipation (P)
This is the maximum power that the diode may dissipate on a continuous basis at a
temperature or 25ºC.
Reverse recovery time (trr)
This is the time that it will take the diode to switch from the conducting state (onstate) to the non-conducting state (off-state).
9.5The diode equation
The electrical characteristics of a diode were given in 9.4 and some of these may be
represented by the following mathematical expression termed the diode equation:
i = IS (e qV / KT) – 1
where
i
IS
q
V
K
T
=
=
=
=
=
=
forward current in ampere
reverse saturation current in ampere
electron charge = 1, 6 ×10-19 coulomb
potential difference across the diode in volt
Boltzmann’s constant = 1, 38 ×10-23 J/K
temperature in Kelvin
It must be noted that Kelvin is termed absolute temperature and is obtained in the
following manner:
K =
273 + º C
N4 Industrial Electronics
Example 9.1
Convert the following temperatures to Kelvin.
(a)
28 º C
(b)
22 º C
(c)
36 º C
Solution:
(a) K
=
=
=
273 + º C
273 + 28
301 Kelvin
(b) K
=
=
=
273 + º C
273 + 22
295 Kelvin
(c) K
=
=
=
273 + º C
273 + 36
309 Kelvin
Example 9.2
Determine the forward current for a diode at 28 º C should the diode have a 0,
03 volts potential difference across it and a reverse saturation current of
0, 5 µ-ampere.
Solution:
i
=
=
=
IS(e qV / KT) -1
0, 5 × 10-6 × e(1,6 × 10-19 × 0,03 / 1,38 × 10-23 × 301) - 1
1,088 µ-ampere
131
132
Semi-conductor Diodes
Example 9.3
Determine the reverse saturation current for a rectifier diode given the
following data:
Potential difference across diode
=
0, 25 volt
Temperature=
17 º C
Forward current
=
42,6 m-ampere
Solution:
i
=
IS(e qV / KT) -1
i
=
IS
qV
e / KT - 1
42,6 × 10-6
=
(1,6 × 10-19 × 0,25 / 1,38 × 10-23 × 290)
e
=
1,944 µ-ampere
-1
Example 9.4
A silicon diode produces a forward current of 6 m-ampere and a reverse
saturation current of 21 µ-ampere. Determine the voltage across the junction
at a temperature of 25 º C.
Solution:
i
=
IS(e qV / KT) -1
V
=
K×T
q
× ln
i-1
IS
1,38 × 10-23 × 298
=
1,6 × 10-19
=
119,645 m-volt
x In
6 × 10-3
21 × 10-6
- 1
N4 Industrial Electronics
Example 9.5
The following data concerning a rectifier diode is given:
Junction potential
=
0,126 volt
Forward current
=
2 m-ampere
Reverse saturation current
=
15 µ-ampere
Determine from this given information the temperature of this junction in º C.
Solution:
i
=
IS(e qV / KT) -1
q ×V
T
=
K × ln i + 1
IS
=
1,6 ×10-19 × 0,126
× ln
1,38 × 10-23
=
298,136 K
ºC =
289,136 - 273
=
25,136 ºC
9.6
2 × 10-3
15 × 10-6
+ 1
Forward resistance of a diode
All diodes are manufactured from semi-conductor materials and will have a resistance
caused by its atomic structure. This resistance is given by:
K×T
R =
where
q×i
K
T
q
=
=
=
Boltzmann’s constant = 1,38 × 10-23 J/K
temperature in Kelvin
electron charge = 1,6 × 10-19 coulomb
Example 9.6
A rectifier diode has a current of 300 m-ampere flowing through it at a
temperature of 26ºC. Determine the forward resistance of this diode.
133
134
Semi-conductor Diodes
Solution:
K×T
R =
q×i
1,38 × 10-23 × 299
=
1,6 × 10-19 × 300 × 10-3
=
0,0859 ohms
Example 9.7
A rectifier diode has a forward resistance of 0,103 ohm at a temperature of
25ºC. Calculate the forward current that will flow through the diode.
Solution:
R =
i
=
=
=
K×T
q×i
K×T
q×R
1,38 × 10-23 × 299
1,6 × 10-19 × 0,103
249,54 m-ampere
Example 9.8
At what temperature in ºC will a diode operate should it have a forward
resistance of 1,428 ohm and allows a forward current of 18 m-ampere to flow
through it?
Solution:
K×T
R =
q×i
q×i×R
T =
K
1,6 × 10-19 × 18 × 10-3 × 1,428
=
1,38 × 10-23
ºC
=
=
=
298 Kelvin
289 - 273
25 ºC
N4 Industrial Electronics
9.7The diode load-line
The load-line for a diode is obtained by considering the maximum values of forward
current and the maximum value of the forward bias for a particular rectifier diode.
This load-line is then drawn on the forward characteristic of a germanium or silicon
diode and the point where these two intersect one another is termed the Q-point or
operating point. Should we assume a 2 volt forward bias and a maximum forward
current of 20 m-ampere then the loadline will be as indicated by the graphic
representation in figure 9.3.
Since the loadline is always constructed using the maximum parameters of the diode it
will be required to limit and protect the diode by using a series resistor with the diode
in order to limit the current to an acceptable level. The reason for this is that the diode
should never be operated at its maximum value since it will reduce the life expectancy
of the diode. The network in figure 9.4 illustrates this concept.
The value of the limiting resistor RS is mathematically given by:
RS
=
where
VS - VDF
IDF
RS
VS
VDF
IDF
=
=
=
=
value of series limiting resistor in ohm
forward bias voltage in volt
forward voltage value of diode in volt
maximum forward value of diode current in ampere
I (mA)
20
10
Germanium
Silicon
Q-point
Q-point
V
1
Figure 9.3
2
135
136
Semi-conductor Diodes
RS
D
VR
VDF
VS
Figure 9.4
Example 9.9
Silicon as well as a Germanium diode needs to be protected by a series limiting
resistor. The forward bias voltage has a value of 3 volt and the maximum
forward current for both types of diodes is 850 m-ampere. Determine the value
of a series limiting resistor for both diodes.
Solution:
SiliconGermanium
RS =
=
=
VS - VDF
RS =
IDF
3 - 0,6
=
850 × 10-3
VS - VDF
IDF
2,824 ohm
3,176 ohm
=
3 - 0,3
850 × 10-3
Exercise 9.1
1.
A diode may be best described by its characteristic curve. This curve
defines four distinct areas or points. Give such a characteristic curve,
indicate these distinct areas and then describe in detail the essence of
these areas.
2.
Tabulate the range as well as rating commonly used for diodes.
N4 Industrial Electronics
3.
Clarify the following electrical characteristic abbreviations of a diode and then also give a brief description of each one:
3.1 VRM;
3.2 VBR;
3.3 IO or IF;
3.4 IFM (SURGE);
3.5 IR;
3.6 VF;
3.7 P; and
3.8 trr.
4.
Give the expression for a diode and explain the meaning of each symbol used in the expression.
5.
The following data concerning rectifier diodes are given.
5.1 Potential difference across the diode
=
0,96 volt
Temperature
=
31ºC
Reverse saturation current =
0,0045 m-ampere
Determine the forward current of the diode.
5.2 Potential difference across the diode
=
0, 96 volt
Temperature
=
31ºC
Forward current
=
55,56 m-ampere
Calculate the reverse saturation current for the diode.
5.3 Potential difference across the diode
=
0,96 volt
Forward current
=
1,6 m-ampere
Reverse saturation current
=
0,0045 m-ampere
Calculate the temperature of this diode in ºC.
6.
7.
Give the expression for the forward resistance of a diode and explain the meaning of each symbol used in the expression.
The following data concerning rectifier diodes are given.
7.1 Temperature
=
Forward current
=
Determine the forward resistance of the diode.
7.2 Forward resistance
=
Temperature
=
Calculate the forward current for the diode.
7.3 Forward current diode
=
Forward resistance
=
Calculate the temperature of this diode in ºC.
21ºC
0,045 m-ampere
0,916 ohm
26ºC
25 m-ampere
1,625 ohm
137
138
Semi-conductor Diodes
8.
Explain how you would draw the loadline for diode given the following values.
• forward bias
=
3 volt
• maximum forward current
=
15 m-ampere
Draw your own forward characteristic for a diode and then indicate clearly the Q-point on your characteristic.
9.
Explain using a suitable network diagram the importance of a series limiting resistor used in conjunction with a diode.
10. Give the expression for a series limiting resistor of a diode and explain the meaning of each symbol used in the expression.
11. The following data concerning a diode is given:
• forward bias voltage
=
6,2 volt
• forward current for Ge diode
=
24 m-ampere
• forward current for Si diode
=
30 m-ampere
Determine a suitable value for a series limiting resistor for a:
11.1 Silicon diode; and
11.2 Germanium diode.
Diode Applications
N4 Industrial Electronics
CHAPTER
10 Diode Applications
Learning Outcomes
On completion of this module you will be able to:
•
•
•
•
Explain using suitable circuit diagrams as well as input and output waveforms the
application of a diode in:
– Series clippers; and
– Parallel clippers.
Explain using suitable circuit diagrams, calculations as well as input and output
waveforms the application of a diode in:
– Half-wave rectification and filters (C, RC and LC);
– Full-wave rectification (two diodes) and filters (C, RC and LC); and
– Full-wave rectification (bridge circuit - four diodes) and filters (C, RC and LC).
Explain using suitable circuit diagrams the application of a diode in:
– Full-wave voltage doubling circuits; and
– Other voltage multiplication circuits.
Explain the concept and do calculations of the following diode rectifier concepts:
– Transformer ratio;
– Average dc voltage;
– Ripple voltage;
– Ripple factor; and
– Peak-inverse-voltage (PIV) ratings.
10.1 Introduction
Diodes have many applications and in this module we will cover the most common
applications. It is however very important that you must understand the principle of
operation of a diode as was discussed in the pervious module otherwise the concepts
that will be discussed in this module will be futile.
139
140
Diodes APPLICATIONs
10.2 Application of diodes
10.2.1 Clippers
The term clipper is used to represent a diode-resistor circuit or network which will
change or separate a given waveform into two components or parts and may be
interchanged with the term limiter. These terms are interchangeable since it will depend
on the point where the output is taken from. As a rule of thumb a limiter is referred to
as a circuit whose output is below a certain value whereas a clipper refers to a circuit
whose output is always above a certain value. In this module we will only refer to
clippers as it is most commonly referred to and these clippers are divided into series and
parallel clippers.
10.2.1.1 Series clippers
A series clipper may be described as a circuit or network that will remove or clip one
half of the input signal and are divided into two main groups.
A negative series clipper is illustrated in figure 10.1 (a), the input waveform is illustrated
in figure 10.1 (b) and the output waveform is illustrated in figure 10.1 (c).
A positive series clipper is illustrated in figure 10.2 (a), the input waveform is illustrated
in figure 10.2 (b) and the output waveform is illustrated in figure 10.2 (c).
•
Negative series clipper
D
R
Input
Output
(a)
+
-
t
+
-
(b)
t
(c)
Figure 10.1
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141
The circuit is so constructed that the diode D is connected in series with the resistor
R and that the output is taken across the resistor R. The input voltage varies centrally
around the zero volt line and has equal positive and negative values. During the positive
half of the input cycle diode D is forward-biased and will conduct producing a current
to flow through resistor R causing a voltage drop across the resistor R. During the
negative half of the input cycle diode D becomes reverse-biased and does not conduct
and no current can flow through resistor R. Since no current can flow through resistor
R during the negative half of the input cycle there can be no voltage drop across resistor
R and no output can be obtained. In this fashion the negative half of the input cycle is
removed.
Now, should the input voltage have a magnitude of VS then the output voltage will have
a magnitude of VO = VS - VF where VF is equal to the forward voltage drop across the
diode during the conduction cycle. The output voltage will therefore be slightly less
than the input voltage. Account must be taken of the type of diode used, Germanium
or Silicon in order to determine the value that needs to be subtracted. Note again that
these values are ± 0,3 volt for Germanium and ± 0,6 volt for Silicon.
• Positive series clipper
The circuit is so constructed that the diode D is again connected in series with the
resistor R and that the output is again taken across the resistor R. The input voltage
varies centrally around the zero volt line and has equal positive and negative values.
During the positive half of the input cycle diode D is reverse-biased and will not
conduct producing no current to flow through resistor R causing no voltage drop
across the resistor R. During the negative half of the input cycle diode D becomes
forward-biased and conducts and a current can flow through resistor R. Since no
current can flow through resistor R during the positive half of the input cycle there can
be no voltage drop across resistor R and no output can be obtained. In this fashion the
positive half of the input cycle is removed.
D
R
Input
Output
(a)
+
-
t
+
-
(b)
t
(c)
Figure 10.2
142
Diodes APPLICATIONs
However, we now find that should the input voltage have a magnitude of VS then
the output voltage will have a magnitude of VO = - [VS - VF] where VF is equal to the
forward voltage drop across the diode during the conduction cycle. The output voltage
will therefore be slightly less than the input voltage. Account must be taken of the type
of diode used, Germanium or Silicon in order to determine the value that needs to be
subtracted. Note again that these values are ± 0,3 volt for Germanium and ± 0,6 volt for
Silicon. An interesting observation concerning series clippers, positive or negative, is
that the output will always consist of that part of the input cycle during which the diode
conducts.
10.2.1.2 Parallel clippers
A parallel clipper may also be described as a circuit or network that will remove or clip
one half of the input signal and are divided into two main groups.
A negative parallel clipper is illustrated in figure 10.3 (a), the input waveform is
illustrated in figure 10.3 (b) and the output waveform is illustrated in figure 10.3 (c).
A positive parallel clipper is illustrated in figure 10.4 (a), the input waveform is
illustrated in figure 10.4 (b) and the output waveform is illustrated in figure 10.4 (c). It
should be noted that parallel clippers are also referred to as shunt clippers.
• Negative parallel clipper
The circuit is so constructed that the diode D is again connected in series with the
resistor R but that the output is now taken across the diode D. The input voltage varies
centrally around the zero volt line and has equal positive and negative values. During
the positive half of the input cycle diode D is reverse-biased and will not conduct
producing no current to flow through resistor R causing no voltage drop across the
resistor R. But, diode D is in parallel with the input and will appear as such across the
output. During the negative half of the input cycle diode D becomes forward-biased and
conducts and current can flow through resistor R. But a short circuit is produced when
diode D conducts and no output is obtained. In this fashion the negative half of the
input cycle is removed. We will now find that should the input voltage have a magnitude
of VS then the output voltage will have a magnitude of VO = [VS - (IO × R)] where IO is
equal to the output current. The output voltage will therefore be slightly less than the
input voltage.
Account must be taken of the type of diode used, Germanium or Silicon in order to
determine the value that needs to be subtracted. Note again that these values are ± 0,3
volt for Germanium and ± 0,6 volt for Silicon. Note that during the negative half of the
input cycle a small output is obtained and that will be equal to the forward voltage of
the diode since it will only become a short circuit once this value has been exceeded.
N4 Industrial Electronics
143
IO
R
Input
D
Output
(a)
+
-
t
+
-
t
VF
(b)
(c)
Figure 10.3
• Positive parallel clipper
The circuit is so constructed that the diode D is again connected in series with the
resistor R but that the output is now taken across the diode D. The input voltage
varies centrally around the zero volt line and has equal positive and negative values.
During the positive half of the input cycle diode D is forward-biased and will conduct
producing current to flow through resistor R causing a voltage drop across the resistor
R. But, diode D is in parallel with the input and no output will appear across the output.
During the negative half of the input cycle diode D becomes reversed-biased and will
not conduct and no current can flow through resistor R. But an open circuit is produced
when diode D does not conduct and an output is obtained. In this fashion the positive
half of the input cycle is removed.
We will now find that should the input voltage have a magnitude of VS then the output
voltage will have a magnitude of VO = - [VS – (IO × R)] where IO is equal to the output
current. The output voltage will therefore be slightly less than the input voltage. Account
must be taken of the type of diode used, Germanium or Silicon in order to determine
the value that needs to be subtracted. Note again that these values are ± 0,3 volt for
Germanium and ± 0,6 volt for Silicon. Note that during the positive half of the input
cycle a small output is obtained and that will be equal to the forward voltage of the
diode since it will only become a short circuit once this value has been exceeded.
An interesting observation concerning parallel clippers is that the output will always
consist of that part of the input cycle during which the diode does not conduct. The
arrow point of the diode also serves as a good indication of the type of clipper. Should
it points downward the positive half of the input is clipped and only the negative half of
the input is obtained as an output. On the other hand, should it point upward then the
negative half of the input is clipped and only the positive half of the input is obtained as
an output.
144
Diodes APPLICATIONs
IO
R
Input
D
Output
(a)
+
-
t
+
-
t
VF
(b)
(c)
Figure 10.4
10.2.2The diode as a rectifier
One of the other main uses of a diode is as a rectifier is that it will convert an alternating
quantity into a pulsating direct quantity and depending on the efficiency of the filter
circuit may supply an almost ripple-free output.
10.2.2.1
Half-wave rectifier
A half-wave rectifier is a network so constructed using one diode plus additional
components to perform the function of rectification and is illustrated in figure 10.5 (a)
and (b).
A half-wave rectifier derives its name from the fact that the output voltage VO only
consist of one half of the input voltage Vi. This can be seen on the waveforms illustrated
in figure 10.6 (a), (b) and (c). A half-wave rectifier circuit is illustrated in figure 10.5 and
consists of a mains step-down transformer T1, a diode D1 and load resistor RL connected
in series with the diode across the secondary side of the mains transformer T1. The
input voltage Vi is supplied to the primary side of the mains transformer T1 and the
output voltage Vo is taken across the load resistor RL. There is no fundamental difference
between the two circuits except for the fact that different halves of the input cycle will
be rectified and this is illustrated in figure 10.6 (a), (b) and (c).
During the first half of the input voltage Vi, that is between A and B in figure 10.6(a),
the diode D1 is reverse-biased and no current will flow through the load resistor RL as
illustrated in figure 10.5. The output voltage Vo taken across the load resistor RL is equal
to zero since there will be no current flowing. The above statements of operation will
be reversed for figure 10.5 (b). During the second half of the input voltage Vi, that is
between B and C in figure 10.6 (b), the diode D1 is forward-biased and a current will
N4 Industrial Electronics
145
flow through the load resistor RL as indicated by the dotted line with the polarities as
indicated in figure 10.5. The output voltage Vo taken across the load resistor RL is almost
equal to Vi. The only difference there is will be the forward voltage drop across the diode
D1 since the circuit on the secondary side operates as a series voltage divider with the
load resistor RL. The above statements of operation will be reversed for figure 10.5 (b).
The same pattern will be repeated for every alternate positive and negative half cycle
and the output voltage thus obtained is termed a pulsating direct quantity. But this
is not really what is required since the magnitude of the output is not constant. The
way in which this pulsating direct quantity is smoothed is by means of a smoothing
capacitor which will remove the varying quantity and an output as constant as possible
will be obtained. This aspect will be discussed under the section that will deal with filter
networks.
The applicable waveforms for the network in figure 10.5 are illustrated in figure 10.6 (a)
and (b) where figure 10.6 (a) are those for figure 10.5 (a) and figure 10.6 (b) are those
for figure 10.5 (b). Figure 10.6 [a (i)] and [b (i)] are the inputs Vi for both networks and
figure 10.6 [a (ii)] and [b (ii)] are the outputs obtained from the secondary side of the
mains transformer for both networks. Figure 10.6 [a (iii)] and [b (iii)] are the outputs
obtained across RL for both networks and note must be taken of the polarity of the
respective outputs.
10.2.2.2
Full-wave rectification (two diodes)
As the name would imply, full-wave rectification will make use of both halves of the
input voltage Vi. Such a network diagram is illustrated in figure 10.7 (a), (b) and (c). The
associated waveforms are given in figure 10.8 (a) and (b).
A circuit diagram of a full-wave rectifier, using two diodes is illustrated in figure 10.7
(a), (b) and (c) and consists of a centre-tap mains step-down transformer T1, two
diodes D1 and D2 and a load resistor RL connected in series with the diodes across the
secondary side of the centre-tap mains step-down transformer T1. The input voltage Vi
is supplied to the primary side of the centre-tap mains transformer T1 and the output
voltage Vo is taken across the load resistor RL.
146
Diodes APPLICATIONs
D1
T1
+
D1
T1
-
-
+
+
VF
VF
RL
Vi
-
Vi
VO
RL
+
+
VO = Vi - VF
-
-
(a)
D1
T1
+
D1
T1
-
-
-
+
VF
VF
RL
-
VO = Vi - VF
Vi
+
+
RL
+
VO
-
(b)
Figure 10.5
(i)
+
-
t
A
( ii )
( iii )
B
C
+
-
D
+
-
E
t
A
t
B
C
+
-
t
t
(a)
(b)
Figure 10.6
E
t
-
+
D
N4 Industrial Electronics
147
During the first half of the input voltage Vi, that is between A and B in figure 10.8,
the diode D1 is reverse-biased and diode D2 is forward-biased and a current will
flow through the load resistor RL as indicated by the dotted line with the polarities as
indicated in figure 10.7 (b). The output voltage Vo taken across the load resistor RL is
almost equal to Vi. The only difference there is will be the forward voltage drop across
the diode D2 since the circuit on the secondary side operates as a series voltage divider
with RL.
T1
D1
Vi
RL
VO = Vi - VF
D2
+
T1
+
+
D1
Vi
-
T1
+
-
D1
+
-
D2
+
Vi
-
+
RL
VO = Vi - VF
D2
+
-
RL
VO = Vi - VF
-
Figure 10.7
During the second half of the input voltage Vi, that is between B and C in figure 10.8,
the diode D2 is reverse-biased and diode D1 is forward-biased and a current will
flow through the load resistor RL as indicated by the dotted line with the polarities as
indicated in figure 10.7 (c). The output voltage Vo taken across the load resistor RL is
almost equal to Vi. The only difference there is will be the forward voltage drop across
the diode D1 since the circuit on the secondary side operates as a series voltage divider
with RL.
The same pattern will be repeated for every alternate positive and negative half cycle
and the output voltage thus obtained is termed a pulsating direct quantity. But this
is not really what is required since the magnitude of the output is not constant. The
way in which this pulsating direct quantity is smoothed is by means of a smoothing
capacitor which will remove the varying quantity and an output as constant as possible
will be obtained. This aspect will be discussed under the section that will deal with filter
networks.
148
Diodes APPLICATIONs
The applicable waveforms for the network in figure 10.7 are illustrated in figure 10.8.
Figure 10.8 (a) is the input Vi for the network and figure 10.7 (b) is the output obtained
from the secondary side of the centre-tap mains transformer for the network. Figure
10.7 (c) is the output obtained across RL for the network.
10.2.2.3 Full-wave rectification (bridge circuit-four diodes)
As the name would imply, full-wave rectification will also make use of both halves of
the input voltage Vi. Such a network diagram is illustrated in figure 10.9 (a), (b) and (c).
The associated waveforms for this type of network is given in figure 10.10 (a) (b) and
(c). The only difference between a bridge rectifier and a rectifier using two diodes is
that a bridge rectifier uses two diodes per half cycle which would increase your current
delivering capabilities if compared to only using one diode per half cycle.
A great advantage in using a bridge rectifier is that does not require a centre-tap mains
transformer which is quite expensive and this type of circuit may be fed directly from
the mains supply line but is not very often done since the transformer does supply some
sort of protection.
(a)
+
-
t
A
(b)
B
C
D
E
+
-
F
G
H
I
t
+
(c)
t
Figure 10.8
A network diagram of a full-wave rectifier, using four diodes commonly known as the
bridge rectifier is illustrated in figure 10.9 (a), (b) and (c) and consists of a mains stepdown transformer T1, four diodes D1 to D4 and a load resistor RL connected in series
with the diodes across the secondary side of the mains step-down transformer T1. The
input voltage Vi is supplied to the primary side of the mains transformer T1 and the
output voltage Vo is taken across the load resistor RL.
During the first half of the input voltage Vi, that is between A and B in figure 10.10,
the diodes D1 and D4 are reverse-biased and diodes D2 and D3 are forward-biased and
N4 Industrial Electronics
149
a current will flow through the load resistor RL as indicated by the dotted line with the
polarities as indicated in figure 3.9 (b). The output voltage Vo taken across the load
resistor RL is almost equal to Vi. The only difference there is will be the forward voltage
drop across the diodes D2 and D3 since the circuit on the secondary side operates as a
series voltage divider with the load resistor RL.
During the second half of the input voltage Vi, that is between B and C in figure 10.10,
the diodes D2 and D3 are reverse-biased and diodes D1 and D4 are forward-biased and
a current will flow through the load resistor RL as indicated by the dotted line with the
polarities as indicated in figure 10.9 (c). The output voltage Vo taken across the load
resistor RL is almost equal to Vi. The only difference there is will be the forward voltage
drop across the diode D1 and D4 since the circuit on the secondary side operates as a
series voltage divider with the load resistor RL.
The same pattern will be repeated for every alternate positive and negative half cycle
and the output voltage thus obtained is termed a pulsating direct quantity. But this
is not really what is required since the magnitude of the output is not constant. The
way in which this pulsating direct quantity is smoothed is by means of a smoothing
capacitor which will remove the varying quantity and an output as constant as possible
will be obtained. This aspect will be discussed under the section that will deal with filter
networks. The applicable waveforms for the network in figure 10.9 are illustrated in
figure 10.10. Figure 10.10 (a) is the input Vi for the network and figure 10.10 (b) is the
output obtained from the secondary side of the centre-tap mains transformer for the
network. Figure 10.10 (c) is the output obtained across RL for the network.
T1
D1
(a)
Vi
+
(b)
D2
T1
D3
D4
D1
D2
Vo = Vi - 2VP
RL
Vo = Vi - 2VP
-
Vi
-
RL
+
D3
D4
150
Diodes APPLICATIONs
T1
-
+
D1
(c)
D2
Vi
Vo = Vi - 2VP
RL
+
D3
-
D4
Figure 10.9
(a)
+
-
t
A
(b)
B
C
D
+
-
E
F
G
H
I
t
+
(c)
t
Figure 10.10
10.3 Rectifier concepts
It was mentioned earlier that the output obtained from a rectifier network is deemed
to be a pulsating dc. The ideal output however should be a steady dc but that will
be explained at a later stage. There are a number of concepts that will determine the
magnitude of the output obtained from a rectifier. The first and most important factor is
that of the transformer ratio. A very brief description of that aspect will be given.
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151
10.3.1 Transformer ratio
Although different types of transformers are available on the market today we will
concentrate mainly on the step-down transformer which is the most common
transformer used in conjunction with rectifier networks. Such a transformer is
illustrated in figure 10.11.
VO
Secondary side
Vi
Primary side
Figure 10.11
A transformer consists of two sets of windings, a primary winding or side across which
the input is applied (Vi) and a secondary winding or side across which the output is
taken (VO) and these two sets of winding are electrically isolated from one another.
Energy is transferred from the primary side to the secondary side by means of a
magnetic coupling or more commonly known as electromagnetic coupling. Altough all
transformers are specifically designed to perform a specific task we need to have a look
at the different transformers that are found in industry and these are given below in
figure 10.12 (a), (b), (c), (d) and (e).
• Step-down transformer
These are transformers that have more primary windings than secondary winding
which will result in a lower secondary voltage but with an increase in secondary
current. This type of transformer consists of a single primary winding as well as a single
secondary winding.
(a)
152
Diodes APPLICATIONs
• Step-up transformer
These are transformers that have more secondary windings than primary winding
which will result in an increased secondary voltage but with a decrease in secondary
current. This type of transformer consists of a single primary winding as well as a single
secondary winding.
(b)
• Centre-tap transformer
These are transformers that consist of a single primary winding and a split secondary
winding which actually amounts to two secondary windings. Each set of secondary
windings has its own specified output.
(c)
• Multi-tap transformer
These are transformers much like the centre-tap transformer but actually has more than
one tap on the secondary side in order to obtain different voltages from the secondary
side of the transformer.
(d)
• Auto transformer
The auto transformer consists of a single inductor with the facility to change the ratio
of the primary to the secondary by means of a tapping on the transformer. These
transformers may be used as step-up or a step-down transformer.
N4 Industrial Electronics
153
(e)
Figure 10.12
The magnitudes of the different quantities are mathematically related to one another
and are mathematically expressed by the transformer equation.
NP
VP
IS
=
=
where
NS
VS
IP
NP
VP
IS
NS
VS
IP
=
=
=
=
=
=
number of primary windings
magnitude of primary voltage in volt
magnitude of secondary current in ampere
number of secondary windings
magnitude of secondary voltage in volt
magnitude of primary current in ampere
Important to note is that the secondary current is inversely proportional to the primary
number of winding as well as the primary voltage whereas the primary current is also
inversely proportional to secondary number of windings as well as the secondary
voltage. Also of great importance is that the above values for the primary and secondary
current as well as voltage are always given in RMS-values unless otherwise stated. These
RMS-values are converted to maximum or peak values as follows:
VP (MAX)
VS (MAX)
=
=
1,414 × VP
1,414 × VS
IP (MAX )
IS (MAX)
= 1,414 × IP
= 1,414 × IS
154
Diodes APPLICATIONs
Example 10.1
The following data concerning a transformer is given:
• Primary voltage =
250 volt
• Turns ratio
=
10 : 1
• Secondary current =
5 ampere
Determine from the above information the:
(a)
Peak primary voltage;
(b)
Peak primary current; and
(c)
Peak secondary voltage.
Solution:
(a)
VP (MAX)
=
1,414 × VP
=
1,414 × 250
=
352,5 volts
NP
IS
=
IP (MAX)
(b)
NS
IP
IS × NS
=
IP
NP
5×1
=
10
=
0,5 ampere
=
=
=
NP
VP
NP
=
or
=
NS
VS
NS
VP × NS
VS
=
VS (MAX)
=
N
P
250 × 1
=
=
10
=
25 volt
=
(c)
VS (MAX) =
1,414 × VS
=
1,414 × 25
=
35,35 volt
1,414 × IP
1,414 × 0,5
0,707 ampere
VP (MAX)
VS (MAX)
VP (MAX) × NS
NP
353,5 × 1
10
35,35 volt
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10.3.2 Average dc-voltage
Refer to the two waveforms illustrated in figure 10.13 (a) and (b).
Maximum value
(a)
+
Average dc-value
t
Maximum value
(b)
Average dc-value
+
t
Figure 10.13
The waveform in figure 10.13 (a) illustrates the output obtained from a half-wave
rectifier network and figure 10.13 (b) illustrates the output obtained from a full-wave
rectifier network. Note the average dc value that will be obtained between the two
types of rectifier networks. These values can be determined by using the following
mathematical expressions for full-wave and half-wave rectification respectively.
Half-wave
VDC (AVE)
=
Full-wave
0,318 × VMAX
VDC (AVE)
=
0,637 × VMAX
Example 10.2
A rectifier has the following peak voltage outputs:
(a)
55 volt
(b)
16 volt
Determine the average dc-output in each instance for a half-wave and fullwave rectifier.
Solution:
(a)
Half-wave
VDC (AVE)
=
=
=
0,318 × VMAX
0,318 × 55
17,49 volt
155
156
Diodes APPLICATIONs
VDC (AVE)
=
=
=
0,318 × VMAX
0,318 × 16
5,088 volt
Full-wave
VDC (AVE)
=
=
=
0,637 × VMAX
0,637 × 55
34,485 volt
VDC (AVE)
=
=
=
0,637 × VMAX
0,637 × 16
10,192 volt
(b)
10.3.3
Ripple voltage
Refer to the two waveforms illustrated in figure 10.14 (a) and (b).
A
B
(a)
+
Ripple
t
A
(b)
B
Ripple
+
t
Figure 10.14
During the period indicated by A the filter capacitor will charge and during the period
indicated by B the capacitor will discharge and the output will have a ripple and not
be a constant value. This will be discussed in more detail when we deal with filter
networks. The ripple voltage is a variation of voltage values around the average dcvalue and is always measured as an RMS-value and can be determined by the following
mathematical expressions.
Half-wave
Vr (RMS) = 0,385 × VMAXVr (RMS)
Full-wave
= 0,305 × VMAX
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157
Example 10.3
The maximum output on the secondary side of a power supply is measured at
10 volt. Determine the ripple voltage should this value be measured across a:
(a)
Half-wave rectifier; and
(b)
Full-wave rectifier.
Solution:
It is very important that the maximum value given be converted to an RMSvalue before the calculation is done!
VS (MAX)
(a)
Vr (RMS)
=
=
=
(b) Vr (RMS)
1,414 × VS
1,414 × 10
14,14 volt
=
0,385 × VMAX
=
0,385 × 14,14
=
5,444 volt
=
=
=
0,305 × VMAX
0,305 × 14,14
4,313 volt
A very important observation concerning the above answers obtained is that it
should be obvious that the ripple voltage will be greater for a half-wave rectifier
than it would be for a full-wave rectifier.
10.3.4
Ripple factor
The ripple factor of a rectifier, half-wave as well as full-wave is mathematically given by:
Vr (RMS)
r
=
V
DC (AVE)
The ideal situation would be that the ripple factor should be as close as possible to zero
and would therefore give us an indication of the performance of the rectifier.
158
Diodes APPLICATIONs
Example 10.4
Calculate the ripple factor for a:
(a)
Half-wave rectifier; and
(b)
Full-wave rectifier.
Solution:
(a) r
=
But
Vr (RMS)
VDC (AVE)
r
=
=
(b) r =
But Vr (RMS)
VDC (AVE)
r =
=
Vr (RMS)
VDC (AVE)
=
=
0,385 × VMAX and
0,318 × VMAX
0,385 × VMAX
0,318 × VMAX
1,21
Vr (RMS)
VDC (AVE)
=
=
0,305 × VMAX and
0,637 × VMAX
0,305 × VMAX
0,637 × VMAX
0,48
The above answers would therefore indicate to us that a full-wave rectifier is more
suitable since the ripple factor is much closer to zero than that of the half-wave
rectifier.
10.3.5PIV-rating (peak-inverse-voltage)
The PIV-rating for a diode is probably the most important aspect when diodes are used
in rectifier networks. This value will specify the maximum reverse voltage that may be
applied across the diode or diodes and exceeding this value may render the component
useless since the junction will be damaged.
Half-wave
PIV
= VMAX
Full-wave
PIV
=
2 × VMAX
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Example 10.5
A value of 16,2 volt is measured across the secondary side of a rectifier.
Determine the PIV-rating for the diodes should the rectifier be used as a:
(a)
Half-wave device; and
(b)
Full-wave device.
Solution:
(a) PIV
=
=
VMAX
16,2 volt
(b) PIV
=
=
=
2 ×VMAX
2 × 16,2
32,4 volt
Example 10.6
A step-down transformer has a turns ratio of 30:1 and is fed from a 300 V/50
Hz supply. Determine for a half-wave rectifier network the:
(a)
Average dc voltage output;
(b)
Ripple voltage;
(c)
Ripple factor; and
(d)
PIV-rating for the diode.
Solution:
NP
(a) VDC (AVE) =
0,318 × VMAX
where
=
NS
=
0,318 × 14,14
=
4,497 volt
VS
=
=
=
VS (MAX)
= 1,414 × VS
= 1,414 × 10
= 14,14 volt
VP
VS
VP × NS
NP
300 × 1
30
10 volt
159
160
Diodes APPLICATIONs
(b) Vr (RMS)
=
=
=
0,385 × VMAX
0,385 × 14,14
5,444 volt
Vr (RMS)
(c) r
=
V
DC (AVE)
5,444
=
4,497
= 1,21
(d) PIV = VMAX
= 14,14 volt
Example 10.7
A diode has a PIV rating of 17,675 volt. Determine the average dc output
voltage for a half-wave rectifier.
Solution:
VDC (AVE)
=
=
=
0,318 × VMAX
0,318 × 17,675
5,62 volt
where VMAX = PIV = 17,675 volt
Example 10.8
A centre-tap step-down transformer has a turns ratio of 10 : 1 and is fed from a
100V/50 Hz supply. Determine for this rectifier network the:
(a)
Average dc voltage output;
(b)
Ripple voltage;
(c)
Ripple factor; and
(d)
PIV-rating for the diodes.
N4 Industrial Electronics
Solution:
A point that should be noted here is that the above rectifier network is that of a fullwave rectifier since use is made of a centre-tap transformer.
NP
(a) VDC (AVE) =
0,637 × VMAX
where
=
NS
=
0,637 × 14,14
=
9 volt
VS
=
=
=
VS (MAX)
= 1,414 × VS
= 1,414 × 10
= 14,14 volt
(b) Vr (RMS)
=
=
=
0,308 × VMAX
0,308 × 14,14
4,355 volt
Vr (RMS)
(c) r
=
V
DC (AVE)
4,355
=
9
= 0,484
(d) PIV
=
=
=
2 × VMAX
2 × 14,14
28,28 volt
VP
VS
VP × NS
NP
100 × 1
10
10 volt
161
162
Diodes APPLICATIONs
10.4 Filter networks
It was mentioned earlier that the output obtained from a rectifier circuit is a pulsating
dc which means that it has a varying value but is not the type of output we require. The
ideal output is a steady dc-value and that is obtained by using a filter with the rectifier.
Definition 10.1 Filter
A filter can therefore be defined as a component that will remove the ripple
(pulsating) component from the output of a rectifier circuit.
There are various types of filters that can be used and each one has its own distinct
characteristics and each type will be explained. Before we actually commence with
the explanation of the filter circuits it is important that we look at the concept of filter
circuits. You are therefore requested to refer to figure 10.15 (a) and (b) where this
principle is illustrated.
Half-wave
rectifier
Filter
Input
Output
(a)
Full-wave
rectifier
Filter
Input
Output
(b)
Figure 10.15
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163
10.4.1The capacitor filter
10.4.1.1
Half-wave rectification
A half-wave rectifier circuit with a capacitor filter is illustrated in figure 10.16. Note that
the capacitor used is an electrolytic type which has a fixed polarity.
D1
T1
+
-
IDC
VF
+
Vi
C
-
+
-
RL
+
VO
Figure 10.16
The load voltage VDC across RL is mathematically expressed by:
V
VDC = Vp - r (p - p)
or
VDC = Vp 2
IDC
Vm
and
IDC =
2×f×C
RL
The RMS ripple voltage Vr (RMS) across RL is mathematically expressed by:
Vr (RMS)
=
Vr (p - p)
or Vr (RMS)
=
2√3
IDC
2√3 × f × C
The ripple r of the output voltage VDC across RL is mathematically given by:
r
=
Vr (RMS)
or
VDC IDC
or
VDC × 2√3 × f × C
1
RL × 2√3 × f × C
Consider the graphical representation in figure 10.17 for a clarification of the quantities
used in the various expressions.
164
Diodes APPLICATIONs
Capacitor discharge time
VC
Vr (p-p)
VDC
VP
t
Capacitor charge time
Figure 10.17
Although a capacitor filter is extensively used it does have the following disadvantages:
•
•
Its current delivering capabilities are limited in that it cannot supply a large
current;
Should there be an increase in current the ripple voltage will increase likewise;
Example 10.9
A half-wave rectifier is fed from a 220V/50 Hz supply and the following
additional data is given:
• Peak secondary voltage
=
15 volt
• Value of filter capacitor
=
30 µF
• Load current
=
15 mA
Use the above data and determine the:
(a)
Output dc voltage (VDC);
(b)
Output ripple voltage (Vr (RMS)); and
(c)
Ripple factor expressed as a percentage (r %).
Solution:
15 -
10 volt
=
Vp -
IDC
2×f×C
(a) VDC
=
=
15 × 10-3
2 × 50 × 30 × 10-6
N4 Industrial Electronics
IDC
(b)Vr (RMS)
=
2 x √3 × f × C
15 × 10-3
=
2 ×√3 × 50 × 30 × 10-6
=
Vr (RMS)
× 100
VDC
2,887
× 100
10
(c) r
=
=
=
10.4.1.2
•
2,887 volt
28,87 %
and
It does not have good regulation capabilities.
Full-wave rectification
T1
D1
+
Vi
-
D2
RL
VO = Vi - VF
C
Figure 10.18
T1
D1
D2
+
Vi
D3
Figure 10.19
D4
RL
C
Vo = Vi - 2VP
165
166
Diodes APPLICATIONs
A full-wave rectifier circuit using two diodes with a capacitor filter is illustrated in figure
10.18 and a bridge rectifier using four diodes is illustrated in figure 10.19.
The load voltage VDC across RL is mathematically expressed by:
V
IDC
Vm
VDC = Vp - r (p - p)
or
VDC = Vp and
IDC =
2
4×f×C
RL
The RMS ripple voltage Vr (RMS) across RL is mathematically expressed by:
Vr (RMS)
=
Vr (p - p)
or Vr (RMS)
=
2√3
IDC
4√3 × f × C
The ripple r of the output voltage VDC across RL is mathematically given by:
r
=
Vr (RMS)
or
VDC IDC
or
VDC × 4√3 × f × C
1
RL × 4√3 × f × C
Example 10.10
Consider the following full-wave rectifier circuit diagram.
Use the given information and determine the following quantities for a 20 µF
and 35 µF capacitor.
(a)
Average value;
(b)
RMS-value;
(c)
Output dc voltage;
(d)
Output RMS ripple voltage; and
(e)
Ripple percentage.
T1
D1
10 mA
D2
+
220 V/50 Hz
15 V
D3
Solution:
(a)
VAVE
=
=
=
Vm × 0,637
15 × 0,637
9,555 volt
D4
RL
C
1 500 ohms
N4 Industrial Electronics
(b)VRMS
=
=
=
(c) 20 µF capacitor
167
Vm × 0,707
15 × 0,707
10,605 volt
(d)
I
Vr (RMS) =
VDC = Vp - DC
4×f×C
10 × 10-3
= 15 =
4 × 50 × 20 × 10-6
= 12,5 volt
=
IDC
4√3 × f ×C
10 × 10-3
3 × √3 × 50 × 20 × 10-6
1,925 volt
35 µF capacitor
IDC
IDC
VDC = Vp Vr (RMS) =
4×f×C
4√3 × f ×C
10 × 10-3
10 × 10-3
= 15 =
4 × 50 × 35 × 10-6
3 × √3 × 50 × 35 × 10-6
=
13,571 volt
=
1,099 volt
(e) 20 µF capacitor
r =
=
Vr (RMS)
× 100
r =
VDC
1,925
× 100
12,5
=
=
15,4= 11,55%
r =
=
=
1
× 100
RL × 4√3 × f × C
1 × 100
1 500 × 4√3 × 50 × 20 × 10-6
9,623%
IDC
× 100
VDC × 4√3 × f × C
10 × 10-3
× 100
12,5 × 4 × √3 × 50 × 20 × 10-6
168
Diodes APPLICATIONs
35 µF capacitor
r =
=
Vr (RMS)
IDC
× 100
r =
× 100
VDC
V × 4√3 × f × C
DC
1,099
× 100
10 × 10-3
13,571
=
× 100
13,571 × 4 × √3 × 50 × 35 × 10-6
=
8,098 = 6,077%
r =
=
=
1
RL × 4√3 × f × C
× 100
1 × 100
1 500 × 4√3 × 50 × 35 × 10-6
5,499%
10.4.2The RC-filter
10.4.2.1
Half-wave rectification
A half-wave rectifier circuit with a resistor-capacitor (RC) filter is illustrated in figure
10.20. This type of filter is very often referred to as a π-type RC-filter. The RC-type filter
has the following advantages when compared to the capacitor filter:
•
•
•
A decrease in the ripple voltage;
A slight increase in current delivering capabilities; and
An increase in the voltage regulation capabilities.
D1
T1
+
+
Vi
C1
-
IDC
RS
VF
-
+
+
+
C2
-
RL
VO
Figure 10.20
10.4.2.2
Full-wave rectification
A full-wave rectifier circuit using two diodes with a resistor-capacitor filter is illustrated
in figure 10.21 and a bridge rectifier using four diodes is illustrated in figure 10.22.
N4 Industrial Electronics
+
T1
+
+
Vi
+
-
IDC
RS
D1
-
D2
+
+
RL
C1
VO
C1
-
Figure 10.21
+
T1
D1
+
+
Vi
-
D3
+
+
RS
D2
RL
C1
D4
Figure 10.22
The load voltage V’DC across RL is mathematically expressed by:
RL
V’DC = VDC ×
R + RS
L
The RMS ripple voltage V’r (RMS) across RL is mathematically expressed by:
V’r (RMS)
=
Vr (RMS) ×
XC
2
(RS + XC2)½
where XC =
1
2×π×f×C
1
XC = 4×π×f×C
for half-wave and
for full-wave.
The ripple r of the output voltage VDC across RL is mathematically given by:
V’r (RMS)
r
=
V’
DC
Vo
C2
-
169
170
Diodes APPLICATIONs
Example 10.11
The following data concerning a power supply is given:
Supply voltage
=
250 V/ 0 Hz
Voltage across first capacitor
=
120 V
Ripple voltage across first capacitor
=
10 V
Series resistor=
1 k-ohm
Second capacitor
=
150 µF
Load resistor=
5 k-ohm
Determine for half-wave- and full-wave rectifier the:
(a)
DC output voltage;
(b)
Ripple voltage; and
(c)
Ripple factor.
Solution:
(a) V’DC
=
=
VDC ×
100 volt
=
(b) Half-wave
120 ×
RL
RL + RS
5 × 103
5 × 103 + 1 × 103
Full-wave
1
1
XC =
XC =
2×π×f×C
4×π×f×C
=
1
1
=
-6
2 × 3,142 × 50 × 150 × 10
4 × 3,142 × 50 × 150 × 10-6
= 21,2 ohm
=
10,6 ohm
V’r(RMS) = Vr(RMS) ×
XC
V’r(RMS) = Vr(RMS) ×
XC
2
2 ½
2
(RS + XC2)½
(RS + XC )
= 10 ×
21,2
= 10 ×
10,6
3 2
2 ½
[(1 × 103)2 + 10, 62]½
[(1× 10 ) + 21, 22 ]
= 0,212 volt
= 0,106 volt
N4 Industrial Electronics
(c) Half-wave
r =
=
=
171
Full-wave
V’r(RMS)
V’r(RMS)
r =
V’DC
V’DC
0,212
0,106
=
100
100
2,212 × 10-3 = 1,06 × 10-3
10.4.3The LC-filter
10.4.3.1
Half-wave rectification
A half-wave rectifier circuit with an inductor-capacitor (LC) filter is illustrated in figure
3.23. This type of filter is very often referred to as a π-type LC-filter. The inductor has a
fairly large value which will tend to smooth the current flowing through it.
The LC-type filter has the following advantages when compared to the capacitor filter:
•
•
•
A decrease in the ripple voltage;
A slight increase in current delivering capabilities; and
An increase in the voltage regulation capabilities.
D1
T1
+
-
L
VF
+
Vi
C1
-
IDC
RS
+
+
-
C2
+
-
RL
VO
Figure 10.23
10.4.3.2
Full-wave rectification
A full-wave rectifier circuit using two diodes with an inductor-capacitor filter is
illustrated in figure 10.24 and a bridge rectifier using four diodes is illustrated in figure
10.25.
172
Diodes APPLICATIONs
T1
+
+
D1
+
Vi
+
-
IDC
RS
L
D2
-
+
+
RL
C1
VO
C1
-
Figure 10.24
T1
+
D1
L
D2
+
+
Vi
-
D3
+
+
RS
C2
D4
Figure 10.25
The load voltage V’DC across RL is mathematically expressed by:
RL
V’DC = VDC ×
R + RS
L
The RMS ripple voltage V’r (RMS) across RL is mathematically expressed by:
V’r(RMS)
=
Vr(RMS) (2πf )2L ×C
for half-wave and
V’r(RMS)
=
Vr(RMS) (4πf )2L × C
for full-wave.
The ripple r of the output voltage VDC across RL is mathematically given by:
r
=
V’r(RMS)
V’DC
Vo
RL
C1
-
N4 Industrial Electronics
Example 10.12
Consider the following filter circuit and waveform for a full-wave rectifier.
L
+
RS
+
+
100 V
-
-
C1
RL
C2
V
10 V
t
Supply voltage to power supply
=
=
250 µF
C1
=
250 µF
C2
L
=
3H
=
10 ohm
RS
=
5 k - ohm
RL
Use this information and determine the:
(a)
DC output voltage; and
(b)
Ripple factor.
250 V/50 Hz
Solution:
RL
(a)
V’DC
=
VDC ×
RL + RS
5 × 103
=
100 ×
5 × 103 + 10
=
99,8 volt
173
174
Diodes APPLICATIONs
(b) Vr(RMS) =
=
=
r
=
Vr(p - p)
V’r(RMS)
=
Vr(RMS)
2√3
(4 × π ×f )2 L × C
10
=
2,886
2√3
(4 × 3,142 × 50)2 × 3 × 250 × 10-6
2,886 volt
= 9,745 × 10-3 volt
V’r ( RMS )
V’DC
9,745 × 10-3
=
99,8
=
9,76 × 10-5
10.5No-load voltage
The no-load voltage of any power supply may be defined as that voltage which is
supplied by the secondary winding of the transformer when the load to that power
supply is not connected. Consider figure 10.26.
+
T1
+
-
S
+
Vi
+
+
D1
D2
-
VO
C
RL
Figure 10.26
During the conduction half-cycles of the input cycle the capacitor C1 will charge up to
a value of ± VP and will discharge during the period when the input cycle goes through
zero degrees. Since there is no discharge path for the filter capacitor according to figure
10.26 the voltage is measured across C1 and is termed the no-load voltage since the
switch S is open. This voltage contains for all practical purposes no ripple. Should we
however close the switch S then the capacitor C1 will have a discharge path through
the load resistor RL since the current will be flowing through it. This output will then
contain a ripple and is termed the full-load voltage. An important aspect that should
also be noted is that the no-load voltage will always have a higher value than the fullload voltage.
N4 Industrial Electronics
175
10.6 Voltage regulation
Definition 10.2 Voltage
Voltage regulation may be defined as that change in the output voltage (full-load)
for varying load conditions.
This concept is illustrated in figure 10.27.
O
u
t
p
u
t
No-load voltage
Voltage
variation
Full-load voltage
v
o
l
t
a
g
e
Load current
Figure 10.27
The following conclusions can be made concerning the graphical representation in
figure 3.26:
•
•
The output voltage of a power supply will decrease with an increase in load
current; and
The smaller the change in no-load voltage and full-load voltage, the more effective
the power supply will be.
This voltage regulation is given by the following mathematical expression:
Voltage regulation =
No-load voltage - Full-load voltage
Full-load voltage
176
Diodes APPLICATIONs
Example 10.13
A full-wave power supply delivers 15,6 volt under no-load conditions.
Determine the percentage regulation should the following voltages be
delivered under full-load conditions.
(a)
13,4 volt; and
(b)
12 volt.
Solution:
(a) % regulation
=
=
=
(b) % regulation
=
=
=
VNL - VFL
× 100
VFL
15,6 - 13,4 × 100
15,6
14,1%
VNL - VFL
× 100
VFL
15,6 -12
× 100
15, 6
23,07%
10.7 Voltage multiplication
Rectifier circuits, full-wave or half-wave, are limited to an output that will never be
greater than the peak input voltage to that circuit. It may sometimes be required that
greater voltages are required and for this purpose we make use of voltage multiplication
circuits which may, depending on the design, supply two or more times the peak input
value as an output. Two such circuits will be discussed.
10.7.1 Voltage doubler
The circuits in figure 10.28 (a) and (b) illustrates a voltage doubler where figure 10.28
(a) illustrates one half of the input cycle and figure 10.28 (b) illustrates the other half of
the input cycle.
N4 Industrial Electronics
Electron
flow
D1
+
C1
Vi = 240 V
-
D1
-
-
340 V
C1
Vi = 240 V
+
C2
680 V
+
+
D2
+
+
+
177
340 V
D2
-
Electron
flow
(a)
C2
-
-
(b)
Figure 10.28
Assume the polarities of the input voltage Vi as illustrated in figure 10.28 (a) then diode
D1 will conduct and capacitor C1 will charge up to a value equal to the peak voltage VP
which in this instance will be equal to:
VP
= 1,414 × VRMS
= 1,414 × 240
= 340 V with polarities as indicated.
Now assume the polarities of the input voltage Vi as illustrated in figure 10.28 (b) then
diode D2 will conduct and capacitor C2 will charge up to a value equal to the peak
voltage VP which in this instance will be equal to:
VP
= 1,414 × VRMS
= 1,414 × 240
= 340 V with polarities as indicated.
10.7.2 Voltage multiplication
The circuit illustrated in figure 10.28 only has the capability to double whatever the
input to the circuit is. What are we going to do should we require multiples of the input?
The circuit in figure 10.29 (a) illustrates such a voltage multiplier circuit. The circuit in
figure 10.29 (b) illustrates the effect of the positive half cycle of the input whereas figure
10.29 (c) illustrates the effect of the negative half cycle of the input.
Assume the polarities as indicated in figure 10.29 (b) then the diode D1 will be forward
biased and the diode D2 will be reverse biased. Since diode D1 is forward biased it will
conduct charging capacitor C1 to a value of VP and polarities as indicated.
178
Diodes APPLICATIONs
Now assume the polarities as indicated in figure 10.29 (c) then the diode D2 will be
forward biased and the diode D1 will be reverse biased. Since diode D2 is forward biased
it will conduct charging capacitor C2 to a value of VP and polarities as indicated. The
sum of the combination of capacitors C1 and C2 will now result in a magnitude of 2 VP.
In this manner we will be able to obtain magnitudes of 3 VP and 4 VP.
4 VP
2 VP
-
+
-
C2
C4
D1
-
C1
+
D2
+
D3
-
C3
D4
+
VP
3 VP
(a)
+
-
-
C2
D1
-
-
C1
+
+
C2
D2
D1
+
(b)
-
C1
D2
+
(c)
Figure 10.29
Exercise 10.1
1.
Clippers may be sub-divided into series and parallel as well as positive
and negative types. Draw the network diagrams as well as input and
output waveforms that will describe the principle of operation of each of
these types of clippers. The magnitude of the input waveform has a peakto-peak value of 10 volt and is centrally clamped around the zero-
volt line. In your output waveforms you are required to make provision for
Germanium as well as Silicon diodes.
N4 Industrial Electronics
2.
Diodes are commonly used for rectification purposes. Make use of suitable network diagrams as well as input and output waveforms to illustrate the following types of rectification.
2.1 Half-wave;
2.2 Full-wave (two diodes); and
2.3 Bridge rectifier.
Indicate on your network diagrams the direction of all currents that may flow in the network.
3.
The transformer of a bridge type rectifier has a turns ratio of 15:1 and is supplied from a 220 V/100 Hz alternating supply. Calculate the:
3.1 Average dc output voltage;
3.2 Ripple voltage;
3.3 Ripple factor; and
3.4 PIV-ratings for the diodes.
4.
A 220 V/50 Hz supply rated at 15 ampere feeds a step-down transformer with 220 windings on the primary side. Should 10 volt be measured on the secondary side then determine for full-wave as well as half-wave the:
4.1 Average dc output voltage;
4.2 Ripple factor; and
4.3 RMS secondary current.
5.
Explain in your own words what is meant by a filter network used in conjunction with rectifier circuits. Enhance your explanation with suitable sketches.
6.
Draw the following fully labelled circuit diagrams:
6.1 Half-wave rectifier with:
(a) An RC-filter; and
(b) A capacitor filter.
6.2 Full-wave rectifier (two diodes) with:
(a) LC-filter; and
(b) An RC-filter.
6.3 Full-wave rectifier (bridge circuit) with:
(a) LC-filter; and
(b) A capacitor filter.
7.
A half-wave rectifier is fed from a 220 V/60 Hz supply and the following additional data is given:
• Peak secondary voltage
=
25 volt
• Value of filter capacitor
=
100 µF
• Load current
=
30 mA
Use the above data and determine the:
179
180
Diodes APPLICATIONs
7.1 Output dc voltage (VDC);
7.2 Output ripple voltage (Vr(RMS)); and
7.3 Ripple factor expressed as a percentage (r %).
8.
Use a graphical representation to illustrate the concept of voltage regulation and then use your own values to show how voltage regulation can be determined.
9.
Use suitable circuit diagrams and illustrate the concept of voltage doubling as well as voltage multiplication.
10. Use suitable sketches or graphic representations to illustrate the concept of voltage regulation.
Use a graphical representation to illustrate the concept of voltage regulation and then use your own values to show how voltage regulation can be determined.
11. Use suitable circuit diagrams and illustrate the concept of voltage doubling as well as voltage multiplication.
12. A full-wave bridge rectifier is constructed and a capacitor filter is used. The following additional data concerning the rectifier is given:
• Input to transformer
=
280 V/60 Hz
• Voltage measured across secondary winding =
18 volt
• Value of load resistor
=
1,2 k-ohm
• Load current
=
15 mA
Use the above information and determine the following quantities for a
150 µF and 90 µF capacitor.
12.1 Average value;
12.2 RMS value;
12.3 Output dc voltage;
12.4 Output RMS ripple voltage; and
12.5 Ripple percentage.
13. The following data concerning a power supply with an RC-filter is given:
• Supply voltage
=
250 V/100 Hz
• Voltage across first capacitor
=
90 V
• Ripple voltage across first capacitor
=
10 V
• Series resistor
=
1,2 k-ohm
• Second capacitor
=
100 µF
• Load resistor
=
2,5 k-ohm
Determine for half-wave and full-wave the:
13.1 dc-output voltage;
13.2 Ripple voltage; and
13.3 Ripple factor.
N4 Industrial Electronics
14. An LC-filter consists of an inductor having a value of 1,4 H and a resistance
of 120 ohm. The capacitor has a value of 120 µF and the load resistor has a value of 80 k-ohm. A voltage of 12 volt is measured across the first capacitor and has a ripple of 1,6 volt. The primary of the transformer is fed from a 180 V/50 Hz supply. Use the given information to calculate the:
14.1 dc-output voltage; and
14.2 Ripple factor.
Before you commence with the calculations you are required to draw a neat
labelled circuit diagram of the power supply inserting all relevant values as is given.
181
182
Diodes APPLICATIONs
Special Diodes and Applications
183
CHAPTER
11 Special diodes and applications
Learning Outcomes
On completion of this module you will be able to:
•
•
Discuss with the aid of circuit diagrams the operation, characteristics and
application of the following diodes:
– Zener diodes;
– Varactor diodes;
– Tunnel diodes;
– Photo-diodes:
– Photo-sensitive diodes; and
– Photo-emissive diodes.
Calculate:
– Current values;
– Voltage values;
– Resistor values; and
– Wattage ratings with reference to a Zener diode.
11.1 The Zener diode
Should an ordinary diode be reverse-biased then a small leakage current will flow which
for all practical purposes are normally disregarded. Should we however increase the
magnitude of the reverse-bias then the diode will be destroyed since it will not be able
to handle this increase since the PIV of the diode will be exceeded. When this PIVpoint is controlled with the aid of a series resistor we end up with a Zener diode voltage
regulator. In order to understand the operation of this type of diode we need to have
a look at the characteristic curve for a Zener diode. A point that must be mentioned
is that a Zener diode operates in the reverse characteristic of the curve and very good
voltage regulation capabilities are obtained and will behave like a normal diode when
forward-biased. The point at which the Zener diode displays the voltage regulation
capabilities are specifically manufactured as such.
184
Special Diodes and Applications
11.1.1 Electrical characteristics of a Zener diode
The following characteristics are the most common and can be found in all specification
sheets and are indicated on the characteristic curve illustrated in figure 11.1 (b) whereas
figure 11.1 (a) illustrates the IEC circuit symbol.
Zener current (IZM)
This is the maximum allowable current when the diode is used as a reference source at
the Zener point.
Reference voltage (VZ)
This is the Zener breakdown voltage and specifies the maximum voltage when the diode
is used as a reference source. (Also termed the Zener knee)
Average forward current (IF (AV))
This is the average forward current permissible when the diode is forward biased but is
not important when the Zener diode is used as a reference source.
Repetitive reverse current (IZ (RM))
This is the maximum specified reverse current that is permissible when the Zener diode
is used as a reference source.
Zener dissipation rating (PZ)
This is the maximum power dissipation rating of the Zener diode.
Junction temperature (Tj)
This is the maximum operating temperature of the Zener diode.
Zener current near the knee (IZK)
This is the minimum current required to sustain breakdown.
Important to note that this characteristic curve of the Zener diode only indicates the
reverse bias characteristics since it is in this region that the diode is used.
IF
Forward bias region
VZ
VR
VF
IS
IZK
Cathode
IR
IZT
Anode
Reverse bias region
(a)
(b)
Figure 11.1
IZM
185
11.1.2 Application of the Zener diode
All Zener diodes are manufactured in order to conduct at a given voltage rating in the
reverse-bias area. Zener diodes are mainly used as voltage regulators and are available
from 1, 5 volt up to 30 volt but higher values can be obtained. The voltage rating for
which it is designed is termed the Zener voltage and has thát value as voltage reference.
What this basically means is that the design value is thát value that will be measured
across the Zener diode when it is conducting and is used as a voltage reference source.
An elementary circuit diagram in figure 11.2 illustrates the typical application of a
Zener diode. The circuit is termed a voltage regulator since the Zener diode will supply
a constant voltage (regulated) to the load depending on what the design value is. The
resistor RS is connected in series with the Zener diode to limit the current flow through
the Zener diode as well as the load RL which is connected in parallel across the Zener
diode and will therefore supply a constant voltage to the load.
Using the concepts of Kirchhoff ’s current and voltage laws we may formulate the
following expressions:
VO = VZ
VRS = Vi - VZ
IRS = IZ + IL
Should the load RL not be connected then IZ = 0 and VRL = VZ
Various component values can be determined by the following mathematical
expressions:
+
VRS
IRS
+
RS
Unregulated
dc Vi
IZ
IL
VZ
RL
DZ
Regulated dc
VO
-
Figure 11.2
PZ
VZ
IZ =
RL
=
RS
=
VZ
IL
where
VO = Output voltage
VZ = Zener voltage
VRS
IRS
Vi = Input voltage
IRS = Current through series resistor
186
Special Diodes and Applications
VRS = Voltage across series resistor
PZ = Zener power dissipation
VRS = Voltage across series resistor
IZ = Zener current
IL = Load current
Example 11.1
Consider the following circuit diagram.
+
IRS
RS
10 V
+
IZ
IL
20 mA
RL
6 V at 3 W
VO
DZ
-
Use the available data and determine the:
(a)
Value of the load resistor;
(b)
Value of the series resistor; and
(c)
Current magnitude through the Zener diode.
Solution:
VZ
(a)
RL
=
IL
6
=
20 × 10-3
=
300 ohm
VRS
(b)
RS =
VRS = Vi - VZ
IRS
IRS
= 10 – 6
4
=
= 4 volt
0,52
= 7,69 ohm
PZ
=
(c)
IZ
VZ
3
=
6
=
0,5 ampere
=
=
=
IZ + IL
0, 5 + 20 × 10-3
0,52 ampere
187
A more practical circuit diagram of a Zener diode voltage regulator is illustrated in
figure 11.3. This circuit consists of two sections as indicated. The rectifier section will
not be discussed since it was done in the previous chapter. The output from the rectifier
section is applied to the regulator (voltage reference) section and ZD and RS form a
voltage divider section. The voltage drop across the Zener diode will always be its design
value and the rest of the voltage will appear across the series resistor. This will only be
so if the voltage applied to the regulator section exceeds the Zener diode design value.
Should this value be less than the Zener diode design value there will be no output.
+
-
T1
+
D1
+
D2
+
RS
+
-
+
C1
DZ
RL
VOB
VOA
Rectifier section
Regulator section
Figure 11.3
Since the load resistor RL is in parallel with the Zener diode, the Zener diode voltage
will appear across the load as well as VOA which is termed the regulated output. Should
the output be taken across the Zener diode and the series resistor an output VOB will be
obtained which is termed the unregulated output.
11.2 The varactor diode
These diodes are also known as epicaps, varicaps and/or voltage variable capacitance
diodes and are also operated in the reverse-bias area of the characteristic curve. You will
recall from the PN-junction theory that when such a junction is reverse-biased that the
depletion layer increases. Since a PN-junction is not an electrical junction this depletion
layer will have the characteristics of a capacitor.
188
Special Diodes and Applications
11.2.1 Circuit symbol and characteristic curve
The IEC circuit symbol and characteristic curve is illustrated in figure 11.4 (a) and (b).
Important to note that the characteristic curve in figure 11.4 (b) is not current versus
voltage as with many other characteristic curves but capacitance versus voltage and in
this instance it is reverse voltage.
11.2.2 Application of a varactor diode
The main application for a varactor diode is that of tuning circuits in FM-receivers
as well as television receivers and replaces the variable tuning capacitor. Such an
application is illustrated in figure 11.5.
Anode
C
a
p
a
c
I
t
a
n
c
e
Cathode
Reverse bias
(a)
(b)
Figure 11.4
Since the magnitude of the reverse-bias is proportional to the width of the depletion
layer a high reverse-bias will result in result in a wider depletion layer and a lower
reverse bias will result in a narrower depletion layer. This will then in turn determine
the capacitance CPN of the varactor diode which is inversely proportional to the
magnitude of the reverse-bias.
189
T1
+ VCC
f out
L1
D1
R2
CC2
CC1
R3
Q1
RV1
Bias control for
D1 capacitance
R1
- VCC
Figure 11.5
In the diagram illustrated in figure 11.5 a tuned circuit is formed by an inductor L of the
transformer T1 and the capacitance of the varactor diode D1 which is further dependant
on the magnitude of the reverse-bias which is controlled by the setting of RV1.
The range of frequencies that the tuned circuit will operate is mathematically given by:
where
1
=
fr
2 × π (L × C) ½
fr
π
L
C
=
=
=
=
frequency in hertz
3,142
value of inductor in henry
value of capacitance in farad
It should therefore be obvious that should the reverse-bias on D1 be changed that
its capacitance will also change and therefore the frequency of the tuned circuit will
likewise change.
190
Special Diodes and Applications
Example 11.2
A variation in reverse bias will cause a capacitance variance of 15 pF to 100 pF
in the circuit diagram illustrated in figure 11.5. Determine the frequency range
of the tuned circuit if it is given that the inductor L has a constant value of 100
µH.
Solution:
fr =
1
2 × π (L × C) ½
=
1
2 × 3,142 (100 × 10-6 × 15 × 10-12)½
= 4,1 MHz
fr =
1
2 × π (L × C) ½
=
1
2 × 3,142 (100 × 10-6 × 100 × 10-12)½
= 1,6 MHz
The frequency fout will therefore vary between 1,6 MHZ and 4,1 MHz.
11.3 The tunnel diode
The tunnel diode is also termed an Esaki diode. It is also a two-terminal device and is
almost exclusively used as a high-frequency component in the following applications:
•
•
•
An amplifier;
An oscillator; and
A switch.
Its very fast response it displays makes it ideal for use in the above applications. The
main advantage of this type of diode is that it requires a smaller bias voltage in the
region of mV if compared to other types of diodes.
11.3.1Circuit symbol and characteristic curve
The IEC circuit symbol and characteristic curve is illustrated in figure 11.6 (a) and (b).
191
The tunnel diode is manufactured from heavily doped semi-conductor materials to
form a PN-junction and is used in the forward-bias region as indicated in figure 11.6.
The large number of impurities in the material ensures that the depletion layer to be
so thin that electrical charges moves through the junction very easily and is termed
‘tunnelling’, hence the name Tunnel diode.
The operating principle and characteristic curve can best be described in the following
manner.
At switch-on there is a sudden and steep increase in current ID up to the point IP then
there is a sudden decrease in current ID up to the point IV with an increase in forwardbias voltage. This area in which the sudden decrease takes place is termed the ‘negative
resistance area’ of the diode and the negative resistance RD of the diode is the most
important characteristic of the Tunnel diode. This is completely opposite to a ‘resistance’
in which there is an increase in current when the supply voltage is increased. A further
increase in forward-bias will have a corresponding increase in ID and will almost follow
the characteristic of an ordinary diode.
mA
Negative
resistance
region
IP
ID
Anode
Cathode
IV
mV
VR
(a)
VM
VV
VF
(b)
Figure 11.6
11.3.2Application of a tunnel diode
The main application of a Tunnel diode is that of a parallel amplifier and such a
practical circuit diagram is illustrated in figure 11.7.
192
Special Diodes and Applications
IB
IL
ID
VB
D
RL
VR
VS
ac
Figure 11.7
This circuit must, for operation as an amplifier, be biased to the centre of its negative
resistance region. This parallel diode amplifier has the following characteristics:
•
•
•
Current gain;
Power gain; and
No voltage gain.
11.4Photo-diodes
Photo-diodes are classified in two main categories namely photo-sensitive components
and photo-emissive components.
11.4.1Photo-sensitive diodes
The photo-diode is a semi-conductor component that will depend on light energy in
order to become effective. But it is important to note that the photo-diode be connected
to a supply voltage and reverse-biased in order to become operative. Should the PNjunction be exposed to a source of light electron-hole pairs will be created which will
cause a small reverse current to flow. The magnitude of this current will be determined
by the intensity of the light source. The more intense this light source the greater the
magnitude of the reverse current.
11.4.1.1Circuit symbol and characteristic curve
The IEC circuit symbol and characteristic curve is illustrated in figure 11.8 (a) and (b).
Photo-diodes are always used in the reverse-bias region of the characteristic curve and
the incident falls onto the PN-junction through a window in the casing of the diode.
This source of light energy may be from the sun or it may be artificially generated light.
A small current, termed the ‘dark current’ will flow in the absence of a light source and
typical values are 10 µA for germanium diodes and 1 µA for silicon diodes. It can be
seen that as the light intensity increases so does the reverse current also increase.
193
I
1 200 lumen
Radiant light
energy
800 lumen
IR
400 lumen
0 lumen
V
Anode
VR
Cathode
(a)
(b)
Figure 11.8
11.4.1.2Application of a photo-diode
A photo-diode may switch a device ‘on’ or ‘off ’ depending on how it is connected in the
circuit. Two such applications are illustrated in figure 11.9 (a) and (b). Figure 11.9 (a)
will switch the device ‘on’ whereas figure 11.9 (b) will switch the device ‘off ’.
The circuit is constructed in such a manner that the device being controlled is in the
‘off ’ state (figure 11.9 (a)) and need to be energised. In the absence of incident light
the current through diode D is very small (dark current) and the transistor Q1 cannot
conduct.
However, should incident light now fall onto the PN-junction a current will flow
dependant on the intensity of the incident light. Typical values of sensitivity lie between
10 mA and 50 mA/lumen. Since the current will increase because of the presence of
incident light the transistor will switch on and start conducting. The transistor is also
an amplifier and should the collector current increase sufficiently the relay contact will
be energised, providing that the holding current value of the relay is exceeded, into the
N/C position and the device coupled to it will commence operation.
The circuit is constructed in such a manner that the device being controlled is in the ‘on’
state (figure 11.9 (b)) and need to be de-energised. Should incident light fall onto the
PN-junction producing current to flow the diode will become a virtual short circuit and
the transistor will switch off.Since there will be no collector current the relay will go into
the N/O position because the holding current value of the relay has dropped below its
specified value. In this manner the device to be controlled will be switched ‘off ’.
194
Special Diodes and Applications
N/O (off)
N/O (off)
Relay contacts
N/C (on)
Relay contacts
To supply and device
to be controlled
To supply and device
to be controlled
+ VCC
N/C (on)
+ VCC
Incident light
Relay coil
RB
Relay coil
D
QI
Q1
Incident light
RB
RE
RE
D
- VCC
- VCC
(a)
(b)
Figure 11.9
11.4.2Photo-emissive diodes
Electrons and holes will continually combine in a PN-junction when there is a flow of
electrical current through the device. What is however important is that the electrons
are in the conduction band and the holes are in the valence band. When an electrical
current now flows through the diode the electrons have to give off energy in order to
drop from the conduction band into the valence band and this energy is given off in
the form of an electromagnetic radiation of different colours depending on the doping
level of the semi-conductor material. Silicon and Germanium however radiate mainly
in the Infra-red spectrum which is not detectable by the human eye. To overcome this
problem Gallium Arsenide Phosphide (Cds) and Gallium Phosphide is used and the
radiation will now take place within the visible spectrum. This radiation takes place at
the area of re combination and the PN-junction is placed in a translucent encapsulation
and the energy given off is now visible as light. This component is termed a ‘Light
Emitting Diode’ (LED) and the colours available include red, green, amber, yellow
and blue. For this radiation to take place the LED need to be forward-biased. There
are however dual-colour LED’s available where forward-bias produces one colour and
reverse-bias produces another colour.
11.4.2.1Circuit symbol and characteristic curve
The IEC circuit symbol, construction and characteristic curves are illustrated in figure
11.10 (a), (b), (c) and (d).
195
Figure 11.10 (a) illustrates the IEC circuit symbol and figure 11.10 (b) illustrates the
physical construction of an LED. Figure 11.10 (c) illustrates the characteristic curve for
an LED and it must be noted that these diodes normally operate at a voltage of between
1,4 V and 1,7 V and at currents ranging from 50 mA to 100 mA. The curve illustrated
in figure 4.10 (d) indicates the intensity or brightness of the radiated light depending on
the magnitude of the forward current. The higher the forward current the more intense
the radiation is.
I
Radiant light
energy
IF
Anode
V
Cathode
VF
(a)
(c)
I
n
t
e
n
s
I
t
y
Cathode
Lumen
I
Anode
IF
(b)
(d)
Figure 11.10
11.4.2.2Application of an LED
Light emitting diodes are mainly utilised as indicating components or devices such as
seven-segment displays, dot-matrix displays or panel indicators. A seven-segment LED
display and a 7 × 5 dot-matrix are illustrated in figure 11.11 (a) and (b).
196
Special Diodes and Applications
a
f
b
g
e
c
d
Decimal points
(a)
(b)
Figure 11.11
• Seven-segment display
The seven-segment display (figure 4.11 (a)) consists of seven LED’s arranged in a figure
eight. Should the applicable segments be energised the decimal numbers from 0 to 9
can be displayed as well as a limited number of alphabetical letters. Typical applications
include the screen of your pocket calculator.
• 7 × 5 dot-matrix display
The main disadvantage of not being able to display all the alphabetical letters using a
seven-segment display is overcome by using a 7 × 5 dot-matrix (figure 4.11 (b)) display.
By applying bias to the correct LED’s any alpha-numerical display can be obtained.
A single LED can be mounted offset to the matrix which is utilised for the decimal
point when displaying numerical numbers.
11.4.2.3Circuit diagram for the application of an LED
A typical circuit diagram for use with an LED is illustrated in figure 11.12. Since LED’s
are current operated devices one is required to limit the current through the LED and
that is acquired by inserting resistor R2 in the given circuit.
197
+ VCC
LED
R2
Q1
Vi
R1
- VCC
Figure 11.12
Exercise 11.1
1.
Describe with the aid of a characteristic curve and circuit symbol the
operation and electrical characteristics of a Zener diode.
2.
You are supplied with the following components:
• A step-down mains transformer.
• Four diodes.
• Two resistors.
• One electrolytic capacitor.
• One other component.
Use the above components and draw the circuit diagram of a device that
will be able to deliver a constant 6,4 volts to a load of 150 ohm.
3.
The following values were measured in a rectifier and regulator circuit:
• Load current
=
50 mA
• Zener diode specifications
=
12 V/3 W
• dc input to regulator section
=
16 V
Use this data and determine the:
3.1 Value of the series resistor;
3.2 Current magnitude through the Zener diode; and
3.3 Value of the load resistor.
You are now further required to draw a neat labelled circuit diagram of
this regulator section and insert all relevant values that were supplied and
calculated.
4.
Draw the IEC circuit symbol and then describe in your own words the difference between a photo-sensitive component and a photo-emissive
component.
198
Special Diodes and Applications
5.
A device can be controlled by using a photo-diode which can cause the device to be controlled to either switch ‘on’ or ‘off ’. Draw two neat labelled circuit diagrams and explain the operation thereof in detail.
6.
Explain with the aid of the characteristic curves the operation of a photo-
emissive component.
7.
Explain in your own word why certain semi-conductor material emits light in the visible spectrum.
8.
What advantage does the 7 × 5 dot-matrix display have over the seven-
segment display?
9.
Show by means of a neat labelled circuit diagram how an LED can be protected against excessive current.
TRANSISTORS
199
CHAPTER
12 Transistors
Learning Outcomes
On completion of this module you will be able to:
•
•
•
•
•
•
•
Describe the operation of a transistor with the aid of schematic block diagrams and
circuit symbols;
Understand the concept of transistor voltages;
Calculate the values of:
– Base currents;
– Collector currents; and
– Emitter currents.
Understand and explain the transistor as an amplifier and switching device taking
note of the areas of operation as well as the switching time of a transistor;
Explain the characteristics and the basic operation of a transistor as a:
Switch;
– Basic common emitter amplifier;
– Basic common collector amplifier; and
– Basic common base amplifier.
Do development of a common emitter amplifier, fixed bias and automatic bias
amplifiers; and
Do the graphical analysis of a common emitter amplifier.
12.1 Introduction
The name transistor is derived from the term ‘transfer resistor’ and you will notice that
the parts of the words printed in Italics spells out the word transistor. The principle of
operation of the transistor is an extension of the PN-theory dealt with when the diode
was discussed.
The transistor is a three-terminal two junction component and is commonly referred
to as a ‘junction transistor’ but it must be noted that there are other types of transistors
also available but will be covered at a later stage.
200
TRANSISTORS
12.2 The transistor
The junction transistor consists of two types of extrinsic or doped semi-conductor
material. Since it contains only two junctions it however has three terminals and
one type of doped material (N- or P-type) is sandwiched between two types of the
other type of doped material (N- or P-type). This arrangement provides us with the
opportunity of obtaining two types of transistors namely an NPN- or PNP transistor.
Their basic operation is virtually the same but their bias arrangement is different.
The schematic block diagrams and IEC circuit symbols are illustrated in figure 12.1 (a)
and (b). Figure 12.1 (a) illustrates an NPN transistor and figure 12.1 (b) illustrates a
PNP transistor. Take note of the voltage polarities indicated on the IEC circuit symbols
since this will be your guide for the future in electronics.
It must also be noted, and this is a very important aspect, that when we talk about
current flow, that we refer to electron flow and that is from negative to positive. We
will further refer to hole flow which is the opposite to electron flow, from positive to
negative. It may not be correct to refer to hole flow as conventional current flow.
Emitter
Base
Collector
Emitter
Base
Collector
N
P
N
P
N
P
Emitter-base
junction
Collector-base
junction
Emitter-base
junction
VCE
Emitter
VCE
Collector
-
Emitter
++
Base
Collector
+
+
VBE
Collector-base
junction
--
VCB
VBE
(a)
Base
VCB
(b)
Figure 12.1
In order to grasp and understand the operation of a transistor we need to familiarise
ourselves with the concept of the depletion layers again. This is illustrated in figure 12.2
(a) and (b).
201
Emitter
Base
Collector
Emitter
Base
Collector
N
P
N
P
N
P
Depletion layers
(a)
(b)
Figure 12.2
The following points need to be taken note of.
•
•
•
The base region is much thinner than the two outer regions.
The two outer reasons are more heavily doped than the base region.
Because of the doping level of the two outer regions the depletion layer at both
junctions will penetrate deeper into the base region.
This deeper penetration of the depletion layers at both junctions therefore
minimises the distance between the two junctions.
•
These are the conditions that will exist when there is no voltage supply connected to
the transistors. Should we now connect a voltage supply to the transistor a number of
significant and important changes will take place. This concept is illustrated in figure
12.3 (a) and (b).
IE
+
-
Forward bias
IB
Emitter
+
Reverse bias
+
IE
-
+
Forward bias
IC
Collector
IB
Emitter
-
Reverse bias
IC
Collector
Holes emitted
N
P
N
P
Base
N
P
Base
Depletion layers
(a)
(b)
Figure 12.3
In both transistors the emitter-base is forward-biased and the collector-base is reversebiased. This is a fundamental condition to exist should a transistor be used as an
amplifier. This will be further explained when the operating areas of a transistor is
discussed. (Refer to the section on PN-junctions since the transistor may be seen as two
PN-junctions connected back-to-back).
202
TRANSISTORS
You will notice that for both types of transistors the current directions are indicated
to be in the same direction but there is a subtle difference. In the NPN transistors we
have a current flow which is by means of electron flow as majority carriers. However, in
the PNP transistors we have a current flow which is by means of hole flow as majority
carriers.
Important to note is that the width of the depletion layers have changed since we added
the supply voltages. It would therefore appear that the width will have an influence on
the conduction properties but this is not the case. Should you refer back to figure 12.1
(a) and (b) that in the case of the NPN transistor the high positive potential will pull
the electrons through the base region and in the case of the PNP transistor the high
negative potential will pull the holes through the base region. Note that the current
directions are indicated as external direction.
The current magnitudes for both transistors are given by the following mathematical
expression:
=
IB + IC
where IE
=
emitter current in ampere
IE
=
base current in ampere
IB
=
collector current in ampere
IC
Important to note is that the magnitude of the base current will seldom exceed ± 2% of
the emitter current. To obtain a clearer view of the above mathematical expression refer
to figure 12.4 (a), (b) and (c).
You will recall the transistor is derived from ‘transfer resistor’ and we can now apply
Ohm’s Law to this schematic diagram.
A resistor variable R, acting as the base, is placed between the two semi-conductor
materials and we will find that the current between the emitter and collector will be
regulated by the variable resistor R. Should we now vary the resistance of the variable
resistor R the electrons that will be emitted will not be the same as those collected
by the collector. The resistance of R will thus have an influence on the magnitude of
the current flow. We can therefore safely assume that the conduction of a transistor is
mainly dependant upon the increase and decrease of current between the base-emitter.
It will be found that the higher the current between the base-emitter junctions the lower
the resistance of R and the lower the current between base-emitter junctions the higher
the resistance of R.
We can therefore safely assume that the collector current is controlled by the base
current and the collector current is almost equal to the emitter current.
203
Base
Emitter
R
NPN
Collector
C
IC
B
Electron or hole flow
IE
IE
VS
(a)
C
IC
B
IB
Amperemeter
PNP
E
IB
IE
IE
E
Electron flow
Hole flow
(b)
(c)
Figure 12.4
Example 12.1
It was found by measurement that the current flow in the emitter of a
transistor is 80 mA and that causes a current flow of 0,8 mA in the base of the
transistor. Determine the magnitude of the collector current.
Solution:
IE
IC
=
=
=
=
IB + IC
IE - IB
80 × 10-3 - 0,8 × 10-3
79,2 mA
12.3 Operating regions of a transistor
There are three regions in which a transistor is normally operated and that is illustrated
in the graphical representation in figure 12.5.
204
TRANSISTORS
IC
Saturation region
IB
IB
IB
Active region
IB
IB
Cut-off region
VCE
Figure 12.5
Three main operating regions can be identified for a transistor as illustrated in figure
12.5. The output characteristics illustrated is that for the common emitter configuration
in which most transistor applications are used. Important to note is that these three
regions are termed ‘static characteristics’ since they are determined by dc conditions
which are static.
There are also another condition which will be covered later namely ‘dynamic
characteristics’ which takes into consideration and ac (signal) application which on the
other hand is a changing quantity.
• The cut-off region
The cut-off region is thát region where the emitter-base as well as the collector-base
junctions is reverse biased. There is no major current that will flow besides a minute
leakage current and the transistor is ‘off ’ since there is no collector current that can flow.
• The active region
The active region is thát region where the emitter-base junction is forward biased and
the collector-base junction is reverse biased. Majority carriers, electrons or holes, are
allowed to flow as a current. The transistor is now ‘on’ and typical voltage values that
will be measured are:
• ± 0,7 V between emitter and base for silicon transistors;
• ± 0,3 V between emitter and base for germanium transistors; and
• ± 20 V
between collector and base and may in some instances even exceed this
value. In this region a transistor may be used as an amplifier.
205
• The saturation region
The saturation region is thát region where the emitter-base as well as the collectorbase junctions is forward biased. The transistor is conducting heavily and the voltage
between collector and emitter is less than 0, 6 V. In this region a transistor may be used
as a switch.
12.4 The switching speed of a transistor
The switching time of a transistor is a very important characteristic. In order to explain
this concept we need to study the circuit diagram and waveforms illustrated in figure
12.6 (a) and (b).
On
Base current
Off
+ VCC
90 %
IC
RC
Collector current
Q1
IB
10 %
- VCC
(a)
td
tr
ts
ton
tf
toff
(b)
Figure 12.6
Like may things in life response is not always immediate and when an input current is
applied to the base of a transistor it does not respond immediately since the electrons
have, what is termed a ‘transit time’, to move across the junction as well as the junction
capacitance to overcome. This time delay is the time it takes between the applications
of the base current and the instance that the collector current starts flowing. This is
indicated on the graphical representation as td and is termed the delay time. However,
even if the current flow commences the collector current does not reach maximum
value immediately and a time period lapses termed the rise time. This rise time tr may
be defined as the time it takes for the collector current to rise from 10% of its maximum
value to 90% of its maximum value. It should be obvious that the time taken for the
collector current to reach 90% of its maximum value is the sum of the delay time td and
the rise time tr which gives us a total time ton as illustrated in figure 12.6.
206
TRANSISTORS
Should the input base current now be removed from the transistor will not switch off
immediately. The collector current will only cease flowing once the turn-off time toff
has lapsed. This turn-off time also consists of two time periods namely the storage
time ts and the fall time tf and these times may be defined in the following manner.
Storage time is thát time that lapses since the majority charge carriers are trapped in
the depletion layer of the junction when the junction polarity is reversed. The fall time
on the other hand is thát time taken for the collector current to reduce from 90% of
its maximum value to 10% of its maximum value. These times are clearly indicated
on the graphical representation in figure 12.6. It must be noted that these time delays
mentioned, ton and toff are in the order of micro-seconds and even nano-seconds for fast
switching transistors.
12.5 The transistor as a switch
A transistor can be utilised as an electronic switch by merely applying the principles
that we have discussed up to this point. We have seen the effect that the base current
IB has on the operation of a transistor and this will be the principle that we will use to
operate a transistor as an electronic switch.
Such a circuit diagram of a transistor used as an electronic switch is illustrated in figure
12.7.
Supply voltage to device or
load to be controlled
+ VCC
N/C
( on )
D1
N/O
( off )
Relay coil
and contacts
Q1
Battery
S1
- VCC
Figure 12.7
Device or
load to be
controlled
207
The circuit is illustrated with the switch S1 as well as the relay contacts in the N/O (off)
position. The load, which could for instance be a borehole pump, is also off. When
switch S1 is activated a high positive voltage will appear on the base of transistor Q1 and
the transistor will start conducting heavily which will pull the relay contacts into the
N/C (on) position, provided that the holding current value of the relay is exceeded, and
the remote load will also switch on. Please note that the load to be controlled must have
its own supply.
In order to protect the transistor when switch S1 is de-activated a diode D1 is placed in
parallel with the relay coil. The reason for this is that during the time that the transistor
is conducting there will be a current flowing through the relay coil which will keep
the relay contacts in the N/C (on) position and a magnetic field is set up around the
relay coil. At switch-off the magnetic field around the relay coil will collapse causing a
back-emf which could damage the transistor. During this period of back-emf the diode
will conduct thus short-circuiting the relay coil and no damage will be done to the
transistor.
12.6 The transistor as an amplifier
By virtue of a transistor having three terminals, emitter, base and collector, will allow us
to use a transistor in three configuration or modes as an amplifier. But we first need to
clarify the term amplifier. An amplifier may de defined as an electronic device that will
increase the value of whatever is fed into the device. The three configurations or modes
of amplifier operation are:
•
•
•
Common emitter;
Common base; and
Common collector (emitter follower).
These amplifiers are termed common on the grounds of reference to the terminal which
is common to both the input as well as the output. In the case of the common emitter
the emitter is common to the input as well as the output, in the common base the base
is common to the input as well as the output, in the common collector the collector is
common to the input as well as the output. Each configuration is unique having its own
characteristics as well as application which will be dealt with in this section. Only the
very basic configurations will be dealt with and a more detailed approach will follow in
latter sections.
12.6.1
The common emitter amplifier
The circuit diagram in figure 12.8 illustrates a transistor configured in the common
emitter mode.
208
TRANSISTORS
+ VCC
RC
CC2
CC1
Q1
Output
Input
- VCC
Figure 12.8
The common factor, emitter and other terminals are circled with a dotted line. The
input signal is applied between the base and emitter and the output signal is obtained
between the emitter and collector and the common denominator here being the emitter.
This is the most frequently and commonly used configuration and has the following
characteristics:
•
•
•
•
•
12.6.2
High current gain;
High voltage gain;
High input impedance;
High output impedance; and
180º phase shift between input and output.
The common collector amplifiere
The circuit diagram in figure 12.9 illustrates a transistor configured in the common
collector mode.
The common factor, collector and other terminals are circled with a dotted line. The
input signal is applied between the base and collector and the output signal is obtained
between the emitter and collector and the common denominator here being the
collector. Note that the -VCC line becomes the collector under dynamic conditions.
209
+ VCC
CC1
Q1
CC2
C
Input
RE
Output
- VCC
Figure 12.9
This configuration is mainly used for matching purposes and has the following
characteristics:
•
•
•
•
•
12.6.3
High current gain;
Low voltage gain;
High input impedance;
Low output impedance; and
0º phase shift between input and output.
The common base amplifier
The circuit diagram in figure 12.10 illustrates a transistor configured in the common
base mode.
+ VCC
RC
CC2
CC1
Q1
Output
Input
- VCC
Figure 12.10
210
TRANSISTORS
The common factor, base and other terminals are circled with a dotted line. The
input signal is applied between the base and emitter and the output signal is obtained
between the base and collector and the common denominator here being the base.
This configuration is also mainly used for matching purposes and has the following
characteristics:
•
•
•
•
•
Low current gain;
High voltage gain;
Low input impedance;
High output impedance; and
0º phase shift between input and output.
All the circuit diagrams given up to this point have made use of NPN transistors but
PNP transistors may also be used. All that need to be done is to substitute the NPN
transistor with a PNP transistor as well change the +VCC line and -VCC line around. The
rest of the circuit including the input and output stay exactly the same. The only real
difference between the two circuits is the majority carriers for the different types of
transistors.
All of these characteristics mentioned here, especially those for the common emitter
and common base will be proved when the h-parameters are discussed at a later stage in
your career. For simplicity a summary of these characteristics is given in table 12.1.
Characteristics
Current gain
Voltage gain
Input impedance
Output impedance
Phase shift
Type of amplifier
Common emitter Common base
High
High High
High
High
180º
Low
High
Low
High
0º
Common Collector
High
Low
High
Low
0º
Table 12.1
12.7 Common emitter amplifier development
The first concept that needs to be discussed is that of transistor biasing since this
concept will determine the operation of the transistor. This concept is illustrated by the
schematic block diagram (a) and circuit diagram (b) in figure 12.11.
211
VCE
IE
IC
VBE
VCB
IB
IC
Emitter
IB
VCE
Collector
VBE
N
P
IE
N
Base
(a)
(b)
Figure 12.11
Two important concepts are identified in figure 12.11.
•
•
The emitter-base is forward-biased by VBE; and
The collector-base is reverse-biased by VCE.
Upon careful inspection of figure 12.11 (a) you will notice that two batteries are used in
order to obtain a voltage on the base of the transistor. Since these batteries are in series
forming a voltage divider they may be replaced by a voltage divider resistive network
and the transistor may then be operated off a single voltage source. The following two
methods of obtaining transistor biasing are widely used for a single voltage source.
12.7.1
Fixed bias common emitter amplifier
The common emitter amplifier circuit diagram illustrated in figure 12.12 is that of a
fixed bias configuration. This means that only one supply is used bias both the base and
the collector.
+ VCC
RB
RC
CC2
Q1
Vi
CC1
VBE
Figure 12.12
VCE
VO
- VCC
212
TRANSISTORS
This circuit will operate as an amplifier when and only when the base-emitter is
forward-biased and the base-collector junction is reverse-biased. In order to function
as an amplifier the collector-emitter voltage VCE must always be positive with respect to
ground. Should this not be the case and the collector-emitter voltage VCE reduces to a
fraction of the base-emitter voltage VBE the base-collector junction will become forward
biased and the transistor will then not function as an amplifier since the transistor will
then move into the saturation area of its characteristic.
To ensure satisfactory operation as an amplifier RB and RC must be so calculated so as to
prevent the transistor from going into saturation which means that the collector-emitter
voltage VCE must always remain greater than the base-emitter voltage VBE. The biasing of
the amplifier will also be dependant upon the current - or β-gain (beta) of the transistor
and for the transistor to always operate in the linear region of the characteristic the
following principle shall apply:
=
β × IB
where IC = collector current in ampere
IC
β = current gain of transistor
IB = base current in ampere
This type of fixed bias however does produce problems with stability since the collector
current IC will vary with a change in temperature because of the following factors:
•
•
•
Leakage current will double for every 10ºC change in temperature;
The base-emitter voltage VBE will decrease by 2,5mV/º C; and
The current gain β which increases with an increase in temperature.
The following mathematical expressions can be used for the design of a fixed bias
amplifier.
VCC - VCE
RC =
where
IC
VCC - VBE
RB =
IB
RC
VCC
VCE
RB
VBE
=
=
=
=
=
value of collector resistor in ohm
supply voltage in volt
collector-emitter voltage in volt
value of base resistor in ohm
emitter-base voltage in volt
213
Example 12.2
The following data for the fixed bias amplifier in figure 12.12 is given:
=
10 V
• VCC
• β
=
100
=
10 mA
• IC
=
5V
• VCE
Use the above information and calculate the resistive values of the base resistor
and the collector resistor. Use is made of a Silicon resistor and VBE may be used
as 0,7 volts.
Solution:
RC =
RC =
VCC - VCE
IC
10 - 6
10 × 10-6
= 400 k-ohm
V -V
RB = CC BE
IC =
IB
IB =
= 10 - 0,7
-6
=
0, 1 × 10 = 9,3 M-ohm
=
β × IB
IC
β
10 × 10-6
100
0,1 µA
The stability for the amplifier may be improved by adding an emitter resistor RE as well
as an emitter by-pass capacitor CE and for the calculation of these component values
one may assume that IE is approximately equal to IC and VE is equal to 1/10 of VCC. The
value of the emitter by-pass capacitor must have a reactance value of 1/10 of the emitter
resistor value. Such a circuit diagram is illustrated in figure 12.13. Therefore, for a fixed
bias amplifier with emitter resistor and emitter by-pass capacitor the component value
expressions are mathematically given by:
VE
IE
=
=
CE
=
1/10 × VCC
IC
10
2 × π × f × RE
214
TRANSISTORS
RE
=
RC
=
RB
=
VE
V
= E
IC
IE
VCC - VCE - VE
IC
VCC-VBE - VE
IB
+ VCC
RB
RC
CC2
Q1
CC1
VO
Vi
RE
CE
- VCC
Figure 12.13
Example 12.3
The following data for the fixed bias amplifier in figure 12.13 is given:
=
15 V
• VCC
• β
=
200
=
5 mA
• IC
=
6V
• VCE
• f
=
60 Hz
Use the above information and calculate the resistive values of the base resistor,
collector resistor, emitter resistor and the capacitance value of the emitter bypass capacitor. Use is made of a Silicon resistor and VBE may be used as 0,7
volts.
Solution:
VE = 1/10 × VCC
= 1/10 × 15
= 1,5 volt
215
V
VE
10
RE =E
=
CE =
IC
IE
2 × π × f × RE
10
=
1,5
1,5
2 × 3,142 × 60 × 300
=
=
5 × 10-3
5 × 10-3
= 88,4 µF
= 300 ohm
VCC - VCE - VE
IC = β × IB
RC =
IC
IB = IC
β
15 – 6 – 1,5
=
5 × 10-3
5 × 10-3
=
200
= 1, 5 k-ohm
= 25 µA
VCC-VBE - VE
RB =
IB
15 – 0,7 – 1,5
=
25 × 10-6
=
12.7.2
512 k-ohm
Voltage divider bias common emitter amplifier
The common emitter amplifier circuit diagram illustrated in figure 12.14 is that of a
voltage divider bias configuration. This means that only one supply is used to bias both
the base and the collector.
+ VCC
RB1
RC
CC2
CC1
Q1
Output
Input
RB2
RE
CE
- VCC
Figure 12.14
216
TRANSISTORS
The fixed bias amplifier discussed has a serious disadvantage in that the values of bias
current as well as voltage is dependant upon the β-gain of the transistor. Since this gain
is extremely sensitive to temperature changes especially with Silicon transistors it is
required that we design a dc-bias network which will be independent of the β-gain and
thus not so sensitive to temperature variations.
It should be mentioned that our greatest enemy with amplifier circuits is that of
temperature variations.
The use of a voltage divider network for our dc-bias arrangement overcomes to some
extent this serious problem and use is made of such a network indicated as RB1 and RB2
as illustrated in figure 12.14.
The values of the voltage divider resistors RB1 and RB2 is mathematically given by:
RB1
=
RB2 × (VCC – VB)
VB
and
RB2
=
1/10 (RE × β)
The magnitude of the base voltage VB is mathematically given by:
RB2
× VCC
where
RB1 + RB2
VB =
VB
RB1
RB2
VCC
=
=
=
=
value of base voltage in volt
resistive value of base resistor in ohm
resistive value of base resistor in ohm
supply voltage in volt
Referring to the mathematical expression it must be noted that the voltage drop across
RB2 is the voltage that will be measured on the base of the transistor.
The value of the collector load resistor RC is mathematically given by:
RC
=
VRC
where
IC
RC
IC
=
=
resistive value of load resistor in ohm
magnitude of collector current in ampere
Also VCC = VRC + VCE
where
VCC
= magnitude of supply voltage in volt
= magnitude of voltage drop across load resistor in volt
VRC
VCE = magnitude of voltage between emitter and collector in volt
Having obtained an output from the transistor we now encounter a more serious
problem termed ‘thermal runaway. It was mentioned at an earlier stage that a small
leakage current flows across the reverse biased base-collector junction and that it could
be ignored. But this is not altogether true. An increase in this leakage current causes
217
an increase in collector current which increases the temperature of the transistor
which decreases its resistance which increases the leakage current which increases the
collector current and in this manner it will snowball until such time that the transistor
base-collector junction is damaged and the transistor becomes unserviceable. Once an
amplifier comes into operation this leakage current affects the stability of the amplifier
if not countered. In order to counter this thermal runaway and keep the transistor at
a stable operating point we add to the circuit an emitter resistor as illustrated in figure
12.14.
Although the addition of an emitter resistor RE has a stabilising affect on the transistor
it causes negative feedback which will affect the gain of the amplifier which is an
unwanted situation. To counter this reduction in gain we add an emitter capacitor CE in
parallel with the emitter resistor RE as illustrated in figure 12.15. A further precaution
that can be used is to mount the transistor on a ‘heat sink’ which will dissipate a lot of
the heat generated by the current flowing through the transistor. This is often done with
power transistors.
If you compare the circuit diagram illustrated in figure 12.14 with the circuit diagram
illustrated in figure 6.8 you will notice quite a difference although the characteristics
and operation is exactly the same.
The circuit diagram in figure 12.15 illustrates an automatic bias common base amplifier.
+ VCC
RB1
RC
CC1
Q1
CC2
Output
RB2
CB
RE
Figure 12.15
Input
- VCC
The circuit diagram in figure 12.16 illustrates an automatic bias common collector
amplifier.
218
TRANSISTORS
+ VCC
RB1
RC
C
Q1
CC1
CC2
Input
RB2
RE
Figure 12.16
Output
- VCC
12.8 Common emitter graphical analysis
The common emitter has three main characteristic curves which defines its behaviour.
It will be possible to determine the transistor operation for static as well as dynamic
conditions.
12.8.1
Input characteristics
The input characteristic curve is obtained by plotting the obtained base current
IB against the base-emitter voltage VBE and this is illustrated in the graphical
representation in figure 12.17.
The input characteristic curve will allow us to calculate the static impedance ZS as well
as the dynamic impedance ZD. Before we continue we need to differentiate between the
terms dynamic and static.
219
IB (µ-ampere)
6
4
2
VB (mV)
0,3
0,4
0,5
Figure 12.17
• Static
Static conditions imply that we only consider fixed parameters such as dc-values.
• Dynamic
Dynamic conditions imply that we only consider variable parameters such as ac-values.
The static- and dynamic impedances are given by:
ΔVBE
where
ΔIB
ZS =
VBE
IB
ZD =
ZS
ZD
IB
VBE
Δ
=
=
=
=
=
static impedance in ohm
dynamic impedance in ohm
base current in ampere
base voltage in volt
difference between
Example 12.4
Use the data in the graph in figure 6.18 and determine the:
(a)
Static impedance; and
(b)
Dynamic impedance.
220
TRANSISTORS
Solution:
VBE
(a) ZS =
IB
0,4 × 10-3
=
4 × 10-6
=
100 ohm
ΔVBE
(b) ZD =
ΔIB
(0,5 - 0,3) × 10-3
=
(6 - 2) × 10-6
12.8.2
=
50 ohm
Transfer characteristics
The transfer characteristic curve is obtained by plotting the obtained base current IB as
input against the collector current IC as output and this is illustrated in the graphical
representation in figure 12.18.
The transfer characteristic gives a clearer indication that a small change in base current
IB causes a substantial change in collector current IC and since the base current IB is
our input current and the collector current IC our output current it is now possible that
we can calculate our current gain of the amplifier. Note that the collector current IC is
controlled by the base current IB and this is termed the beta (β) gain.
IC (mA)
30
20
10
IB (µ-ampere)
10
Figure 12.18
20
30
221
The static- and dynamic current gain is given by:
I
IC
ΔIO
ΔIC
AS
= O =
AD
=
=
Ii
IB
ΔIi
ΔIB
where AS = static current gain
AD = dynamic current gain
IB = base current in ampere
IC = collector current in ampere
IO = output current in ampere
Ii = input current in ampere
Δ=difference between
Example 12.5
Use the data in the graph in figure 12.18 and determine the:
(a)
Static current gain; and
(b)
Dynamic current gain.
Solution:
IC
(a) AS =
IB
30 × 10-3
=
20 × 10-6
=
1 500
ΔIC
(b) AD =
ΔIB
(50 - 10) × 10-3
=
(30 - 10) × 10-6
12.8.3
=
2 000
Output characteristics and dc-loadline
The output characteristics are obtained by plotting the collector current IC to the
emitter-collector voltage VCE and also indicating the base current IB and this is
illustrated in the graphical representation in figure 12.19. In order to get a clearer
222
TRANSISTORS
picture of the output characteristic curve we need to use a circuit with this curve since
we need to draw a dc-loadline on the graph and then analyse the quantities. Refer to
figure 12.20.
IC
(mA)
IB
IB
IB
µ-ampere
IB
IB
VCE
Figure 12.19
We now need to draw the dc-loadline on the graph in figure 12.19. This dc-loadline
follows the equation y = mx + c, which is a straight line and only two points need to be
plotted.
Point A VCE =
VCC - IC × RC
Should the transistor not be conducting, that is, no signal is being applied, then IC = 0.
VCE
= 20 - 0
= 20 volt
VCC
Point B IC =
R
C
20
=
10 × 103
=
2 mA
223
+ 20 V
RB1
RC
10 k-ohms
CC2
Q1
CC1
Output
Input
RB2
RE
- 20 V
Figure 12.20
These two points A and B can now be plotted on the graphical representation of the
output characteristic curve illustrated in figure 12.21.
IC
(mA)
Point B
IB
2
IB
IB
1
IB
dc bias point
(Q-point)
IB
Point A
VCE
10
20
Figure 12.21
On this dc-loadline a point is now selected as the dc-bias point or Q-point and must
indicate the collector current IC and collector-emitter voltage VCE when no signal
is applied. Should a signal now be applied the base current IB will vary causing the
collector current IC to also vary according to the amplitude of the input signal resulting
in an output voltage variation across the collector resistor RC. This concept is illustrated
in figure 12.22.
224
TRANSISTORS
2
90
1,8
80
1,6
70
1,4
1,2
Collector
output
current
60
Base input
current
50
1
40
0,8
30
0,6
20
0,4
10
0,2
0
2
4
6
8
10
12
14
16
18
IB (µ-ampere)
20
Collector output
voltage
Figure 12.22
The dc-operating or Q-point in this instance is taken at 10 volt. But how will this point
be determined. Refer back to the mathematical expression of:
RB2
VB
=
R + RB2
B1
× VCC
and this is the expression we will use to determine the dc-operating or Q-point on
the base of the transistor. This is a very important concept since this dc-operating or
Q-point will determine our class of amplification which will be dealt with at a later
stage. The use and interpretation of these graphical representations allow us to make
quite a number of observations and calculations concerning the amplifier. These will
now be discussed.
225
12.8.3.1
Gain
Definition 12.1 Gain
The gain of an amplifier may be defined as that capability of an amplifier to
increase the magnitude of the input signal ideally without distortion.
The output must be a replica but larger in amplitude than the input signal.
12.8.3.1.1 Current gain
The current gain has been discussed but it will be repeated here. The static- and
dynamic current gain is given by:
IO
AS =
=
I
i
where
AS
AD
IB
IC
IO
Ii
Δ
IC
ΔIO
ΔIC
AD
=
=
IB
ΔIi
ΔI
B
=
=
=
=
=
=
=
static current gain
dynamic current gain
base current in ampere
collector current in ampere
output current in ampere
input current in ampere
difference between
12.8.3.1.2 Voltage gain
The static- and dynamic voltage gain is given by:
VO
IO × ZO
ΔVO
AVS =
=
AVD
=
V
I × Zi
ΔV
i
i i
where
AVS
AVD
ZO
Zi
VO
Vi
=
=
=
=
=
=
static voltage gain
dynamic voltage gain
output impedance in ohm
input impedance in ohm
output voltage in volt
input voltage in volt
=
ΔIO × ZO
ΔIi × Zi
226
TRANSISTORS
12.8.3.1.3 Power gain
AP = VO
=
Vi
IO2 × ZO
Ii2 × Zi
where AP
=
power gain in watt
12.8.3.1.4 Decibel gain
It is more than often required to express the gain of an amplifier in decibels (dB). All
the previous expressions have no unit and is merely a figure and the dB is a convenient
method of expressing gain.
Current gain
Ai =
Voltage gain
AV =
Power gain
AP =
IO
where
PO
Ii
Pi
V
20 log O
Vi
PO
10 log
Pi
20 log
= output power in watt
= input power in watt
Example 12.6
Refer to figure 12.22 and use the indicated information to determine the:
(a)
Static current gain;
(b)
Magnitude of the input signal;
(c)
Voltage magnitude of the output signal; and
(d)
Dynamic current gain.
Solution:
IC
(a) AS =
IB
1 × 10-3
=
40 × 10-6
=
25
(b)
20 µA (p-p)
(c)
4 V (p-p)
227
ΔIC
(d) AD =
ΔIB
0,4 × 10-3
=
20 × 10-6
=
20
Example 12.7
An amplifier has the following specifications:
• Input power
=
3 500 watt
• Output power
=
1 000 watt
• Input current
=
200 ampere
• Input voltage
=
200 volt
• Output impedance
=
20 ohm
Use the above data and determine in dB the:
(a)
Current gain;
(b)
Voltage gain; and
(c)
Power gain.
Solution:
I
(a) Ai
=
20 log O
PO
=
Ii
=
I2O
50
=
20 log
200
=
=
- 12,04 dB
=
(b) AV =
=
=
(c) AP
=
=
=
VO
VO
Vi
1 000
20 log
200
20 log
13,979 dB
PO
Pi
1 000
10 log
3 500
10 log
- 5,44 dB
=
=
=
I2O × ZO
PO
ZO
1000
20
50 ampere
IO × ZO
50 × 20
1 000 volt
228
TRANSISTORS
An interesting point to note here is the minus sign in front of the current gain and the
power gain. What does it mean? It merely means that the amplifier operates at a loss
and not a gain!
Exercise 12.1
1.
Use suitable schematic diagrams that will illustrate a:
1.1 Biased NPN-transistor; and
1.2 Biased PNP-transistor.
2.
Determine the magnitude of the base current of a transistor if it was found
by measurement that a current of 80 mA flowed from the emitter and
78,5 mA was measured in the collector of the transistor.
3.
Draw a neat labelled graphical representation that will illustrate the three
regions of operation of a transistor and then give a detailed description of
the three regions with a possible application.
4.
Draw a neat labelled graphical representation that will illustrate the
switching speed of a transistor and then give a detailed description of the
times indicated.
5.
Complete the following table with reference to three amplifier configurations.
Characteristics
Current gain
Voltage gain
Input impedance
Output impedance
Phase shift
Type of amplifier
Common emitter Common base Common Collector
High
High
High
Low
0º
6.
Explain with the aid of a suitable circuit diagram the operation of a
transistor when used as a switch.
7.
Draw neat labelled circuit diagrams with input - and output waveforms of
a PNP transistor used as an amplifier in the following configurations:
7.1 Common collector;
7.2 Common emitter; and
7.3 Common base.
229
8.
Use suitable graphs to explain the following characteristics of a transistor:
8.1 Input characteristics;
8.2 Transfer characteristics; and
8.3 Output characteristics.
9.
Consider the following circuit diagram and output characteristic.
9.1 Use the available information on the circuit diagram and output characteristic curve and draw the dc loadline showing all relevant calculations.
9.2 Draw the collector output current waveform as well as the base input current waveform.
IC
(mA)
IB
2
IB
IB
1
IB
IB
VCE
10
20
30
+ 30 V
RB1
RC
15 k-ohms
CC2
CC1
Q1
Output
Input
RB2
RE
- 30 V
µ-ampere
230
TRANSISTORS
9.3
Should the amplifier have an input impedance of 300 ohm and an output impedance of 400 ohm then calculate the:
9.3.1
Static current gain;
9.3.2
Dynamic current gain;
9.3.3
Static voltage gain;
9.3.4
Dynamic voltage gain;
9.3.5
Power gain;
9.3.6
Current gain in dB;
9.3.7
Voltage gain in dB; and
9.3.8
Power gain in dB.
Amplification Classes, Coupling
Methods and Feedback
N4 Industrial Electronics
231
CHAPTER
13
Amplification Classes, Coupling
Methods and Feedback
Learning Outcomes
On completion of this module you will be able to:
•
•
•
•
Understand and explain the concept of amplification classes;
Understand and explain with the aid of circuit diagram and input- and output
waveforms:
– RC-coupling of transistors as well as frequency response;
– Transformer coupling of transistors as well as frequency response;
– Direct coupling of transistors as well as frequency response;
– Push-pull amplification; and
– Complementary symmetric amplification;
Explain using waveforms the concept of distortion and methods of overcoming
such distortion; and
Understand and explain the concept of negative feedback.
13.1 Introduction
All amplifiers fall into a specific category of amplification which is dependant upon
the application of the amplifier as well as the coupling method used. The class of
amplification refers to the conduction period of the output signal compared to the input
signal and there are four types of amplification classes that will be discussed.
13.2 Classes of amplification
13.2.1 Class A amplification
The graph in figure 13.1 illustrates the concept of class A amplification
232
Amplification Classes, Coupling Methods and Feedback
IC
(mA)
Output
collector
current
Base
current
VCE
dc-bias point
Input
voltage
Figure 13.1
Definition 13.1 Class A Amplifiction
Class A amplification may be defined as a process whereby an output collector
current will flow for the full 360 º of the input cycle.
The advantage of this type of amplification is that it gives a fairly distortion free output
in that the output is an almost exact replica of the input signal. The disadvantage is that
it has a very low efficiency, in the region of 25%. `
Typical application includes pre-amplifiers and general purpose amplifiers. Studying
figure 13.1 one can see that the transistor is biased in the linear part of the output
characteristic on the dc-loadline.
13.2.2
Class B amplification
The graph in figure 13.2 illustrates the concept of class B amplification.
N4 Industrial Electronics
IC
(mA)
Output
collector
current
Base
current
VCE
dc-bias point
Input
voltage
Figure 13.2
Definition 13.2 Class B Amplifiction
Class B amplification may be defined as a process whereby an output collector
current will flow for only 180º of the input cycle.
The advantage is that it has a higher efficiency if compared to class A, in the region of
78,5 %. Typical applications include power amplifiers. Studying figure 13.2 one can see
that the transistor is biased at cut-off on the output characteristic on the dc-loadline.
13.2.3
Class AB amplification
The graph in figure 13.3 illustrates the concept of class AB amplification.
Definition 13.3 Class AB Amplifiction
Class AB amplification may be defined as a process whereby an output collector
current will flow for more than 180º but less than 360º of the input cycle.
233
234
Amplification Classes, Coupling Methods and Feedback
IC
(mA)
Output
collector
current
Base
current
VCE
dc-bias point
Input
voltage
Figure 13.3
The efficiency is also quite high and is often used in the push-pull configuration of
amplification which will be dealt with later. Studying figure 13.3 one can see that the
transistor is biased above cut-off on the output characteristic on the dc-loadline.
13.2.4
Class C amplification
The graph in figure 13.4 illustrates the concept of class C amplification.
IC
(mA)
Output
collector
current
Base
current
VCE
Input voltage
dc-bias point
Figure 13.4
N4 Industrial Electronics
235
The efficiency is very high and typical applications include waveform shaping circuits
used in television receivers as well as oscillator circuits. Studying figure 13.4 one can see
that the transistor is also biased beyond cut-off on the output characteristic on the dcloadline.
Definition 13.4 Class C Amplifiction
Class C amplification may be defined as a process whereby an output collector
current will flow for less than 180º of the input cycle.
13.3 Amplifier coupling methods
It is sometimes required to connect two or more transistors in cascade in order to
increase the gain since one transistor may not supply the required gain on its own.
There are several methods that may be employed and they will now be discussed.
13.3.1
RC-inter-stage coupling
An RC (resistive-capacitive) inter-stage coupling is illustrated in figure 13.5.
+ VCC
RC1
RB1
RC2
RB3
CC2
CC3
Q1
CC1
Q2
RB2
RB4
RE1
CE1
RE2
CE1
- VCC
Stage 1
Stage 2
Figure 13.5
236
Amplification Classes, Coupling Methods and Feedback
Two common emitter amplifiers are connected in cascade (series) by means of RCcoupling compromising of RC1 and CC2 which are enclosed in dotted lines in figure 13.5.
The coupling from stage 1 to stage 2 is by means of CC2 which also serves to isolate the
dc-biasing between the two stages from one another.
Also of importance is the frequency response of an amplifier. All amplifiers are
designed to cover a pre-determined range of frequencies and this response is given by
the frequency response curve for RC-coupled amplifiers which is illustrated in figure
13.6. All frequency response curves are interpreted in the same manner and it does not
matter whether we have a single stage or more than one stage.
The bandwidth, between fLCO (lower cut-off frequency) and fHCO (higher cut-off
frequency), is that range of frequencies that will be reproduced by the amplifier in a
fairly constant manner. You will notice that the bandwidth is taken at a point termed the
half-power points. That is when the output is at its RMS-value and is also termed the
-3 dB point. Also important to note is that the frequency is indicated on a logarithmic
scale and not a linear scale since it will be rather difficult to fit such a scale.
VO
0,707 or - 3 dB
(half-power
points)
Bandwidth
100 Hz
1 kHz
10 kHz
100 kHz
f
fHCO
fLCO
Figure 13.6
13.3.2Direct inter-stage coupling
A direct inter-stage coupling is illustrated in figure 13.7. Two common emitter
amplifiers are connected in cascade (series) by means of direct coupling which are
enclosed in dotted lines in figure 13.7. A simpler direct coupled amplifier is illustrated
in figure 13.8
N4 Industrial Electronics
+ VCC
RC1
RC2
RB1
RB3
CC2
Q1
CC1
Q2
RB2
RB4
RE1
CE1
CE1
RE2
- VCC
Stage 1
Stage 2
Figure 13.7
+ VCC
RB
Q2
Q1
RE
- VCC
Figure 13.8
Another form of direct coupling, the Darlington pair is illustrated in figure 13.9
237
238
Amplification Classes, Coupling Methods and Feedback
+ VCC
RB
Q1
CC
Darlington
pair
Q2
RE
- VCC
Figure 13.9
The latter two applications of direct coupling are normally associated with low
frequency applications. The dc-levels are related to one another for the complete circuit
and the bias arrangement is so designed to cater for that. All variations in bias voltages
will affect both transistors which will have an effect on the stability of the amplifier as
a whole. The advantage of the Darlington pair is that it is an extremely good current
amplifier and is available in a single package.
Again the bandwidth, between fLCO (lower cut-off frequency) and fHCO (higher cut-off
frequency), is that range of frequencies that will be reproduced by the amplifier in a
fairly constant manner. Again it is important to note is that the frequency is indicated
on a logarithmic scale and not a linear scale. The frequency response curve for a direct
coupled amplifier is illustrated in figure 13.10.
VO
0,707 or - 3 dB
(half-power
points)
Bandwidth
100 Hz
fLCO
Figure 13.10
1 kHz
10 kHz
100 kHz
fHCO
f
N4 Industrial Electronics
239
13.3.3Transformer inter-stage coupling
A transformer inter-stage coupling is illustrated in figure 13.11.
Two common emitter amplifiers are connected in cascade (series) by means of
transformer coupling which are enclosed in dotted lines in figure 13.11. Important to
note is the phase shifts that takes place.
+ VCC
T1
RC1
RB1
RB3
CC2
Q1
CC1
Q2
RB2
RB4
RE1
CE1
CE1
RE2
- VCC
Stage 1
Stage 2
Figure 13.11
The frequency response curve for a transformer coupled amplifier is illustrated in figure
13.12.
VO
0,707 or - 3 dB
(half-power
points)
Bandwidth
100 Hz
1 kHz
10 kHz
100 kHz
fHCO
fLCO
Figure 13.12
f
240
Amplification Classes, Coupling Methods and Feedback
Again the bandwidth, between fLCO (lower cut-off frequency) and fHCO (higher cut-off
frequency), is that range of frequencies that will be reproduced by the amplifier in a
fairly constant manner. Again it is important to note is that the frequency is indicated
on a logarithmic scale and not a linear scale since it will be rather difficult to fit such a
scale in on a linear scale. The disadvantage of this type of coupling is a poorer frequency
response and increased size and cost. We do however have the advantage that power
consumption is improved in that the transformer can be used for impedance matching
as well.
13.3.4Push-pull amplifiers
13.3.4.1Two similar transistors
In a previous section we discussed the classes of amplification and there we mentioned
that class B has a very high efficiency. This concept is often used in push-pull amplifiers
but it must be noted that the amplifier does not use class B but rather use class AB
as will be discussed at a later stage. There are two methods employed for push-pull
amplifiers and the first method is illustrated in figure 13.13.
+ VCC
RB1
T2
Q1
T1
Q2
RB2
RE
Figure 13.13
- VCC
The input transformer T1 has a centre-tapped secondary winding which will act as a
phase splitter for the transistors Q1 and Q2. The transistors Q1 and Q2 will only conduct
when their respective bases become positive with respect to their collectors. It is
therefore obvious that only one transistor can conduct at any given time.
The output transformer T2 which has a centre-tapped primary will now combine
the two halves of the signal to produce an output. Worthy of note is that the output
N4 Industrial Electronics
241
transformer T2 also acts as impedance matching. The resistors RB1 and RB2 are so chosen
that the amplifier operates in class AB operation to minimise distortion. That means
that before one transistor switches off the other transistor already switches on and in
this manner cross-over distortion is minimised or in some cases negated.
A different circuit may also be utilised as a phase splitter and such a circuit is illustrated
in figure 13.14.
+ VCC
RC
Output 1
Q1
Output 2
RE
- VCC
Figure 13.14
This particular circuit has its own advantages and disadvantages in that should RC and
RE have the same resistive values then the magnitude of the two outputs will be exactly
the same. The problem however is that the output impedance from the collector is high
and the output impedance from the emitter is low which will cause a mismatch to the
two transistors of the push-pull amplifier.
13.3.4.2
Complementary pair
The complementary pair is also termed a complementary symmetric push-pull
amplifier and is illustrated in figure 13.15.
242
Amplification Classes, Coupling Methods and Feedback
+ VCC
RB1
Q1
CC1
RV1
CC3
CC2
Q2
RB2
Figure 13.15
- VCC
In this type of circuit use is made of an NPN- and a PNP transistor. The NPN transistor
Q1 will amplify only the positive half of the input cycle to it whereas the PNP transistor
Q2 will only amplify the negative half of the input cycle to it. These two halves are
then combined to give an output. The base biasing resistors RB1 and RB2 are once again
so chosen so as to provide for class AB amplification thereby minimising cross-over
distortion.
13.4 Cross-over distortion
Cross-over distortion falls into the category of distortion and distortion may be defined
as a condition that will occur when the output waveform is not an amplified version
of the input waveform. This is particularly so with push-pull amplifiers since the two
halves of the input waveform is amplified separately and then need to be put together
again and it is here where the problem comes in. We have already discussed what need
to be done to overcome this problem, that is selecting the base bias resistors in such
a fashion that the switching times of the two transistors actually overlap. Thus, when
one transistor is still conducting the other transistor already switches on before the
other transistor switches off. The graphical representation in figure 13.16 illustrates the
concept of cross-over distortion.
N4 Industrial Electronics
Cross-over
distortion
Figure 13.16
13.5The transistor as a switch
As can be seen in figure 13.16 there is a distortion of the waveform as it approaches the
zero volt line. The desired output is that indicated by the dotted line on the graphical
representation.
13.6Feedback
Definition 13.5 Feedback
Feedback can be defined as a process whereby a part of the output signal is fed
back to the input in anti-phase so as to stabilise the gain of the amplifier or similar
circuit.
This process is termed negative feedback. In order to derive at an expression for
negative feedback you are referred to the illustration in figure 13.17 (a) and (b).
243
244
Amplification Classes, Coupling Methods and Feedback
Vi
VS
VO
Amplifier
Vf
Feedback network
(Beta network)
(a)
+
=
Vf
VS
VO
(b)
Figure 13.17
Figure 13.17 (a) illustrates and amplifier with a feedback network Beta. A portion of the
output waveform VO is fed back in anti-phase to the input waveform VS and is termed
Vf. In this manner the input waveform Vi to the amplifier is reduced in that Vi = VS Vf which results in the overall gain of the amplifier stage being reduced and kept at a
constant level. This can be observed from the waveforms illustrated in figure 13.17 (b).
The gain with feedback of an amplifier is mathematically given by:
A
where
1–A×β
A’
=
A’ =
A =
β =
overall gain with feedback
overall gain without feedback
feedback fraction
The product A is termed the loop gain or feedback factor and when this factor is
negative we have achieved negative feedback. Important to note that negative feedback
is also termed de-generative feedback.
N4 Industrial Electronics
Example 13.1
An amplifier has an open loop gain of -50 with no feedback. Should a negative
feedback fraction of 0, 04 be applied to the amplifier determine what the new
gain would be.
Solution:
A
A’ =
1-A×β
- 50
=
1 - (- 50 × 0,04)
= - 16,667
13.6.1Types of feedback
Although feedback is a very important concept we will not go into detail with circuit
description but merely investigate such feedback by means of types of feedback and
block diagram descriptions.
13.6.1.1Series voltage feedback
It should be mentioned that our greatest enemy with amplifier circuits is that of
temperature variations.
A block diagram for series voltage feedback is illustrated in figure 13.18.
Vi
Amplifier
VO
Figure 13.18
13.6.1.2Shunt (parallel) voltage feedback
A block diagram for shunt (parallel) voltage feedback is illustrated in figure 13.19.
245
246
Amplification Classes, Coupling Methods and Feedback
Amplifier
Vi
VO
Figure 13.19
13.6.1.3Series current feedback
A block diagram for series current feedback is illustrated in figure 13.20.
Vi
Amplifier
VO
Figure 13.20
13.6.1.4Shunt (parallel) current feedback
A block diagram for shunt (parallel) current feedback is illustrated in figure 13.21.
Vi
Amplifier
Figure 13.21
VO
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247
Worthy of note concerning the effects of the different types of negative feedback
includes:
•
•
Series feedback will tend to increase the input impedance to the amplifier;
Shunt (parallel) feedback will tend to decrease the input impedance to the
amplifier;
Voltage feedback will tend to decrease the output impedance to the amplifier; and
Current feedback will tend to increase the output impedance to the amplifier
•
•
The effects of negative feedback will lead to:
•
•
•
•
•
Reduced but stabilised gain;
Reduction in frequency distortion;
Reduction of noise generation particularly in the early stages of amplification;
Reduction of non-linear distortion; and
Increase in bandwidth particularly in the higher frequency band.
Exercise 13.1
1.
Describe with the assistance of waveforms and applicable descriptions:
1.1 Class A amplification;
1.2 Class AB amplification
1.3 Class B amplification; and
1.4 Class C amplification.
Mention at least one application for each class of amplification.
2.
An amplifier has a feedback fraction of 0,02 and an open loop gain of -80.
Determine the gain should negative feedback be applied to the amplifier.
3.
Two common emitter amplifier stages using PNP transistors need to be
connected in tandem so as to increase the gain. Draw neat labelled circuit
diagrams with waveforms of:
3.1 Transformer coupling;
3.2 Direct coupling; and
3.3 RC-coupling.
4.
Draw neat fully labelled graphic representations of the frequency response
of:
4.1 Transformer coupling;
4.2 Direct coupling; and
4.3 RC-coupling.
5.
Explain using a suitable circuit diagram what is meant by a Darlington
pair.
248
Amplification Classes, Coupling Methods and Feedback
6.
Draw neat labelled circuit diagrams of a push-pull amplifier using:
6.1 Complementary symmetry; and
6.2 Two similar resistors.
Indicate on both circuit diagrams all the relevant waveforms.
7.
Phase splitting required for push-pull amplifiers may be obtained from
a centre-tap transformer. A transistor may also be used but has its own advantages and disadvantages. Draw a neat labelled circuit diagram using an NPN transistor and indicate the input- and the output waveforms. What
is the main disadvantage of this circuit?
8.
Push-pull amplifiers suffer from phenomena termed cross-over distortion.
Use an applicable sketch to illustrate this concept and indicate how this problem can be overcome.
9.
Use a suitable block diagram, waveforms and mathematical expression that will illustrate the concept of negative feedback.
10. Negative feedback can either be series or parallel, current or voltage. Draw neat labelled block diagrams that will illustrate:
10.1 Parallel current feedback;
10.2 Series voltage feedback;
10.3 Series current feedback; and
10.4 Parallel voltage feedback.
11. Negative feedback can either be series or parallel, current or voltage but will
have an effect on the input- or output impedance of the amplifier. Mention these effects.
12. Mention five effects negative feedback will have on the characteristics of an amplifier.
Hybrid-Parameters
N4 Industrial Electronics
249
CHAPTER
14 Hybrid-Parameters
Learning Outcomes
On completion of this module you will be able to:
•
•
•
•
•
•
Explain using a simple sketch what is meant by matching of amplifier circuits;
Convert an NPN and PNP transistor to a two-port device in the:
– Common emitter mode; and
– Common base mode.
Clarify the meaning of the following h-parameters:
– hi;
– ho;
– hr; and
– hf.
Explain the subscript of ‘e’ and ‘b’.
Convert common base and common emitter amplifier circuits to an equivalent
circuit;
Do calculations using the approximate method the:
– Input impedance;
– Reverse voltage gain;
– Forward current gain and;
– Output conductance for common emitter and common base amplifiers.
14.1 Introduction
Any transistor has a relation to current, voltage and impedance and is very often
referred to the parameters of the transistor. In the previous module a number of
activities were done with reference to these parameters, but was not a very accurate
of determining specific gains of an amplifier and in most cases only took into
consideration the dc-conditions of the amplifier.
The question may now be raised whether the amplifier will display the same
characteristics when a signal is applied to the amplifier. We therefore have to investigate
the operation and characteristics of a transistor amplifier when we apply ac-conditions
250
Hybrid-Parameters
and for that we need to look at the principle of ‘hybrid-parameters’ which literally
means ‘of mixed origin’ which will now include the ac-conditions for that specific
amplifier. It is of utmost importance that circuits need to be ‘matched’ to one another in
order to obtain maximum power transfer.
This matching of circuits specifically refers to the impedances to be matched. This
concept of matching is illustrated in figure 14.1
Amplifiers
Z
A
H
I
g
h
Z
B
L
o
w
Mismatch.
Input and output
impedances not the
same
Z
Z
H
I
g
h
H
I
g
h
Z
C
H
I
g
h
Matched.
Input and output
impedances the
same
Figure 14.1
According to figure 14.1 there is a mismatch of impedances between amplifiers A and
B and maximum power transfer will not take place. On the other hand the impedances
between amplifiers B and C are matched and maximum power transfer will take place.
For the purpose of this section only the common emitter and common base amplifiers
will be discussed.
Our first step will therefore be to reduce an amplifier circuit to its equivalent ac-format
and use a ‘black-box’ approach or more commonly known in electronic terms ‘a
two-port device’ in order to eliminate the dc- circuit and to allow for small signal acanalysis to be done.
14.2The transistor as a two-port device
The treatment of ac-analysis will be done in a manner that makes no distinction
between an NPN and PNP transistor. The transistor in figure 14.2 illustrates a transistor
as a two-port device in the common emitter configuration.
N4 Industrial Electronics
IC
IB
C
B
B
C
VBE
VCE
E
IB
E
E
(a)
(b)
Figure 14.2
The transistor in figure 14.3 illustrates a transistor as a two-port device in the common
base configuration.
E
IE
C
IC
E
C
VBE
B
B
VBC
IB
B
(a)
(b)
Figure 14.3
Figure 14.3 (a) illustrates an NPN transistor and figure 14.3 (b) illustrates the same
transistor as a four-terminal network. The hybrid equivalent network for the input
and output is illustrated in figure 14.4. Important to note is that the input circuit
is represented by a Thevenin equivalent circuit and the output circuit by a Norton
equivalent circuit. The complete hybrid equivalent circuit is illustrated in figure 14.5
(a) and (b) where figure 14.5 (a) is a common emitter network and figure 14.5 (b) is a
common base network.
I1
I2
hi
hr
V1
Input
hf
Figure 14.4
ho
Output
V2
251
252
Hybrid-Parameters
IC
IB
B
hi
V1
IE
C
hr
ho
IC
E
V2
C
hi
V1
hr
ho
hf
IE
V2
hf
IB
E
(a)
B
(b)
Figure 14.5
What do the different h-parameters mean? Refer to table 14.1
Parameter
hi
Meaning
input resistance
e
Common emitter
b
Common base
hr
reverse transfer voltage ratio
Common emitter
Common base
hf
forward transfer current ratio
Common emitter
Common base
ho
output conductance
Common emitter
Common base
Therefore, hie would mean: input resistance, common emitter and hib would mean: input
resistance, common base.
14.3Small signal analysis (common emitter)
All amplifiers are basically two-port devices in that it will consist of two input terminals
and two output terminals.
In order to arrive at mathematical expression for the solving of the above quantities you
are referred to the illustration in figure 14.6
I2
I1
B
RS
Z1
V1
C
hi
hr
hf
VS
E
Figure 14.6
ho
V2
Z2
ZL
N4 Industrial Electronics
The following mathematical expressions are applicable:
Δ Vin
hie
=
=
Δ Iin
Δ Vbe
Δ Ib
Δ Vin
hre
=
=
Δ Vo
Δ Vbe
Δ Vce
Δ Io
hfe
=
=
Δ Iin
Δ Ic
Δ Ib
Δ Io
hoe
=
=
Δ Vo
Δ Ic
Δ Vce
Important to note is that 1/hoe = output impedance
The best way of explaining small signal analysis is by doing a number of activities and
then to indicate the different and various approaches that need to be followed.
Example 14.1
Consider the fixed bias amplifier and obtain the equivalent h-parameter
network.
+ VCC
RB
RC
CC2
CC1
Q1
ZO
Vi
VO
Zi
- VCC
253
254
Hybrid-Parameters
Solution:
Step 1
Short circuit all dc-supplies and capacitors and obtain the equivalent network for the
amplifier.
+ VCC
RB
RC
CC2
Q1
CC1
Vi
ZO
VO
Zi
- VCC
Step 2
Equivalent network
Ii
I1
I2
B
RB
Zi
C
hie
V1
IO
hre
Z1
hfe
hoe
E
Example 14.2
Consider the following characteristics and determine:
(a)
The input impedance hie;
(b)
The output impedance 1/hoe; and
(c)
The forward current gain Ai = hfe.
V2
Z2
RC
ZO
N4 Industrial Electronics
IB (µ-ampere)
8
6
4
VB (mV)
0,2
0,3
0,4
Ic (mA)
8
6
4
Vce
4
6
Solution:
Δ Vin
(a) hie =
Δ Iin
0,2
=
4 × 10-6
= 50 k-ohm
8
255
256
Hybrid-Parameters
Δ Vo
Δ Io
(b) 1/hoe =
=
4
4 × 10-3
=
1 k-ohm
(c) Ai = hfe =
=
=
Δ Io
Δ Iin
4 × 10-3
4 × 10-6
1000
14.4Small signal analysis (common base)
All amplifiers are basically two-port devices in that it will consist of two input terminals
and two output terminals.
In order to arrive at mathematical expression for the solving of the above quantities you
are referred to the illustration in figure 14.7. All the expressions already given are also
applicable to a common base amplifier. A few subtle changes however to occur but will
be taken care of with the activities that will be done.
I1
I2
E
RS
Z1
C
hi
V1
hr
hf
VS
B
Figure 14.7
The following mathematical expressions are applicable:
Δ Vin
hib
=
=
Δ Iin
Δ Vbe
Δ Ie
Δ Vin
hrb
=
=
Δ Vo
Δ Vbe
Δ Vcb
ho
V2
Z2
ZL
N4 Industrial Electronics
Δ Io
hfb
=
=
Δ Iin
Δ Ic
Δ Ie
Δ Io
hob
=
=
Δ Vo
Δ Ic
Δ Vcb
257
Important to note is that 1/hob = output impedance
When observing figure 14.7 you will notice that it is similar to figure 14.6 with the
exception that the base and emitter terminals have changed around. This implies that
the same process that was followed to obtain the equivalent circuit for a common
emitter amplifier is followed for a common base amplifier. To further note is that for all
the quantities that need to be calculated that the ‘e’ is substituted with a ‘b’.
Example 14.3
Consider the amplifier and then obtain the equivalent h-parameter network.
Ii
CC1
Q1
RE
Vi
IO
Zi
CC2
RC
ZO
RL
VO
Solution:
Step 1
Short circuit all dc-supplies and capacitors and obtain the equivalent network
for the amplifier.
258
Hybrid-Parameters
Ii
IO
CC1
CC2
Q1
RE
Vi
RC
Zi
ZO
RL
VO
Step 2
Equivalent network
Ii
I1
E
RE
C
hib
hrb
Vi
hfb
hob
V1
Zi
Z1
IO
I2
RC
RL
V2
B
VO
Z2
ZO
ZL
Example 14.4
Consider the following characteristics and determine:
14.4.1 The input impedance hib;
14.4.2 The output impedance 1/hob; and
14.4.3 The forward current gain Ai = hfb.
N4 Industrial Electronics
Ic (mA)
8
6
4
VCE
4
6
8
IB (µA)
8
6
4
0,2
Solution:
(a) hib =
=
=
(b)1/hob =
=
=
Δ Vin
Δ Iin
0,2 × 10-6
4 × 10-6
50 ohm
Δ Vo
Δ Io
4
4 × 10-3
1 k-ohm
0,4
VBE (mV)
259
260
Hybrid-Parameters
Δ Io
Δ Iin
(c) Ai = hfb =
=
4 × 10-3
4 × 10-6
=
1000
Exercise 14.1
1.
Explain the meaning of ‘h-parameters’.
2. 2.1
You are given the following amplifier circuit. You are required to
obtain the equivalent circuit for this amplifier circuit.
+ VCC
RB
RC
CC2
CC1
Q1
ZO
Vi
VO
Zi
- VCC
2.1
2.3
Give the mathematical expression you would use to calculate the
following quantities:
2.2.1
The input impedance hie;
2.2.2
The output impedance 1/hoe; and
2.2.3
The forward current gain Ai = hfe.
You are given the following amplifier circuit. You are required to
obtain the equivalent circuit for this amplifier circuit.
Ii
CC1
Q1
RE
Vi
IO
Zi
CC2
RC
ZO
RL
VO
N4 Industrial Electronics
2.3
3.
Give the mathematical expression you would use to calculate the following quantities:
2.3.1
The input impedance hib;
2.3.2
The output impedance 1/hob; and
2.3.3
The forward current gain Ai = hfb.
The following characteristics for an amplifier are given. Determine from the
characteristics the following:
3.1 The input impedance hie;
3.2 The output impedance 1/hoe; and
3.3 The forward current gain Ai = hfe.
IB (µ-ampere)
10
8
6
VB (mV)
0,3
0,4
0,5
Ic (mA)
7
5
3
Vce
5
7
9
261
262
Hybrid-Parameters
4.
The following characteristics for an amplifier are given. Determine from the
characteristics the following:
4.1 The input impedance hib;
4.2 The output impedance 1/hob; and
4.3 The forward current gain Ai = hfb.
Ic (mA)
7
5
3
VCE
2
4
6
IB (µA)
6
4
2
0,2
0,4
VBE (mV)
263
Uni-Junction- and Field Effect Transistors
N4 Industrial Electronics
CHAPTER
15 Uni-Junction- and Field Effect Transistors
Learning Outcomes
On completion of this module you will be able to:
•
•
•
•
•
•
•
Discuss the Uni-Junction Transistor with the aid of suitable sketches indicating the
construction, IEC symbol and characteristic curve;
Explain the operation of the Uni-Junction Transistor with the aid of its equivalent
circuit;
Give a circuit diagram with output waveforms and operation for the application of
the Uni-Junction Transistor;
Calculate the frequency of operation of a Uni-Junction Transistor;
Mention the characteristics of Field Effect Transistors;
Give with the aid of suitable sketches the construction, IEC symbol and
characteristic curve of an N-channel and P-channel Junction Field Effect
Transistors; and
Give with the aid of suitable sketches the construction, IEC symbol and
characteristic curve of an N-channel and P-channel Metal Oxide Semi-conductor
Junction Field Effect Transistors.
15.1 Introduction
The Uni-Junction Transistor is mainly used in digital circuits and for the firing
circuits in SCR-Control (Silicon Controlled Rectifier). Before we can commence on
the application of a UJT as a relaxation oscillator we need to explain the principle of
operation of a UJT.
15.2 The UJT-Transistor
The construction of a UJT-transistor is illustrated in figure 15.1 (a) and consists of a
piece of N-type silicon material on which a heavily doped P-type material is attached
and is termed the emitter. The UJT is a three-terminal device but only contains one PNjunction. The IEC-circuit symbol is illustrated in figure 15.1 (b) and the characteristic
curve for a UJT-transistor is illustrated in figure 15.1 (c). The two ends of the N-type
264
Uni-Junction- and Field Effect Transistors
material are termed base 1 and base 2. During normal operation base 2 is held positive
with respect to base 1 and is reverse biased. Should the emitter voltage now increase
current will flow from the emitter to the base 1 region and the potential of the emitter
will be reduced and goes into the negative resistance area of the characteristic curve.
This is mainly caused by the reduction in resistance between the emitter and the base 1
region.
B2
Negative
resistance
region
Cut-off region
Base 2
Saturation
region
Emitter
E
P
Valley
point
N
Base 1
IE
0
B1
(a)
(b)
(c)
Figure 15.1
In order to fully understand the operation of a UJT is required that we familiarise
ourselves with an equivalent circuit thereof and this is illustrated in figure 15.2.
The silicon bar possesses a relatively high resistance when no forward bias is applied to
the emitter resulting in a small current that will flow between base 1 and base 2. When a
forward bias is now applied on the emitter-base 1 region the internal resistance between
base 1 and the emitter will decrease. This in turn causes an increase in current between
base 1 and base 2. Should a signal now be applied the voltage across the device will vary
in sympathy with the input signal. The intrinsic stand-off ratio of the UJT is taken as the
ratio of the resistance RB1/RB2.
+ B2
RB1
IE
E
VBB
D1
VE
VBE
RB2
Figure 15.2
B1
A typical UJT-transistor relaxation oscillator is illustrated in figure 15.3.
N4 Industrial Electronics
265
The output from the UJT-transistor is obtained via base 1 and base 2 as indicated on
the circuit diagram. When the UJT fires a surge of current through base 1 causes a
voltage drop across resistor R1 and a positive-going spike is produced. At the same time
a reduction in the voltage across the emitter-base 2 junction causes the current flow
towards base 2 to increase rapidly and produces a negative-going pulse from base 2
across resistor R2.
The circuit compromising resistor RE and the capacitors C1, C2 and C3 allows for a
variable frequency outputs to be obtained. These three capacitors may also be replaced
by a single variable capacitor in order to get an adjustable or variable output.
+ VBB
R2
RE
Negative going pulses
UJT
Output spikes
S1
C1
C2
C3
Positive going pulses
R1
- VBB
Figure 15.3
The main application of a UJT-transistor is for the firing circuits in power control
which will be dealt with at a later stage. The frequency of oscillation of a UJT-transistor
oscillator is mathematically given by:
f
= 1,5
where
RE × C
f =
C =
RE =
frequency of oscillator in hertz
value of capacitor (C1, C2 or C3)
value of emitter resistor in ohms
Example 15.1
The following component values for the circuit diagram illustrated in figure
15.3 are given:
=
10 k-ohm
• RE
=
5 nF
• C1
=
100 nF
• C2
=
25 pF
• C3
Determine in each instance the frequency of operation of the oscillator.
266
Uni-Junction- and Field Effect Transistors
Solution:
1,5
1,5
f
=
f =
R × C1
R × C2
E
E
1,5
1,5
=
=
3
-9
10 × 10 × 5 ×10
10 × 103 × 100 × 10-9
=
30 kHz = 1,5 kHz
1,5
f
=
RE × C3
=
1,5
10 × 103 × 25 × 10-12
=
6 MHz
15.3 Field Effect Transistors
The main difference between the bi-polar transistors and Field Effect Transistors is that
bi-polar transistors are voltage operated and that Field Effect Transistors are current
operated. Field Effect Transistors are also three terminal devices like bi-polar transistors
with two PN-junctions. The Field Effect Transistor is a channel of semi-conductor
material of which the resistance can be controlled which results in the current through
the device being controlled.
Field Effect Transistors however have the following characteristics over bi-polar
transistors:
•
•
•
•
•
•
No off-set voltage when used as a switch;
Small gain-bandwidth;
Low noise level;
Relatively immune to radiation;
Extremely high input impedance; and
Good thermal stability.
Having indicated the characteristics of the Field Effect Transistor we need to mention
that it has one major disadvantage when compared to bi-polar transistors in that it is
very susceptible when it comes to handling them in that the acid present in the human
hand may lead to damage.
Field Effect Transistors are mainly divided into Junction Field Effect Transistors (JFET)
and Metal Oxide Semi-conductor Field Effect Transistor (MOSFET).
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15.3.1
267
Junction Field Effect Transistors (JFET)
The JFET is constructed either of N-type silicon or P-type silicon. The N-type has
two P-type regions diffused into it to form the gate region whereas the P-type has two
N-type regions diffused into it to form the gate region. These concepts are illustrated in
figure 15.4 (a) which is an N-type channel with its IEC circuit symbol and characteristic
curves and figure 15.4 (b) is a P-type channel with its IEC circuit symbol and
characteristic curves. The terminals associated with a FET are termed the gate, drain
and source.
G
ID
VGS = 0 V
D
P
VGS = -1 V
S
N
D
G
VGS = -2 V
P
S
VDS
(a)
G
ID
D
N
VGS = 0 V
VGS = 1 V
S
P
D
G
VGS = 2 V
N
S
VDS
(b)
Figure 15.4
The question may now be raised on how the JFET operates. In order to do this we need
to look at the illustration in figure 15.5 (a), (b) and (c).
268
Uni-Junction- and Field Effect Transistors
D
ID
ID
D
Pinch-off
point
VGS = 0 V
IDSS
G
P
N
P
VDD
G
P
N
P
VGS = -1 V
VGS = -2 V
VGS
S
ID
(a)
S
VP
(b)
VDS
(c)
Figure 15.5
The most important factor that need to taken into account is that the correct polarities
must be maintained when connecting a JFET into a circuit. Current will be able to flow
from source to drain in the absence of no gate-source reverse bias. Should we now apply
a reverse bias across the gate-source a depletion layer will start forming around the
two P-type materials (figure 15.5 (b) and will extend into the N-type material thereby
increasing the resistance resulting in that less area for current to flow and the current
flow will decrease. Should the reverse bias increase further a point will be reached when
the current will stop flowing and a cut-off point will be reached and current will cease
flowing. This cut-off point is also termed the pinch-off point and is clearly indicated in
figure 15.5 (c).
15.3.2 Metal Oxide Semi-conductor Field Effect Transistors (MOSFET’s)
Metal Oxide Semi-conductor Field Effect Transistors (MOSFET’s) are found in two
main types namely the enhancement mode type and the depletion mode type. Worthy
to take note of is that MOSFET’s are often referred to as IGFET’s (Insulated Gate Field
Effect Transistors). The main difference between a JFET and a MOSFET lies therein that
the gate terminal is isolated from the channel which is created between the source and
the drain in the absence of a gate potential. Current will only flow when a potential is
applied to the gate.
15.3.2.1 The Enhancement Mode (MOSFET)
The channel is not part of the original construction but is obtained when a potential is
applied across the gate-source terminals for both the N-and P-channels. Figure 15.6 (a)
(b) and (c) illustrates the construction of a P-channel enhancement MOSFET (a), the
IEC circuit symbol (b) and the characteristic curve (c).
N4 Industrial Electronics
S
G
D
ID
Silicon Oxide layer
VGS = 0 V
D
N
N
P-substrate
269
Induced Nchannel
P-substrate
G
VGS = -1 V
S
VGS = -2 V
VDS
(a)
(b)
(c)
Figure 15.6
Figure 15.7 (a) (b) and (c) illustrates the construction of an N-channel enhancement
MOSFET (a), the IEC circuit symbol (b) and the characteristic curve (c).
S
G
D
ID
Silicon Oxide layer
VGS = 0 V
D
P
P
N-substrate
Induced Pchannel
N-substrate
G
VGS = 1 V
S
VGS = 2 V
VDS
(a)
(b)
(c)
Figure 15.7
15.3.2.2 The Depletion Mode (MOSFET)
In the depletion mode the channel already exists between the drain and the source
and a current will be produced when a potential is applied across the source and
drain terminals. Figure 15.8 (a) (b) and (c) illustrates the construction of a P-channel
depletion MOSFET (a), the IEC circuit symbol (b) and the characteristic curve (c).
S
G
D
ID
Silicon Oxide layer
VGS = 0 V
D
P
P
P-channel
N-substrate
G
N-substrate
VGS = 1 V
S
VGS = 2 V
VDS
(a)
(b)
Figure 15.8
(c)
270
Uni-Junction- and Field Effect Transistors
Figure 15.9 (a) (b) and (c) illustrates the construction of an N-channel depletion
MOSFET (a), the IEC circuit symbol (b) and the characteristic curve (c).
S
G
D
ID
Silicon Oxide layer
VGS = 0 V
D
N
N
N-channel
P-substrate
G
P-substrate
VGS = -1 V
S
VGS = -2 V
VDS
(a)
(b)
Figure 15.9
(c)
Exercise 15.1
1.
Explain by means of an equivalent circuit the operation of a Uni-Junction
Transistor.
2.
The following component values for the circuit diagram illustrated in
figure 15.3 are given:
=
20 k-ohm
•
RE
=
3 nF
•
C1
=
80 nF
•
C2
=
40 pF
•
C3
Determine in each instance the frequency of operation of the oscillator.
3.
Refer to figure 15.3 and discuss its operation.
4.
What is the main difference between a FET and a bi-polar transistor?
5.
Mention six characteristics that a FET possesses over a bi-polar transistor.
6.
Mention the main difference between an enhancement MOSFET and
depletion MOSFET.
POWER CONTROL
N4 Industrial Electronics
CHAPTER
16 Power Control
Learning Outcomes
On completion of this module you will be able to:
•
•
•
•
•
•
•
Give the IEC circuit symbols of:
– A Silicon Controlled Rectifier (SCR);
– A Diac;
– A Triac; and
– A Quadrac.
Explain the operating principle of an SCR by means of:
– Two transistor analogy; and
– Characteristic curve.
Explain using graphical representation where applicable:
– Conduction angle;
– Delay angle;
– Phase control;
– Cycle control;
– Cyclotronic control;
– Commutation; and
– Duty cycle.
Draw a circuit diagram with the applicable waveforms to illustrate the application
of an SCR in:
– Half-wave control;
– Full-wave control; and
– dc-circuits.
Explain using characteristic curves the operating principle of:
– A Diac;
– Triac; and
– A Quadrac.
Draw a circuit diagram with the applicable waveforms to illustrate the application
of a Diac and Triac in a speed control circuit; and
Draw a circuit diagram with the applicable waveforms to illustrate the application
of a Quadrac in a speed control circuit.
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272
Power Control
16.1 Introduction
Power control forms an integral part in the electronic as well as electric field in
modern industry today. This module will explain the concepts of different electronic
components in use today to accomplish power control.
16.2 The Silicon Controlled Rectifier (SCR)
Definition 16.1 SCR
An SCR may be defined as an ordinary diode with a control element namely the gate.
The forward current that the device will experience will only flow once the control
element (gate) has been activated. Once the gate has been activated the voltage on the
gate can be removed and the SCR will stay in conduction provided certain parameters
are adhered to. The SCR will therefore act as a high-speed switch.
An SCR may be represented by a four-layer, three junction PNPN device and by two
transistor analogy. This concept is illustrated in figure 16.1 (a), (b), (c) and (d). Figure
16.1 (a) illustrates a four-layer, three junction PNPN device and figure 16.1 (b) is
an equivalent representation of an SCR. Figure 16.1 (c) illustrates the two transistor
analogy around which the operating principle will be discussed. Figure 16.1 (d) is the
IEC circuit symbol for an SCR.
The four-layer PNPN device in figure 16.1 (a) has three junctions: J1 between P1 and N1,
J2 between N1 and P2 and J3 between P2 and N2 and has three terminals namely an anode
connected to P1, a cathode connected to N2 and the control element (gate) connected to
P2. This four layer PNPN device is now split according to figure 16.1 (b) and now forms
two transistors Q1 and Q2 of which Q1 is a PNP-transistor and Q2 is an NPN-transistor
and is illustrated in figure 16.1 (c). It is shown in figure 16.1 (c) that the gate of the
SCR is connected to the base of transistor Q2 and the emitter of transistor Q1 and the
collector of transistor Q2 is connected to the base of transistor Q1. The anode is taken
from the emitter of transistor Q1 and the cathode is taken from the emitter of transistor
Q2. This concept is termed two transistor analogy and is in most cases used to explain
the principle of operation of an SCR.
Should the device now be forward-biased, anode positive with respect to the cathode
no forward current will flow since the gate needs to be triggered by means of a positive
pulse. This positive pulse will forward-bias the base of transistor Q2 thereby causing
a current to flow in the collector of Q2 which in turn is connected to the base of
N4 Industrial Electronics
Anode
Anode
P1
P1
J1
N1
N1
N1
P2
P2
J2
Gate
Gate
P2
J3
N1
N2
Cathode
Cathode
(a)
(b)
Anode
Q1
Gate
Anode
Q2
(c)
Cathode
Cathode
Gate
(d)
Figure 16.1
transistor Q1 which will cause transistor Q1 to also switch on. The collector current of
both transistors will provide more than enough base current and both transistors will
therefore remain in a state of conduction even if the gate pulse is removed and only a
small voltage across the anode-cathode of the SCR. Under these conditions the SCR is
said to have latched.
The characteristics of an SCR may be summarised in the following fashion:
•
•
•
•
•
A positive pulse is required to switch an SCR on.
Once the SCR is conducting the gate has no further control provided that the
forward current remains above the holding current value for the device.
The SCR will continue conducting until the anode-cathode voltage is reduced to
near-zero.
An SCR is an open-circuit device and will only conduct should the gate be
triggered. It therefore behaves like an on-off switch.
The gate may be triggered by either an ac or dc source.
The characteristics of the SCR, as with most semi-conductor components, are best
described by means of a characteristic curve which is illustrated in figure 16.2.
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274
Power Control
Forward conduction
region
IF
Holding
currents IH
+V
IG1
IG2
IG3
Figure 16.2
The characteristic curve illustrated in figure 16.2 only indicates the forward conduction
characteristics. The reverse characteristics are the same as that of an ordinary diode.
Also indicated is the different gate triggering values as well as the holding current
values.
Definition 16.2 Holding Current
The holding current may be defined as that minimum value of current required to
keep the SCR in the state of conduction and should this value be reduced below the
specified value the SCR will switch off.
There is also another type of SCR available namely the Light Activated Silicon
Controlled Rectifier (LASCR) and will operate in exactly the same manner as an
ordinary SCR with the exception the SCR is triggered by means of a light source instead
of a positive voltage on the gate. The characteristics are the same as that illustrated in
figure 16.2. Figure 16.3 illustrates the IEC-circuit symbol of a LASCR.
Anode
Cathode
Gate
Figure 16.3
Another concept that must be understood is that of commutation.
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275
Definition 16.3 Commutation
Commutation is a process that will switch an SCR from its conducting state to its
non-conducting state.
This in short means it is a method to switch an SCR off and there are two methods in
which this can be accomplished.
• Line commutation
An SCR is switched off every time that the input waveform moves through zero going
negative and this is termed line commutation and line commutation may therefore be
defined as switching an SCR off by means of the input alternating supply.
• Forced commutation
In dc circuits a problem arises in that there is no alternating supply that can switch an
SCR off and the SCR must be forced to switch off. This is done by placing a capacitor
in series with the load and thereby forcing a current through the SCR in the opposite
direction of normal conduction hence the term forced commutation.
16.3 Delay angle and conduction angle
Definition 16.4 Delay Angle
The delay angle may be defined as that part of the waveform for which no
conduction will take place whereas the conduction angle may be defined as that part
of the waveform for which conduction will take place.
Consider the waveforms illustrated in figure 16.3 (a) and (b). The waveform in figure
16.3 (a) illustrates that of half-wave control and the waveform in figure 16.3 (b)
illustrates that of full-wave control. The concept of delay angle and conduction angle
is the same whether it is half-wave or full-wave. You will further notice that half-wave
only utilises only one half of the waveform whereas full-wave utilises both halves of the
waveform. In both waveforms the area indicated by A is termed the delay angle and the
area indicated by B is termed the conduction angle.15.3 Field Effect Transistors.
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Power Control
+
(a)
t
-
+
(b)
t
A
B
A
B
A
B
A
B
Figure 16.4
A point that must be noted is that once an SCR conducts that it then behaves like an
ordinary diode which will actually deliver a pulsating dc and when averaged will have a
value proportional to the conduction angle. Since an SCR could be used for controlling
the speed of a motor, drilling machine or as a light dimmer it will be obvious that the
conduction angle will determine either the speed of the motor, drilling machine or
brightness of the lamp. Having this in mind the following conclusions may be made:
•
•
The smaller the delay angle the more of the input waveform is made available
therefore the higher the voltage will be across the load. Should we therefore apply
this voltage to a motor for instance it will have a relative high speed.
The larger the delay angle less of the input waveform is made available therefore
the lower the voltage will be across the load. Should we therefore apply this
voltage to a motor for instance it will have a relative low speed.
Consider a basic circuit diagram of a half-wave SCR control and waveforms illustrated
in figure 16.4 (a) and (b).
•
•
•
The gate of the SCR is triggered by the leading edge of the train of pulses.
The delay angle and therefore the conduction angle is the same for every positive
half cycle of the input waveform.
Should you graphically add the shaded areas of the voltage across the load and the
voltage across the SCR then it will equal the input waveform.
The waveforms illustrated in figure 16.5 indicate the concept of full-wave control.
277
N4 Industrial Electronics
Input voltage
t
VSCR
Gate trigger pulses
t
VIN
VL
RL
(b)
Voltage across load
t
Voltage across SCR
(a)
t
Figure 16.5
Input voltage
t
Gate trigger pulses
t
Voltage across load
t
Voltage across SCR
t
Figure 16.6
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Power Control
16.4 SCR applications
It has already been mentioned that an SCR may be used for speed control or for a light
dimmer. No matter what application is used, the operating principle will be the same.
16.4.1
Half-wave control
The circuit diagram illustrated in figure 16.6 indicates the concept of half-wave control.
Only the circuit diagram is given and the waveforms are the same as that illustrated in
figure 16.4 (b).
During the positive half cycle of the input waveform the anode of the SCR is positive
with respect to the cathode but the SCR will not conduct until such time that the gate is
triggered. This is determined by the delay angle to the gate terminal of the SCR by the
time constant of RV1 and C. This part of the circuit is also termed the triggering circuit.
As soon as the time constant lapses a positive pulse will be applied to the gate of the
SCR via the diode D1 and the SCR will commence conduction. The main purpose of
the diode D1 is to prevent a pulse on the gate when the input cycle goes negative since
repeated negative triggering may damage the SCR.
Should the polarity of the input waveform change the anode of the SCR will now
be negative with respect to the cathode and the SCR will switch off. This process of
changing polarity will continue for as long as there is an input waveform. Note the
position of the different loads.
Universal
load
R
SCR
VIN
RV1
D
dc load
C
Figure 16.7
16.4.2
Full-wave control
The circuit diagram illustrated in figure 16.7 indicates the concept of full-wave control.
N4 Industrial Electronics
279
Only the circuit diagram is given and the waveforms are the same as that illustrated in
figure 16.5.
In order to obtain full-wave control two SCR’s are connected in parallel (back-to-back)
as illustrated in figure 16.7. During the positive half of the input waveform SCR1 is
forward-biased and will commence conduction as soon as the delay angle has been
reached. At the same time SCR2 is reversed biased and will not commence conduction.
During the negative half of the input waveform the roles will be reversed in that SCR1
becomes reverse-biased and will not conduct whereas SCR2 now becomes forwardbiased and will commence conduction as soon as the delay angle has lapsed.
Universal
load
SCR1
SCR2
Triggering
circuit
VIN
dc load
Figure 16.8
16.4.3
An SCR in a DC-circuit
The circuit diagram illustrated in figure 16.8 indicates the concept of an SCR in a dccircuit. The SCR is connected across the output terminals of a full-wave bridge rectifier
and in this manner a single SCR can be used to obtain operation from both halves of
the input waveform. The current through the SCR will be in the same direction for both
halves of the input waveform.
280
Power Control
Universal
load
D1
D2
R1
SCR
VIN
D3
D4
dc load
R2
Figure 16.9
16.5 SCR control methods
16.5.1
Phase control
Since an alternating quantity has a cycle of 360º it is possible to trigger such a waveform
at any desired angle and thereby controlling the amount of voltage applied to the SCR
and hence the load. The waveforms in figure 16.9 (a) and (b) illustrate this principle.
+
(a)
t
-
+
(b)
t
-
Figure 16.10
The waveform in figure 16.9 (a) illustrates half-wave control and the waveform in
figure 4.9 (b) illustrates full-wave control. It must however be noted that the angle for
conduction (shaded area in both waveforms) are determined by the triggering circuit.
Also to be noted is that the larger the delay angle the smaller the conduction angle and
the smaller the conduction angle the lower the voltage and hence the speed of the motor
or brightness of the lamp whatever our application is. It must however be indicated that
this type of control generates unwanted transients which is not desirable.
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281
Definition 16.5 Transient
A transient may be defined as unwanted voltage and/or current peaks which occur
at the point (angle) where the triggering takes place.
16.5.2
Cycle control
+
t
On time
Off time
On time
Figure 16.11
The waveform in figure 16.10 illustrates the principle of cycle control.
Cycle control is obtained by applying a number of full cycles to the load for a number
of pre-determined periods of time and then removing the supply for a pre-determined
period of time. Cycle control has the advantage that very little or no transients are
generated since the control takes place at 0º.
16.5.3
Cyclotronic control
The waveform in figure 16.11 illustrates the principle of cyclotronic control. Cyclotronic
control is a combination of the previous two methods discussed and is obtained by
applying phase control for a number of cycles where the conduction angle is increased
by a pre-determined amount until full cycle control is obtained where it is then kept for
a number of pre-determined cycles. The conduction angle is then decreased by a predetermined amount until such time that zero is reached. The supply is then switched
of completely for a pre-determined number of cycles and the whole process starts all
over again. Typical applications include for instance a situation where a motor has to
start running but at a very low speed with a gradual increase in speed until full speed is
reached and then needs to be slowed down again.
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Power Control
t
Phase control
Fully on
Phase control
Off
Phase control
Figure 16.12
16.5.4
Duty cycle
The waveforms in figure 16.12 (a), (b), (c) and (d) illustrate the principle of duty cycle.
The waveform illustrated in figure 16.12 (a) has a reference time of 2 seconds, whereas
the reference time in figures 16.12 (b), (c) and (d) has a duty cycle of 75%, 50% and
25% compared to that of figure 16.12 (a).
Definition 16.6 Duty Cycle
Duty cycle may therefore be defined as the ratio of on time to the reference time and
the reference time being the sum of the on time and the off time.
+
100 %
(a)
t
1
2
3
4
5
6
+
7
8
9
10
75 %
(b)
t
+
50 %
(c)
t
+
25 %
(d)
t
Figure 16.13
N4 Industrial Electronics
283
16.6 The Diac
A Diac is a two terminal bi-directional semi-conductor component which is normally
used in conjunction with a Triac. The IEC circuit symbol is illustrated in figure 16.13 (a)
and the characteristics in figure 16.13 (b). A Diac can conduct in both directions, hence
the term bi-directional meaning two, and may be seen as two diodes connected backto-back. The use of the Diac will be discussed under the next section that will deal with
the Triac.
+I
Forward bias
Conducting
region
Anode 1
Anode 2
-V
+V
Reverse bias
Conducting
region
(a)
-I
(b)
Figure 16.14
16.7 The Triac
A Triac is a three terminal bi-directional gate-controlled semi-conductor component
which is normally used in conjunction with a Diac. The IEC circuit symbol is illustrated
in figure 5.14 (a) and the characteristics in figure 16.14 (b). This component has the
advantage that it may be triggered by a positive pulse on the gate for one half of the
input waveform as well as a negative pulse on the other half of the input waveform
thereby giving us full-wave control in a single component.
+I
Forward bias
Conducting
region
Holding currents IH
MT1
MT2
-V
+V
Holding currents IH
Gate
(a)
Reverse bias
Conducting
region
-I
(b)
Figure 16.15
284
Power Control
16.8 Power control using a Diac and Triac
The circuit diagram illustrated in figure 16.15 is that of a Diac and Triac used for speed
control or a light dimmer. The waveforms are not given since it will be the same as that
for figure 16.5.
Universal
load
R
Triac
VIN
RV1
Diac
dc load
C
Figure 16.16
Using a Diac and a Triac gives us the opportunity to make use of full-wave control. The
purpose of the diac is mainly to allow a negative pulse on the gate during the negative
half of the input waveform and to allow a positive pulse on the gate during the positive
half of the input waveform.
16.9 The Quadrac
A Quadrac is a three terminal bi-directional semi-conductor component and consists
of a Diac and a Triac in one package and has the characteristics of each component on
its own. The characteristic curve of the Quadrac will therefore not be given and neither
will a circuit diagram illustrating its application be given. The application is the same
as that for figure 16.15 with the exception that a Quadrac is used as a single component
instead of a Diac and Triac as indicated. The IEC circuit symbol is illustrated in figure
16.16.
Anode
Gate
Cathode
Figure 16.17
N4 Industrial Electronics
16.10
285
Control systems
There are two types of main control systems available in industry today but it must be
mentioned that both types are not always used.
16.10.1
Open-loop control systems
In this type of control the result cannot be controlled since the output is not monitored.
The following scenario will illustrate this concept. Assume a borehole is used to fill a
water tank. The borehole pump will pump water from the borehole into the tank for
as long as the pump is activated. Since there is no indication when the tank is full the
pump will keep on pumping which will result in a waste of a very scares commodity.
The only time when the flow of water will be interrupted is when the pump is manually
deactivated.
This action can be rectified by using a closed-loop system.
16.10.2
Closed-loop control systems
Output
Comparator
Input
Quantity to
be controlled
Sensor
Reference
section
supplies a
constant
signal
Correcting signal
Feedback signal
Correcting
circuit
Amplifier
Figure 16.18
Error signal
286
Power Control
A quantity need to be controlled to alleviate the scenario that was mentioned in 16.10.1.
A sensor is used to determine the state of the output to comply with the reference we
supply. As soon as this reference is exceeded a signal will be sent to the comparator
which will compare the received signal and the reference signal and if they are different
it will generate an error signal. This error signal will be amplified and sent as a feedback
signal to the correction section which will send a correcting signal which will adjust
the quantity to be controlled to the required level in order to obtain a pre-set output
specified.
Exercise 16.1
1.
Define an SCR.
2.
An SCR may be represented by two transistor analogy in order to explain
its operating principle. Draw such a neat labelled circuit diagram and
explain the operating principle of an SCR.
3.
Use a suitable characteristic curve to illustrate the principle of operation
of an SCR and then tabulate its characteristics.
4.
Differentiate between line commutation and forced commutation.
5.
Use a graphic representation and illustrate the difference between a delay
angle and conduction angle with reference to an SCR. In your illustration
you must make reference to full-wave as well as half-wave control.
6.
Draw a neat fully labelled circuit diagram with input and output waveforms to illustrate the application of an SCR in:
6.1 Full-wave control;
6.2 A dc circuit; and
6.3 Half-wave control.
7.
Use suitable graphic representation to illustrate the principle of:
7.1 Cyclotronic control;
7.2 Phase control; and
7.3 Cycle control, and mention at least one advantage or disadvantage of
each.
8.
Use your own reference time and a square wave to explain the concept of
duty cycle.
9.
Compare the characteristics by means of a characteristic curve of a Diac
and a Triac.
N4 Industrial Electronics
10. Full-wave control may be obtained by using a Quadrac. Draw such a neat labelled circuit diagram and indicate the following waveforms:
10.1 An input waveform;
10.2 An output waveform across the load; and
10.3 An output waveform across the Quadrac.
11. Explain the difference between an open-loop control system and a closed-
loop control system.
12. Draw a neat labelled sketch of a closed-loop control system and then give a detailed description of the operation thereof.
287
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Power Control
Operational Amplifiers
N4 Industrial Electronics
289
CHAPTER
17 Operational Amplifiers
Learning Outcomes
On completion of this module you will be able to:
•
•
•
•
•
Describe the characteristics of an operational amplifier;
Draw the IEC circuit symbol and pin-layout of an operational amplifier;
Give the uses for the following operational amplifier applications:
– Inverting amplifier;
– Non-inverting amplifier;
– Summing amplifier (inverting);
– Voltage follower;
– Integrator; and
– Differentiator.
Draw the circuit diagram indicating the input and output waveforms of an
operational amplifier used as a (n):
– Inverting amplifier;
– Non-inverting amplifier;
– Summing amplifier (inverting);
– Voltage follower;
– Integrator; and
– Differentiator.
Do calculations of resistor values, capacitor values, input and output voltages for the
following operational amplifier applications:
– Inverting amplifier;
– Non-inverting amplifier;
– Summing amplifier (inverting); and
– Integrating amplifier.
17.1 Introduction
In the previous modules on transistor applications it was found that all circuits had to
be constructed using discreet components (transistors, resistors, capacitors, inductors,
diodes, etc.) and that it is quite a cumbersome effort. The operational amplifier
290
Operational Amplifiers
solves this problem to quite an extent in that an amplifier is now available in a single
integrated circuit (IC) package. Other types of IC-packages are also available but are not
covered by the scope of this module. These IC-packages are fairly cheap to manufacture
and has the following characteristics:
•
•
•
•
•
High input impedance;
Low output impedance;
A high voltage gain in the region of 106;
Wide bandwidth; and
The ability to handle both ac and dc signals.
17.2The operational amplifier
Various modes of operation can be obtained from an operational amplifier and the
operation is based on the concept of a differential amplifier which we will not discuss at
this point in time. The IEC-circuit symbol and pin-layout for a 741 dual-in-line (DIL)
8-pin operational amplifier is illustrated in figure 17.1 (a) and (b).
8
-
7
6
5
3
4
+V
Vi
+
-V
Vi
+
VO
1
(a)
2
(b)
Figure 17.1
The following list supplies the purpose of each pin on the 741 dual-in-line (DIL) 8-pin
operational amplifiers:
1
Null offset
This pin is utilised together with pin 5 in order to obtain a zero output when no signal is
present on any of the inputs.
2
Inverting input
This is the pin on which an input may be applied and that will give an amplified but
inverted output which means that the output is 180º out of phase with the input.
3
Non-inverting input
The pin on which an input may be applied and that will give an amplified but an in
phase output and the phase angle is 0º.
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291
4
-V
The pin on which the negative polarity of the bias supply is connected to the operational
amplifier.
5
Null offset
This pin is utilised together with pin 1 in order to obtain a zero output when no signal is
present on any of the inputs.
6
Output
This is the pin from which the amplified output is taken.
7
+V
The pin on which the positive polarity of the bias supply is connected to the operational
amplifier.
8
NC
This pin is unconnected.
In order to get a clearer view of an operational amplifier and how it functions you are
referred to the following illustrations. An operational amplifier has two input terminals
namely an inverting input marked - and a non-inverting input marked +. A single
output can be obtained depending on which of the above terminals have been utilised
as the input.
Figure 17.2 illustrates the concept of a signal being applied to the inverting input
terminal of an operational amplifier and the non-inverting input grounded.
+V
-
+
-V
Figure 17.2
The terminal marked - is the inverting input and a signal is applied between the
inverting input and ground. The output will therefore be the inverted (180 ºout of
phase) with the input.
Figure 17.3 illustrates the concept of a signal being applied to the non-inverting input
terminal of an operational amplifier and the inverting input grounded.
292
Operational Amplifiers
+V
+
-V
Figure 17.3
The terminal marked + is the non-inverting input and a signal is applied between the
non-inverting input and ground. The output will therefore be in phase (0 º) with the
input. Figure 17.4 illustrates the concept of a signal being applied to both the input
terminals of an operational amplifier.
+V
-
+
-V
VO = 0
Figure 17.4
Here we have a situation where two signals, equal in magnitude and in phase with
one another are applied to both the inverting input as well as the non-inverting input
and the output is zero. But why is the output zero? The input on the inverting input
will cause a phase shift of 180º at the output and the input on the non-inverting input
will have a 0º phase shift at the output and since these outputs are out of phase with
one another they will cancel one another out and no output will be obtained. It must
however be noted that this condition will only exist when the null-offset has been set
to zero and that the operational amplifier have no output in the absence of any input
signal.
Figure 17.5 and figure 17.6 illustrates the concept of a signal being applied to both the
input terminals of an operational amplifier but is different from the circuit illustrated
in figure 17.4. Please observe the phases of the signals to the inverting input and noninverting input of the operational amplifier since it will have an effect on the output of
the operational amplifier.
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293
+V
-
+
-V
Figure 17.5
Here we have a situation where two signals, equal in magnitude and out phase with
one another are applied to both the inverting input as well as the non-inverting input
and the output is double the magnitude of a single input. But why is the output double?
The input on the inverting input will cause a phase shift of 180º at the output and the
input on the non-inverting input will have a 0º phase shift at the output and since these
outputs are in phase with one another they will add together out and double the output
will be obtained. It must however be noted that this condition will only exist when the
null-offset has been set to zero and that the operational amplifier have no output in the
absence of any input signal.
+V
-
+
-V
Figure 17.6
17.3Modes of operation
The good characteristics, size and cost of operational amplifiers makes them very
versatile in a number of applications.
17.3.1The inverting amplifier
An operational amplifier connected in the inverting mode is illustrated in figure 17.7.
This type of configuration is normally used for amplification purposes since the main
characteristic of an operational is its good gain that can be obtained. It is also very
useful in applications where a phase inversion is required.
294
Operational Amplifiers
Rf
RIN
Vi
+
VO
Figure 17.7
The output of this mode of operation is mathematically given by:
-Rf × Vi
VO =
where
RIN
VO
Rf
Vi
RIN
=
=
=
=
magnitude of output voltage in volt
value of feedback resistor in ohm
magnitude of input voltage in volt
value of input resistor in ohm
The gain of the operational amplifier in this particular mode is mathematically given by:
- Rf
A
=
R
IN
VO
=
A × Vi
Example 17.1
The values for an inverting operational amplifier are given as follows:
• Input voltage
=
3 volt
• Input resistor (RIN)
=
150 ohm
Determine the value of the output voltage when the following resistance values
are inserted for the feedback resistor:
(a)
300 ohm
(b)
500 ohm
(c)
1 500 ohm
(d)
Having obtained the output voltage values for the different feedback
resistors, what conclusions can be made concerning the answers?
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Solution:
- Rf × Vi
(a) VO
=
RIN
- 300
× 3
=
150
=
- 6 volt
- Rf × Vi
=
(b) VO
RIN
- 500
=
× 3
150
=
- 10 volt
(c) VO
=
=
=
- Rf × Vi
RIN
-1 500
×
150
3
- 30 volt
(d) The higher the resistive value of the feedback resistor the higher the output voltage. This is so because the portion of the voltage which is fed back negatively becomes smaller and therefore the gain of the amplifier increases.
A point that needs to be mentioned is that for the inverting mode of operation the
answer is preceded by a minus sign and this minus sign indicates that there is a 180º
phase shift between the input and the output. However, should any other value be
determined it is important that this minus sign then be used as a mathematical sign
since it would be impossible to determine the value of a resistor and ending up with a
minus value.
17.3.2The non-inverting amplifier
An operational amplifier connected in the inverting mode is illustrated in figure 17.8.
This type of configuration is normally used for amplification purposes since the main
characteristic of an operational is its good gain that can be obtained.
The overall gain of this mode (non-inverting) is higher than that of the inverting mode
of operation.
The output of this mode of operation is mathematically given by:
295
296
Operational Amplifiers
R
VO =
1 + f × Vi where
RIN
VO
Rf
Vi
RIN
=
=
=
=
magnitude of output voltage
value of feedback resistor in ohm
magnitude of input voltage
value of input resistor in ohm
The gain of the operational amplifier in this particular mode is mathematically given by:
Rf
RIN
A
=
1+
VO
A × Vi
=
+
Vi
Rf
VO
RIN
Figure 17.8
Example 17.2
The values for a non-inverting operational amplifier are given as follows:
• Input voltage
=
3 volt
• Input resistor (RIN)
=
150 ohm
Determine the value of the output voltage when the following resistance values
are inserted for the feedback resistor:
(a)
300 ohm
(b)
500 ohm
(c)
1 500 ohm
(d)
Having obtained the output voltage values for the different feedback
resistors, what conclusions can be made concerning the answers? In
your conclusion you are also to refer to the answers you obtained in
example 17.1 above.
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Solution:
(a) VO
=
=
=
(b) VO
=
=
=
Rf
× Vi
RIN
300
× 3
1+
150
1+
9 volt
Rf
× Vi
RIN
500
1+
×3
150
1+
13 volt
Rf
(c) VO
=
1+
× Vi
RIN
1 500
=
1+
×3
150
=
33 volt
(d) The higher the resistive value of the feedback resistor the higher the output voltage. This is so because the portion of the voltage which is fed back negatively becomes smaller and therefore the gain of the amplifier increases.
However, should we make use of an operational amplifier in the non-inverting mode the gain increases further.
Worthy to take not of is that whether we are connecting an operational amplifier in the
inverting mode or non-inverting mode, that the input resistor and the feedback resistor
form a voltage divider network and the amount of feedback is determined by the ratio
between these two resistors.
17.3.3The inverting summing amplifier (adder)
An operational amplifier connected in the inverting summing mode (adder) is
illustrated in figure 17.9. Typical application is that of data collection where it is often
required to add several output voltages from various sources such as sensors etc.
298
Operational Amplifiers
Rf
RIN1
RIN2
+
Vi1
VO
RIN3
Vi2
Vi3
Figure 17.9
The output of this mode of operation is mathematically given by:
VO =
-Rf
-Rf
-Rf
× Vi1 +
× Vi2 +
× Vi3
RIN1
RIN2
RIN3
where
VO
Rf
Vi
RIN
=
=
=
=
=
magnitude of output voltage in volt
value of feedback resistor in ohm
magnitude of input voltage in volt
value of input resistor in ohm
V
Vi2
V
Rf ( i1 +
+ i3 )
( RIN1
RIN2
RIN3)
should
RIN1 ≠ RIN2 ≠ RIN3
Should RIN1 = RIN2 = RIN3 = R then:
Rf
VO =
(Vi1 + Vi2 + Vi3)
R
The gain of the operational amplifier in this particular mode is mathematically given by:
Rf
Rf
Rf
A1 =
A2
=
A3
=
R
RIN2
R
IN3
IN1
VO
=
AT × (Vi1 + Vi2 + Vi3)
where
AT
=
A1 + A2 + A3
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Example 17.3
The values for a non-inverting summing operational amplifier are given as
follows:
10 k-ohm
Vi1
=
2,5 m-volt
RIN1 =
15 k-ohm
Vi2
=
7,5 m-volt
RIN2 =
25 k-ohm
Vi3
=
12,5 m-volt
RIN3 =
Determine the output voltage for the following value feedback resistors.
(a)
100 k-ohm;
(b)
50 k-ohm; and
(c)
150 k-ohm.
Solution:
(a)VO =
=
=
=
-Rf
-Rf
× Vi1 +
× Vi2 +
RIN1
RIN2
-Rf
× Vi3
RIN3
100 × 103 (2,5 × 10-3 + 7,5 × 10-3 + 12,5 × 10-3)
25 × 103 )
(10 × 103 15 × 103
3
-6
-6
100 × 10 (0,25 × 10 + 0,5 × 10 + 0,5 × 10-6 )
-125 m-volt
-Rf
-Rf
-Rf
(b) VO
=
× Vi1 +
× Vi2 +
× Vi3
R
RIN2
RIN3
IN1
=
50 × 103 (2,5 × 10-3 + 7,5 × 10-3 + 12,5 × 10-3)
25 × 103 )
(10 × 103 15 × 103
=
50 × 103 (0,25 × 10-6 + 0,5 × 10-6 + 0,5 × 10-6 )
=
-62,5 m-volt
-Rf
-Rf
-Rf
(c) VO
=
× Vi1 +
× Vi2 +
×V
RIN1
RIN2
RIN3 i3
=
150 × 103 (2,5 × 10-3 + 7,5 × 10-3 + 12,5 × 10-3)
25 × 103 )
(10 × 103 15 × 103
3
-6
-6
=
150 × 10 (0,25 × 10 + 0,5 × 10 + 0,5 × 10-6 )
=
-187,5 m-volt
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300
Operational Amplifiers
Example 17.4
The values for a non-inverting summing operational amplifier are given as
follows:
RIN2
=
RIN3
=
R
=
15 k-ohm
RIN1 =
Vi2 = 15 m-volt;
Vi3 = 20 m-volt
Rf = 60 k-ohm;
Determine the output voltage for this operational amplifier.
Solution:
(a) VO
=
=
=
-Rf (Vi1 + Vi2 + Vi3)
R
60 × 103 (12,5 × 10-3 + 15 × 10-3 + 20 × 10-3)
15 × 103
-190 m-volt
Example 17.5
The output of an operational amplifier is given as:
=
- (A + B + C) volts
VO
Draw a neat fully labelled circuit diagram of an operational amplifier that will
conform to the given output expression. NO component values are required
but input and output wave forms must be indicated.
Solution:
Rf
RIN1
RIN2
+
A
RIN3
B
C
VO
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17.3.4The voltage follower
An operational amplifier connected as a voltage follower in the inverting mode is
illustrated in figure 12.10. It has a gain of unity and VO = Vi and it is sometimes referred
to an amplifier where the output follows the input hence the name ‘voltage follower’.
Its main application is that of and impedance matching circuit since it has a high input
impedance and a low output impedance. It is further used as a ‘buffer circuit’ when two
circuits need to be isolated from one another and the first circuit need to be protected.
17.4Mathematical functions using operational amplifiers
+
VO
Vi
Figure 17.10
Over and above the application already mentioned for operational amplifiers it may also
be used for mathematical decision making applications.
17.4.1The differentiator
An operational amplifier used as a differentiator and will be applied where
differentiation of an input waveform is required and is illustrated in figure 17.11 (a), the
input waveform is illustrated in figure 17.11 (b) and the output waveform is illustrated
in figure 17.12 (c).
302
Operational Amplifiers
Rf
CIN
VIN
+
VO
(a)
Input
Output
(b)
(c)
Figure 17.10
17.5.2The integrator
An operational amplifier used as an integrator and will be applied where integration of
an input waveform is required and is illustrated in figure 17.12 (a), the input waveform is
illustrated in figure 17.12 (b) and the output waveform is illustrated in figure 17.12 (c).
Cf
RIN
VIN
+
VO
(a)
Input
Output
(b)
(c)
Figure 17.11
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303
The rate of change of the output voltage is mathematically given by:
Vi
Vo =
V/sec
where Vo
C×R
Vi
IN
C
RIN
=
=
=
=
Rate of change of output voltage
Value of input voltage in volt
Value of feedback capacitor in Farad
Value of input resistor in ohm
Example 17.6
The following data concerning an integrating operational amplifier is given:
Feedback capacitor
=
100 µF
Input resistor
=
10 k-ohm
Input voltage
=
5 volt
Use this information to determine the rate of change of the output voltage and
then represent this change determined graphically.
Solution:
Vi
Vo
=
V/sec
C × RIN
Vo
=
5
100 ×10-6 × 10 ×103
=
- 5 V/sec
1 sec
t
Slope
-
Output voltage
5
V
304
Operational Amplifiers
The following set of waveforms in figure 17.13 illustrates the input- as well as different
output waveforms of the different applications of operational amplifiers that have been
discussed.
+
-
t
Square wave input to
all circuits
+
-
t
Output for inverting
amplifier and voltage
follower
+
-
t
Output for noninverting amplifier
t
Output for integrator
circuit
t
Output for
differentiator circuit
-
+
-
Figure 17.13
Exercise 17.1
1.
Name five characteristics of an operational amplifier.
2.
Draw a neat labelled sketch of an 8-pin dual-in-line representation of a
741 operational amplifier and describe the purpose of each pin.
3.
Draw neat labelled circuit diagrams of an operational amplifier complete
with input- and output waveforms for the following conditions:
3.1 Input on non-inverting input with inverting input grounded;
3.2 Equal magnitude in-phase input on both inputs;
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3.3 Input on inverting input with non-inverting input grounded; and
3.4 Equal magnitude out of phase inputs on both inputs.
4.
An operational amplifier is marked with a positive and a negative sign. What is the meaning of these signs?
5.
The output of an operational amplifier can either be in-phase or out of phase with the input signal. Draw neat labelled circuit diagrams that will illustrate this principle.
6.
The output from three sensors is given as:
= 4,5 m-volt
• Vi1
= 6,5 m-volt
• Vi2
= 2,5 m-volt
• Vi3
and is applied to the non-inverting input of an operational amplifier. Determine the output voltage of this amplifier if it uses a 50 k-ohm feedback resistor and the following input resistors:
6.1 15 k-ohm;
6.2 25 k-ohm; and
6.3 40 k-ohm.
7.
Consider the following input and output waveforms.
(a)
(b)
(c)
+
-
+
-
t
t
t
Waveform (a) is an input waveform and waveforms (b) and (c) are output waveforms. Draw neat labelled circuit diagrams of an operational amplifier that will have as an output:
7.1 Waveform (b): and
7.2 Waveform (c).
305
306
Operational Amplifiers
8.
The following data is given:
• Feedback resistor
=
100 k-ohm
• Input voltage
=
125 m-volt
Determine the output voltage for:
8.1 A non-inverting operational amplifier; and
8.2 An inverting operational amplifier.
The following resistors are connected as input resistors for the two modes of
operation.
(a) 15 k-ohm;
(b) 50 k-ohm; and
(c) 75 k-ohm.
9.
A network with a high input impedance need to be matched to a network having low impedance by means of an operational amplifier. Draw a neat labelled circuit diagram of an operational amplifier that will conform to this
specification.
10. The following data concerning an integrating operational amplifier is given:
• Feedback capacitor
=
200 µF
• Input resistor
=
10 k-ohm
• Input voltage
=
4 volt
Use this information to determine the rate of change of the output voltage and then represent this change determined graphically.
Function generator and oscilloscope
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307
CHAPTER
18 Function Generator and Oscilloscope
Learning Outcomes
On completion of this module you will be able to:
•
Explain by means of a block diagram a(n):
– Function generator; and
– Oscilloscope.
• Indicate the uses of a(n):
– Function generator; and
– Oscilloscope.
• Explain using suitable sketches the operating principle of a cathode ray oscilloscope
(CRO);
• Discuss the following focusing concepts:
– Electrostatic; and
– Electromagnetic.
• Discuss the following deflection concepts:
– Electrostatic; and
– Electromagnetic.
• Explain using suitable sketches the concept of synchronisation;
• Explain, describe and do calculations of signal analysis for:
– Amplitude measurement;
– Time measurement; and
– Frequency.
In the field of Electronics it is required that use is made of different test instruments to
perform our task. Some of these instruments are used to provide specific quantities ie
function generator and other instruments are used to be able to measure, but more so to
observe ie oscilloscope.
A function generator is an instrument that is capable of delivering a choice of
waveforms and whose frequencies are adjustable over a fairly large range. The output
wave forms available include:
308
Function Generator and Oscilloscope
•
•
•
•
Sine waves;
Triangular waves;
Square waves; and
Saw-tooth waves from terminals A, B and C.
Depending on where and how the function generator is utilised it is of utmost need and
importance in an electronic workshop. The block diagram in figure 18.1 illustrates such
a function generator.
It must be noted however that the function generator is not deemed a measuring
instrument but deemed a test instrument and will be used together with certain
measuring instruments, digital voltmeter and oscilloscope, to perform specific tasks.
The purpose of each block illustrated in figure 18.1 may be summarised as follows:
•
Frequency control network
This section is controlled from the front panel of the function generator and will
control the upper current source and the lower current source.
Upper current
source
C
Frequency
control network
Voltage
comparator
A
Intrgrator
Lower current
source
Output amplifier 1
B
Shaping circuit
C
Output
Output amplifier 2
Figure 18.1
•
•
•
Lower current source
The lower current source supplies a constant negative current to the integrator of
which the output voltage will increase in a linear fashion with time.
Upper current source
The upper current source supplies a constant positive current to the integrator of
which the output voltage will increase in a linear fashion with time.
Integrator
A triangular wave is obtained from the integrator output and its magnitude is
determined by the constant current source and this triangular waveform is supplied
to the comparator and wave-shaping circuits.
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•
309
Comparator
The main purpose of the comparator is to deliver a square wave at the same
frequency of the triangular waveform.
Wave shaping circuit
This circuit will use the triangular waveform slope changes to convert the waveform
to a sine-wave.
Output amplifiers
Two amplifiers supplies individually and independent outputs simultaneously.
•
•
An oscilloscope is a measuring instrument which is not only capable of giving the
magnitude of a measurement but also the shape of the variable. Such a block diagram is
illustrated in figure 18.2 and an oscilloscope may be used in the following applications:
•
•
•
•
•
•
Studying waveforms;
Indicating phase relationships of waveforms;
ac- and dc-voltage measurement;
Amplification gain;
Frequency determination; and/or
Distortion indication.
The block diagram in figure 18.2 indicates all the main controls of an oscilloscope. The
following list will give an indication of the main purpose of the indicated controls.
Vertical
position
V/div
VIN
Attenuator
Vertical amplifier
CRT
Focus
Z
Trigger circuit
External
trigger
Time-base
Horisontal
amplifier
+
Level
-
Slope
T/div
Figure 18.2
CRT control
Horisontal
position
Intensity
310
Function Generator and Oscilloscope
Focus and intensity
Ensures a well-focused display at a convenient
intensity.
Vertical- and horisontal position
The vertical- and horisontal position will allow
for movement of the trace left and/or right and/
or up and/or down.
V/div
The V/div will indicate the voltage per division
on the oscilloscope display screen.
T/div
The T/div will indicate the time per division on
the oscilloscope display screen.
Slope and level
The slope and level controls will select the point
on the waveform where the display is to commence.
The main component in an oscilloscope is the cathode ray tube (CRT) and is illustrated
in figure 18.3 and consists of an electron gun, electron lens, Y-deflection plates,
X-deflection plates and an oscilloscope screen. The operation is quite elementary and
can be described in the following manner:
•
•
•
•
•
The cathode termed the electron gun which is at a negative potential and emits
electrons;
A focusing and accelerating section which provides for acceleration of the
electron beam as well as a well defined focus;
A deflection system which may be electrostatic or magnetic controlling the
movement of the electron beam onto the screen;
A screen at a very high positive potential covered in phosphor which will glow
when the electron beam strikes the screen; and
The above components are covered by an evacuated glass tube.
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311
X-deflecting
plates
Electron gun
CRT screen
Anode
Cathode
Electron lens
Y-deflecting
plates
Electron beam
Figure 18.3
Focusing of the electron beam can either be done magnetically or electrostatically.
Figure 18.4 illustrates the concept of electromagnetic focusing.
There are two coils placed around the neck of the glass tube and the magnetic thus
created will extend into the tube and it will affect the focusing of the electron beam.
The magnitude of the current in the external coils will determine the magnetic field
intensity and therefore the sharpness or focus of the spot produced on the CRT screen.
Coil
Electron gun
CRT screen
Anode
Cathode
Coil
Figure 18.4
Figure 18.5 illustrates the concept of electrostatic focusing.
This is the most commonly used concept of focusing used in oscilloscopes and employs
the use of intermediate anodes. The most acceleration takes place between the cathode
and the first anode which is at a fairly high positive potential. The second anode is
at a lower positive potential than the first and third and controls the focusing. The
third anode will provide for further acceleration providing enough energy to produce
a spot on the screen since the screen anode is at a higher positive potential than the
preceeding anodes.
312
Function Generator and Oscilloscope
Deflection of the electron beam can either be done electromagnetically or
electrostatically.
Anodes
CRT screen
Anode
Electric field
Electron beam
Figure 18.5
Two pairs of coils are placed symmetrically on the outside of the glass tube. The
X-deflector coils are mounted horizontally below and above the tube and the
Y-deflector coils are mounted vertically one on either side of the tube.
Electrostatic deflection uses two pairs of parallel plates which operate independently
from one another. Should a potential difference between the two horizontally mounted
Y-deflection plates the electrons are deflected in the vertical plane. Should a potential
difference between the two vertically mounted X-deflection plates the electrons are
deflected in the horisontal plane.
The latter of the above two concepts are illustrated in figure 18.6.
Electrons will move from X1 towards X2 at a constant speed and in the middle of the
screen provided that the voltages on all four plates (X1, X2, Y1 and Y2) are equal in
magnitude. The velocity of the electrons will be determined by the time base signal.
The waveform to be studied is applied to the vertical amplifier and from there to the
vertical deflection plates (Y1 and Y2) and is controlled by the V/div control whereas the
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313
Y1
Beam
movement
X1
X2
Input signal
Y2
Time-base
signal
Figure 18.6
time base signal is applied to the horisontal amplifier and from their to the horisontal
deflection plates (X1 and X2) and is controlled by the T/div control and it is this signal
that will display a trace across the screen. When all four plates have the same magnitude
voltage the electron beam will be displayed exactly in the middle of the screen.
Increasing the voltage on the Y1-plate will cause the beam to move up. The converse to
this is also true. Increasing the voltage on the X1-plate will cause the beam to move left.
The converse to this is also true. Note that this will be the case whether we apply a signal
or not. Should we now apply a signal to the Y-plates we want our display to be as stable
as possible. This concept is termed synchronisation.
18.1Synchronisation
The signal to be studied and the time base signal must be synchronised in order to
ensure a stable signal on the oscilloscope screen. The graphical representation in figure
18.7 illustrates this concept.
Should the signal to be viewed and the time-base frequency be synchronised then
the display on the oscilloscope screen will be stable as illustrated in figure 18.7 (a).
However, should the time-base frequency be too low then the display will drift to the
left as illustrated in figure 18.7 (b). The converse is also true in that should the time-base
frequency be too high then the display will drift to the right as illustrated in figure 18.7 (c).
314
Function Generator and Oscilloscope
+
-
(a)
+
+
+
+
+
-
+
-
(c)
Vertical input
signal
t
Horisontal input
signal
t
-
(b)
t
+
+
-
Stable display
t
Vertical input
signal
t
Horisontal input
signal
t
Display drifts
to the left
t
Vertical input
signal
t
Horisontal input
signal
t
Display drifts
to the right
Figure 18.7
18.2Signal analysis
In order to analyse signals we need to understand the layout of the CRT-screen and this
is illustrated in figure 18.8.
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Vertical
axis
1 cm
1 cm
Horisontal
axis
Figure 18.8
18.2.1 Amplitude measurement
The amplitude (peak-value) of the displayed signal is governed by the V/div control.
This control is further calibrated in terms of V/cm and since the screen is normally
divided into 1 cm squares it converts to each block representing a given voltage. The
amplitude (VP) measurement is determined by the following mathematical expression.
VP
=
height × V/div
and VP – P
=
2 × VP
Example 18.1
The following waveform is displayed on an oscilloscope screen and the V/div
control is set at 2,5 V/div. Determine the peak-to-peak value of the quantity.
Solution:
VP
=
=
=
height × V/div
VP – P
3 × 2,5
7,5 volt
=
=
=
2 × VP
2 × 7,5
15 volt
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Function Generator and Oscilloscope
Example 18.2
Calculate the peak-to-peak value of an alternating quantity for a V/div setting
of 1,5 V/cm when the amplitude of this alternating quantity is 4 cm on the
oscilloscope screen.
Solution:
VP
=
=
=
height × V/div
VP – P
4 × 1,5
6 volt
=
=
=
2 × VP
2 × 6
12 volt
18.2.2 Time Measurement
The time measurement, also termed periodic time of the displayed signal is governed
by the T/div control. This control is further calibrated in terms of seconds/cm and since
the screen is normally divided into 1 cm squares it converts to each block representing a
given time period. The time measurement is determined by the following mathematical
expression.
t
=
length of wave × T/div
Example 18.3
The following waveform is displayed on an oscilloscope screen and the T/div
control is set at 2, 5 µ-sec/div. Determine the time duration of the signal.
Solution:
t
=
=
=
length of wave × T/div
8 × 2,5 × 106
10 µ-sec
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Example 18.4
Calculate the periodic time of an alternating quantity for a T/div setting of
2 m-sec/cm when the distance of this alternating quantity is 10 cm on the
oscilloscope screen.
Solution:
t
18.2.3
=
=
=
length of wave × T/div
10 × 2 × 103
20 m-sec
Frequency measurement
Although we term this section frequency measurement it is actually the calculation of
frequency since frequency can only be measured by means of a frequency counter. Since
frequency may be related to time we may define frequency calculation by the following
mathematical expression.
1
f
=
t
where
t
=
periodic time
Example 18.5
Determine the frequencies of example 18.3 and 18.4 above.
Solution:
f
=
1 f
=
1
t
t
=
1 =
1
6
20 × 103
10 × 10 =
400 MHz
=
50 Hz
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Function Generator and Oscilloscope
Example 18.6
Refer to the following oscilloscope display.
The oscilloscope is adjusted to the following settings.
V/div = 7, 5 V/cm and V/div = 25 µ-sec/cm
Determine the:
(a)
Peak-to-peak value; and
(b)
Frequency of the alternating quantity.
Solution:
(a) VP
=
=
=
height × V/div
VP - P
5 × 7,5
37,5 volt
=
=
=
2 × VP
2 × 37,5
75 volt
(b) t
= length of wave × T/div
f = 1
-6
= 12 × 25 × 10 t
= 300 µ-sec =
1
300 × 10-6
= 3,333 kHz
18.3
Waveforms
During the process of utelising test instruments ie oscilloscopes and function generators
one will meet up with various forms of waves which need to be described and
understood.
18.3.1 Sine-waves
The development and obtaining of a sine wave was done in previous studies and only
a brief background will be given so as to recap the principles and concepts that were
discussed at that stage.
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A sine-wave is illustrated in figure 18.9 with all important aspects noted.
The most important concepts will be recapped.
V or I
Pos
Peak-to-peak
value
Peak
value
t
Neg
One cycle
V or I
(b)
Figure 18.9
Frequency
The frequency of this alternating quantity is related to time as well as the number of
times the reversal of direction takes place. Since the same repetitive action takes place
every 360º and one cycle is completed in this time.
In order to determine the frequency we make use of the expression:
Definition 18.1 Frequency
Frequency may therefore be defined as the number of full cycles that moves past a
given point in one second.
f
=
1
where
t
t
f
=
=
time in seconds
frequency in hertz
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Function Generator and Oscilloscope
Example 18.7
Determine in each instance the frequency for the alternating quantity:
(a)
1 second;
(b)
1 m-second; and
(c)
10 m-seconds.
Solution:
1
(a) f
=
t
1
=
1
=
1 Hz
1
(b) f
=
t
=
1
1 × 10-3
=
1 kHz
(c) f
=
1
t
=
1
10 × 10-3
=
100 Hz
Figure 18.10 (a) and (b) illustrates the frequency of an alternating quantity over the
same time period.
Figure 18.10 (a) illustrates a frequency of one cycle over a given time period and figure
18.10 (b) illustrates a frequency of two cycles over the same time period.
+
+
t
-
t
(a)
Figure 18.10
(b)
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Periodic time
Definition 18.2 Periodic Time
The periodic time of an alternating quantity is that time it takes to complete one
full cycle of 360°.
Example 18.8
The following frequencies of an alternating quantity were measured:
(a)
150 kHz;
(b)
10 Hz
(c)
1 MHz
Calculate in each instance the periodic time.
Solution:
1
(a) f
=
t
1
t
=
f
=
1
150 × 103
=
6,667 µ-seconds
1
(b) f
=
t
1
t
=
f
=
1
10
=
100 m-seconds
1
(c) f
=
t
1
t
=
f
=
1
1 × 106
=
1 µ-second
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Function Generator and Oscilloscope
Peak value
The maximum- or peak value will be reached when the conductor lies at right angles
(90º) to the magnetic field. You will notice that this value is reached twice during a full
cycle, at 90º and once at 270º where the first is during the positive half-cycle and the
second during the negative half-cycle.
Peak-to-peak value
The peak-to-peak value is the total value measured between the two peak values. This is
mathematically expressed as:
VP-P = 2 × VP
where
VP = peak value of alternating quantity
Example 18.9
Calculate the peak-to-peak value of the following alternating quantities:
(a)
35 V;
(b)
60 V; and
(c)
120 V
Solution:
(a) VP-P
=
=
=
2 × VP
2 × 35
70 volt
=
(b) VP-P
=
=
2 × VP
2 × 60
120 volt
(c) VP-P
=
=
=
2 × VP
2 × 120
240 volt
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323
Example 18.10
An alternating quantity has the following peak-to-peak values:
(a)
240 V;
(b)
100 V; and
(c)
180 V
Determine in each instance the peak value of the alternating quantity.
Solution:
(a) VP-P
=
2 × VP
VP-P
VP =
2
240
=
2
=
120 volt
(b) VP-P
=
2 × VP
VP-P
VP =
2
100
=
2
=
50 volt
(c) VP-P
=
2 × VP
VP-P
VP =
2
180
=
2
=
90 volt
RMS- and Average values
The root mean square (RMS) value of an alternating quantity is also termed the effective
value and is that value that actually does the work.
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Function Generator and Oscilloscope
Definition 18.3 RMS-value
The RMS-value may be defined as that value of the alternating quantity that will
produce the same heat energy as that of a direct quantity when applied to the same
resistive component.
This RMS-value is given by the following mathematical expression:
VRMS
=
0,707 × VP
where
VRMS
=
0,707 × IPIRMS
IRMS
VP
IP
=
=
=
=
RMS-value in volt
RMS-value in ampere
peak value of voltage
peak value of current
The average value of an alternating quantity is mathematically given by:
VAVE = 0,637 × VP
where
IAVE = 0,637 × IP
VAVE
IAVE
VP
IP
=
=
=
=
average value in volt
average value in ampere
peak value of voltage
peak value of current
Example 18.11
An alternating quantity has a maximum voltage value of 120 V at a peak-topeak current of 10 A. Determine in each instance the:
(a)
RMS-value of the voltage and current; and
(b)
Average value of the voltage and current.
Solution:
(a) VRMS =
0,707 × VPIRMS
=
0,707 × IP
=
0,707 × 120
=
0,707 × 10
=
84,84 volt
2
=
3,535 ampere
0,637 × VP
I AVE
=
0,637 × IP
(b) VAVE =
=
0,637 × 120
=
0,637 × 10
=
76,44 volt
2
=
3,185 ampere
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Figure 18.11 illustrates a square wave.
A square wave can either be positive going or negative going and are mainly used in
computer circuitry as well as logic circuits since it will either be on or off.
On
Off
Peak value
Leading edge
Mark
Space
Trailing edge
One cycle
Figure 18.11
Figure 18.12 illustrates a sawtooth wave.
The sawtooth waveform is also termed a triangular wave corresponding to its shape. It
is mainly used in oscilloscope time-base circuits in order to deflect the electron beam
across the oscilloscope screen. You will further notice that the positive ramp and the
Positive ramp
Negative ramp
Rise time
Fall time
Figure 18.12
negative ramp have different slopes and that the rise time is greater than the fall time
which is also termed the fly-back time,
Exercise 18.1
1.
Draw neat labelled block diagrams of a (n):
1.1 Function generator; and
1.2 Oscilloscope and indicate all the main controls.
Give in each instance as many uses as possible for all the measuring
instruments and test equipment above.
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Function Generator and Oscilloscope
2.
There are eight major controls on an oscilloscope. Discuss each of these controls.
3.
Use a suitable sketch to describe the operation of a cathode ray tube with specific reference to the display of a signal.
4.
Use suitable graphic representations to illustrate the concept of synchronisation.
5.
Use suitable sketches to illustrate signal analysis with special reference to:
5.1 Amplitude measurement; and
5.2 Frequency determination.
6.
The following data concerning an oscilloscope display is given:
Vertical height =
4 cm
Horisontal width =
3 cm
V/div =
25 µV/cm
T/div =
20 µsec/cm
Determine the:
6.1 Peak-to-peak voltage of the signal; and
6.2 Frequency of the signal.
7.
Consider the following displayed oscilloscope signal.
7.1 How many cycles are displayed?
7.2 Calculate the:
7.2.1
Peak-to-peak value of the displayed signal; and
7.2.2
Frequency of the displayed signal.
The following additional data is given.
V/div = 10 mV/cm
T/div = 10 m-sec/div
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8.
Draw neat labelled graphic representation with all relevant characteristics
of the following waveforms:
9.1 Square wave;
9.2 Sine wave; and
9.3 Triangular wave.
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TRANSDUCERS
N4 Industrial Electronics
CHAPTER
19 Transducers
Learning Outcomes
On completion of this module you will be able to:
•
•
•
•
•
•
•
Define:
– A transducer;
– Temperature coefficient of resistance;
– A Thermistor; and
– A strain gauge.
Discuss the selection of transducers;
Discuss transducer applications;
Describe with the aid of sketches and graphic representations where applicable the
following mechanical transducers:
– Pressure transducer; and
– Temperature transducer.
Describe with the aid of sketches and graphic representations where applicable the
following electrical transducers:
– Resistive transducers;
– Potentiometer application of the Wheatstone bridge;
– Strain gauge application of the Wheatstone bridge; and
– Thermistor application of the Wheatstone bridge.
Describe with the aid of sketches and graphic representations where applicable the
following capacitive transducers:
– Pressure measurement;
– Flow measurement;
– Displacement measurement; and
– Level measurement.
Describe with the aid of sketches and graphic representations where applicable the
following inductive transducers:
– Speed measurement;
– Pressure measurement;
– Displacement measurement; and
– The Linear Variable Differential Transformer (LVDT).
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TRANSDUCERS
•
Describe with the aid of sketches and graphic representations where applicable the
following photosensitive transducers:
– Photovoltaic cells;
– Photoconductor (LDR);
– Photodiode;
– Phototransistor;
– Light Emitting Diode (LED);
Describe with the aid of sketches the following:
– Opto-isolators
– Thermocouples
•
19.1 Introduction
In the world of the modern Electronics we are at times required to observe specific
changes that have taken place and in most instances we cannot do this should the
changes be very small. Take for example the difference between 19 °C and 20 °C.
Without the proper equipment or measuring device this will be an impossible
task. Transducers are normally associated with measuring instruments in the
instrumentation field in that it will enable us to measure quantities such as:
•
•
•
•
Liquid level;
Pressure;
Temperature; and
Light intensity.
Definition 19.1 Transducer
A transducer may be defined as a device that will convert one form of energy into
another form of energy.
Although there are various types of transducers available and which will be impossible
to mention here the following list is not exhaustive.
•
•
•
•
•
•
Mechanical.
Electrical.
Optical.
Pressure.
Thermal
Chemical.
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331
Transducers are mainly divided into mechanical and electrical transducers where the
output of a mechanical transducer will be indicated on an analogue calibrated scale.
A typical example of such a device is a bathroom scale. An electrical transducer will
on the other hand supply us with an electrical signal which then need to be calibrated
in terms of the change that had taken place and it is these principles that we will be
discussing in this module. Electrical transducers are however divided into two distinct
groups.
Passive transducers
These transducers will produce a variation in electrical parameters such as resistance,
inductance, capacitance, etc which will be measured in current and/or voltage.
Important to note is that these types of transducers will require an external excitation
supply.
Active transducers or self-generating transducers
These are transducers which will generate a voltage and/or current when subject to
some form of quantity and therefore do not require an external excitation supply. Table
18.1 will indicate the type of transducer, the principle of operation of the transducer and
the typical application of that transducer.
All transducers are classified under the headings of application, nature of the output
signal obtained and the method used for the energy conversion. This module will be
devoted to discussing the following types of transducers.
•
•
•
•
•
Mechanical.
Resistive.
Capacitive.
Inductive.
Photosensitive.
19.2 Selection of transducers
Before we discuss the criteria for the selection of a transducers we need to determine
the major elements of a measuring system and this concept is illustrated in the block
diagram in figure 19.1.
Input device
Process device
Figure 19.1
Output device
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TRANSDUCERS
The input device will receive a quantity, normally non-electrical. This quantity under
measurement is converted to an electrical quantity and will be passed onto the process
device in the form of a proportional electrical signal. The process device will then
modify the quantity to an acceptable format for display purposes.
Since there are a multitude of transducers available on the market today that may be
utelised for virtually any type of measurement that need to be observed it is always
required that the following criteria need to be taken into account when selecting a
transducer.
•
•
•
Determine the physical quantity that needs to be measured.
Which transducer principle is best suited to measure a particular quantity?
What is the level of accuracy that will be required for this measurement?
Table 19.1 indicates the classification of a variety of transducers. This classification is
according to electrical principles and is divided into two groups.
Type of transducer
Operating principle
Passive transducers
Resistive types
Resistance strain gauge Resistance of wire changes when subjected to elongation or compression
due to external stress being applied.
Resistance
Resistance wire with a high positive
thermometer
temperature coefficient changes with
a variance in temperature.
Thermistor
Resistance of certain metal oxides
with a negative temperature coefficient changes with a variance in
temperature.
Resistance
Resistance of conductive material
hygrometer
changes with a variance in moisture
content.
Photoconductive cell Resistance of cell changes with a variance in incident light.
Application
Measurement of force, torque
or displacement.
Measurement of temperature
and radiant heat.
Measurement of temperature.
Relative humidity.
Photosensitive relay.
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Type of transducer
Variable capacitance
pressure gauge
Capacitor
microphone
Dielectric gauge
Type of transducer
Differential
transformer
Magnetostriction
gauge
Type of transducer
Photoemissive cell
Photomultiplier
Operating principle
Passive transducers
Capacitive types
Distance between two parallel plates
is varied by an externally applied
force.
Sound pressure varies the
capacitance between a fixed plate
and a moveable diaphragm.
Variation in capacitance by variation
of the dielectric medium.
Operating principle
Passive transducers
Inductive types
The difference voltage across two
series opposed secondary
transformer windings is varied by
the positioning of an iron core.
Magnetic properties are varied.
Operating principle
Passive transducers
Voltage and Current types
Electron emission will occur due to
incident light on photoemmissive
surface.
Secondary electron emission will
occur due to incident light on photoemmissive surface.
Application
Displacement or pressure.
Music, noise or speech.
Liquid level and thickness
gauging
Application
Force, direction, pressure,
position and displacement.
Displacement and rotational
speed.
Application
Radiant light, radiation and
photosensitive relay.
Radiant light, radiation and
photosensitive relay.
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TRANSDUCERS
Type of transducer
Thermocouple
Piezoelectric
Photovoltaic cells
Operating principle
Active Transducers
An emf will be generated across the
junction of two dissimilar metals when
the junction temperature is varied.
An emf will be generated when an
external force is applied across certain
crystalline materials.
An emf will be generated when incident
Application
Temperature, heat flow and
radiation.
Pressure changes, vibration,
acceleration and sound.
Solar cell and light meter.
light strikes the surface of a semi-conductor junction.
Table 19.1
Exercise 19.1
1.
Define a transducer.
2.
Differentiate between an active- and passive transducer.
3.
Name five types of transducers in use.
4.
What criteria need to be taken into account when selecting a transducer?
19.3 Transducer applications
19.3.1
Mechanical transducers
It must be noted that when we discuss mechanical transducers that no electrical
variations take place. The input is of a mechanical nature and the output is also of a
mechanical nature.
19.3.1.1
Pressure transducer
The transducer illustrated in figure 19.2 is one of many that are used for the
measurement of pressure.
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Pressure in
335
Calibrated scale
Movement
Bellows capsule
Figure 19.2
When pressure is exerted via the inlet the bellows capsule will be compressed and since
the pointer is mechanically linked to the bellows capsule there will be a movement
of the pointer on the calibrated scale. Using a suitable conversion ratio the scale will
indicate the magnitude of the pressure exerted.
19.3.1.2
Temperature transducer
The transducer illustrated in figure 19.3 is one of many that are used for the
measurement of temperature.
Calibrated scale
Metal 1
Bi-metal strip
Metal 2
Figure 19.3
Temperature measurement is accomplished by using a bi-metal strip. This bi-metal
strip is manufactured from two different materials that have different coefficients of
expansion and this means that one material will expand at a quicker tempo than the
other one causing the bi-metal strip to bend. Since the pointer is mechanically linked
to the bi-metal strip there will be movement on the calibrated scale. Using a suitable
conversion ratio the scale will indicate the magnitude of the temperature measured.
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TRANSDUCERS
19.3.2
Electrical transducers
19.3.2.1
Resistive transducers
As a basis for resistive transducers we will commence with an ordinary potentiometer
as illustrated in figure 19.4.
V1
VS
RV1
V2
Figure 19.4
Assuming that the potentiometer is a linear device we know that VS = V1 + V2. Should
the slider of the potentiometer be exactly in the middle we will find that V1 = V2.
Movement of the slider away from the centre position will result in V1 < V2 or V1 >
V2 depending on the direction of the movement. It is therefore now possible that the
direction as well as magnitude of the change that had occurred be determined by
suitable calibration.
A second very popular resistive transducer is the strain gauge which can be used where
space or inaccessible areas are at the order of the day. These strain gauges are very often
used to measure stress on:
•
•
•
Concrete pylons;
Aircraft wings and structures; and
Bridging structures.
Although these strain gauges are available in various patterns the principle of operation is
exactly the same and the basic construction of a strain gauge is illustrated in figure 19.5.
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337
Connecting wire
manufactured from
resistance wire
Paper
Figure 19.5
The principle of operation of a strain gauge is determined by the following
mathematical expression.
ρ×ℓ
where
a
R=
R
ρ
ℓ
a
=
=
=
=
resistance of strain gauge in ohms
specific resistance of strain gauge in ohms/m
length of conductor in m
cross-sectional area of conductor in m2
Definition 19.2 Strain Gauge
A strain gauge is a component that will have a change in resistance due to
dimensional changes of the conductor from which the strain gauge is manufactured.
When stress is applied to the strain gauge and according to the mathematical expression
there are two variables that will undergo a change in that the length (ℓ) will increase and
the cross-sectional area (a) will decrease. Since the length, which directly proportional
to the resistance, will increase the resistance will also increase but at the same time the
cross-sectional area, which is inversely proportional to the resistance, will decrease
resulting in the resistance to increase. Important to note is that the specific resistance of
the strain gauge will remain the same.
This principle of operation of the strain gauge is best explained in the following
example.
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TRANSDUCERS
Example 19.1
The following data concerning a strain gauge is given.
Condition 1
Condition 2
Before stress application
After stress application
R = Unknown
R = unknown
−6
ρ = 0, 0177 × 10−6 ohm/m
ρ = 0,0177 × 10 ohm/m
−6
2
a
= 2 × 10−6 m2
a = 4 × 10 m
ℓ = 1,4 m
ℓ
= 1,5 m
Determine in each instance the resistance of the strain gauge.
Solution:
Condition 1
Condition 2
R =
ρ × ℓ
R = ρ×ℓ
a
a
=
0,0177 × 10−6 × 1,4 = 0,0177 × 10−6 × 1,5
2 × 10−6
4 × 10−6
=
6,195 m-ohm = 13,275 m-ohm
It can be seen from the solutions that were obtained for condition 1 and 2 that the
resistance increased for condition 2 since:
•
•
The length increased;
The cross-sectional area decreased.
A third very popular resistive transducer is the thermistor. The first two types we
discussed both had the ability to measure the magnitude of stress and the thermistor on
the other hand has the ability to measure a change in temperature.
Definition 19.3 Thermistor
A thermistor is a component that will have a change in resistance due to
temperature changes of the material from which the thermistor is manufactured.
The IEC-circuit symbol and characteristic curve for a thermistor is illustrated in figure
19.6 (a) and (b).
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339
R
e
s
I
s
t
a
n
c
e
C
(a)
Temperature
(b)
Figure 19.6
Not all materials, however, respond to temperature in the same manner. Some materials
have a positive temperature coefficient (PTC) whereas other materials have a negative
temperature coefficient (NTC). What does this mean in practice? Refer to figure
19.7. This characteristic illustrates the variation of resistance with temperature for
conductors, insulators or non-conductors and semi-conductors. Conductors having a
PTC or NTC have that particular characteristic since it is manufactured from a material
which possesses that particular characteristic.
R
e
s
I
s
t
a
n
c
e
Insulator
Conductor
(NTC)
Semi-conductor
Conductor
(PTC)
Temperature
Figure 19.7
This phenomenon is termed the temperature coefficient of resistance.
Definition 19.4 Temperature Coefficient of Resistance
Temperature coefficient of resistance may be defined as the relationship between
the increases in resistance per °C temperature increase to the increase in the actual
resistance.
The temperature coefficient of resistance is given in ohm/ohm ˚C. Typical values
include 0,00428 ohm/ohm ˚C for copper and 0,0072 ohm/ohm ˚C for mercury. For the
purpose of our discussion we only need to look at the following two aspects: should the
temperature increase then the resistance of the conductor will also increase and should
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TRANSDUCERS
the temperature decrease the resistance of the conductor will also decrease. Sintered
mixtures of metallic oxides that are used in the manufacturing of thermistors include:
•
•
•
•
•
Manganese;
Nickel;
Cobalt;
Iron; and
Uranium.
The resistance of these materials varies between 0,5 ohms to 75 M-ohms. An important
point that need to be mentioned is the fact that the abovementioned materials
exhibit and excellent characteristic of NTC (negative temperature coefficient) in
that the resistance will decrease with an increase in temperature as illustrated in the
characteristic curve in figure 1.6 (b). The operation of a thermistor is mathematically
given by:
Rt = RO(1 + α∆t) where
Rt
RO
α
∆t
=
=
=
=
Resistance at the new temperature in ohms
Resistance at initial temperature in ohms
Temperature coefficient of material in ohm/ohm °C
Change in temperature in °C
Example 19.2
Calculate the difference in resistance of a copper conductor should it have a
resistance of 28 ohm at 10 ˚C and a temperature coefficient of 0,00428 ohm/
ohm ˚C at a final temperature of:
(a)
145˚C; and
(b)
263˚C
Solution:
(a) Rt1
=
=
=
RO [1 + αΔt]
28 [1 + 0,00428 × 35]
32,19 ohm
(b) Rt2
=
=
=
RO [1 + αΔt]
28 [1 + 0,00428 × 53]
34,35 ohm
Difference in resistance
=
=
=
Rt2 - Rt1
34,35 - 32,19
2,16 ohm
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341
Having discussed the operation of the different resistive transducers we need to
also discuss the use of the Wheatstone bridge concept which is very widely used in
transducer principles. This concept is illustrated in figure 19.8. The Wheatstone bridge
is made up of two equal resistors R1 and R2 connected between AD and AB respectively.
Resistor R3 is a variable resistor and resistor R4 may be a resistor of which the value
is unknown connected between DC and BC respectively. The supply VS is connected
across AC and a galvanometer is connected across BD. In order to determine the value
of the unknown resistor R4 the variable resistor R3 is adjusted until there is no deflection
on the galvanometer. Since there is no deflection on the galvanometer it results in the
bridge being balanced and you will find that the currents I1 and I2 are equal since the
resistance ADC = ABC and the voltage across ABC = ADC. To determine the value of
the unknown resistor R4 we need to find a mathematical expression and we will derive
such expression in the following manner.
A
VS
I1
I2
R1
R2
R3
R4
D
B
0
Galvanometer
C
Figure 19.8
I1 × R1 = I2 × R2
I1 × R3 = I2 × R4
…… (1)
…… (2)
Divide equation (1) by equation (2)
I1 × R1
I2 × R2
=
I1 × R3
I2 × R4
R1
R2
=
R3
R4
R2 × R3
R4 =
R1
Any deviation from the balanced condition will indicate a change that has occurred.
This phenomenon can now be utilised with great ease in resistive transducers.
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TRANSDUCERS
Example 19.3
The following Wheatstone bridge is constructed.
A
VS
R1
R2
R3
R4
D
B
C
The bridge has the following parameters.
=
100 volt
R1
=
10 k-ohm
VS
=
10 k-ohm
R3
=
10 k-ohm
R2
=
Unknown
R4
The bridge is adjusted by means of R3 until it is balanced again and the value
of R3 is then measured at 5 k-ohm. Use the data and determine the value of the
unknown resistor R4.
Solution:
R2
× R3
(a) R4
=
R1
=
10 × 103 × 5 × 103
10 × 103
=
5 k-ohm
19.3.2.2
Potentiometer application of the Wheatstone bridge
Assume we need to measure the pressure in a pipe by using resistive potentiometers.
The layout in figure 19.9 illustrates such an application. Referring back to figure 19.4
you will recall that the voltage drops across the two parts of the potentiometers are
equal when they are situated exactly in the centre of the potentiometers. The slider of
RV2 is locked into the centre position by a set screw resulting in two equal voltage drops
across RV2. The slider of RV1 is mechanically linked to the bellows capsule and also has
two equal voltage drops across the potentiometer in the absence of any pressure being
applied. When pressure is now applied the bellows capsule will expand thereby moving
the slider of RV1 downward and the voltage drop across the two halves will no longer be
the same and will now be proportional to the pressure applied and the bridge becomes
imbalanced. This imbalance causes an output voltage to be obtained and can now be
calibrated in terms of the magnitude of the pressure applied ie 1 V = 5 kPA, 1 V = 10
kPA, etc. Such a potentiometer as illustrated in figure 18.9 has the advantage that it can
be utelised on ac- or dc-supplies.
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Pressure
RV2
RV1
Bellows
capsule
VS
VO
Figure 19.9
19.3.2.3
Strain gauge application of the Wheatstone bridge
Assume we need to measure the stress on a concrete pylon by using a strain gauge. The
layout in figure 19.10 illustrates such an application. When using the Wheatstone bridge
configuration for this application the following points need to be taken into account.
Active
gauge
Temperature
compensation gauge
VS
R1
R2
A
V
Figure 19.10
•
•
•
Two strain gauges need to be used, one as the active gauge and the other one for
temperature compensation;
The two strain gauges must be mounted at right angles (90°) to one another; and
The Wheatstone bridge must be balanced before stress is applied.
The question may be raised on why two strain gauges are used when one would
be sufficient to measure the stress applied. One of the main criteria for using the
Wheatstone bridge configuration is that it needs to be balanced before measurements
344
TRANSDUCERS
are done and we need to compensate for any temperature changes that may occur
since resistance will vary with changes in temperature. Should any stress now be
applied to the active gauge it will have a change in resistance which will cause the
bridge to become unbalanced and an output voltage will be obtained in proportion
to the magnitude of the stress applied. An operational amplifier can now be used in
conjunction with the bridge in order to obtain a readable output.
The strain gauge has one major disadvantage in that it is not useable for a second time
since it has undergone dimensional changes and will have a different resistance in
comparison to its design value.
19.3.2.4
Thermistor application of the Wheatstone bridge
Assume we need to measure the temperature in an oven by using a thermistor.
A thermistor is also termed a resistance thermometer hence the application of
determining temperature changes. The layout in figure 19.11 illustrates such an
application and the thermistor is placed in the area where the temperature needs to be
measured like the inside of an oven. Before the oven is switched on you must ensure
that the bridge is balanced. When the thermistor now heats up caused by an increase
in the oven temperature an out of balance reading will be obtained on the measuring
device which is caused by the resistance of the thermistor decreasing with the increase
in temperature.
This imbalance causes an output voltage to be obtained and can now be calibrated in
terms of the magnitude of the temperature change ie 1 V = 5 °C, 1 V = 10 °C, etc.
R1
R2
VS
R3
-t°
C
A
V
Figure 19.11
A more practical application for a thermistor is that of a control mechanism for a heater
element of a geyser and is illustrated in figure 19.12.
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N/C
Relay
-t°
N/O
Heater
element
C
Relay coil
Rheostat
Figure 19.12
Exercise 19.2
1.
Differentiate by definition the difference between a strain gauge and a
thermistor.
2.
The temperature need to be measured in an oven and you are requested to
design a transducer circuit that will be able to do such measurement.
3.
Explain the construction of a thermocouple
4.
An ordinary potentiometer may be used to illustrate the principle of
operation of a transducer. Refer to the sketch given and explain the operation thereof.
V1
VS
RV1
V2
5.
Refer to the following sketch and describe the operation thereof.
Pressure
RV1
Bellows
capsule
RV2
VS
VO
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TRANSDUCERS
19.4 Capacitive transducers
The basic construction of a capacitor is illustrated in figure 19.13. The capacitor is
constructed by means of two conducting materials electrically isolated from one
another by a dielectric material. This dielectric material is a non-conductive material
and the most commonly used include the following:
• Air;
• Mica;
• Paper;
• Ceramic; and
• Electrolytic and each material will have its own given dielectric constant.
+
S1
S1
+
+
Dielectricum
-
(a)
(b)
Figure 19.13
Before the switch S1 is closed (see figure 19.13 (a)) each plate of the capacitor has a
net charge of zero, there is no electric field between the plates resulting in no voltage
existing between the plates. The capacitor is therefore uncharged. Should the switch
S1 be closed (see figure 19.13 (b)), free electrons will leave the upper plate and move
to the positive side of the battery terminal and at the same time free electrons leave
the negative battery terminal and will accumulate on the lower plate of the capacitor.
Since there was a movement of electrons we now find that the upper plate has no
free electrons and will be positively charged whereas the lower plate has an excess of
electrons and will be negatively charged and this will produce an electric field which
will in turn produce a voltage. This action will continue until such time that the voltage
across the two capacitor plates equal the supply voltage. Only now will the capacitor be
fully charged. Should the voltage source now be removed the charges will be trapped on
the respective plates thus storing the energy accumulated. There are three main factors
that influence the capacity of a capacitor.
Dielectric constant
It was mentioned earlier that the two plates constituting a capacitor is separated by a
dielectric material and all such materials will have by virtue of the type of material their
own dielectric constant. Important to note is that dielectric used is of a non-conductive
nature.
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347
Cross-sectional area of plates
By virtue of the cross-sectional are being large or small will determine the amount of
energy being stored. The smaller the cross-sectional area the less space there will be for
storing energy and the smaller the capacitance will be . Should we however have a large
cross-sectional area more space will be available for storing energy and the higher the
capacitance will be.
Distance between the plates
The closer the two capacitor plates are to one another the greater its capacitance would
be and the further they are apart the smaller the capacitance would be. The capacitance
of a capacitor can be expressed by the following mathematical expression.
C = 2,24 × 10-13 [k × A (N - 1)]
d
where
k
A
N
d
C
=
=
=
=
=
dielectric constant
cross-sectional area of one plate in metres2
number of plates
distance separation of plates in meters
capacitance value in Farad
Example 19.4
The following information concerning a two-plate capacitor is given:
Cross-sectional area of one plate
=
0,5 m2
Distance apart=
900 mm
Dielectric constant=
1,006
Applied voltage
=
30 volt
Determine from the given information the capacitance of the capacitor.
Solution:
C = 2,24 × 10-13 [k × A (N - 1)]
d
= 2,24 × 10-13 × [1,006 × (0,5)2 × (2 -1)]
900 × 10-6
= 6,26 × 10-11 Farad
= 62,6 Pico Farad
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TRANSDUCERS
It will be found that the only variable that will change in capacitive transducers is the
distance between the capacitor plates or the cross-sectional area of the capacitor plates
depending on the application. Another important factor that needs to be noted is that
capacitive transducer needs to be exited from an alternating current source. Capacitive
transducers are suitable to measure the following quantities.
•
•
•
•
Pressure.
Level.
Flow rate.
Displacement.
In order to fully grasp the operation of a capacitive transducer we need to look at the
principle of frequency. It is quite normal for a capacitive transducer to form part of an
oscillator circuit to obtain the required measurement. An oscillator network consists of
an inductor and a capacitor normally arranged in parallel which will then determine the
frequency at which the network will operate. This frequency of operation is given by the
following mathematical expression.
f
= 1
2 π (L × C)½
where f
π
L
C
=
=
=
=
Frequency of oscillator in Hz
3,142
Value of inductance in Henry
value of capacitance in Farad
Observing the above expression you must realise that the only constants that will vary
is L and C and since we are discussing capacitive transducers it should be obvious
that only C will vary and L will stay constant. Therefore, with a variation in C a new
frequency will be obtained which will be in relation to the quantity that was applied to
the capacitive transducer. This concept will be illustrated by the following activity.
Example 19.5
The following information concerning a capacitive transducer is given.
Condition 1Condition 2
No pressurePressure applied
L
=
120 mH
L
=
120 mH
C
=
100 pF
C
=
140 pF
Determine in each instance the frequency of operation.
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349
Solution:
Condition 1
f
=
1
2 π (L × C)½
=
1
2 × 3,142 (120 × 13−3 × 100 × 10−12)½
=
695,1 Hz
Condition 2
f
=
1
2 π (L × C)½
=
1
2 × 3,142 (120 × 13−3 × 140 × 10−12)½
=
2,554 kHz
From the above it can be seen that there is a difference in frequency between no
pressure applied and pressure applied and can now be calibrated in terms of the
quantity being measured irrespective of the type of quantity being measured ie 100 Hz
= 1 kPA; 300 Hz = 1 kPA, etc. Our next objective will be to discuss the different types of
capacitive transducers available as well as their principle of operation.
19.4.1
Pressure measurement
The capacitive transducer illustrated in figure 19.14 is that for measuring pressure and is
the basis on which a capacitor microphone also operates.
Flexible metal
foil diaphram
Fixed plate
Pressure
in
d
Output
Dielectric
Figure 19.14
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TRANSDUCERS
When pressure is applied the flexible metal foil diaphragm will vibrate at the tempo at
which the pressure is applied thereby decreasing the distance between the moveable
plate and the fixed plate. This new value of capacitance can now be calibrated in terms
of the pressure applied.
19.4.2
Flow measurement
The capacitive transducer illustrated in figure 19.15 is that for measuring flow rate or
pressure.
Gas flow
Movement
d
To oscillator
Bellows capsule
Figure 19.15
This transducer consists of a bellows capsule attached to the pipe in which the flow rate
or pressure needs to be measured. The capacitor plate attached to the bellows capsule
is the moveable plate whereas the plate on the right-hand side is the fixed plate and is
coupled to an oscillator network. As soon as a quantity is moved through the pipe the
bellows capsule will decrease and the distance between the two plates will be reduced
thereby causing a new value for the capacitor and hence a new frequency as was
indicated in activity 19.5. This type of capacitive transducer is mainly used to measure
magnitude of a quantity but not the direction of the quantity.
19.4.3
Displacement measurement
In order to measure direction and magnitude we make use of a capacitor transducer as
illustrated in figure 19.16 (a) and (b).
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Fixed plate
R1
Movement
Moveable plate
To oscillator
351
Supply
R2
Fixed plate
Rotation
(a)
Fixed plate
R1
Movement
Rotation
To oscillator
Supply
R2
Fixed plate
(b)
Figure 19.16
You will notice that the capacitive transducers illustrated in figure 19.16 (a) and (b) once
again make use of the Wheatstone bridge configuration. Note that when the moveable
plate is exactly in the middle then the bridge will be balanced and no output is obtained.
Moving away from the centre will result in an out of balance output which could be
positive or negative indicating the direction of displacement and the magnitude thereof
will indicate the change that is taking place. It should also be noted that the resistive
arm of the Wheatstone bridge can be replaced with inductors and the same effect will
be obtained.
19.4.4
Level measurement
In order to measure level we make use of a capacitor transducer as illustrated in figure
19.17. It was noted that there are three effects that will determine the capacitance of a
capacitor and in the previous discussions we had a change of capacitance caused by a
change in the distance between the capacitor plates. In order to measure level we make
use of the change in the capacitor plate area since the liquid level will determine the area
of the plates that will be submersed in the liquid.
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TRANSDUCERS
Electronic circuitry
and read-out
devices
Container
Liquid feeder pipe
Liquid level
Liquid serving
as dielectric
medium
Capacitor plates
Figure 19.17
The graphic representation in figure 19.18 will clearly illustrate this concept.
Capacitance
Capacitance
of air as
dielectric
Fluid level
Figure 19.18
Exercise 19.3
1.
Draw neat labelled diagram of a capacitive transducer that will be able to
measure the following quantities.
1.1 Flow; and
1.2 Level.
You are further required to discuss the operation of each one with the aid
of a mathematical expression and example.
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19.5 Inductive transducers
The operation of inductive transducers is somewhat different from resistive- and
capacitive transducers in that with the aforementioned transducers the resistance or
capacitance changed with the application of a varying quantity. Inductive transducers
make use of a change in magnetic field characteristics since it is impossible to alter the
inductance value of the inductor. In order to more clearly understand this concept you
are referred to the illustration in figure 19.19 (a) and (b).
(a)
(b)
Figure 19.19
Observing figure 19.19 (a) the soft iron core is outside the inductor and will have no
effect on the magnetic field which exists around the inductor. When the soft iron core
is now inserted into the inductor without coming into contact with the inductor the
existing magnetic lines become more concentrated and it this effect which is utelised in
inductive transducers as illustrated in figure 19.19 (b).
19.5.1
Speed measurement
The illustration in figure 19.20 shows a method used to determine the rotational speed
of a gear, wheel or axle.
Secondary windings
&
A
7
7
7
7
Digital readout
Toothed
wheel
Direction of
rotation
Primary windings
Figure 19.20
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TRANSDUCERS
The primary windings are exited from a dc-source thereby creating a steady magnetic
field around the primary winding. As soon as the toothed wheel is rotated the tooth will
cause an increase in the magnetic field strength across the secondary windings thereby
creating a pulse every time a tooth penetrates deeper into the primary magnetic field.
By suitable methods of calibration the number of pulses can now be counted for a given
time period and the speed can be calculated.
A second concept makes use of a tacho-generator and is illustrated in figure 19.21.
Read-out unit
Direction of
rotation
Armature
Figure 19.21
A tacho-generator is basically a two-phase induction motor and has two coils mounted
at right-angles to one another. An alternating current is supplied to the one winding
and a read-out unit normally a calibrated voltmeter is connected to the other winding.
When the armature is now rotated it will induce a voltage into the coil to which the
calibrated voltmeter is connected and the reading obtained will be in proportion and
relation to the speed at which the armature is rotated.
19.5.2
Pressure measurement
The inductive transducer illustrated in figure 19.22 is that for measuring flow rate or
pressure.
Pressure
Movement
Soft iron core
Inductor wound around a
cardboard holder
Bellows capsule
Figure 19.22
Supply and
measuring
device
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This transducer consists of a bellows capsule attached to the pipe in which the flow
rate or pressure needs to be measured. Attached to the bellows capsule is the moveable
piston that will move in and out of the inductor coil thereby creating a varying magnetic
field. This type of inductive transducer is mainly used to measure magnitude of a
quantity but not the direction of the quantity.
19.5.3
Displacement measurement
The inductive transducer illustrated in figure 19.23 will measure direction as well as
magnitude of any variance applied.
Iron core
L1
Displacement
R1
L2
R2
VS
Figure 19.23
You will once again notice that use is made of a Wheatstone bridge and this Wheatstone
bridge consists of inductors L1 and L2 with resistors R1 and R2. When the iron core is
situated in the middle of the two inductors L1 and L2 then the voltage drops across
resistors R1 and R2 are equal and no output is obtained.
Moving the core in either direction will cause an imbalance in the Wheatstone bridge
and an output will be obtained directly proportional to the magnitude of the change.
19.5.4
The Linear Variable Differential Transformer (LVDT)
The operation and graphic representation of an LVDT is illustrated in figure 19.24 (a)
and (b).
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TRANSDUCERS
The LVDT consists of a primary winding L1 and two secondary windings L2 and L3.
The secondary windings are connected in series opposition and the output will be zero
when the core is exactly in the centre position since the magnetic flux is distributed
evenly between L2 and L3.
Moving away from the centre position will result in an increase in magnetic coupling
depending on the direction of displacement and either a positive voltage or negative
voltage output will be obtained which will indicate direction of the quantity as
illustrated in figure 19.24 (b). The voltage magnitude of the quantity applied will
indicate the magnitude of the quantity.
+V
L1
B
Displacement
Iron core
A
Output
across L3
B
Output
across L2
L2
A
L3
-V
Output
(a)
(b)
Figure 19.24
Exercise 19.4
1.
The rotational speed of a wheel needs to be determined by means of an
inductive transducer. Show by means of a neat labelled sketch how you
would be able to perform this requirement and then discuss the operation
of the circuit you had provided.
2.
Discuss with the aid of a suitable sketch the operation of a tacho-
generator.
19.6 Photo-sensitive transducers
Semi-conductors exhibits the phenomena to change its characteristics when exposed
to light whether it be natural light or artificial light hence the term ‘photosensitive’
which literally translates to being sensitive to light. These components may operate
when they are subjected to a source of light and are then termed ‘photoconductive or
N4 Industrial Electronics
357
photosensitive’ components but in this instance they need to be connected in a network
or circuit that has a supply voltage. On the other hand they may radiate light and is
then termed ‘photo-emmissive’ devices and will radiate light when subjected to a supply
voltage.
19.6.1
Photo-voltaic cells
Photovoltaic cells are also termed solar cells and having already discussed the meaning
of the term ‘photo’ which means light we also need to discuss the term ‘voltaic’. Voltaic
means voltage and we can therefore proclaim that ‘photovoltaic’ means the generation
of a voltage by means of light which still conforms to our definition of a transducer
given earlier. It was mentioned earlier that semi-conductors changes its characteristics
when exposed to a source of light and this change is the combination of electron-hole
pairs in the semi-conductor material by thermal agitation which constitutes a current
flow in the external circuit coupled to the device. The voltage developed by these cells
will be dependant upon the strength and intensity of the light falling onto the cell. The
output obtained from these cells is in the region of ± 0,5 volt per cell and in order to get
a reasonable output depending on the application, these cells are connected in seriesparallel. These cells are used in what is termed a ‘solar panel’ to supply electricity to
devices where the ESKOM network cannot reach. The construction of a selenium cell
and a silicon cell is illustrated in figure 19.25 (a) and (b).
Light
Light
Ring-shaped
electrodes
Glass cover
Glass cover
+
Gold
Selenium
Metal base
+
V
P-type layer
V
-
-
(a)
N-type layer
Metal base
(b)
Figure 19.25
19.6.2
Photo-conductor (LDR)
A photoconductor is a component that will have a change in resistance when subjected
to a source of light. Important to note is that an LDR will not generate a voltage as was
the case with the solar cells and will need a voltage source in the network where it is
installed. The following materials used in the manufacture of LDR’s.
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TRANSDUCERS
•
•
•
•
Selenium.
Lead Sulphide.
Cadmium Sulphide.
Cadmium Selenite
The construction of an LDR is illustrated in figure 19.26.
Conductor
Figure 19.26
The circuit symbol and characteristic curve for an LDR is illustrated in figure 19.27 (a)
and (b).
R
Incident light
(a)
Light intensity in lumen
(b)
Figure 19.27
In the absence of light onto the LDR it has a very high resistance termed the ‘dark
resistance’ and the resistance will decrease when the LDR is exposed to a source of light.
The amount by which the resistance will decrease is dependant upon the intensity of the
light source as illustrated in figure 18.27 (b). The following applications of an LDR are
illustrated in figure 19.28 (a) and (b). The circuit in figure 19.28 (a) is off in the absence
of light and as soon as the LDR is illuminated then the circuit will switch on. The circuit
in figure 19.28 (b) is on in the absence of light and as soon as the LDR is illuminated
then the circuit will switch off.
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359
+ VCC
+ VCC
LDR
RB
Relay
Relay
Q
Q
RB
LDR
- VCC
- VCC
(a)
(b)
Figure 19.28
19.6.3
Photo-diode
A photodiode operates in more or less the same principle as an LDR but the difference
is that use is made of a PN-junction. The photodiode is connected in the reverse bias
mode in the circuit and the junction is exposed to light through a window in the casing
of the photodiode.
The circuit symbol and characteristic curve for a photodiode is illustrated in figure
19.29 (a) and (b).
I
600 lumen
500 lumen
400 lumen
Anode
300 lumen
Cathode
200 lumen
100 lumen
0 lumen
Incident light
-V
(a)
(b)
Figure 19.29
In the absence of incident light a small current, termed dark current, will flow but is too
small to make any significant change in the characteristics of the photodiode. This dark
current is typically in the region of 10 μA for germanium diodes and 1 μA for silicon
diodes.
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TRANSDUCERS
The following applications of a Photodiode are illustrated in figure 19.30 (a) and (b).
The circuit in figure 19.30 (a) is off in the absence of light and as soon as the diode D is
illuminated then the circuit will switch on.
The circuit in figure 19.30 (b) is on in the absence of light and as soon as the diode D is
illuminated then the circuit will switch off.
+ VCC
+ VCC
D
RB
Relay
Relay
Q
Q
RB
D
- VCC
- VCC
(a)
(b)
Figure 19.30
19.6.4
Photo-transistor
A photo-transistor has similar characteristics to a conventional bi-polar transistor.
Should the emitter-base junction be exposed to a source of light a current in the base
will be produced which in turn causes a collector current to flow. The magnitude of the
collector current is therefore dependant upon the magnitude of the base current and is
determined by the intensity of the light source. The circuit symbol and characteristic
IC
600 lumen
500 lumen
400 lumen
C
300 lumen
B
Incident light
200 lumen
100 lumen
E
0 lumen
(a)
V
(b)
Figure 19.31
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A typical application of a phototransistor is illustrated in figure 19.32. In the absence of
incident light the base current is too small to cause any significance change in collector
current and the relay can not be energised since the collector current is below the
holding current value of the relay.
Any incident light on the base-emitter junction of the phototransistor causes an
increase in base current which has a corresponding change in collector current and
since the collector current now has a value above the holding current value of the relay,
the relay will be energised.
Figure 19.32 illustrates the application of a phototransistor.
+ VCC
RB1
Relay
Q
RB2
Figure 19.32
19.6.5
- VCC
Light Emitting Diode (LED)
An LED falls into the category of photo-emmissive transducers since it will give off
light when supplied with a suitable voltage source. Therefore it still conforms to the
definition of a transducer.
In any PN-junction there is a continuous combination of holes and electrons when
a current is passed through the junction. The electrons of the junction are in the
conduction band whilst the holes are in the valence band. In order to drop from the
conduction band into the valence band the electrons have to give off energy for this
transition, and are given off in the form of electromagnetic radiation of different colours
depending on the level of doping of the junction as well as the material used for the
junction.
By placing this PN-junction in a translucent encapsulation the electromagnetic
radiation energy is given off as light and the intensity thereof is determined by the
magnitude of the forward current passed through the junction. These LED’s are then
available in red, green, amber, blue and yellow. It is possible to obtain an LED in
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TRANSDUCERS
dual colours in that forward current will produce one colour and reverse current will
produce a different colour. The construction and circuit symbol for an LED is illustrated
in figure 19.33 (a) and (b).
Anode
Cathode
Cathode
Anode
Radiated light
(a)
(b)
Figure 19.33
Although there are various applications for an LED the most popular ones are those for
digital read-out devices like you would find on the calculator you use. These LED’s are
arranged either as a seven-segment display or as a 7 × 5 matrix arrangement. These two
concepts are illustrated in figure 19.34 (a) and (b).
a
f
b
g
e
c
d
(b)
(a)
Figure 19.34
19.6.6
Opto-isolators
An opto-isolator or opto-coupler utilises an LED and phototransistor in one package as
a six pin dual-in-line and is illustrated in figure 19.35.
1
6
5
2
3
NC
4
Figure 19.35
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363
A current is allowed to flow through the LED and light is emitted which is directed
at the base-emitter junction of the phototransistor which is then switched on and the
device that is to be controlled by the photo-transistor becomes operational.
The opto-isolator is therefore used as a switch with both the LED and phototransistor
initially off. The coupling is therefore optical which allows for a high degree of electrical
isolation between the input and the output of the opto-isolator and this feature enables
us to use the opto-isolator as an interface between a low voltage- and a high-voltage
system.
19.7 Thermo-couples
It is possible to generate a voltage by means of a thermo-couple. In analysing the word
thermo-couple it will be noticed that ‘thermo’ has reference to temperature (heat)
and it is on this principle that a thermo-couple operates. A thermo-couple consists of
two dissimilar metal wires joined at one end termed the sensing or hot junction and
terminated at the other end termed the reference or cold junction. This concept is
illustrated in figure 19.36.
+
G
Hot junction
Cold junction
Figure 19.36
Should a temperature difference exist between the hot and the cold junction a voltage
will be generated equal to the magnitude of the temperature difference. This magnitude
of the voltage developed will also be depending on the type of material used in the
manufacture of the thermocouple. Typical materials used in the manufacture of
thermocouples include:
•
•
•
•
Chromel-Constantan;
Iron-Constantan;
Copper-Constantan; and
Platinum-Platinum/Rhodium.
364
TRANSDUCERS
Exercise 19.5
1.
The following components can be used to switch a device on or off.
1.1 Photo-diodes; and
1.2 Light Dependant Resistors.
Draw two neat labelled circuit diagrams that will illustrate this concept.
2.
Use suitable sketches that will illustrate the construction of:
2.1 A selenium cell; and
2.2 A silicon cell.
3.
Illustrate using suitable sketches two applications of a light emitting diode.
4.
Discuss the operation of a thermo-couple.