2. The use of decibels is widespread throughout the
electronics industry. Many electronic instruments
are calibrated in decibels, and it is common for
data sheets describing various instruments and
devices to give the specifications in terms of
decibels. For these reasons a study of their use is
essential.
3. Let us begin by considering the situation depicted
by Fig. 1. A voltage generator is driving an
amplifier. The input resistance of the amplifier is
R3.
Fig. 1 Developing the decibel concept
4. The amplifier delivers an output voltage to a load
resistor whose value is R2. In describing the gain
of the amplifier we may speak of either the power
gain or the voltage gain. The power gain G is-
defined as the power delivered to R2 divided by
the power delivered to the amplifier resistance R1.
That is,
(1)
2
1
2
2
2
1
2
1
2
2
2
1
2
/
/
R
R
R
R
P
P
G
5. The voltage gain A is defined simply as the output
voltage divided by the input voltage . That is
(2)
It will be recalled from basic electronics courses
that the power gain in decibels is defined as
(3)
where G is the power gain and, Gdb is the decibel
equivalent of G. Note that the base 10 is used in
1
2
A
Gdb = 10 log G
6. If we like, we can obtain an alternate formula for
Gdb by substituting Eq. (1) into Eq. (3) to obtain
(4)
The properties of logarithms allow us to rewrite
this equation as
(5)
2
1
2
1
2
log
10
log
10
R
R
G
Gdb
2
1
1
2
log
10
log
20
R
R
Gdb
7. Historically, decibels were defined strictly for use
with power ratios. However, modern usage of
decibels has developed to the point where it is
now common to speak of either the power gain in
decibels or the voltage gain in decibels. The
voltage gain in decibels is the first term of Eq. (5),
that is,
Adb = 20 log A
(6)
where A is the voltage gain, , and Adb is the
decibel equivalent of A. Thus, we can rewrite Eq.
(5) simply as
2
1
log
10
R
R
A
G db
db
8. Note carefully that if R1/R2 is unity, then Gdb = Adb.
If R1/R2 does not equal unity, then Gdb Adb. It is
very important, therefore, when using decibel
equivalents to specify voltage gain or power gain.
In other words, it is not enough to say that the
gain is so many decibels. We must specify that
the voltage gain is so many decibels or that' the
power gain is so many decibels, whichever the
case may be.
9. Let us summarize the use of decibels up to this
point. In the modern usage of decibels we may
speak of either the decibel equivalent of a power
ratio or the decibel equivalent of a voltage ratio.
The decibel equivalent of a power ratio is given
by
(8)
where G is the ratio of two powers, p2/p1. The
decibel equivalent of a voltage ratio is given by
Adb = 20 log A
(9)
G
Gdb log
10
10. where A is the ratio of two voltages, . In speaking
of the gain of an amplifier, Gdb = Adb only if the
load resistance R2 equals the input resistances
R1. If the two resistances are not equal, it is
essential to specify whether the gain in decibels
is for the voltage gain or for the power gain.
11. In the remainder of this section we develop some
shortcuts that allow rapid conversion between
ordinary numbers and decibel equivalents. Adb is
a function of A, meaning that each value of A
selected, only one value of Adb can be calculated.
For instance,
When A = 1 Adb = 20 log 1 = 0 db
When A = 2 Adb = 20 log 2 = 20(0.3) =
6 db
When A = 4 Adb = 20 log 4 = 20(0.6) =
12 db
When A = 10 Adb = 20 log 10 = 20(1) = 20 db
12. We can continue in this fashion until there are
enough pairs of A and Adb to tabulate as shown in
Table 1.
To bring out certain properties of decibels more
clearly, let us graph the data of Table 1, using
semi-logarithmic paper to compress the A values .
13. A Adb
-40
-20
-18
-12
½ -6
1 0
2 6
4 12
8 18
10 20
100 40
1000 60
100
1
10
1
8
1
4
1
Table 1
Voltage Gain and Its Decibel Equivalent
14. First, note in Fig. 2 a than each time A is
increased by a factor of two.
A = 1, 2, 4, S, …
Adb increases by 6 db
Adb = 0, 6, 12, 18, …
Conversely, each time that A decreases by a factor
of two
A = 1, ½ , ¼ , , …
Adb decreases by 6 db
Adb = 0, -6, -12, -18, …
15. Adb increases by 20db
Adb = 0, 20, 40, 60, ….
Fig. 2Graphs of Adb VS. A.
16. Conversely, each time A is decreased by a factor
of te
Adb decreases by 20 db
Adb = 0, -20, -40, 60, …
These properties of decibels make the
conversion from ordinary numbers into decibels a
simple matter. We need only express the ordinary
number in factors of two and ten and convert
according to the decibel properties described. As
an example, let us convert A= 4000 into its
decibel equivalent.
A = 1, 1000
1
,
100
1
,
10
1
, …
17. A = 4000 = 2 . 2 . 10 . 10 .10
Adb = 6 + 6 + 20 + 20 + 20 = 72db
We have factored A into twos and tens and added
6 or 20 db for each factor of two or ten to obtain
the total of 72 db.
As another example, consider A = 0.004. We
write this as a fraction and then factor into twos
and tens 10
.
10
.
10
2
.
2
1000
4
004
.
0
A
db
Adb 48
20
20
20
6
6
18. In this case we add 6 db for the numerator factors
and subtract 20 db for each denominator factor.
When the value of A is not exactly factorable into
twos and tens, we can obtain an approximate
answer by interpolation. For instance, if A = 60,
we observe that this is a number between A = 40
and A = 80. Hence,
Since A = 40 = 2 - 2 - 10 Adb = 32 db
Since A = 80 =2-2-2-10 Adb = 38 db
19. A = 60 is halfway between 40 and 80, so that Adb
35 db. We have interpolated to find the
approximate decibel equivalent of A = 60. using
the exact formula Adb = 20 log 60, yields a value
of Adb = 35.56db. The error in our approximate
answer is only about 0.5db. Usually, errors of less
than 1 db are acceptable in practice. Only in
those situations where the greatest is required
must we use the exact formula,
Adb = 20 log A
20. Let us summarize our procedure for finding decibel
equivalents.
1-For any ratio A of voltages or currents, express the
number in factors of two and ten If the number is not
exactly factorable into twos and tens, bracket
between the next lower and higher numbers that are
factorable into twos and tens.
2-Add 6 db for every factor of two in the numerator
and 20 db for every factor of ten. Subtract 6 db for
every factor of two in the denominator and 20 db for
every factor of ten.
3-Interpolate, if necessary, to obtain the decibel
equivalent.
4-When dealing with a power ratio G, proceed as in
21. EXAMPLE 1
Find the decibel equivalent of A = 2000.
SOLUTION
A = 2000 = 2.10 . 10 . 10
Adb = 6 + 20 + 20 + 20 = 66 db
22. EXAMPLE 2
Find the decibel equivalent of A = 3000
SOLUTION
This number is not factorable into twos and tens, but it
lies between 2000 and 4000, numbers which are so
factorable.
A = 2000 = 2 . 10 . 10 . 10
Adb = 6 + 20 + 20 + 20 = 66db
A = 4000 implies that we add 6 db to obtain Adb = 72 db.
For A = 3000, we interpolate to obtain Adb = 69 db.
(The exact answer is 69.5 db. Whenever we interpolate
the maximum error possible is about 0.5 db.)
23. EXAMPLE 3
Find the decibel equivalent of P2/P1 = 2000
SOLUTION
This is a ratio of two powers. The decibel
equivalent of a power ratio is one-half the decibel
equivalent of a voltage ratio of the same numer-
ical value. We need only proceed in our usual
manner and divide the answer by 2.
2000 = 2 . 10 . 10 . 10
6 + 20 + 20 + 20 = 66db
Hence, Gdb = 33 db
24. EXAMPLE 4
An amplifier has an input voltage of 1 my and an
output voltage of 1.6 volts. Express the voltage
gain of the amplifier in decibels.
SOLUTION
The voltage gain of the amplifier is the output
voltage divided by the input voltage.
Adb = 6+6+6+6+20+20 = 64db
10
.
10
.
2
.
2
.
2
.
2
1600
10
6
.
1
3
A
25. EXAMPLE 5
Find the decibel equivalent of A =
SOLUTION
Adb = -6 –20 –20 = 46 db
200
1
10
.
10
.
2
1
200
1
A
26. EXAMPLE 6
Voltages are often expressed in decibel
equivalents by comparing their value to a
reference voltage. In Fig.3, suppose that we use
a reference voltage of 0.5 volt. Form the ratio of
each given voltage to 0.5 volt and find the decibel
equivalent of these ratios.
Fig. 3
27. SOLUTION:
For 0.1 volt, we have
which has a decibel equivalent of -14 db. Hence,
we would say that the first voltage, 0.1 volt, is -14
db with respect to 0.5 volt.
In a similar way, the ratio of the second voltage to
the reference voltage is
10
2
5
.
0
1
.
0
3
5
.
0
5
.
1
28. which has a decibel equivalent of about 9 db.
Finally, the ratio of 10 volts to 0.5 volt is
which has a decibel equivalent of 26 db.
Hence, our system can be labeled with the
decibel equivalents as shown in Fig. 3b. It is
important to realize that these decibel values
have the correct meaning only for a reference
voltage of 0.5 volt. Had we chosen a different
reference voltage, the decibel equivalents would
all be different from those shown.
20
5
.
0
10
29. Decibel Gain of a System
One important reason for the use of decibels is
that for a system consisting of many stages, the
overall gain in decibels is the sum of the stage
gains expressed in decibels.
Fig.4 The decibel gain of a cascade of stages.
A1, A2, and A3 are the voltage gains of each stage
expressed in ordinary numbers, that is, as ratios. For
instance, the first stage may have a voltage gain of
100, so that A1 is 100, meaning that the output voltage
30. Decibel Gain of a System
To find the ordinary voltage gain of the entire
system we already know that the gains are
multiplied
A = A1 A2 A3
where A is the overall gain.
Let us find the decibel equivalent of the overall
gain.
Adb = 20 log A = 20 log A1 A2 A3
Recall that the logarithm of a product of numbers is
equal to, the sum of the logarithms of each
number.
31. Adb = 20 (log A1 + log A2 + log A3)
= 20 log A1 + 20 log A2 + 20 log A3
Each term on the right-hand side of the last equation
is merely the decibel gain of each stage. Hence,
Adb = A1 (db) + A2(db) + A3(db) (10)
Equation (10) tells us that the overall decibel gain is the
sum of the decibel gains of the individual stages. This
property is another reason for the popularity of decibels. If
we work with decibel gains, we add the stage gains to find
the overall gain. This is considerably easier than working
with ordinary gains, where it is necessary to multiply to find
the overall gain.
32. In practice, we will find that voltmeters often have
a decibel scale, so that the gain of a stage can be
measured in decibels. For instance, on some
voltmeters a reference voltage of 0.77 volt is
used. A decibel scale is provided on the meter
face, so that all voltages can be. read in decibels
with respect to 0.775 volt.
33. We might find, for example, that the input to a
stage reads -10 db and the output reads +20 db.
The gain of the stage is the algebraic difference
between these two values, or 30 db. In this way,
the decibel gains of different stages are easily
found. Once they are known, they can be added
to find the overall gain of a system in decibels.
34. EXAMPLE 7
Find the overall gain for the system of the
following Fig.
SOLUTION
Adb = 20 – 10 + 35 = 45 db
35. EXAMPLE 8
A data sheet for an amplifier specifies that the
voltage gain is 40 db. If we cascade three
amplifiers of this type, what is the overall gain
expressed as an ordinary number?
SOLUTION
Adb = 40 + 40 + 40 = 120db
For every 20db we know that there is a factor of ten
in A. Hence,
Adb = 20 + 20 + 20 + 20 + 20 + 20
A = 10 . 10 .10 . 10 . 10 . 10 = 106
36. EXAMPLE 8
A voltmeter is calibrated in decibels with a
reference voltage of 0.775 volt. What does the
voltmeter read in decibels for a voltage of 3.1
volts?
SOLUTION
The voltmeter will read 12 db, meaning that given
voltage is four times greater than the reference of
0.775 volt.
4
775
.
0
1
.
3
ref
