Download Power Descriptions

APPENDIX B
Power Descriptions
B.1 Logarithm
Logarithms are used in RF engineering to express the ratio of powers using
reasonable numbers. Communication and radar receivers can generally
handle power in the milliwatts to tens of femtowatt range with background
noise over the bandwidth of a communication channel typically being in the
femtowatt range, the actual value depends on the bandwidth of the signal.
This is a very large range of powers, around a factor of 1012 .
Logarithms are taken with respect to a base b such that if x = by then
y = logb (x). In engineering log(x) is the same as log10 (x), and ln(x) is
the same as loge (x) and is called the natural logarithm (e = 2.71828 . . . ).
Unfortunately in physics and mathematics log x (and programs such as
MATLAB) is often taken to mean ln x, so be careful. Common formulas
involving logarithms follow.
Table B-1 Common logarithm formulas.
Description
product
Formula
logb (x) ←→ x = 10b
logb (xy) = logb (x) + logb (y)
ratio
power
root
change
of base
logb (x/y) = logb (x) − logb (y)
logb (xp ) = p logb (x)
√
logb ( b x) = 1p logb (x)
logk (x)
logb (x) =
logk (b)
equivalence
Example
log(1000) = 3 and 103 = 1000
log(0.13 · 978) = log(0.13) + log(978)
= −0.8861 + 2.990 = 2.104
log2 (8/2)
=
log2 (8) − log2 (2) = 3 − 1 = 2
ln 32 √
= 2ln (3) = 2 · 1.0986 = 2.197
log10 3 20 = 13 log10 (20) = 0.4337
2
log10 (100)
=
= 6.644
log2 (100) =
log10 (2)
0.30103
838
MICROWAVE AND RF DESIGN: A SYSTEMS APPROACH
B.2 Decibels
At radio and microwave frequencies signal levels are usually expressed in
terms of the power of a signal. While power can be expressed in absolute
terms such as watts (W) or milliwatts (mW), it is much more useful to use a
logarithmic scale. The ratio of two power levels P and PREF in bels1 (B) is
P
,
(B.1)
P (B) = log
PREF
where PREF is a reference power. Here log x is the same same as log10 x. Now
human senses have a logarithmic response and the minimum resolution
tends to be about 0.1 B, so it is most common to use decibels (dB); 1 B =
10 dB. Common designations are shown in Table B-2. Also, 1 mW = 0 dBm
is a very common power level in RF and microwave power circuits.
Working on the decibel scale enables convenient calculations using power
numbers ranging from 10’s of dBm to −90 dBm to be used rather than
numbers ranging from 0.01 W to 0.00000000000001 W. A transmitter, e.g.
a cell phone, transmits in the range of up to 1 W and and base station could
transmit 100’s of watts. Generally dBm is used as 1 mW is a common power
level in RF circuits. As well dBW is used and this is the power ratio with
respect to 1 W which is convenient for transmitters such as base station and
radar with 1 W = 0 dBW = 30 dBm.
Table B-2 Common power designations.
(a)
PREF
1W
1 mW = 10−3 W
1 fW = 10−15 W
(b)
Power ratio
10−6
0.001
0.1
1
10
1000
106
1
(c)
Bell
units
BW
Bm
Bf
Decibel
units
dBW
dBm
dBf
in dB
−60 dB
−30 dB
−20 dB
0 dB
10 dB
30 dB
60 dB
Power
in dBm
Absolute
power
−120 dBm
−90 dBm
−60 dBm
−30 dBm
−20 dBm
−10 dBm
0 dBm
10 dBm
20 dBm
30 dBm
40 dBm
50 dBm
10−12 mW = 10−15 W = 1 pW
10−9 mW = 10−12 W = 1 pW
10−6 mW = 10−9 W= 1 nW
0.001 mW = 1 µW
0.01 mW = 10 µW
0.1 mW = 100 µW
1 mW
10 mW
100 mW = 0.1 W
1000 mW = 1 W
104 mW = 10 W
105 mW = 100 W
Named to honor Alexander Graham Bel, a prolific inventor and major contributor to RF
communications.
POWER DESCRIPTIONS
EXAMPLE B. 1
839
Gain Calculations
A signal with a power of 2 mW is applied to the input of an amplifier that increases the
power of the signal by a factor of 20.
G
Pout
Pin
(a) What is the input power in dBm?
2 mW
Pin = 2 mW = 10 · log
= 10 · log(2) = 3.010 dBm ≈ 3.0 dBm.
1 mW
(B.2)
(a) What is the gain, G, of the amplifier in dB?
The amplifier gain (by default this is power gain) is
G = 20 = 10 · log(20) dB = 10 · 1.301 dB = 13.0 dB
(B.3)
(b) What is the output power of the amplifier?
G=
Pout
Pin
(B.4)
so the gain in decibels is
G|dB = Pout |dBm − Pin |dBm .
(B.5)
So the output power in dBm is
Pout |dBm = G|dB + Pin |dBm = 13.0 dB + 3.0 dBm = 16.0 dBm.
(B.6)
Note that dB and dBm are dimensionless but they do have meaning, dB indicates a
power ratio but dBm refers to a power. We can add quantities in dB and one quantity
in dBm together to yield dBm, and the difference of two quantities in dBm yields a
power ratio in dB.
840
MICROWAVE AND RF DESIGN: A SYSTEMS APPROACH
EXAMPLE B. 2
Power Calculations
The output stage of an RF front end consists of an amplifier followed by a filter and then an
antenna. The amplifier has a gain of 33 dB, the filter has a loss of 2.2 dB, and of the power
input to the antenna, 45% is lost as heat due to resistive losses. If the power input to the
amplifier is 1 W, calculate the following:
PR
AMPLIFIER FILTER
Pin
=1W
η = 55%
33 dB
−2.2 dB
(a) What is the power input to the amplifier expressed in dBm?
Pin = 1 W = 1000 mW,
PdBm = 10 log(1000/1) = 30 dBm
(b) Express the loss of the antenna in dB.
45% of the power input to the antenna is dissipated as heat.
The antenna has an efficiency, η, of 55%. P2 = 0.55P1
Loss = P1 /P2 = 1/0.55 = 1.818 = 2.60 dB
(c) What is the total gain of the RF front end (amplifier + filter + antenna)?
Total gain = (amplifier gain)dB + (filter gain)dB − (loss of antenna)dB
= (33 − 2.2 − 2.6) dB = 28.2 dB
(B.7)
(d) What is the total power radiated by the antenna in dBm?
PR = Pin |dBm + (total gain)dB = 30 dBm + 28.2 dB = 58.2 dBm
(e) What is the total power radiated by the antenna?
PR = 1058.2/10 = (661 × 103 ) mW = 661 W
B.2.1 Decibels and Voltage Gain
Figure B-1(a) is an amplifier with input and output resistances that could be
different. If Av is the voltage gain of the RF amplifier then
V2 = Av V1
(B.8)
and the input and output powers will be
Pin =
V12
2R1
Pout =
V22
.
2R2
(B.9)
The ‘2’ in the denominator arises because V1 and V2 are peak amplitudes of
POWER DESCRIPTIONS
841
AMPLIFIER
V1
R1
Av
R2
V2
V1
R1
OUTPUT
INPUT
5Ω
R2
V2
200 Ω
(a)
(b)
Figure B-1 Amplifiers with an input resistance R1 and output resistance R2 : (a)
general amplifier; and (b) with R1 = 200 Ω and R2 = 5 Ω.
sinusoids in RF engineering. Thus the power gain is
G=
V 2 2R1
R1 2
Pout
= 22
=
A .
Pin
V1 2R2
R2 v
(B.10)
The power gain depends on the input and utput ratio of the amplifiers and
this is commonly used to realize significant power gain even if the voltage
gain is quite small. If the input and output resistances of the amplifier are
the same then the power gain is just the voltage gain squared.
In handling this situation some authors have used the unit dBV (decibel
as a voltage ratio). This should not be used, decibels should always refer to
a power ratio and it is needlessly confusing to use dBV in RF engineering
and its usage is the source of many errors.
EXAMPLE B. 3
Voltage Gain to Power Gain
Figure B-1(b) is a differential amplifier with a large input resistance but low output resistance.
If the voltage Av is 0.6, what is the power gain of the amplifier in dB?
The input and output powers are
Pin = 21 V12 /R1
Pout = 12 V22 /R2 =
1
2
(Av V1 )2
R2
(B.11)
Thus the power gain is
2
(Av V1 )
Pout
=
G=
Pin
R2
V12
R1
−1
=
200 2
R1 2
A =
0.6 = 14.4 = 11.58 dB.
R2 v
5
(B.12)
The surprising result is that even with a voltage gain of less than 1 a significant power gain
can be obtained if the input and output resistances are different. A result used in many RF
amplifiers. The numbers here are typical of cell phone amplifiers. Impedance transformation
is required to match the input and output impedances of stages.