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Transcript
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.