* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download College of Southern Nevada
Power inverter wikipedia , lookup
Spectral density wikipedia , lookup
Power factor wikipedia , lookup
Voltage optimisation wikipedia , lookup
Pulse-width modulation wikipedia , lookup
Standby power wikipedia , lookup
Wireless power transfer wikipedia , lookup
History of electric power transmission wikipedia , lookup
Opto-isolator wikipedia , lookup
Electrification wikipedia , lookup
Buck converter wikipedia , lookup
Power over Ethernet wikipedia , lookup
Electric power system wikipedia , lookup
Amtrak's 25 Hz traction power system wikipedia , lookup
Alternating current wikipedia , lookup
Mains electricity wikipedia , lookup
Power electronics wikipedia , lookup
Power engineering wikipedia , lookup
Power supply wikipedia , lookup
Switched-mode power supply wikipedia , lookup
College of Southern Nevada Decibels and dBm’s Professor Joe Miller THE DECIBEL SYSTEM The decibel (dB) is a logarithmic unit that is used in electronics, acoustics, optics, video & digital imaging, and telecommunications industries to indicate a ratio. For example, power measurements for rf circuits are usually indicated in terms of decibels (dB), electronic amplifier gain in terms of power or voltage often is expressed in terms of dBs, initial and final transmitter power when there is an upgrade can be expressed as a power gain in dBs, and other examples can be found in the fields of acoustics and optics. ⁄ Equation dB-1 The basic unit of measurement in the system is not the decibel; it is the bel. The bel is a unit that expresses the logarithmic ratio between the input and the output of any given component, circuit, or system. It may be ratio of voltages, currents, powers, sound pressure level in µPascals, or the intensity of the radiation (light). In many cases, it is used to show the ratio between input and output power otherwise known as gain. REPRESENTING POWER GAIN IN DECIBELS You can express the power gain of the amplifier (N) in bels by dividing the output (P1) by the input (P2) and taking the base 10 logarithm of the resulting quotient. For a review of logarithms see the linked WISC-Online module Log-Review (note the first equation in the module lacks superscript format on the “y” ) The formula for determining this gain is: Equation dB-2 If an amplifier doubles the input power, the quotient of POut to PIn will be 2. If you consult a logarithm table, you will find that the base 10 logarithm of 2 is 0.3, making the power gain of the amplifier 0.3 bel. Experience has shown that because the bel is a rather large unit, it is difficult to apply. A more practical unit, and one that can be used more easily, is the decibel (1/10 bel). You can convert any figure expressed in bels to decibels by multiplying that figure by 10 or simply by moving the decimal point one place to the right. Applying this rule, we find that the above ratio of 0.3 bel is equal to 3 decibels. Equation dB-3 The decibel (dB) cannot be used to represent actual power; only the ratio of one power compared to another. To say that an amplifier has a 3 dB gain means that the output power is twice the input power. This gives no indication of the actual power represented. You must be able to state the input power for it to be meaningful. Example DB.1 Given: An amplifier has input power level of 50 mW and a output power level of 2.75 W Find: AP and APdB AP= PO/PI=2.75W/50 mW = 55 APdB= 10 Log AP = 17.4 dB The power gain of any amplifier can be calculated in dB using the same steps shown above. Example DB.2 Given: An amplifier has output power level of 680 mW and is reconfigured to have an output power level of 1.2 W Find: Find the change in power level in decibels AP= Pfin/Pint=1.2W/680 mW = 1.76 AP dB= 10 Log AP = 2.47 dB Reverse the calculation. If you are given the gain in dBs you can find the gain as a ration (AP). Start with the following simple equation for APdB. AP dB= 10 Log AP = 17.4 dB Divide both sides of the equation by 10. AP dB/10 = Log AP Take the antilog of both sides. 10 AP dB/10 = AP Equation dB-5 After you have derived AP, you can find either the input or output powers if the other is known. Example DB.3 Given: An amplifier has a power gain of 26 dB and an input power of 70 uW Find: AP and PO AP = 10 APdB/10 = 10 26/10 = 398 AP = 398 = PO/70uW PO=398 x 70uW = 27.9 mW ABSOLUTE POWER GAIN EXPRESSED USING DECIBELS If a standard reference value is used for PIn , AP dB and even AP represent POut as a ratio to the standard reference and a specific power. If the standard reference value is one watt then the gain in dB becomes AP dBm or if the standard reference for telecommunications, 1 mW, is used then the gain in dB becomes AP dBm. In acoustics pref is used and it is the sound pressure of 20 micro-pascals., specifically - dB SPL (sound pressure level) – for sound in air and other gases (Note: 20 micro-pascals (µPa) = 2×10-5 Pa, approximately the quietest sound a human can hear). The standard reference that is used in electronics and telecommunications the dBm. The dBm is an abbreviation used to represent power levels above or below 1 milliwatt. Negative dBm (−dBm) represents power levels below 1 milliwatt, and positive dBm (+dBm) represents power levels above 1 milliwatt. In other words, a dBm value is a specific amount of power; 0 dBm is equal to 1 milliwatt. Briefly stated, the amount of power in a given value of dBm is the power which results if 1 milliwatt is amplified or attenuated by that dB value. For example, 40 dBm represents an actual power level (watts or milliwatts) that is 40 dB above 1 milliwatt, whereas −10 dBm represents a power level that is 10 dB below 1 milliwatt. The formula for finding dBm is a variation of the dB power formula: Equation dB-6 You do not need to use the formula in many applications. The following shows conversions of dBm to mW: a. 20dBm = 100mW f. 3dBm = 2mW b. 10dBm = 10mW g. 0dBm = 1mW c. 7dBm = 5mW h. -3dBm = .5mW d. 6dBm = 4mW i. -10dBm = .1mW e. 4dBm = 2.5mW For a +10 dBm level, start with the 1 milliwatt reference and move the decimal point one place to the right (+10 dBm = 10 mW). Another 10 dB increment brings the power level to +20 dBm, thereby moving the decimal point another place to the right (+20 dBm 100 mW) For a −10 dBm level, again start with 1 milliwatt, but this time move the decimal point one place to the left (−10 dBm = .1 mW). An additional 10 dB decrease results in another decimal point shift to the left (−20 dBm 01 mW) For a 3 dB increase, you double the power. For a 3 dB decrease, you reduce the power by one-half (+3 dBm 2 mW and −3 dBm 5 mW) A +6 dBm level is an additional 3 dB change from +3 dBm. In this case, you just double the power level of the +3 dBm (+6 dBm = 4 mW). The dB change can be made in either direction. For example, +7 dBm is a decrease from +10 dBm. Reducing the +10 dBm power by one-half, we have +7 dBm, or 5 mW. A +4 dBm power level is a 3 dB decrease from +7 dBm (+4 dBm − 2 5 mW) By using this simple method, you can quickly find any power level that corresponds to a given dBm. Example DB.4 Given: A 2.4 W signal Find: The expression of the signal as APdBm By manipulating equation dB-6 the way dB-3 was manipulated to yield equation dB5 you can develop the follow equation for finding a power level in mW that is originally expressed in dBm’s 0.001W{10 AP dB/10} = Power Equation dB-7 Example DB.5 Given: A 10 Mhz test tone measures -16 dBm on a spectrum analyzer. Find: The power level of the signal 0.001W{10 AP dB/10} = 0.001W{10-1.6} 25 1 μW RELATIVE VOLTAGE GAIN Many times it is preferable to analyze a circuit in terms of voltage or current and if we use the following with equation dB-3. PO=VO2 /RO APdB = 10 Log { VO2 /RO/ VI2/RI } PI = VI2/RI … Then we assume that RO = RI APdB = 10 Log { VO2 / VI2 } = 10 Log { VO / VI }2 = 20 Log { VO / VI } Thus AVdB= 20 Log { VO / VI } Equation dB-8 and in the case of the circuit above we substitute in the values for the above circuit APdB = 20 Log 3.0/0.3 = 20dB Equation DB-8 can be used to calculate voltage gain even if the input and output resistance are not equal, but the result wouldn’t represent the gain in power Example DB.6 Given: The output of a telecommunications amplifier is 2.00 VRMS for an input of 144 mVRMS Find: AV and the gain in dBs AV =VO/VI = 2.00V/144mV = 13.89 AVdB= 20 Log 13.89 = 22.85 dB Example DB.7 Given: The voltage gain of a network amplifier is 52 dB Find: Voltage gain as a ratio, VO if VI = 35mV, the drop of the output signal in dBs if the input decreases from 35mV to 5 mV. a. AV =VO/VI =10AVdB/20 =1052/20 =398 b. AV =VO/VI => 398 = VO/35mV => VO= 13.9V c. AVdB = 20 Log Vfin/ VInit = 20 Log 5mV/35mV = -16.9 dB For a review of decibels and a few more examples please review the linked animation module from WISC-Online Decibel-Review. Links: 1. Log-Review – http://www.wisc-online.com/Objects/ViewObject.aspx?ID=SSE2503 2. Decibel-Review - http://www.wisc-online.com/Objects/ViewObject.aspx?ID=SSE2603 Note: You may need to cut and paste the links into a separate browser window.