The ebook FEEE  Fundamentals of Electrical Engineering and Electronics is based on material originally written by T.R. Kuphaldt and various coauthors. For more information please read the copyright pages. 
Home Semiconductors Amplifiers and Active Devices Attenuators Decibels  
Search the VIAS Library  Index  
Decibels10 dB + 6 db = 16 dB Changing sound levels are perceptible roughly proportional to the logarithm of the power ratio (P_{I} / P_{O}). sound level = log_{10}(P_{I} / P_{O}) A change of 1 dB in sound level is barely perceptible to a listener, while 2 db is readily perceptible. An attenuation of 3 dB corresponds to cutting power in half, while a gain of 3 db corresponds to a doubling of the power level. A gain of 3 dB is the same as an attenuation of +3 dB, corresponding to half the original power level. The power change in decibels in terms of power ratio is:dB = 10 log_{10}(P_{I} / P_{O})Assuming that the load R_{I} at P_{I} is the same as the load resistor R_{O} at P_{O} (R_{I} = R_{O}), the decibels may be derived from the voltage ratio (V_{I} / V_{O}) or current ratio (I_{I} / I_{O}): P_{O} = V _{O} I_{O} = V_{O}^{2} / R = I_{O}^{2} R P_{I} = V_{ I} I_{I} = V_{I}^{2} / R = I_{I}^{2} R dB = 10 log_{10}(P_{I} / P_{O}) = 10 log_{10}(V_{I}^{2} / V_{O}^{2}) = 20 log_{10}(V_{I}/V_{O}) dB = 10 log_{10}(P_{I} / P_{O}) = 10 log_{10}(I_{I}^{2} / I_{O}^{2}) = 20 log_{10}(I_{I}/I_{O}) The two most often used forms of the decibel equation are: dB = 10 log_{10}(P_{I} / P_{O}) or dB = 20 log_{10}(V_{I} / V_{O}) We will use the latter form, since we need the voltage ratio. Once again, the voltage ratio form of equation is only applicable where the two corresponding resistors are equal. That is, the source and load resistance need to be equal. Example: Power into an attenuator is 10 watts, the power out is 1 watt. Find the attenuation in dB. dB = 10 log_{10}(P_{I} / P_{O}) = 10 log_{10} (10 /1) = 10 log_{10} (10) = 10 (1) = 10 dB Example: Find the voltage attenuation ratio (K= (V_{I} / V_{O})) for a 10 dB attenuator. dB = 10= 20 log_{10}(V_{I} / V_{O}) 10/20 = log_{10}(V_{I} / V_{O}) 10^{10/20} = 10^{log10(VI / VO)} 3.16 = (V_{I} / V_{O}) = A_{P(ratio)} Example: Power into an attenuator is 100 milliwatts, the power out is 1 milliwatt. Find the attenuation in dB. dB = 10 log_{10}(P_{I} / P_{O}) = 10 log_{10} (100 /1) = 10 log_{10} (100) = 10 (2) = 20 dB Example: Find the voltage attenuation ratio (K= (V_{I} / V_{O})) for a 20 dB attenuator. dB = 20= 20 log_{10}(V_{I} / V_{O} ) 10^{20/20} = 10^{ log10(VI / VO )} 10 = (V_{I} / V_{O} ) = K


Home Semiconductors Amplifiers and Active Devices Attenuators Decibels 