Voice communication receiver sensitivity is often stated in terms of the radio-frequency voltage level at the input necessary to produce a particular signal-to-noise ratio of the desired signal at the audio-frequency output. Conversion of this voltage level to a power level in decibels referenced to one-milliwatt (dBm) can be useful when analyzing radio circuits. The formula for circuit analysis predicts a certain received power level. Conversion of rated receiver sensitivity in micro-volts to a power level in dBm will help assess the receiver performance for a particular predicted level of received signal. We must also assume a particular resistance in the antenna, R, normally about 50-ohms.
Power (P), voltage (E), and resistance (R) are related by Ohm's Law; (1) P = E2 / R If we express power in terms of dB relative to one milliwatt, we get: (2) dBm = 10 LOG ( E2 / R ) + 30 Separating the first term into components we have (3) dBm = 10 LOG ( E2 ) + 10 LOG ( 1/R ) + 30 Further simplifying we get (4) dBm = 20 LOG E - 10 LOG (R) + 30 Here E is in volts. If Eµ is the same voltage in micro-volts, then (5) E = Eµ X 10-6 Substituting into (4) we get (6) dBm = 20 LOG (Eµ X 10-6) - 10 LOG (R) + 30 Now simplifying we get (7) dBm = 20 LOG Eµ - 120 - 10 LOG (R) + 30 (8) dBm = 20 LOG Eµ -10 LOG (R) - 90 For the common situation where R=50, this simplifies to (9) dBm = 20 LOG Eµ - 107 To convert the other way we find (10) (dBm + 107)/20 = LOG Eµ (11) Eµ = 10(dBm + 107)/20
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Last modified: Monday, 25-May-2009 09:28:01 EDT
Author: James W. Hebert
This article first appeared January 8, 2005.