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From Microvolts to dBm

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 = E^{2}/ RIf we express power in terms of dB relative to one milliwatt, we get:

(2) dBm = 10 LOG ( E^{2}/ R ) + 30Separating the first term into components we have

(3) dBm = 10 LOG ( E^{2}) + 10 LOG ( R^{-1}) + 30Further simplifying we get

(4) dBm = 20 LOG E - 10 LOG (R) + 30Here E is in volts. If E

_{µ}is the same voltage expressed in units of micro-volts, then to convert to volts we calculate

(5) E = E_{µ}X 10^{-6}Substituting into (4) we get

(6) dBm = 20 LOG (E_{µ}X 10^{-6}) - 10 LOG (R) + 30Now simplifying we get

(7) dBm = 20 LOG E_{µ}- 20 LOG (10^{-6}) - 10 LOG (R) + 30

(8) dBm = 20 LOG E_{µ}-10 LOG (R) - 90For the common situation where R=50, this simplifies to

(9) dBm = 20 LOG E_{µ}- 107To convert the other way we find

(10) (dBm + 107)/20 = LOG E_{µ}

(11) E_{µ}= 10^{(dBm + 107)/20}

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Copyright © 2005 by James W. Hebert. Unauthorized reproduction prohibited!

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Author: James W. Hebert

This article first appeared January 8, 2005.