Conversion of signal levels from dBm measurement to voltage measurement is demonstrated.
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 ( R-1 ) + 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 - 20 LOG (10-6) - 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 if voltage is in microvolts and resistance is 50-Ohm, we find
(10) (dBm + 107)/20 = LOG Eµ
(11) Eµ = 10(dBm + 107)/20
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Author: James W. Hebert This article first appeared January 8, 2005.