## Conversion of Receiver Sensitivity

From micro-volts to dBm

### by James W. Hebert

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} / R
If we express power in terms of dB relative to one milliwatt,
we get:
(2) dBm = 10 LOG ( E^{2} / R ) + 30
Separating the first term into components we have
(3) dBm = 10 LOG ( E^{2} ) + 10 LOG ( R^{-1} ) + 30
Further simplifying we get
(4) dBm = 20 LOG E - 10 LOG (R) + 30
Here E is in volts. If E_{µ} is the same voltage in micro-volts, then
(5) E = E_{µ} X 10^{-6}
Substituting into (4) we get
(6) dBm = 20 LOG (E_{µ} X 10^{-6}) - 10 LOG (R) + 30
Now 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) - 90
For the common situation where R=50, this simplifies to
(9) dBm = 20 LOG E_{µ} - 107
To convert the other way we find
(10) (dBm + 107)/20 = LOG E_{µ}
(11) E_{µ} = 10^{(dBm + 107)/20}

DISCLAIMER: This information is believed to be accurate but there is no
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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.