continuousWave --> Radio-->dBm to Microvolt Conversion

## Conversion of Receiver Sensitivity from micro-volts to dBm

### by James W. Hebert

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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