Signal Power Converter
Convert signal power between watts, milliwatts, kilowatts, dBm, and dBW. Calculate RMS voltage and current from power and impedance for RF.
Signal Power Converter
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Formula: dBm = 10 * log10(P_mW) | P = V^2 / Z
Worked example โ 20 dBm = 100 mW = 0.1 W | Vrms = 2.236 V at 50 ohms
Formula
dBm = 10 * log10(P_mW) | P = V^2 / Z
Convert between linear power (watts) and logarithmic power (dBm, dBW) using decibel formulas. dBm uses 1 mW as reference, dBW uses 1 W. Voltage = sqrt(Power * Impedance).
Worked Examples
Example 1: WiFi Router Power Conversion
Problem:A WiFi router transmits at 20 dBm. Convert to watts and find the voltage across 50 ohms.
Solution:Power = 10^((20-30)/10) = 10^(-1) = 0.1 W = 100 mW Voltage RMS = sqrt(0.1 * 50) = sqrt(5) = 2.236 V Current RMS = sqrt(0.1 / 50) = 0.04472 A
Result:20 dBm = 100 mW = 0.1 W | Vrms = 2.236 V at 50 ohms
Example 2: Cell Phone Signal Strength
Problem:A received signal measures -80 dBm. What is the power in watts?
Solution:Power = 10^((-80-30)/10) = 10^(-11) Power = 1e-11 W = 10 picowatts dBW = -80 - 30 = -110 dBW Vrms = sqrt(1e-11 * 50) = 7.07e-5 V
Result:-80 dBm = 10 pW | -110 dBW | Vrms = 70.7 uV
Frequently Asked Questions
What is dBm and how does it relate to watts?
dBm is a unit of power expressed in decibels relative to 1 milliwatt. The conversion formula is dBm = 10 * log10(power in mW). For example, 0 dBm equals 1 mW, 10 dBm equals 10 mW, 20 dBm equals 100 mW, and 30 dBm equals 1 watt. The logarithmic scale is preferred in RF engineering because it simplifies gain and loss calculations to simple addition and subtraction.
What is the difference between dBm and dBW?
Both dBm and dBW are logarithmic power units, but they use different reference levels. dBm uses 1 milliwatt as the reference (0 dBm = 1 mW), while dBW uses 1 watt as the reference (0 dBW = 1 W). The conversion is simple: dBW = dBm - 30. dBm is more common in telecommunications and RF engineering for lower power levels, while dBW is often used in satellite communications and broadcasting where power levels are higher.
Why is impedance needed to convert between power and voltage?
Power, voltage, and current are related through impedance by the formulas P = V^2/Z and P = I^2*Z, where Z is the impedance. Without knowing the impedance, you cannot determine the voltage or current from power alone. Standard impedances in RF systems are 50 ohms (most test equipment and transmitters), 75 ohms (cable TV and video), and 600 ohms (traditional telephone lines). The choice of impedance affects signal levels significantly.
What are typical signal power levels in communications?
Signal power levels vary enormously across applications. WiFi routers transmit at about 20 dBm (100 mW). Cell phones transmit up to 23 dBm (200 mW) for 4G. FM radio stations broadcast at 50-100 kW (47-50 dBW). GPS satellite signals arrive at Earth at about -130 dBm (0.1 femtowatts). Cell tower receivers can detect signals as weak as -110 dBm. This vast range is why logarithmic units like dBm are essential in RF engineering.
How do I add gains and losses along an RF chain in decibels?
Because decibels are logarithmic, cascaded gains and losses simply add rather than multiply. Starting from a source at 0 dBm, an amplifier of 20 dB followed by 3 dB of cable loss and a 6 dB splitter loss gives 0 + 20 - 3 - 6 = 11 dBm at the output. This additive property is the main reason link budgets are worked in dBm and dB rather than watts.
What does a change of 3 dB or 10 dB mean in linear power terms?
Power in decibels is ten times the logarithm of the power ratio, so 3 dB is a factor of about two and 10 dB is exactly a factor of ten. That makes 20 dB a hundredfold change and 30 dB a thousandfold. Halving power is therefore about minus 3 dB, which is why the half-power points of a filter response are called the 3 dB points.
Why is 50 ohms the standard impedance for most RF systems?
Fifty ohms is a compromise for coaxial line, sitting between the impedance that minimises attenuation and the one that maximises power handling for a given cable size. Standardising on it means sources, cables, attenuators and instruments all match, so reflections stay small and power transfers efficiently. Video and cable television use 75 ohms instead, chosen for lowest loss, which is why the two families are not interchangeable.
How do I convert a dBm reading into RMS voltage across a 50 ohm load?
First return to watts with P = 10 raised to the power of (dBm minus 30) divided by 10, then take V = sqrt(P x Z). At 0 dBm the power is 0.001 watts, so the RMS voltage across 50 ohms is the square root of 0.05, about 0.2236 volts, and the peak is 1.414 times that. Changing the impedance changes the voltage even though the power is unchanged.
What is EIRP and how does antenna gain enter a link budget?
EIRP is effective isotropic radiated power, the power an ideal isotropic antenna would need to match the signal radiated in the main beam. In decibel form it is the transmitter power in dBm plus the antenna gain in dBi minus feeder and connector losses in dB. Because gain is expressed logarithmically, the whole budget stays a simple sum, and regulatory limits on transmitted power are usually written as EIRP.
When should the decibel formula use 20 log instead of 10 log?
Use 10 log for power quantities such as watts, and 20 log for field quantities such as volts or amps. The factor of two appears because power goes as voltage squared into a fixed impedance, and the logarithm of a square is twice the logarithm. Doubling the voltage into a fixed load quadruples the power, and the two formulas agree on that one event: 20 log of 2 is 6 dB and 10 log of 4 is also 6 dB. Applying the wrong multiplier is a common way to end up a factor of two out in decibels.
References
Reviewed for accuracy by Manoj Kumar, Mathematics Educator ยท Editorial policy
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