Electrical & Electronics

dB, dBm & Watts Converter

Convert power and RMS voltage levels, or calculate gain and loss ratios.

Your inputs

Your results

Power: —

dBm: —

dBW: —

RMS voltage: —

dBV: —

dBµV: —

Current: —

Formula—

Convert absolute signal levels and gain ratios correctly

Convert power and RMS voltage levels, or translate a gain or loss between decibels and linear ratios. Absolute level mode uses a positive resistive impedance to connect voltage, current and power. Gain mode assumes equal input and output impedances. Keeping level and ratio distinct helps compare measurements without confusing a referenced unit such as dBm with a relative change in dB.

Step by step

Choose power and voltage levels when you have an absolute measurement. Enter its value, choose W, mW, V, mV, dBm, dBW, dBV or dBµV, and enter impedance in Ω. Voltage entries represent RMS voltage, not peak or peak-to-peak amplitude.
Read power, RMS voltage, current and the four logarithmic level results. A negative logarithmic value is allowed: it means the level is below that unit's reference. Linear power and voltage entries must be nonnegative; impedance must remain positive.
Choose gain and loss when comparing output with input. Enter dB, a power ratio or a voltage ratio, then read the corresponding ratios. Positive dB represents gain and negative dB represents loss; the mode does not calculate an absolute output level without an input level.

Settings and limits

Reference levels: dBm references 1 mW and dBW references 1 W. dBV references 1 V RMS and dBµV references 1 µV RMS. Power levels use 10 log10 of the referenced ratio; voltage levels use 20 log10. Select the actual measurement unit rather than treating all logarithmic values as interchangeable.
Impedance and RMS: The model uses P = Vrms² / Z and I = Vrms / Z for a resistive impedance. Changing impedance at fixed power changes voltage and current; changing it at fixed voltage changes power. Reactive loads, impedance mismatch and transmission-line reflections need a different model.
Gain and loss ratios: Power ratio is 10 raised to dB/10; voltage ratio is 10 raised to dB/20. The paired ratios assume equal impedances. If input and output impedances differ, a voltage ratio alone does not determine power gain; calculate each power using its corresponding impedance first.

Worked example

In level mode, enter 0 dBm into 50 Ω. The result is 0.001 W, about 0.223607 V RMS and 4.47214 mA. Change the level to 30 dBm at the same impedance: power becomes 1 W, voltage 7.07107 V RMS and current 141.421 mA.

Expected result
dBm	P	Vrms	Irms	dBW
0	0.001 W	0.223607 V	4.47214 mA	−30
30	1 W	7.07107 V	141.421 mA	0

The 30 dB increase multiplies power by 1000 while voltage increases by the square root of that factor. In gain mode, enter 6 dB to obtain power ratio 3.98107 and voltage ratio 1.99526. A rounded 6 dB gain is therefore close to doubling voltage under the equal-impedance assumption.

Questions and troubleshooting

Does 0 dBm mean there is no signal?

It means power equals the 1 mW reference. Zero linear power instead produces zero voltage and current and a logarithmic result of −∞. A finite negative dBm value still represents positive power below the reference.

Why did changing impedance leave dBm unchanged?

When the input is a power unit, the entered power stays fixed and impedance changes the derived voltage and current. With a voltage-unit input, voltage stays fixed and power changes. Check which input unit is selected before comparing the two cases.

Can I enter a peak sine-wave voltage directly?

Convert it to RMS before using the level mode: Vrms = Vpk / √2 for a sine wave. Peak-to-peak amplitude must first be halved to obtain peak amplitude. The calculator's V and mV choices do not perform these waveform conversions automatically.