FundamentalsSignals & Interfaces

Decibels in electronics: dB, dBm, dBV, and dBc

Understand power and voltage ratios, dBm, dBV and dBu references, impedance, conversions, and measurement pitfalls in electronics and RF.

Level

Intermediate

Reading time

15 min

Primary standard or source

IEC 80000-3, IEC 60268-3, and NIST — logarithmic units and signal levels

Last verified

13 September 2026

Associated tool

Decibel converter

Written and technically reviewed byElectroDesignForge Engineering Team

View the editorial process

Key point: dB always expresses a ratio. Suffixes such as dBm, dBV, dBu, dBc, and dBFS state the reference that makes that ratio an actionable level. Without a reference—and an impedance where needed—a decibel value is incomplete.

Quick reference

Decibels compress very large ratios into numbers that are easy to add. A chain of RF gains and losses can therefore be handled with simple algebra in dB.

dB is a logarithmic ratio

For two powers:

L(dB) = 10 × log10(P2 / P1)

For two voltages or two currents, only when both impedances are the same:

L(dB) = 20 × log10(V2 / V1) = 20 × log10(I2 / I1)

The factor of 20 follows from power being proportional to the square of voltage at fixed resistance: P = V² / R.

Useful shortcut: +3 dB approximately doubles power, but increases voltage by only about 41%. +6 dB nearly doubles voltage and quadruples power.

Gain, loss, and cascade addition

An amplifier of +12 dB followed by a cable of −2.5 dB and a filter of −1 dB gives:

total gain = +12 − 2.5 − 1 = +8.5 dB

With an input of −20 dBm, ideal output is −11.5 dBm. This works for power gains and losses using a consistent convention, provided devices stay below compression and saturation.

Do not mistake a 3 dB insertion loss for 3% loss: it passes only about 50% of the power.

dBm and dBW: referenced power levels

dBm is referenced to 1 milliwatt:

P(dBm) = 10 × log10(P(mW))
P(mW) = 10^(P(dBm) / 10)

dBW is referenced to 1 watt:

P(dBW) = P(dBm) − 30
P(dBm) = P(dBW) + 30

dBm describes real power without assuming impedance. Impedance matters when translating that power into voltage or current.

dBV and dBu: referenced voltage levels

dBV is referenced to 1 V RMS:

V(dBV) = 20 × log10(VRMS / 1 V)
VRMS = 10^(V(dBV) / 20) V

dBu is referenced to 0.775 V RMS, the voltage that once dissipated 1 mW in 600 Ω:

V(dBu) = 20 × log10(VRMS / 0.775 V)

The difference between dBu and dBV is about 2.21 dB: 0 dBu = −2.21 dBV. dBu is now a voltage reference; it does not imply a 600 Ω load is present.

Power-to-voltage conversion: impedance is essential

For a resistive load R and RMS voltage:

P = V² / R
V = √(P × R)

Combined with dBm and dBV references:

dBm = dBV + 10 × log10(1000 / R)
dBV = dBm + 10 × log10(R / 1000)

RF example: +10 dBm into a matched 50 Ω load is 10 mW and 0.707 V RMS. With a mismatched load, measured voltage can differ because power is reflected.

dBc and dBFS: specialist relative levels

dBc means dB relative to carrier. A harmonic at −40 dBc is 10,000 times less powerful than the carrier. With a +10 dBm carrier, that harmonic is ideally at −30 dBm:

−40 dBc + 10 dBm = −30 dBm

dBFS means dB relative to digital full scale. 0 dBFS is the maximum representable level, so useful signal values are negative, such as −18 dBFS. Conversion to dBu or dBV depends on the converter maker’s specified analogue alignment.

A reliable calculation method

1. Identify the quantity: power, voltage, current, or digital level.
2. State the reference: 1 mW, 1 V RMS, 0.775 V RMS, carrier, or full scale.
3. For power-to-voltage conversion, state impedance and check that it is resistive and matched.
4. Use 10 × log10 for power, and 20 × log10 for voltage/current at equal impedance.
5. Keep RMS, peak, and peak-to-peak separate. For a sinewave, Vpeak = √2 × VRMS.
6. Add gains and losses in dB only after checking compatible levels.

Use the decibel converter to convert dB, dBm, dBW, dBV, and dBu with the appropriate reference.

Common laboratory pitfalls

• Using 20 log for power. Power requires 10 log10.
• Converting dBm to volts without an impedance. There is no single voltage for 0 dBm: it is 224 mV RMS in 50 Ω and 775 mV RMS in 600 Ω.
• Mixing RMS, peak, and peak-to-peak. Confirm the instrument setting and waveform.
• Forgetting 50 Ω termination. A generator specified for a 50 Ω load may appear to deliver twice its set voltage when open-circuit; check its display convention.
• Confusing dB and dBm. +6 dB is a relative gain; +6 dBm is an absolute power of 3.98 mW.
• Directly adding uncorrelated sources. Add their powers linearly, then convert back to dB. Two equal uncorrelated sources give +3 dB, not +6 dB.

Worked RF power budget

A source provides +5 dBm. The signal passes through a cable with 1.2 dB loss, a 0.8 dB filter, then a 15 dB amplifier.

Amplifier input = +5 − 1.2 − 0.8 = +3 dBm
Ideal output = +3 + 15 = +18 dBm

+18 dBm is about 63.1 mW. In a matched 50 Ω system, that is about 1.78 V RMS. Then verify amplifier maximum rating, filter power handling, and any reflections: a link-budget calculation does not replace those checks.

Bibliography

• NIST — Units outside the SI: bel and decibel
• IEC 80000-3 — Quantities and units, Part 3: Space and time
• IEC 60268-3 — Sound system equipment, Part 3: Amplifiers
• National Instruments — Understanding dB and dBm