dB Gain Calculator (Power, Voltage and Current Ratios)

Background.

Feed 0.2 V into a small audio amplifier and measure 6.32 V across its 8 Ω speaker. The textbook rule, 20 times the logarithm of the voltage ratio, says 30 dB of gain. In fact the amplifier multiplies power by almost 75,000, which is 48.74 dB. The missing 18.75 dB comes from the impedances: the input voltage was measured across 600 Ω and the output across 8 Ω, and the factor-of-20 rule silently assumes they are equal.

That assumption is written into the standard. ITU-R Recommendation V.574-5, which sets out how the decibel is used in telecommunications, gives the factor 20 for voltages and currents "assuming that the impedances are equal", and warns that using the decibel when they are not "is not appropriate unless adequate information is given concerning the impedances involved". It also prints the fix, an extra term of ten times the logarithm of the impedance ratio. Both impedances are asked for here, and that correction is reported as its own number so you can see how much of the answer it supplies.

Underneath, the decibel is simple. A bel is the base-ten logarithm of a ratio of two powers and a decibel is a tenth of a bel, so every 10 dB multiplies power by ten. Being a logarithm, gains and losses along a signal chain add instead of multiplying. Being a ratio, it never says how much power is actually present: 20 dB turns a microwatt into a hundred microwatts just as it turns a watt into a hundred watts.

What is db gain calculator?

Decibel gain is the ratio of output power to input power expressed as 10·lg(P₂/P₁), where lg is the base-ten logarithm. Positive values are gain, negative values loss or attenuation, and 0 dB means the power is unchanged. Voltage, current and sound pressure are field quantities whose square is proportional to power, so their ratios use a factor of 20 instead of 10; NIST SP 811 notes that the two levels agree only when the power ratio equals the square of the field ratio, which in a circuit means equal impedances. The neper is the natural-logarithm counterpart: 1 Np is 8.686 dB.

How to use this calculator.

1. 01Choose what you measured: two powers, two voltages, two currents, or a decibel figure you want turned back into ratios.
2. 02For powers, use any unit as long as both readings share it; milliwatts and watts give the same answer because only the ratio counts.
3. 03For voltages or currents, take both readings as RMS. Mixing an RMS reading with a peak or peak-to-peak one makes the ratio meaningless.
4. 04Enter the impedance each reading was taken across: 50 Ω is usual for RF, 75 Ω for video and antenna feeds, 600 Ω for legacy line-level audio and 4 to 8 Ω for a speaker. Equal values make the correction zero.
5. 05In the decibels mode, enter a loss as a negative number, so a 20 dB attenuator is −20.
6. 06For a chain of stages, work out each stage and add the decibel figures, gains positive and losses negative.

The formula.

For powers the definition is used directly: L = 10·lg(P₂/P₁). For voltages, substitute P = V²/Z. The power ratio becomes (V₂/V₁)² × (Z₁/Z₂), and ten times its logarithm splits into 20·lg(V₂/V₁) + 10·lg(Z₁/Z₂). For currents, P = I²Z, so the impedance ratio turns the other way up: 20·lg(I₂/I₁) + 10·lg(Z₂/Z₁), the identity printed in ITU-R V.574-5 §1.2. When Z₁ = Z₂ the logarithm of 1 is zero and the familiar 20·log rule is left. Running backwards, a decibel figure L gives a power ratio of 10^(L/10) and an amplitude ratio of 10^(L/20), the voltage or current ratio that figure would mean across equal impedances. The neper value is L × ln 10 ÷ 20, about 0.1151 Np per decibel. Every reading and impedance must be greater than zero, since the logarithm of zero does not exist, and results beyond ±2000 dB, a power ratio of 10²⁰⁰, are refused as certain typing errors. Impedances are treated as plain resistances: phase and reactive loading are not modelled.

A worked example.

Frequently asked questions.

References& sources.

1. [1]Recommendation ITU-R V.574-5 (08/2015), "Use of the decibel and the neper in telecommunications", Annex 1. Read directly from the ITU PDF on 2026-07-29. Primary source for every rule on this page: §1.1 defines the bel as the decimal logarithm of a power ratio and the decibel as one tenth of it; §1.2 gives the factor 20 for field quantities "assuming that the impedances are equal", warns that "use of the decibel when the impedances are not equal is not appropriate unless adequate information is given concerning the impedances involved", and prints the corrected identity 10 lg(P1/P2) = 20 lg(I1/I2) + 10 lg(R1/R2) that this calculator implements; §3 gives 1 Np ≈ 8.686 dB and 1 dB ≈ 0.1151 Np; §5 defines gain as an increase in power between two points and loss as a decrease.
2. [2]NIST Special Publication 811, 2008 edition, "Guide for the Use of the International System of Units (SI)", §8.7 "Logarithmic quantities and units: level, neper, bel". Read directly from the NIST PDF on 2026-07-29. The independent check on the factor-10 / factor-20 rule, reached through IEC 60027-3 rather than through the ITU: it gives LF = 2 lg(F/F0) B = 20 lg(F/F0) dB and LP = lg(P/P0) B = 10 lg(P/P0) dB, and adds the condition for their equality — "Note that if P/P0 = (F/F0)², then LP = LF" — together with 1 dB ≈ 0.115 129 3 Np.
3. [3]IEC 60027-3, "Letter symbols to be used in electrical technology — Part 3: Logarithmic and related quantities, and their units". The underlying standard that both sources above are built on: ITU-R V.574-5 cites it in its considering clause (b), and NIST SP 811 §8.7 states that its section is based on it. Cited as the definitional authority of record. Access: gated — the IEC sells this standard and it was not opened for this page; every figure attributed to it here is quoted through ITU-R V.574-5 or NIST SP 811, both of which were read in full.
4. [4]Recommendation ITU-R V.574-5 §6.1 and §6.5. Source of the statement on this page that a decibel figure is a ratio and not a level: §6.1 defines absolute power level and reserves dBW and dBm for levels referred to 1 W and 1 mW, and §6.5 defines the absolute voltage level dBu against 0.775 V RMS — the voltage that dissipates 1 mW in 600 Ω — noting that a voltage level equals the corresponding dBm figure only when the impedance really is 600 Ω, and otherwise differs by 10 lg(R/600). This is the same equal-impedance caveat as §1.2, expressed as a unit rather than as a factor. Retrieved 2026-07-29.