dB Gain Calculator (Power, Voltage and Current Ratios)

Background.

The decibel is a ratio, and almost every mistake made with it comes from forgetting that. It says how much bigger one power is than another; it does not say how big either of them is. A stage with 20 dB of gain multiplies whatever arrives by a hundred — a microwatt or a kilowatt, the number is the same. That is why decibels are so useful for chaining stages together (you add them instead of multiplying) and so easy to misread as a level.

The second mistake is the factor. A power ratio uses 10·log; a voltage or current ratio uses 20·log. The reason is not arbitrary. Power goes as the square of a voltage, so squaring inside a logarithm becomes a doubling outside it. The consequence is the fact that catches everyone once: double the power and you gain 3.01 dB, but double the voltage and you gain 6.02 dB. Both describe a real doubling; they are doublings of different quantities.

The third mistake is the one this page exists to fix, and most dB calculators get it wrong by silently ignoring it. The 20·log rule is only valid when the two voltages are measured across the same impedance. ITU-R V.574-5, the recommendation that defines how the decibel is used in telecommunications, says so in as many words: the factor 20 applies "assuming that the impedances are equal", and "use of the decibel when the impedances are not equal is not appropriate unless adequate information is given concerning the impedances involved". It then prints the corrected identity, which adds a term of 10·log of the impedance ratio.

That term is not a rounding detail. Take a small audio power amplifier with a 600 Ω input and an 8 Ω speaker output, driven with 0.2 V and delivering 6.32 V. The voltage ratio is 31.6, so a naive calculator reports 29.99 dB. The real power gain is 48.74 dB, because the impedance term contributes another 18.75 dB. Nearly nineteen decibels — a factor of 75 in power — went missing because two impedances were assumed equal when they were not. This calculator asks for both, applies the term, and reports it separately so you can see exactly how much of the answer it accounts for.

Four modes are on offer. Give it two powers and it uses the factor 10. Give it two voltages or two currents together with the impedances they were measured across and it uses the factor 20 plus the correction. Or give it a decibel figure and it runs backwards, returning the power ratio, the equal-impedance amplitude ratio, the surviving percentage, and the same quantity in nepers for anyone working with propagation constants and natural logarithms.

Two scope limits, stated here rather than buried in an FAQ. This page works in magnitudes: impedances are taken as the resistive values a measurement refers to, and complex impedance, phase and reactive loading are not modelled — for a load that is meaningfully reactive, the power the reactance carries is not delivered and a magnitude-only answer will overstate the gain. And a decibel figure is a ratio, never a level: to turn a gain into an actual output power you need an absolute starting point such as dBm, which is a different page.

What is db gain calculator?

A decibel is one tenth of a bel, and a bel is the base-ten logarithm of the ratio of two powers. That is the whole definition. Ten decibels is a power ratio of ten, twenty decibels is a hundred, thirty is a thousand — each ten decibels multiplies the power by another factor of ten, which is why enormous dynamic ranges compress into two-digit numbers.

Because it is a logarithm, decibels add where ratios multiply. An amplifier of +20 dB followed by a cable of −3 dB followed by another amplifier of +12 dB gives +29 dB overall, and you never have to multiply 100 by 0.501 by 15.8. This is the single reason the unit exists and the reason every link budget in radio, fibre and audio is written in it.

The decibel is also used for quantities whose square is proportional to power — voltage, current, sound pressure, electric field strength. The standards call these field quantities. To make a field-quantity level come out numerically equal to the power level it corresponds to, the logarithm is multiplied by 20 rather than 10. NIST Special Publication 811, restating IEC 60027-3, puts the two definitions side by side and adds the condition explicitly: if the power ratio equals the square of the field ratio, the two levels are equal. When the impedances differ, the power ratio is no longer the square of the voltage ratio, and the two levels part company.

The neper is the same idea built on natural logarithms instead of base ten: one neper is a field-quantity ratio of e, and therefore a power ratio of e². Telecommunications specifications use the decibel exclusively for international exchange, but nepers survive wherever results emerge from Napierian logarithms — propagation constants and attenuation coefficients in particular. One neper is 8.686 dB.

How to use this calculator.

1. 01Choose what you measured. Two powers is the simplest case and needs no impedance at all. Two voltages is the common bench measurement. Two currents appears in current-mode and antenna work. The fourth mode goes the other way: enter decibels, get ratios.
2. 02In the power mode, enter both powers in whatever unit you like — milliwatts, watts, dBm-free microwatts — as long as both are in the same one. Only their ratio is used.
3. 03In the voltage or current mode, enter both readings in RMS. Mixing an RMS reading with a peak or peak-to-peak reading is the fastest way to a wrong answer, and no calculator can detect it for you.
4. 04Then fill in both impedances honestly. If they are the same, leave them the same and the correction term will be zero, giving you the familiar 20·log result. If they differ — a 600 Ω input and an 8 Ω load, a 50 Ω generator and a high-impedance probe — enter the real values.
5. 05Read the impedance correction term as well as the gain. If it is large, any other calculator that only asked for two voltages has given you an answer that is wrong by that amount.
6. 06To chain stages, run each one separately and add the decibel figures. That is the entire arithmetic of a link budget: gains positive, losses negative, sum at the end.
7. 07Use the decibels mode to sanity-check a specification. A filter quoted at −60 dB rejection passes one millionth of the power; a −0.5 dB connector loses about 11 % of it. The percentage output makes both concrete.

The formula.

The base case is the definition. ITU-R V.574-5 §1.1 states that the bel expresses the ratio of two powers by the decimal logarithm of that ratio, and that the decibel is one tenth of a bel. So a power ratio in decibels is 10·lg(P₂/P₁), and nothing else enters — power is already power, and no impedance is needed.

For voltages, substitute P = V²/Z. The output-to-input power ratio becomes (V₂/V₁)² × (Z₁/Z₂). Taking ten times the base-ten logarithm of that product splits it into two terms: 20·lg(V₂/V₁) from the squared voltage ratio, plus 10·lg(Z₁/Z₂) from the impedances. For currents, substitute P = I²Z instead, and the impedance ratio comes out the other way up: 20·lg(I₂/I₁) + 10·lg(Z₂/Z₁). The current form is printed verbatim in ITU-R V.574-5 §1.2, which is the check that the sign convention here is right rather than plausible.

When the two impedances are equal, lg(Z₁/Z₂) = lg(1) = 0 and both expressions collapse to the familiar factor-of-20 rule. That is the only condition under which the familiar rule is correct, and the standard says so directly. This page keeps the term as its own output so that its size is visible rather than assumed away.

The worked example makes the size concrete. A voltage ratio of 31.6 gives 20·lg(31.6) = 29.9937 dB. The impedances are 600 Ω in and 8 Ω out, so the term is 10·lg(75) = 18.7506 dB. The total is 48.7444 dB. Computing the two powers directly — 0.2²/600 in, 6.32²/8 out — gives a power ratio of exactly 74 892, and 10·lg(74 892) is 48.7444 dB. The two routes agree to every digit, which is what confirms the algebra rather than merely restating it.

The reverse mode inverts the definition: a power ratio of 10^(L/10) and an amplitude ratio of 10^(L/20). The neper figure is L × (ln 10)/20, which is the conversion ITU-R V.574-5 §3 gives as 1 dB = (0.05 ln 10) Np ≈ 0.1151 Np and NIST SP 811 §8.7 gives as 1 dB ≈ 0.115 129 3 Np.

All arithmetic runs as exact decimals at forty significant digits and is rounded once, at the return boundary, to ten decimal places. The gain / unity / loss verdict classifies the unrounded decibel value, so a figure that displays as 0.0000000000 but is really a shade negative is still reported as a loss. The page refuses anything beyond ±2000 dB, a power ratio of 10²⁰⁰, because past that the ratio stops being representable and the honest answer is that you mistyped something.

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.