How this instrument works
Gain in decibels compresses a ratio of two signal amplitudes onto a logarithmic scale, because both human hearing and most measurement chains respond to relative rather than absolute changes. An amplifier that turns 1 volt into 10 volts has multiplied the signal tenfold, and 20·log₁₀(10) reports that multiplication as a clean 20 dB rather than the harder-to-compare raw ratio of 10.
The factor of 20, rather than the 10 used for power ratios, comes from how power relates to amplitude. Power is proportional to voltage or current squared, so a power-based decibel figure — 10·log₁₀(Pout ⁄ Pin) — becomes 10·log₁₀((Vout ⁄ Vin)²) once power is rewritten in terms of voltage, and squaring inside a logarithm doubles the coefficient sitting outside it. The 20 exists purely so a voltage-gain figure and a power-gain figure agree across the same amplifier stage.
The formula breaks at the edges every engineer eventually meets. Vin cannot be zero, since log₁₀ of an infinite ratio is undefined, and a phase-inverted or negative ratio has no real logarithm either — only the magnitude of Vout ⁄ Vin belongs inside the log. The result also measures amplitude gain alone: the two readings must be taken across matching impedances for the decibel figure to double as a power gain, which is why datasheets are careful to state the load a gain spec was measured into.
- Enter the Input amplitude (Vin) — the reference signal level, in volts or amps, ahead of the stage you are measuring.
- Enter the Output amplitude (Vout), measured in the same unit as the input, after that stage.
- Read Gain, dB — the instrument applies 20·log₁₀(Vout ⁄ Vin) automatically and shows the working.
- A positive result means amplification; a negative one means the signal lost amplitude.
Worked example — a tenfold voltage step
A preamp stage takes a 1 V input and produces a 10 V output — a clean tenfold increase, the kind of round number a datasheet quotes on purpose. The ratio Vout ⁄ Vin works out to 10 ⁄ 1 = 10, and since log₁₀(10) equals exactly 1, the gain is 20 × 1 = 20 dB. The instrument reports Gain, dB as 20 for these exact inputs, matching the familiar 20 dB figure engineers associate with a tenfold voltage stage.
The logarithmic scale compresses fast: doubling the voltage, a ratio of only 2, is 20·log₁₀(2) ≈ 6.02 dB, while a hundredfold jump is 40 dB, not 100 dB — every extra factor of 10 in the ratio adds a flat 20 dB. That additive behavior is exactly why cascaded gain stages are specified and summed in decibels rather than multiplied as raw ratios.
Questions
Why does voltage gain use 20·log₁₀ instead of 10·log₁₀?
Because power is proportional to amplitude squared. Power gain in decibels is 10·log₁₀(Pout ⁄ Pin); substituting P = V² turns that into 10·log₁₀((Vout ⁄ Vin)²), and squaring inside a logarithm doubles the coefficient outside it, giving 20·log₁₀(Vout ⁄ Vin). The extra factor of 2 keeps voltage-based and power-based decibel figures consistent for the same amplifier stage.
What does 0 dB gain mean?
Unity gain — the output amplitude equals the input exactly, so Vout ⁄ Vin = 1 and log₁₀(1) = 0. A unity-gain buffer or a cable with negligible loss both read close to 0 dB: no amplification and no attenuation, even though a real signal is still passing through the stage.
Can the gain figure come out negative?
Yes — a negative result means attenuation, not amplification. Whenever Vout is smaller than Vin, the ratio is below 1 and its logarithm is negative; an output at half the input amplitude, for example, gives 20·log₁₀(0.5) ≈ −6.02 dB. Passive attenuators and lossy cables commonly show gain figures in this negative range.
Does 20·log₁₀ apply to current ratios as well as voltage?
Yes. Current, like voltage, scales linearly with amplitude while power scales with its square, so the same derivation and the same factor of 20 apply — 20·log₁₀(Iout ⁄ Iin) is exactly as valid as the voltage form, provided both currents are measured across the same impedance.
Why is Vin not allowed to be zero?
Because a ratio with a zero denominator is infinite, and log₁₀ of infinity is undefined. Physically, a zero input paired with any nonzero output has no meaningful finite gain figure at all; a real circuit approaching that condition is better described by its noise floor or output offset than by a gain in decibels.
How is voltage gain in dB different from dBm or dBW?
This gain figure is a pure ratio between two amplitudes and carries no absolute reference, so it stays the same whether the signals are millivolts or kilovolts. dBm and dBW, by contrast, reference power to a fixed absolute level — 1 milliwatt or 1 watt — so a dBm value describes one actual power level, not a ratio between two measurements.