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Amplitude to dB Converter

This free amplitude to dB converter turns any linear ratio into decibels. Use dB = 20·log₁₀(A) for voltage/SPL/dBFS or dB = 10·log₁₀(A) for power. Output: dB value in current mode, equivalent in the other mode, percentage of reference, log₁₀(A), and zone classification.

Input

A
Current value: 0.5 (must be ≥ 0)
Common Amplitude Values

Result

Decibels
dB (voltage)
Other mode
% of reference
log₁₀(A)
Formulas
Voltage / SPL / dBFS: dB = 20 · log₁₀(A)
Power: dB = 10 · log₁₀(A)
A = 0 → −∞ dB (silence); A < 0 is non-physical (phase inversion, not level).

Common Amplitude → dB Reference

Amplitude (A)dB (voltage / 20·log)dB (power / 10·log)Notes
100+40 dB+20 dB100× gain (voltage)
10+20 dB+10 dB10× amplitude
2+6.02 dB+3.01 dBDouble amplitude / double power
√2 ≈ 1.4142+3.01 dB+1.51 dB+3 dB voltage point
1.00 dB (unity)0 dB (unity)No change
1/√2 ≈ 0.7071−3.01 dB−1.51 dB−3 dB voltage / half-power for power mode
0.5012−6.00 dB−3.00 dBHalf amplitude / half power
0.5−6.02 dB−3.01 dBExactly half (linear)
0.1−20 dB−10 dB10% amplitude
0.01−40 dB−20 dB1% amplitude
0.001−60 dB−30 dBAudible noise floor
1.585e-5≈ −96 dB≈ −48 dB16-bit theoretical floor
1e-6−120 dB−60 dBDeep noise floor
6.31e-8≈ −144 dB≈ −72 dB24-bit theoretical floor
0−∞ dB−∞ dBMathematical silence

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About Amplitude & Decibels

Decibels compress a huge linear range into a manageable logarithmic one. A linear amplitude of 100 (100× the reference) becomes +40 dB; an amplitude of 0.00001585 becomes −96 dB. The conversion direction matters: amplitude (voltage, sound pressure, sample value) uses 20·log₁₀(A), while power (watts, energy flux) uses 10·log₁₀(A). The factor of 2 reflects the relationship P = A² — squaring inside a log multiplies the result by 2. This same logarithmic compression is why the decibel scale is indispensable for expressing dynamic range and signal-to-noise ratio (SNR). As an audio dB calculator, this tool handles both the voltage and power cases so you can convert any linear-to-decibels ratio without ambiguity.

When to use ×20 vs ×10

Use the voltage (×20) formula when the input value represents an amplitude-like quantity — voltage, sample value in a digital audio file, sound pressure, current, or any field-strength quantity. This is the formula behind dBFS, dBu, dBV, and dB SPL. Use the power (×10) formula when the input represents a power-like quantity — watts, intensity, or energy. This is the formula behind dBW, dBm, and most antenna gain figures. If in doubt: voltage → 20, power → 10. Note that dBu and dBV share the same 20·log formula but use different reference voltages (0.775 V for dBu, 1 V for dBV).

Peak vs RMS amplitude

Before converting to dB, you need to know which amplitude you have. Peak amplitude is the maximum excursion of the waveform; RMS (root mean square) amplitude is the power-equivalent average, which for a sine wave is peak / √2 ≈ 0.7071 × peak. Loudness meters, VU meters, and broadcast specs typically use RMS or A-weighted RMS (dBA). DAW peak meters show peak or true-peak (inter-sample peak). Mixing decisions are usually made in RMS headroom, mastering levels in LUFS (loudness units relative to full scale) — a perceptual RMS variant.

Why amplitude=0 gives −∞ dB

log₁₀(0) is mathematically undefined, but the limit as A → 0⁺ is negative infinity. Digital systems represent this as the "mathematical silence" — no signal at all. In practice, even "silent" digital files have a noise floor set by their bit depth: 16-bit floor ≈ −96 dBFS, 24-bit floor ≈ −144 dBFS, 32-bit float floor ≈ −192 dBFS. True −∞ dB is only meaningful as a mathematical limit.

Why negative amplitudes are non-physical

A negative amplitude doesn't represent a quieter signal — it represents a phase inversion. The level (magnitude) is the same; only the polarity flips. Since log₁₀ of a negative number is undefined in the reals, this converter requires A ≥ 0. If your signal has positive and negative excursions (every audio signal does), take the absolute value or use the peak or RMS envelope before converting. When combining two out-of-phase signals, use the dB addition calculator with care — phase cancellation reduces the combined level below either source.

Frequently Asked Questions

What does "amplitude 0.5" mean in dB?
Using the voltage formula: dB = 20 · log₁₀(0.5) = 20 · (−0.301) = −6.02 dB. So an amplitude of 0.5 is "half amplitude" or "−6 dB". Using the power formula it would be 10 · log₁₀(0.5) = −3.01 dB ("half power"). The common shorthand "−6 dB = half" refers to voltage/amplitude, not power.
Which formula should I use — ×20 or ×10?
Use ×20 when your input is amplitude-like: voltage, sound pressure, sample value, current, electric or magnetic field. This is by far the most common case in audio and music (dBFS, dBu, dBV, dB SPL). Use ×10 when your input is power-like: watts, intensity, energy flux, antenna gain. The result of the two formulas differs by a factor of 2 for the same linear ratio.
What does +6 dB mean for amplitude?
+6 dB voltage = roughly 2× the amplitude (exactly 10^(6/20) ≈ 1.995). It corresponds to 4× the power. This is why an audio engineer "boosting by 6 dB" doubles the peak sample values. By contrast, +3 dB voltage ≈ 1.41× amplitude (the square root of 2), and +3 dB power = double power.
Why does the converter complain about negative amplitude?
A negative amplitude is a sign flip, not a quieter signal — and log₁₀ of a negative number isn't defined in the real numbers. If you're measuring instantaneous samples (which can be negative), take the absolute value first, or use the peak or RMS amplitude over a window. If you specifically want to express "180° out of phase", that's a separate quantity from level.
What's the dB equivalent of amplitude 0?
Mathematically: −∞ dB. log₁₀(0) tends to negative infinity, so absolute silence is represented as "negative infinity decibels". In practice, every real system has a noise floor set by quantization or thermal noise: 16-bit digital ≈ −96 dB, 24-bit ≈ −144 dB, and analog gear is usually limited by thermal noise around −120 dB.
How do I get from a sample value (−32768 to +32767 for 16-bit) to dBFS?
Step 1: take the absolute value. Step 2: normalize to a 0..1 range by dividing by the maximum (32768 for 16-bit signed). Step 3: apply dBFS = 20 · log₁₀(normalized). Example: a 16-bit sample of −16384 → |−16384| / 32768 = 0.5 → 20·log₁₀(0.5) ≈ −6.02 dBFS. The same logic works for 24-bit (divide by 8388608) and 32-bit float (already in −1..+1 range — skip normalization).
What is the difference between peak dB and RMS dB?
Peak dB reflects the loudest instantaneous sample; RMS dB reflects the perceived average loudness. For a pure sine wave, peak amplitude is √2 (≈ 1.414) times the RMS amplitude, so peak dB = RMS dB + 20·log₁₀(√2) ≈ RMS dB + 3.01 dB. Complex audio (music, speech) has a much higher crest factor — peaks can be 10–20 dB above the RMS level. Broadcast loudness specs (EBU R128, ATSC A/85) target RMS-based LUFS, not peak dBFS.
How do I convert a voltage gain ratio to dB in a circuit?
Use the voltage formula: dB = 20 · log₁₀(Vout / Vin). Enter the ratio Vout/Vin as the amplitude. Example: an op-amp with Vout = 5 V and Vin = 0.1 V has a gain ratio of 50 → 20·log₁₀(50) ≈ +33.98 dB. For a passive attenuator reducing 1 V to 100 mV, the ratio is 0.1 → 20·log₁₀(0.1) = −20 dB. This is the same formula used for dBu and dBV conversions — only the reference voltage differs.
Why does "−3 dB" mean half power but not half amplitude?
Because power scales as the square of amplitude. Halving amplitude (ratio = 0.5) gives 20·log₁₀(0.5) = −6.02 dB. But halving power corresponds to amplitude ratio 1/√2 ≈ 0.7071, which gives 20·log₁₀(0.7071) ≈ −3.01 dB. So the filter cutoff frequency (the −3 dB point) is where amplitude drops to 70.7% of its passband value — not 50%. This is a frequent source of confusion when reading audio and RF specifications.