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Amplitude to dB Converter

Convert any linear amplitude ratio to 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

Amplitude ratio (A / A_ref)

Current value: 0.5 (must be ≥ 0)

Common Amplitude Values

Result

Decibels

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dB (voltage)

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Other mode

—

% of reference

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log₁₀(A)

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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).

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Common Amplitude → dB Reference

Amplitude (A)	dB (voltage / 20·log)	dB (power / 10·log)	Notes
100	+40 dB	+20 dB	100× gain (voltage)
10	+20 dB	+10 dB	10× amplitude
2	+6.02 dB	+3.01 dB	Double amplitude / double power
√2 ≈ 1.4142	+3.01 dB	+1.51 dB	+3 dB voltage point
1.0	0 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 dB	Half amplitude / half power
0.5	−6.02 dB	−3.01 dB	Exactly half (linear)
0.1	−20 dB	−10 dB	10% amplitude
0.01	−40 dB	−20 dB	1% amplitude
0.001	−60 dB	−30 dB	Audible noise floor
1.585e-5	≈ −96 dB	≈ −48 dB	16-bit theoretical floor
1e-6	−120 dB	−60 dB	Deep noise floor
6.31e-8	≈ −144 dB	≈ −72 dB	24-bit theoretical floor
0	−∞ dB	−∞ dB	Mathematical silence

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.

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.

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.

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).

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