Case 1
- Output voltage
- 10V
- Input voltage
- 1V
20.0000dB
Open with these valuesChange any value to update the result.
20.0000dB
Voltage gain: 20.0000 dBYour current valid inputs, with one value varied across the range.
Held fixed: Output voltage 10.000000 V.
| Input voltage (V) | Voltage gain (dB) |
|---|---|
| 0.500000 | 26.0206 |
| 0.750000 | 22.4988 |
| 1.000000Your value | 20.0000 |
| 1.250000 | 18.0618 |
| 1.500000 | 16.4782 |
For voltage the factor is twenty, not ten: power goes with the square of voltage, and the square doubles the logarithm. Ten times the voltage is 20 dB of gain, twice the voltage is 6.02 dB, and an output below the input comes out negative.
20.0000dB
Open with these values6.0206dB
Open with these values-6.0206dB
Open with these valuesgain = 20 × log₁₀(V_out ÷ V_in)
| Output, input (V) | Voltage ratio | Gain (dB) |
|---|---|---|
| 1, 1 | Unity — a buffer | 0 |
| 2, 1 | Twice the voltage | 6.0206 |
| 10, 1 | Ten times | 20 |
| 100, 1 | A hundred times | 40 |
| 1, 2 | Halved — a divider | -6.0206 |
Power is proportional to the square of voltage, and the logarithm of a square is twice the logarithm. The factor 20 is the same rule as 10, applied to a field quantity.
Roughly twice the voltage — 6.020599913 dB exactly. Twelve decibels is four times, eighteen is eight times.
Yes. Current and sound pressure are field quantities like voltage, so the same factor of 20 applies.
The result turns negative, which is attenuation rather than gain. Halving the voltage is −6.02 dB.
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