Decibel Voltage Gain

GdB=20log10 ⁣(V2V1)G_{dB} = 20 \log_{10}\!\left(\frac{V_{2}}{V_{1}}\right)

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Learning zone

Power goes as voltage squared, and the log of a square is twice the log — so a voltage ratio wears a 20 where a power ratio wears a 10. Ten volts out for one volt in is 20 dB; a 40 dB preamp multiplies 50 mV up to 5 V. Handy landmarks: 6 dB is a doubling of voltage, 20 dB is ten times, −3 dB is the 0.707 point that defines a filter's corner frequency.

Strictly, equating a voltage ratio to a power ratio in dB assumes the input and output see the same impedance — an assumption from the days when everything was 600 Ω. Modern audio and instrumentation ignore it and use 20 log of voltage as a convention, which is fine as long as everyone in the conversation agrees. Where absolute level is meant, look for the suffix: dBV references 1 V, dBu references 0.7746 V (1 mW into 600 Ω).

Decibel Voltage Gain
GdB=20log10 ⁣(V2V1)G_{dB} = 20 \log_{10}\!\left(\frac{V_{2}}{V_{1}}\right)
Where
  • GdBG_{dB}= Gain
  • V2V_{2}= Output voltage
  • V1V_{1}= Input voltage
Missing one of these? Work it out first, then come back