Right I can't work out for the life of me what mu I need to increase signal 10dB, is it x10 or my other calculation of 3333.3 using two stages? It's for a feedback relay
Gain in dB's = log(A)*20 where A is the gain
Gain from dB' = 10^(dB/20) (10 powered to dB/20)
10 dB gain is 10^(10/20) = 3.1
Check also wikipedia for dB calculations.
Gain from dB' = 10^(dB/20) (10 powered to dB/20)
10 dB gain is 10^(10/20) = 3.1
Check also wikipedia for dB calculations.
I'm assuming you mean a x10 increase in volume means sound pressure level?
In which case the correct formula are:
10log(Pout/Pin) for power giving 3dBm increase for a doubled sound pressure level or 10dBm for a ten fold increase.
Translating that to a voltage gain into a load to achieve that sound pressure level gain you use:
20log(Vout/Vin) giving 6dB increase for a doubling of voltage and 20dB for a ten-fold increase.
In which case the correct formula are:
10log(Pout/Pin) for power giving 3dBm increase for a doubled sound pressure level or 10dBm for a ten fold increase.
Translating that to a voltage gain into a load to achieve that sound pressure level gain you use:
20log(Vout/Vin) giving 6dB increase for a doubling of voltage and 20dB for a ten-fold increase.
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