10log or 20log, can someone check my thinking?
hi all. If i want to use a resistor to pad a driver or alter Qes, i use 10log for voltage ratio and 20log for acoustic ratio. Is this right? I.e. If i use a 8R resistor in series with an 8R driver, i would reduce acoustic output 3db, but voltage 6db.

ah. I may have answered my own question, but can anyone confirm or correct me?

power is 20log, voltage is 10log. afaik.

inserting the divide by 2 resistor ladder/attenuator reduces both the voltage and the power by 6dB.

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Voltage and SPL are 20 log measures (so 1/2 is 6dB and 1/4 is 12dB) Power, energy, Intensity are 10 log measures (so 1/2 is 3dB and 1/4 is 6dB) 
thanks tarrogan and andrewT.
Ive just realised that db ratio isnt really necessary unless i need say 5db reduction, since i just need to halve it, the absolute ratio is all i need. Gosh i feel a fool... Re reading my post i spotted my error, voltage is 20log. So, to get a halving of acoustic output, i need an absolute ratio of 0.5? Thats correct, isnt it? 
Yes, an absolute ratio of 0.5 is 6dB.
That would be equivalent to having half the voltage (your 0.5 ratio) and also equivalent to 1/4 (one quarter) of the power. 
Gents
dB is a power ratio normally with a reference to a given power. Say we put 4mw into a cable (or a filter) and the output is 1 mw ? dB = 20 Log ( Power in / Power out ) ? dB = 20 Log ( 1/4 ) ? dB = 20 Log ( 4 ) ? dB = 20 * 0.3 ? dB =  6 So our circuit loss is 6 dB If we are measuring voltage, power = volts^2/R: volts squared divided by resistance. ?dB = 10 Log volts in^2/volts out^2 ? dB = 20 Log volts in/volts out Don 
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But there are some gross errors. decibel ratio = 10*log[p1/p2] decibel ratio = 20*log[v1/v2] all the rest is gobbledegook except Quote:
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Oh, your first statement was almost right: dB decibel is a ratio. don't add voltage or power or current or anything else. It is a ratio. 
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