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-   -   10log or 20log, can someone check my thinking? (http://www.diyaudio.com/forums/multi-way/221854-10log-20log-check-thinking.html)

 mondogenerator 19th October 2012 03:29 PM

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.

 mondogenerator 19th October 2012 03:32 PM

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

 powerflux 19th October 2012 04:11 PM

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

 AndrewT 19th October 2012 04:15 PM

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

 WithTarragon 19th October 2012 04:16 PM

Quote:
 Originally Posted by powerflux (http://www.diyaudio.com/forums/multi-way/221854-10log-20log-check-thinking-post3207499.html#post3207499) power is 20log, voltage is 10log. afaik.
No!
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)

 mondogenerator 19th October 2012 04:32 PM

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?

 AndrewT 19th October 2012 08:37 PM

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.

 dlp4341 20th October 2012 06:58 AM

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

 Wavebourn 20th October 2012 07:00 AM

Quote:
 Originally Posted by mondogenerator (http://www.diyaudio.com/forums/multi-way/221854-10log-20log-check-thinking-post3207459.html#post3207459) 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.
On which frequency?

 AndrewT 20th October 2012 10:51 AM

Quote:
 Originally Posted by dlp4341 (http://www.diyaudio.com/forums/multi-way/221854-10log-20log-check-thinking-post3208299.html#post3208299) ........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...........
I don't fully understand your post.

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:
 ? dB = 20 Log volts in/volts out
this is correct
Quote:
 ?dB = 10 Log volts in^2/volts out^2
this one is also correct but why would you ever use it? It is far simpler to use 20times rather than 10times the log[x].

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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