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Old 19th October 2012, 03:29 PM   #1
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Default 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.
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Old 19th October 2012, 03:32 PM   #2
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ah. I may have answered my own question, but can anyone confirm or correct me?
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Old 19th October 2012, 04:11 PM   #3
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power is 20log, voltage is 10log. afaik.
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Old 19th October 2012, 04:15 PM   #4
AndrewT is offline AndrewT  Scotland
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inserting the divide by 2 resistor ladder/attenuator reduces both the voltage and the power by -6dB.
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Old 19th October 2012, 04:16 PM   #5
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Quote:
Originally Posted by powerflux View Post
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)
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Old 19th October 2012, 04:32 PM   #6
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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?
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Old 19th October 2012, 08:37 PM   #7
AndrewT is offline AndrewT  Scotland
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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.
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Old 20th October 2012, 06:58 AM   #8
dlp4341 is offline dlp4341  United States
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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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Old 20th October 2012, 07:00 AM   #9
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Quote:
Originally Posted by mondogenerator View Post
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?


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Old 20th October 2012, 10:51 AM   #10
AndrewT is offline AndrewT  Scotland
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Quote:
Originally Posted by dlp4341 View Post
........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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Last edited by AndrewT; 20th October 2012 at 10:55 AM.
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