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Old 20th October 2012, 12:19 PM   #11
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thanks guys. I had lost my brain yesterday. Im used to taking volt ratios and i use the 20log daily, but for some reason i had persuaded myself that SPL was a power metric and thus a 10log example. Thanks again andrewT and thanks to the others, even if you were even less correct than i!
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Old 20th October 2012, 01:11 PM   #12
sreten is offline sreten  United Kingdom
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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.
Hi,

Very confused.

8R in series with 8R reduces power and acoustic output by 6dB.

It reduces the voltage level expressed as power and dB by 6dB.

Looking at purely voltage (not dB) the voltage level is halfed.
Power is V[squared]/R, 20log converts the ratio to power.
(doubllng a log squares the value).

You can't talk about pure voltage ratios in dB, like half = 3dB,
its just wrong ,10log V2/V1 does not = dB, and is not used.

rgds, sreten.
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Old 20th October 2012, 03:00 PM   #13
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you cant talk of voltage ratios and express them in dB? Op amp gain is expressed in such a way, so i find that remark a little perplexing. The question was written whilst enebriated somewhat. I guess that i shouldve asked if an absolute voltage ratio will also result in an equal acoustic ratio. 3db is half power, 6db half amplitude, If i now recall correctly. I shouldve asked if spl is a amplitude or power type unit.
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Old 20th October 2012, 03:24 PM   #14
sreten is offline sreten  United Kingdom
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Hi,

All dB's are power - related amplitude ratios are 0.707 for -3dB, 0.5 for -6dB.

A x10 voltage gain op-amp stage has 20 (i.e. 100) dB gain.

You can't talk about x10 in dB, as its x100 in dB.

You can of course just talk about voltage gain, e.g. typically 20 to 30
for a power amplifier and it some respects its a lot simpler to most.
(Nobody talks about power amplifier gain in dB much .... )

Same with the gain of preamp stages, but here dB totally
dominates in expressing the gain, not voltage ratios.

rgds, sreten.

10log is used only for comparing power, e.g. 10W vs 1W = +10dB.
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Old 20th October 2012, 05:48 PM   #15
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ah ha. I see where i was mistaken. Thanks.

So, to attenuate 3db, my potential divider of Resistor and driver should have an absolute ratio of root2/2? The op amp thing confuses me still, but nevermind. Reading Capel's book for a minute revealed where my 3db half power misconception originated. Its half power in the sense that 2 drivers crossing at -3db sum to 0db. Sometimes im so daft. Thanks again.
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Old 21st October 2012, 01:48 PM   #16
dlp4341 is offline dlp4341  United States
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Default DB SPL

See Sound pressure
From Wikipedia, the free encyclopedia.

The Winner

10 Log p *p/pr*pr

Don
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Old 21st October 2012, 02:24 PM   #17
AndrewT is online now AndrewT  Scotland
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you can't be serious.

10 Log (power ratio) ^2 ??????
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Old 21st October 2012, 02:41 PM   #18
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there are times in my life when i regret asking questions when im half cut. I just wanted to confirm that a voltage ratio would translate to the same acoustic ratio. I have to admit that to my knowledge a crossover does 'cross' at -6db, even though in my confusion i said the contrary. Im at a loss to explain where the -3db half power bit comes from.
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Old 21st October 2012, 02:49 PM   #19
jwmbro is offline jwmbro  United States
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Quote:
Originally Posted by mondogenerator View Post
Reading Capel's book for a minute revealed where my 3db half power misconception originated.
3dB half power is not a misconception, it is true. if you have something with 100Watts (of anything, electrical or acoustical power or whatever you like) and you reduce it to 50Watts (of anything) you have 3dB less.

Thing is just that in order to get half the power into a given impedance, you need to reduce the voltage not to half its original value, but to 0.707 times its original value.

p (pressure), v (velocity), V (voltage)... all use 20log(out/in).

I (intensity), P (power)... all use 10log(out/in).

reason for this is simple, it's because the lower values are proportional to the second power of the upper values (P=V^2/R, I=p*v, P_ac=p*v*A...)
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Old 21st October 2012, 04:07 PM   #20
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yes. I should have said 'misunderstanding' rather than 'misconception'. My meaning was that i missed the point about half power in a voltage divider. You are quite correct, half power in terms of voltage is .707
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