formula in calculating attenuation in a series configuration?
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 20th July 2002, 06:39 PM #1 jarthel   diyAudio Member   Join Date: Jan 2002 Location: somewhere in Australia formula in calculating attenuation in a series configuration? Well I was using Sheldon's calculator (http://www.quadesl.com/attenuator.html). It gives you attenuation for a particular step. I gave the calculator the ff. values: 24 steps (I think he means positions here), 100k total resistance and total db of -64. It thens gives me several calculated resistance/attenuation for each step. Since the calculated resistor values is different from what's available, the calculated attenuation also changed. I found this site (http://www.enjoythemusic.com/steppedattenuator.htm) the gives a formula for calculating attenuation. Here's the formula Well, I calculated the db for step 23 which has 48.05 as the calculated resistance. here's the calculation: db=20*log(48.05/100000)= -66.xxdb. Sheldon's site gives be 64.xxdb. When I calculated the db for step 22 with 87.94ohm as calculated resistance. here's the calculation: db=20*log(87.94+48.05/100000)= -57.xxdb. Sheldon's site gives be 59.xxdb. I might be using an incorrect formula. So please guide me to the correct one. Thanks Jayel
 21st July 2002, 11:09 AM #2 gromanswe   Account Disabled   Join Date: Jul 2002 Location: North Can you have forgotten that log 20 scale always compare value to zero attenuation? While a "normal" scale have zero at nothing, the log20 have its zero at maximum output. zero attenuation. A high value is set to zero. Values below are-xxdB values above are+xxdB A normal scale have no negatives, as there can not be anything that doesn't exist. Also nothing can be added to what exist. The bible says you can not take way or add something. gro

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