# formula in calculating attenuation in a series configuration?

#### jarthel

##### Member
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

#### gromanswe

##### Disabled Account
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