Doug Self 16 dB passive attenuator

[angelo66]

Hi,
it's not clear to me what could be the values of R57 / R58 resistors of the "16 dB passive attenuator" illustrated by Mr. Self at page 394 of "Small Signal Audio Design".

The only hints reported my Mr. Self is that the input impedance of the attenuator is around 10 Kohm, and output impedance is 1.4 Kohm.
I looked for more details throughout his past articles in Electonics Word + Wireless World, and Linear Audio, but no success.

You guys have any ideas about those values?

Any help would be appreciated!

Angelo

 

[JonSnell Electronic]

R57 9k1
R58 820R
will give you -12dB
There is no need for -16dB but if you want, change the values using expensive special value resistors by all means.
R57 9k2
R58 800R

Here is an example of how to work out your actual values.

 

[angelo66]

R57 9k1
R58 820R
will give you -12dB
There is no need for -16dB but if you want, change the values using expensive special value resistors by all means.
R57 9k2
R58 800R

Here is an example of how to work out your actual values.

Thanks a lot for your super-fast response! So it seems Mr. Self left the solution "open" to reader's needs...

I can happily go with -12 dB as you suggested.

Best,
Angelo

 

[nattawa]

To have 16db attenuation it requires R58 = 1/6 *10kohm.
Hence
R58=1.6k R57=8.2k, using E24 series values.

 

[angelo66]

To have 16db attenuation it requires R58 = 1/6 *10kohm.
Hence
R58=1.6k R57=8.2k, using E24 series values.

Thanks nattawa! For confirmation, I re-did all calculations for 16 dB attenuation, and got quite the same results as yours.
Even Rout matches with Mr. Self's result.

Angelo

 

[nattawa]

I'm glad I was close 🙂 .
I did it the quick dirty way.
-16db is -10db + -6db, which means dividing a voltage by 3 then by 2, and that's our 1/6.
The rest was the same as yours.