Go Back   Home > Forums > Amplifiers > Solid State

Solid State Talk all about solid state amplification.

Please consider donating to help us continue to serve you.

Ads on/off / Custom Title / More PMs / More album space / Advanced printing & mass image saving
Reply
 
Thread Tools Search this Thread
Old 23rd June 2012, 12:35 PM   #1
diyAudio Member
 
luvdunhill's Avatar
 
Join Date: Jul 2006
Default attenuator dB to reistance calculation

Hi!

I have a project in which I'd like to use this pot:

http://www.partsconnexion.com/prod_pdf/tkd_cp600.pdf

There is a list of attenuations in dB that I'd like to convert to a pair of resistances (series and shunt) so I can plug them into a simulator. I don't have a pot on me, but it seems that this can be done using some math. However, I'm not sure how to do it.

Perhaps someone might be able to steer me in the right direction?
  Reply With Quote
Old 23rd June 2012, 12:42 PM   #2
DF96 is offline DF96  England
diyAudio Member
 
Join Date: May 2007
dB = 20 log ( R2 / (R1 + R2) ), where R1 is series resistor and R2 is shunt resistor. Potential divider theory, plus definition of voltage ratio dB.

Starting from dB, divide by 20, take the antilog. This gives you the ratio R2/(R1+R2). You then pick one resistor and some trivial algebra gives you the other resistor.
  Reply With Quote
Old 23rd June 2012, 01:30 PM   #3
diyAudio Member
 
luvdunhill's Avatar
 
Join Date: Jul 2006
Ok, so I know that R1+R2=10K (assuming that value is chosen) but dont I still need some value from the actual pot to accurately simulate it, or does that not really matter as far as the circuit is concerned?
  Reply With Quote
Old 23rd June 2012, 02:32 PM   #4
diyAudio Member
 
jan.didden's Avatar
 
Join Date: May 2002
Location: Great City of Turnhout, Belgium
Blog Entries: 7
The pot value would be your 10k, or whatever. That means R2 = 10k -R1. Plug that into the equation DF96 gave you, and you can calculate R1. R2 of course follows then.

Edit: so for say a 6db step, you know that log (R1/(R1+R2)) = 6/20 = 0.3
W7 calculator tells me that in this case, R1/(R1+R2) = 0.523 or thereabouts.
Since R2 = 10k - R1 (your choice), R1/(R1 + 10k - R1) = 0.523 which gives us R1 = 5.23k. Thus R2 = 4.77k.

Reality check: we know that -6dB is about half voltage -> result seems to be correct.

jan
__________________
Never explain - your friends don't need it and your enemies won't believe you anyway - E. Hubbart
Check out Linear Audio Vol 7
!

Last edited by jan.didden; 23rd June 2012 at 02:40 PM.
  Reply With Quote
Old 23rd June 2012, 02:42 PM   #5
RJM1 is offline RJM1  United States
diyAudio Member
 
Join Date: Jan 2010
Location: Titusville, Fl.
Do you mean something like this?

Just enter the value of the pot in the attenuator value box.

Series

Stepped Attenuator Resistor String Calculator - Neville Roberts

Shunt
http://homepages.tcp.co.uk/~nroberts/shunt.html

Last edited by RJM1; 23rd June 2012 at 02:59 PM.
  Reply With Quote
Old 23rd June 2012, 02:57 PM   #6
DF96 is offline DF96  England
diyAudio Member
 
Join Date: May 2007
Quote:
Originally Posted by janneman
Edit: so for say a 6db step, you know that log (R1/(R1+R2)) = 6/20 = 0.3
W7 calculator tells me that in this case, R1/(R1+R2) = 0.523 or thereabouts.
Since R2 = 10k - R1 (your choice), R1/(R1 + 10k - R1) = 0.523 which gives us R1 = 5.23k. Thus R2 = 4.77k.

Reality check: we know that -6dB is about half voltage -> result seems to be correct.
Jan, I think you need to buy a new battery for your calculator. -6dB is a voltage ratio of 0.5012, not 0.523. As -6dB is almost exactly half voltage we ought to expect resistor values which are almost equal, not 5% up and down. Your reality check needs a reality check!
  Reply With Quote
Old 23rd June 2012, 03:49 PM   #7
diyAudio Member
 
jan.didden's Avatar
 
Join Date: May 2002
Location: Great City of Turnhout, Belgium
Blog Entries: 7
Quote:
Originally Posted by RJM1 View Post
Do you mean something like this?

Just enter the value of the pot in the attenuator value box.

Series

Stepped Attenuator Resistor String Calculator - Neville Roberts

Shunt
http://homepages.tcp.co.uk/~nroberts/shunt.html
That seems correct, yes, in contrast to mine.
Hmmm. I tried 'INV LOG' on my W7 calculator but the INV doesn't work for logarithms it seems....
Thanks DF96!

jan
__________________
Never explain - your friends don't need it and your enemies won't believe you anyway - E. Hubbart
Check out Linear Audio Vol 7
!
  Reply With Quote
Old 23rd June 2012, 03:54 PM   #8
DF96 is offline DF96  England
diyAudio Member
 
Join Date: May 2007
OK, I understand: log 0.3 is -0.523. Maybe the INV key needs its contacts cleaning? On my old Casio fx-570 (nearly 40 years old but still working OK) the INV works for almost all functions.
  Reply With Quote
Old 23rd June 2012, 04:53 PM   #9
diyAudio Member
 
jan.didden's Avatar
 
Join Date: May 2002
Location: Great City of Turnhout, Belgium
Blog Entries: 7
My only 30 years old Casio FX451 doesn't seem to have an INV key...

jan
__________________
Never explain - your friends don't need it and your enemies won't believe you anyway - E. Hubbart
Check out Linear Audio Vol 7
!
  Reply With Quote
Old 24th June 2012, 01:13 AM   #10
diyAudio Member
 
luvdunhill's Avatar
 
Join Date: Jul 2006
Thanks guys! First, so does the 0 degree / 0 decibel on the datasheet in post #1 mean the spot when the knob is fully counter-clockwise and there is 0dB attenuation, or does that mean the resulting signal is 0dB? I'd suspect the values to be negative if it's the former, but I'm a bit confused. This changes how I situate the pot below, so perhaps this is something I need to figure out.

Regardless, I need a little more guidance. I calculated the values and I'm not positive how to situate the attenuator. I'm trynig a shunted pot configuration, so the wiper and one leg are tied together. The goal is as the attenuator is turned clockwise to increase the volume. I also want max volume to result in as little attenuation as possible.

Here's the portion of my circuit we're concerned with. The input signal in the second image is taken to the left of the series resistor R83 and R84.

The values in the A variable are supposed to represent turning the knob to the right. On the graph, the first value in A is the lowest blue curve. As you can see, the green curve is somewhat affected by the loading of the attenuator. Now, if I tie the wiper to the other leg, then I get (10K - A) as the resultant resistance. I'm not sure if this is what I want.

Also, R83 and R84 seem to be important, if you set these too big then your minimum attenuation doesn't equal to the input signal (as desired).

Ideally, the attenuation would be equally spread out over the input curve.
Attached Images
File Type: png atten.png (32.6 KB, 185 views)
File Type: png atten-1.png (41.1 KB, 178 views)
  Reply With Quote

Reply


Hide this!Advertise here!
Thread Tools Search this Thread
Search this Thread:

Advanced Search

Posting Rules
You may not post new threads
You may not post replies
You may not post attachments
You may not edit your posts

BB code is On
Smilies are On
[IMG] code is On
HTML code is Off
Trackbacks are Off
Pingbacks are Off
Refbacks are Off


Similar Threads
Thread Thread Starter Forum Replies Last Post
Calculation of stepped shunt attenuator Mick_F Chip Amps 2 25th January 2014 10:49 AM
Audax PR170M0 100 dB or just *94 dB?? rick57 Multi-Way 44 22nd December 2013 12:50 AM
Stepped attenuator resistor calculation software pmillett Tubes / Valves 5 13th October 2013 03:27 PM
Does anyone know the plate reistance of a 6CW7/ECC84 grhughes Tubes / Valves 2 22nd May 2009 08:01 PM
formula for DB calaculation for a ladder attenuator jarthel Parts 1 24th June 2006 08:58 AM


New To Site? Need Help?

All times are GMT. The time now is 05:58 AM.


vBulletin Optimisation provided by vB Optimise (Pro) - vBulletin Mods & Addons Copyright © 2014 DragonByte Technologies Ltd.
Copyright 1999-2014 diyAudio

Content Relevant URLs by vBSEO 3.3.2