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23rd June 2012, 12:35 PM  #1 
diyAudio Member
Join Date: Jul 2006

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? 
23rd June 2012, 12:42 PM  #2 
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. 
23rd June 2012, 01:30 PM  #3 
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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?

23rd June 2012, 02:32 PM  #4 
diyAudio Member

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
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23rd June 2012, 02:42 PM  #5 
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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. 
23rd June 2012, 02:57 PM  #6  
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Join Date: May 2007

Quote:


23rd June 2012, 03:49 PM  #7  
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Quote:
Hmmm. I tried 'INV LOG' on my W7 calculator but the INV doesn't work for logarithms it seems.... Thanks DF96! jan
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23rd June 2012, 03:54 PM  #8 
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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 fx570 (nearly 40 years old but still working OK) the INV works for almost all functions.

23rd June 2012, 04:53 PM  #9 
diyAudio Member

My only 30 years old Casio FX451 doesn't seem to have an INV key...
jan
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24th June 2012, 01:13 AM  #10 
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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 counterclockwise 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. 
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