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Attached is attenuator calculator spreadsheet. You can insert any resistors values to tailor the range, total resistance and steps. You will notice that multiplying values by two produces the smoothest curve (and truly logarythmic). The input resistance is constant.
Two additonal relays, setting smallest resistor values, extend the range by two more steps (user adjustable).
The spreadsheet was prepared by Algar_emi.
Two additonal relays, setting smallest resistor values, extend the range by two more steps (user adjustable).
The spreadsheet was prepared by Algar_emi.
Peter Daniel said:
Interesting, perhaps I am missing something, but when I used increasing powers of 2 on that spreadsheet I did not see a logarithmic curve (voltage rising logarithmically with steps toward 0 attenuation), but rather a reverse log or anti-log curve. Not exactly what most people are looking for, especially not I, but hey maybe someone is. Maybe I missed something, it would not be the first time.
Constant input impedance is the same behaviour as a standard pot, not a bad thing, but then again not always the most desirable. As I (feable as my mind is) understand it is nearly always better to keep the impedance between stages (especially is you are attenuating into a stage) constant. So that the input to the stage (say an amplifier) sees something better approximating an ideal source. Of course you could add a buffer to either end and solve (or create) a whole host of problems.
Still it is an intersting approach Peter and very similar to one I have seen elsewhere. I may give that approach a wirl too, it is simple enough to implement.
I chose a constant output impedance approach intentionally to feed the XBOSOZ.
Cheers!
Russ
I'm pretty much sure you are missing something. Take a look at this link: http://www.diyaudio.com/forums/attachment.php?s=&postid=466945&stamp=1094088615
The curve produced by the attenuator is similar to 210Y. It needs some figuring out which way to look at resitance value and rotation though to fully understand it.
Constant input impedance has advantage when using attenuator directly at amp's input, exactly what I'm doing now.
The curve produced by the attenuator is similar to 210Y. It needs some figuring out which way to look at resitance value and rotation though to fully understand it.
Constant input impedance has advantage when using attenuator directly at amp's input, exactly what I'm doing now.
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Joined 2002
Russ White said:
That's indeed interesting, because what you are looking on at the curve is steps vs attenuation function (and not actual voltage). The voltage is perfectly linear function: 50/50 split halfway and 25/75 at quater turn from each end. This works perfectly well in real world and small attenuation values are actually preferred at upper half of the range and much bigger steps at lower listening levels.
You should be looking this way at the curve presented in the spreadsheet:
Attachments
Sure, but you could have easily googled it.
http://sound.westhost.com/pots-f4.gif
Notice it is expressed in terms of output, not attenuation, but the curve is the same regardless.
The Log curve is complete opposite your curve, which is antilog(also shown).
Cheers!
Russ
http://sound.westhost.com/pots-f4.gif
Notice it is expressed in terms of output, not attenuation, but the curve is the same regardless.
The Log curve is complete opposite your curve, which is antilog(also shown).
Cheers!
Russ
Well Russ, acording to graph represantation you posted, the graph presented in my spreadsheet is a perfect log function of attenuation vs number of steps (I'm not talking here about percentage of output).
The actual percentage of output is a linear function vs number of steps, in my attenuator.
I never claimed otherwise.
Now, the log function of attenuation vs steps gives you simply more resolution in attenuation at higher volume levels and less resolution at lower volumes, something I already mentioned as well.
So what exactly is your point?
The actual percentage of output is a linear function vs number of steps, in my attenuator.
I never claimed otherwise.
Now, the log function of attenuation vs steps gives you simply more resolution in attenuation at higher volume levels and less resolution at lower volumes, something I already mentioned as well.
So what exactly is your point?
That is incorrect, your attenuation column values would have to be reversed and the graph remain static for that to be true.
No, it is is not log attenuation, it is antilog attenuation in terms of decibles attenuation per step.
A standard audio taper pot will give pseaudo log attenuation in terms of voltage. A true log pot or attenuator gives linear attenuation in terms of decibles. And your attenuator gives antilog performance in terms of decibles, not linear in terms of decibles as a true LOG attenuator would.
My exact point is that your circuit according to the graph and function in
the spreadsheet you provided increases attenuation in terms of decibles as an antilogarithmic function as the progression of the steps occurs.
It is obvious.
I am sure you will think it is marvelous, and that is great!
No, it is is not log attenuation, it is antilog attenuation in terms of decibles attenuation per step.
A standard audio taper pot will give pseaudo log attenuation in terms of voltage. A true log pot or attenuator gives linear attenuation in terms of decibles. And your attenuator gives antilog performance in terms of decibles, not linear in terms of decibles as a true LOG attenuator would.
My exact point is that your circuit according to the graph and function in
the spreadsheet you provided increases attenuation in terms of decibles as an antilogarithmic function as the progression of the steps occurs.
It is obvious.
I am sure you will think it is marvelous, and that is great!
Russ White said:
While you may be sure of that, I don't see a point whatsover what attenuation represantation curve has to do with greatness of an attenuator curcuit (especially a circuit with 130 steps, controlled by a software).
I stand corrected, and indeed, because attenuation is represented in negative numbers it will be actually antilog function, something I didn't notice before. But this still doesn't change anything and my previous point is valid (with the only difference that the word log is replaced by antilog)
:"The actual percentage of output is a linear function vs number of steps, in my attenuator.
I never claimed otherwise.
Now, the antilog function of attenuation vs steps gives you simply more resolution in attenuation at higher volume levels and less resolution at lower volumes, something I already mentioned as well.
So what exactly is your point?"
My point was made very early on, and it remains quite simple.
It is that your circuit (while fine as it may be) is the exact opposite of what most people will choose for volume control. Now that may be desirable for you, and maybe others, but it is in fact opposite what most people would use.
I for one would not give a hoot about a faction difference of change at -10db attenuation. Just give me a nice linear (in terms of DB) change when I change volume, when you do that, what you have is a logarithmic control. Anything else is... well something else.
Anyway. It was fun.
Ciao.
It is that your circuit (while fine as it may be) is the exact opposite of what most people will choose for volume control.
Now that may be desirable for you, and maybe others, but it is in fact opposite what most people would use.I for one would not give a hoot about a faction difference of change at -10db attenuation. Just give me a nice linear (in terms of DB) change when I change volume, when you do that, what you have is a logarithmic control. Anything else is... well something else.
Anyway. It was fun.
Ciao.
Russ White said:It is that your circuit (while fine as it may be) is the exact opposite of what most people will choose for volume control.
And here you are not correct: it is not exactly opposite. You just mentioned that "a true log pot or attenuator gives linear attenuation in terms of decibles". Mine gives antilog, so it's only half way off.
And I find it advantegous, as with a limited number of steps (130 vs 400) I prefer higher steps resolution at upper range of volume levels. So I'm settling on 0.15dB average and not 0.5dB
It surely was fun and learning expierience
Peter Daniel said:
And here you are not correct: it is not exactly opposite. You just mentioned that "a true log pot or attenuator gives linear attenuation in terms of decibles". Mine gives antilog, so it's only half way off.
And I find it advantegous, as with a limited number of steps (130 vs 400) I prefer higher steps resolution at upper range of volume levels. So I'm settling on 0.15dB average and not 0.5dB
It surely was fun and learning expierience
Ha true, on that you are right.
I respect you, and I appreciate the spirited debate.- Status
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