Basic voltage divider design question

I am interested in having the ability of adding attenuation to some of the devices I have been building, but want to make sure I understand voltage divider design first. I'll break up my lack of knowledge into a few questions :green: :

1. As I understand it, a voltage divider is basically a resistor in series on the signal line and a resistor in parallel shunting to ground. The voltage attenuation is equal to the ratio of the series resistor to the total resistance of the two resistors. Audio signal volume is not directly proportional to the voltage atten, but rather follows a log relationship. Is this correct?

2. How do you determine what the total resistor value is, or in another case, whether a 1K or a 1M pot is to be used?

3. I guess the power considerations would follow ohms law? Are there any rules of thumb for line level, mic level, etc. Obviously I can't strap an Alpha pot on the output of my Marshall to turn down the speaker. Does all line level equipment generally have similar current draw through the signal line?

4. I still need to understand input and output impediance better, but is this a factor in attenuation design?

My first use of an attenuator is going to be on the output for my Calrec EQ's to pad the gain caused by additive EQ. I also want to drop the output of my Alembic F2B bass pre, which seems to run about 20dB hotter than a normal line level output.

Thanks again,
Chris

[quote author="Emperor-TK"]1. As I understand it, a voltage divider is basically a resistor in series on the signal line and a resistor in parallel shunting to ground. The voltage attenuation is equal to the ratio of the series resistor to the total resistance of the two resistors.[/quote]

Yup.

[quote author="Emperor-TK"]Audio signal volume is not directly proportional to the voltage atten, but rather follows a log relationship. Is this correct?[/quote]

Well, there's two different things at play here... Any signal (audio or otherwise) going through an attenuator will have a level directly proportional to the ratio of (Rshunt)/(Rshunt+Rseries). Signal volume or loudness, however, must follow a logarithmic attenuation curve for our ears to hear a linear change. In other words, the ratio of the resistors must change rapidly in the first few dB of attenuation, and slow down as you dial in more and more attenuation. This is why we have audio-taper or log-taper pots.

[quote author="Emperor-TK"]2. How do you determine what the total resistor value is, or in another case, whether a 1K or a 1M pot is to be used?[/quote]

[quote author="Emperor-TK"]4. I still need to understand input and output impediance better, but is this a factor in attenuation design?[/quote]

It all depends on the input and output impedance of the circuit you're attenuating. If the output impedance is 10K (possible with tube stages), you want to use an attenuator whose total resistance is at least 10 times that, or 100K, so you don't load things down. At the same time, a 100K attenuator raises the impedance presented to the next stage of the circuit, so you need to make sure that the input impedance of the next stage is high enough...

[quote author="Emperor-TK"]3. I guess the power considerations would follow ohms law? Are there any rules of thumb for line level, mic level, etc. Obviously I can't strap an Alpha pot on the output of my Marshall to turn down the speaker. Does all line level equipment generally have similar current draw through the signal line?[/quote]

Don't use rules of thumb, the math is simple enough: Power equals the square of the maximum signal voltage you're planning to put through your attenuator, divided by the total resistance of it. Whatever figure you get, just go higher. Potentiometer specs give you maximum long-term dissipation and maximum peak dissipation, so check that.

Peace,
Al.

To supplement Al's reply, here's a good resource that covers much of what you've asked about.

http://www.uneeda-audio.com/pads/

OK, I've been studying Dave's link for a day or two, but am still having a little trouble fully putting everything together. Does this sound right?

The Calrec EQ is designed with modern impedance matching, the "bridged" approach which feautures low output impedances feeding high input impedances. Any pad that I put on the output stage of the Calrec needs to look like a high impedance to the Calrec, and a low impedance to the next stage that I am feeding it to. Using an L pad on each of the + and - outputs to ground, the Calrec sees the total af R series and R shunt as the impedance of it's following stage. The next device after the Calrec and it's pad has a shunt resitor on it's input which determines it's input impedance. That device sees only the shunt resistor in the pad as the output impedance of the device feeding it.

So if this all corect, the bugbear I see in using a potentiometer as a pad is in the "shunt resistor" part of the pot. The Calrec would see the input impedance as the pot value, regardless of pot's setting. However, the value that the following device sees as the output impedance will vary from zero to the pot's maximum value depending on how the pot is set, because it sees only the "shunt" part of the pot as it's input impedance. I'm not sure what the implication of this are since I don't fully get impedance matching or bridging yet.

Al, you also mentioned "loading" the output if the pot isn't big enough (I guess like having too low of an input impedance). The Uneeda article mentions loading the source if the series resistor is too small. Can someone elaborate a little more on what loading is? How does having a small series restor adversely affect things, since that is what a fully open potentiomer would look like.

I hope all these questions make sense. You guys have been a big help so far.

Thanks,
Chris

super cool ref dave and lots of excellent info everyone...thanks
thanks for creating the topic too.
i'l be studying this one as well.
later
ts