Paul_B I went back and found the article on the Internet concerning biasing a single supply digital potentiometer for audio operation. The biasing drawing is about halfway down in the article. Why would this applications engineer draw a drawing that is not correct?
Paul_B here is the circuit I am talking about. The Engineer has all these complicated mathematical formulas---then draws a incorrect drawing. For a guy like me trying to understand, things like this make it very difficult. Why couldn't the guy draw a correct diagram?
Well, it's almost correct. 
No, you have missed that the resistor for biasing the input op-amp should come from the "B" point on the potentiometer.
Paul_B I do not understand, could you draw a diagram?
I get the point. This diagram is what you are wanting Paul_B? I understand. This is the same way any op amp is biased for audio operation with a single supply.
What is crazy about digital pots to me is that both the A terminal and B terminal both have to be biased this way.
Paul_B Is this correct? Can I now use this pot in an analog circuit?
Why isn't this biasing of single supply pots in articles or books, for guys like me trying to learn???????????
Looks good. Anyone else have thoughts?
Maybe it's not a common concern.
Seriously, most of these things are used in commercial designs, not so much DIY. 
hmm, in the last circuit, as the top opamp is a unity gain buffer, pins A and B and the W of the pot are at the same potential - with a DC level as defined by the bottom opamp.
What did you find happened when you built and tested the offered circuit?
a7
alto777 I do not know yet, because I have not hooked it up to the offered circuit. I will try to do that soon and report back. We will see if it works.
That is exactly what is needed, no DC bias across the digipot because you are using it as a volume control.
It avoids noise generated by "moving" the "wiper" of the pot, which is particularly important with the mechanical variety. 
I have been hooking up everything to see how it works with audio. I need to point out something that is very important. The audio ground is different than the power ground.
The audio ground is the DC bias level. If you get the grounds mixed up the audio will not come through. Also, remember the wiper can only handle a tiny amount of current. The wiper must run to a buffer. The buffer can be a transistor or a op amp---but just remember they have to be biased for audio also.
If you will notice how they have biased the digital pot AD5116, in this schematic, it is better than the way I did it in my last diagram. Similar but different! The only thing I see wrong with this diagram is the Audio input impedance is really low for something like a guitar's magnetic pickup. The impedance to ground of the Audio input signal would be 33k and 33k, for a total of 66k. Somewhere between 250K and 500K is needed for a guitar's magnetic pickup. Of course the impedance of this circuit would probably be OK if the guitar's magnetic pickup had a buffered output to the input of the circuit.
I still have not breadboarded a bias circuit for a single supply digital pot for audio use to test. I hope to do that soon. In the mean time, trying to find parts to de-bounce the push buttons.
Like Op Amps, digital pots can be single supply or dual supply. The important thing I have learned ---ON MY OWN---is that either single supply or dual supply digital pots can be biased for audio operation.
The impedance of a voltage divider is equal to the parallel combination of the two resistors, 16.5K. So the total impedance to ground for the audio input is 49.5K, resistive, in series with 220 nF, capacitive.
jremington, the audio signal input path to ground does not go through the one 33k resistor that is in parallel with the other 33k resistor. So I respectfully do not agree. Current takes the shortest route, and that is through two 33k resistors in series---to ground.
Just wait till Mike gets on to you. 
To "channel" the ol' Grumpy Mike, current does not take the shortest route at all. Current takes every route in (inverse) proportion to the impedance of each. You can see how stupid the "shortest route" idea is in mains distribution. Are you seriously suggesting that the house lights go out when the electric radiator - clearly the shortest route, or the kettle - is turned on?
Nor does all current go to ground. It goes - or comes from, depending on the circumstances - "VDD" as well.
What is "seen" at the junction of the two 33k resistors, one to "VDD" and one to ground - is two 33k resistors in parallel (as the internal resistance of the power supply between "VDD" and ground is necessarily very small so the two are essentially the same connection when considering resistances); 16.5k. With another 33k resistor in series with that as @jremington points out to total 49.5k.
A bit of study of electrical network theory may be in order. 
Well, you are wrong. You are struggling with this project because you lack some really fundamental concepts of electronics theory and circuit design.
But you have come a long way! Good luck with the project.
Paul_B and jremington, thank you both for your insight and help. In thinking about the circuit I have changed my mind and I now agree with what you both said. Vdd is mixing with the audio signal. I only considered the audio path to ground, not taking into consideration the effect Vdd had on the circuit.
From the point of view of what a magnetic guitar's pickup needs to see in impedance--- there is not much difference in 49.5K and 66K impedance. Both of these number would degrade the guitar's frequency cutting a lot of highs. With that said---I know I am wrong, and should figure the exact impedance. There are situations where incorrect impedance calculations would really matter.
jremington, thank you for saying this: "But you have come a long way!" All I know about electronics has been self taught by reading and experimenting. I am a slow learner, but determination is one of my strong points. The whole point of this post is to learn how to use digital potentiometers. Why? Because a voltage divider is one of the basic building blocks of circuits.
I have not run any audio through my digital potentiometer circuits yet. Stay tuned and I will try to explain what happens when I do run audio through them.
Not correct either. The resistors and the 220 nF capacitor form a high pass filter on the input, and the value of the resistive component of the impedance is chosen to determine the frequency response.
jremington you are correct again. Yes it does form a high pass filter and the values of components make a huge difference in the frequency. For guitar's magnetic pickups I normally use.1uf capacitor and 500K resistor. But--- this is normally done going into a dual polarity op-amp where I don't have to bias the input of the op-amp. Even if I used a buffer op amp to feed the audio into the circuit above the high pass filter would also have an effect on frequency.
In the circuit above, if I did not use a buffer op amp to feed the audio in, I would play with the values of the resistors and the capacitor. The capacitor would be bigger and the resistors would be bigger.