amplifier
I want to make an audio amplifier with digital volume control. I figure that by adjusting the feed back path that I won't change the input impedance and won't affect the frequency response. (other than the gain bandwidth product)
Does the following setup look reasonable?
Would a digital pot be good for this application?
Are there any drawbacks of digital pots that i'm not aware of that would make them undesirable for this application?
simulate this circuit – Schematic created using CircuitLab
Regarding the capacitor, Ignacio beat me to it. shakefist You'll see capacitive coupling all over the place in circuits like these.
As far as what you feed in, actual amplification will depend on the resistor values. The specs say -.3 to vdd (your source voltage) + .3V, so provided your VDD is >= 3 volts, you should be able to amplify to that level.
update
You can simply leave the V01 disconnected. The speaker wire connected to pin 5 in that schematic would be connected to ground instead.
That said, the configuration shown in the schematic is said to have a gain of $$2(\frac{R_i}{R_f})$$ This is because the outputs are 180 degrees out of phase with eachother, so that even though both amplifiers being unity gain (when Ri = Rf), it winds up doubling the input.
If you don't use pin 5(V01), you'll need to adjust your gain. The new formula would be $$\frac{R_i}{R_f}$$
That said, if you haven't already ordered a different chip, I think the LM4880 is more in line with what you want, it's mentioned in the automatic switching circuit in the datasheet, and it's a single output headphone amplifier.
This is not 100.00% complete:
simulate this circuit – Schematic created using CircuitLab
Re: comments
The YouTube lady cut this out of a larger circuit to demonstrate the potentiometer's function, not necessarily how to use them correctly for audio applications. We aren't shown the input or output stages. The pot should be grounded at or below the lowest voltage expected on the input. As per the comments, her diagram also shows input and output swapped. Her setup was "good-enough" to demonstrate the pot doing work.
Take all internet advice with a grain of datasheet-salt.
See figures 13 and 14 in this selection guide.
Figure 13 is referenced to 0V and is decoupled after the pot.
Figure 14 is referenced to -15V and is decoupled immediately.
Best Answer
I can see two things that are not optimal in that design. For the discussion I will use a random digital potentiometer I found from this article: The DS1881. Most are similar in construction.
Wiper resistance
A digital potentiometer has a pretty high wiper resistance, especially compared to a mechanical potentiometer. Not only is it high, possibly in the order of hundreds of ohms, but it is also not very stable or well-defined. The DS1881 lists it as 160 Ω typical, but up to 250 Ω maximum. It can vary with the signal voltage and such.
It is not a problem if it is in series with a high impedance input, for example the input of an op-amp. Then there is negligible current through it, and thus, a negligible voltage drop. In your case, the wiper is in parallel with a section of the potentiometer:
simulate this circuit – Schematic created using CircuitLab
Since some amount of current will flow here, it will affect the signal. How bad? Not much, it is somewhat swamped by the 10 kΩ series resistance and the rest of the potentiometer section, but that leads to the next problem:
Absolute resistance variation
A digital potentiometer has a well-defined and trimmed ratio, but its absolute resistance can vary a lot:
Note the ±20% resistance tolerance, and the 750 ppm/°C temperature coefficient.
Since you want to use it as a rheostat, you are hit by this. If it matters, that is up to you. It will really only make the gain fluctuate a bit by temperature and so on. A digital potentiometer should really be used as a voltage divider, so you can take advantage of the close matching between the steps.
The article I linked to, from Maxim, has a few audio circuits you may want to take a look at, with pros and cons.