I've been reading an article discussing single-supply op-amp design. Regarding Sallen-Key low-pass filter circuits, the author states that:
A few filter topologies, such as the Sallen-Key low pass (see section 1), require a resistive divider to DC bias the input. Addition of this network negates the requirement of a Virtual Ground.
By virtual ground, the author is referring to ICs such as the TLE2426 virtual ground chip. Can someone explain why Sallen-Key requires a resistor divider? Why won't biasing the AC-coupled signal (i.e. put a capacitor in series with the input) to Vcc/2 (provided by e.g. a TLE2426) work?
Best Answer
Let's look at a Sallen-Key low-pass filter:
One thing you will notice: the filter does not introduce any additional DC path to ground. C2 is connected to "ground", but since there is no DC path to it, it doesn't actually matter where it's connected, as long as it's a fixed voltage. We could just as well connect it to \$V_{CC}\$, or any other power rail. It doesn't matter, except for power-on transients.
How about a high-pass filter?
Here, we have a path to ground through R2, but R2 is 10kΩ. The point of a virtual ground IC is to provide a low impedance virtual ground, but here we need a 10kΩ ground. We don't need an IC for that, we just need a voltage divider made of two 20kΩ resistors. Sure, you could use a virtual ground IC and follow it with a 10kΩ resistor, but what's the point? A pair of 20kΩ resistors is a lot simpler.
Look at the Sallen-Key topology in general:
In this topology, there is always some impedance (\$Z_4\$) between the filter and ground. Since the point of a virtual ground IC is to make a low impedance ground, but we would never need that, the Sallen-Key "negates the requirement of a Virtual Ground". In other words, it isn't that you couldn't use a virtual ground IC: it's that you'd never need to use one.