Is it possible to design a single-op-amp schematic to attenuate a high impedance signal? The designs that I know have these limitations:
- Non-inverting with input directly going into op-amp, has gain ≥1
- Inverting op-amp draws current from the input.
The input is a pressure sensor (probably Wheatstone bridge based) but the impedance is in the area of 100kΩ.
The output goes into a low-pass filter (out of my control) which has non-negligible impedance, and ADC, so it needs to be driven hard.
If possible I don't want to use a voltage follower to buffer the signal. My feeling says it should be possible with one op-amp but I can't think of a way, and searching led me to nothing. Inverted-ness doesn't matter.
[Edit]
The resistance of the filter measures 5.1kΩ at DC. The bandwidth is non-existant (pressure changes take multiple seconds). Accuracy is not important as long as it's repeatable, the min and max are defined, and the domain is monotonic. Input min/max ~2V-12V. Output min/max ~1V-5V. Opamp is rail-to-rail, single-rail powered with 0V/12V.
simulate this circuit – Schematic created using CircuitLab
(From 
Best Answer
Normally it's no problem at all if the output signal you want to feed into the low pass filter has some non-zero impedance. It's just important to know what impedance it has.
Therefor you can use a voltage divider after a voltage follower (upper subcircuit).
It's Thévenin equivalent (lower subcircuit) is a voltage source \$v_{th} = v_{in} \frac{R2}{R1+R2}\$ with series resistance \$R_{th} = R_1 || R_2\$.
simulate this circuit – Schematic created using CircuitLab