# High impedance opamp attenuator

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.

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

• You shouldn't have a problem attenuating it with a resistive voltage divider. No point in overcomplicating things, right? Then just use your op amp as a voltage-follower/buffer after the divider. Apr 25, 2017 at 12:17
• Why don't you want to use a voltage follower? You could make a negative gain to attenuate, but that requires an opamp and 2 resistor, but so does a voltage follower and a resistive divider. It is pretty common to use a voltage follower followed by a resistive divider. Apr 25, 2017 at 12:17
• Underconstrained - put some numbers on things like bandwidth, attenuation, input impedance of low pass filter, accuracy required etc... Apr 25, 2017 at 12:20
• Draw a schematic of what you think you need, explain what it needs to do. State the reasons why an opamp is needed. You're already thinking in solutions while you have not even fully described (the constraints) of your problem yet. Don't make presumptions like high impedance and that because of that you need a certain circuit. Apr 25, 2017 at 12:28

It's Thévenin equivalent (lower subcircuit) is a voltage source $v_{th} = v_{in} \frac{R2}{R1+R2}$ with series resistance $R_{th} = R_1 || R_2$. 