# Why do designers use op-amps with fractional gains?

I often find designs like the following simulate this circuit – Schematic created using CircuitLab

Where the gain is less than one ($$\\ R_2/R_1 < 1\$$)

Why not simply use a resistive voltage divider? Beyond the inversion (which tends to be irrelevant in many applications), a divider with $$\\ (R_1-R_2) \& R_2 \$$ would produce the same output with the same input impedance. Plus, it will not have an offset problem, an input bias current problem, a transistor noise problem, or a bandwidth problem (adding a couple of capacitors can make it basically flat in frequency).

Although my first instinct is to regard this as a possible instability (gain beyond the op-amp specification), I see that it is basically a stable trans-conductance amplifier with the input current given by $$\\ V_{in}/R_1 \$$, so that is not a valid objection.

Why not use just a voltage divider? It decouples the input and output circuits.

• The output impedance is low
• The input impedance is R1
• The load has no effect on the input impedance

The last one is particularly important, as changing the load on the previous stage can have unexpected effects, e.g. changing frequency response or nonlinearity.

Why use this rather than a voltage divider and unity gain follower, for applications where phase doesn’t not matter?

• Same number of components
• The opamp’s inputs are at ground which is best for performance.

The opamp functions as a buffer, providing a much lower output impedance than the bare divider would have. This completely eliminates any loading effects created by the downstream circuitry.

A noninverting voltage follower configuration would provide the same benefit (and the same downsides), but if you want the inversion, this is the way to go.

Also, it is sometimes important for the application that the node between R1 and R2 be held at ground potential.

• @EdgarBrown most applications for a buffer in my experience, to counter your experience, are exactly to provide a low-impedance output to a high-impedance input. Kinda feels like the whole buffer idea. – Marcus Müller Nov 12 '18 at 21:31
• @MarcusMüller I have seen the same basic idea with differential amplifiers, in which the case is not as clear cut and impedance matching/buffering is definitively an issue (e.g., for high-speed differential ADCs). But I commonly see this in low-speed designs where a resistive divider would work at least as well. – Edgar Brown Nov 12 '18 at 22:04

If your sensor has high output impedance, yet the ADC needs to grab a bunch of charge and the sensor cannot provide that charge fast enough to support the desired sampling rate, then I could see using this circuit.