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Answer by periblepsis for Sanity Check: Does adding a second voltage divider change the properties of the original?

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First off, you are right (as everyone clearly points out) that a second voltage divider will load down your existing circuit.

You've got a lot of useful responses. I've one additional thought to add. Mostly, because of its simplicity and it allows you to use your voltage divider idea without an opamp.

schematic

simulate this circuit– Schematic created using CircuitLab

\$R_1\$, \$R_2\$, \$Q_1\$, and \$Q_2\$ are arranged so that the \$R_x\$ and \$R_y\$ voltage divider are buffered and therefore won't load down your sensor output. The base of \$Q_1\$ will only draw about \$1\:\mu\text{A}\$. So it does not mess with the sensor much. (On the order of \$10\:\text{mV}\$, though you can alter \$R_1\$ and \$R_2\$ to make this even less, if desired.) The two BJTs are also arranged so that their respective \$V_{_\text{BE}}\$ values compensate for each other (mostly -- I also tried to keep their respective current densities about the same, as well) and this reduces the impact of operating temperature, though there will still be some relative offset differences for unmatched junkbox BJTs.

Regardless, you can now fairly freely set \$R_x\$ and \$R_y\$ as desired to get the voltage into the range you want.

It's a really simple circuit and takes only a few parts. That's its strength. You can cobble it up very quickly and test it. However, how well it satisfies your needs is another issue that you will have to work out.

(I left \$V_{_\text{CC}}\$ unspecified for now as I'm not sure if you want to use \$5\:\text{V}\$ there, or something else. But \$V_{_\text{CC}}\ge 5\:\text{V}\$ here.)


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