Can a 5V regulator connect directly to a voltage divider to obtain a 2V output

By your values, I have to assume that you are measuring off of the 10kOhm resistor on the bottom. I would hazard a guess that your 10kOhm resistor is a little high and the 100kOhm resistor is a little low. This would mean if your 10kOhm resistor is taking more of the voltage in the voltage division than expected.

There are a few problems at work here.

1. The first is the resistor value error. Ohm's Law: V=I*R. So if there is 5% error in R, then for a constant I, there will be a %5 error in the expected value of V. The %error from resistor tolerances are the maximum +/- error. So in reality, there could be a resistor with +5% error and other with -5% error. So there is a much larger possible range in output voltages for a constant current.

   • This can be eliminated by either hard-coding the real resistances in software or adding a reference voltage that is more accurate than the resistor or A/D quantization error. This reference voltage would be used to get the real resistance values as the A/D sees them.

2. A/Ds can have an offset error. This really has to be calibrated out. Some A/Ds have this built in.

3. A/Ds also have quantization error. So if a voltage falls between two consecutive quantization levels, it will have to be rounded to one of those two, introducing error. There are ways of increasing the number of bits for an A/D by oversampling and averaging blocks of samples into one sample.

There are other errors that could be at work, but those three are the main ones I have run into. If the A/D is fairly linear, then measuring two accurate voltages with the A/D circuitry would let you build an affine linear equation to correct the data. This builds off of the problems in 1 and 2.

A/Ds can appear linear, but end up being very nonlinear at a particular range. I have heard of some implementations using a look-up table to correct the nonlinear behavior. But that is getting a little beyond your current problems.

Edit: One more item. It is a good idea to buffer your analog signals to the A/D. It does a few things, like add another device to protect the microcontroller, and limit any possible transient sampling behavior from the A/D go into the analog signal.

Yes, the divider definitely will affect the gain calculation, because its equivalent source resistance (Thévenin resistance) of 6.23 kΩ is in series with one of your 10 kΩ gain-setting resistors. The overall gain will be -0.2325, rather than the -0.375 that you're looking for.

If you want an overall gain of -3/8, it would be simpler to forget about the voltage divider and just set up your opamp with 20 kΩ input and 7.5 kΩ feedback resistors.

Also, be sure to use an opamp whose common-mode input range includes ground.