Precision Voltage Reference

The very highest output voltage you can expect from this circuit with the chip in regulation is about 3.7V, so a 1K pot would be more appropriate. In fact you should allow a bit of margin, so maybe 3.5V maximum.

Using a pot as a rheostat is bad, using it as all the resistance is worse, and using only 20% (or less) of the element is really, really horrible, even if it's a good pot. A 1% of full scale change in that pot means the output voltage will change by 5%, or 165mV if it's 3.3V.

If you're interested in millivolts, you should definitely limit the range of adjustment as much as possible. For example, if you need 3.3V you might use a 100 ohm pot with 787 ohms in series and an 499 ohm resistor for the bottom part of the divider.

For even better performance, shut the pot with a precision resistor of perhaps 1/10 the value and use it as a voltage divider. For example, a 1K pot used with a 110 ohm shunt. Then you could use a 452 ohm resistor for the and a 787 ohm resistor as follows:

^{simulate this circuit – Schematic created using CircuitLab}

A 1% of full scale change in the pot position will change the output by about 0.2%, which is 100x better than your circuit, whilst still using very inexpensive components. The purpose of shunting the pot is two-fold- pot elements have lousy tolerance compared to resistors and this reduces the variation, and they have lousy temperature coefficient so that is proportionally reduced. Using it as a voltage divider virtually eliminates errors due to contact resistance variation (CRV).

You're also putting considerable current through the TLV431, presumably so you can draw a lot of current from it. Consider using a lower current and buffering the reference- it will reduce temperature-related drift of the bandgap reference. Trade that off against the inaccuracy caused by a high impedance in the feedback terminal.

(of course the above example value are just for illustration- substitute your own requirements and do the math for your situation).

I think you've missed some of the limitations of the circuit. In effect, it takes a 5-volt PWM signal, low-pass filters it, then subtracts 2.495 volts to get the output voltage. Since the PWM signal can have a DC component in the range 0 to 5 volts, the output range is -2.5 to 2.5 volts. Its advantage over a simple RC filter on the PWM output is that a) it allows bipolar output, and b) with a 2-pole response it allows a better tradeoff of PWM switching frequency rejection vs. response time for changes in PWM duty cycle. Oh yes, and it's also cheap.

Note, though, that the accuracy of the output is only as good as the accuracy of both the 5-volt and 0-volt levels of the PWM signal. If you want the output to be accurate to 1 mV, the PWM high and low levels must both be accurate to within 1 mV.

There's no free lunch here. For what you're doing, I'd guess you'd be better off using the LM431 as a simple 2.495 reference, then use an analog switch driven by the PWM followed by a multi-pole filter. The more poles in the filter the sharper the cutoff frequency and the faster the response to duty cycle changes for a given level of switching frequency ripple. More poles means more complexity in the filter, but that's the price you pay. The filter could also incorporate the 0.8 gain you want to produce a 0 to 2 volt swing.