Electronic – Low Voltage Shunt Regulator

The problem is that your 500mV output requirement is rare, so that very few regulators use a reference that is lower than ~1.25V (a few use 800mV).

One way to get this to work is to amplify the feedback signal using an op-amp. A gain of 3 would allow you to use any regulator with a reference voltage up to 1.5V.

Edit:

Say your regulator has a 1.25V reference, so you'd normally use a voltage divider RA/RB on the output such that Vout = 1.25V (1+ \$\frac {R_A}{R_B})\$.

Add a suitable op-amp to amplify the output voltage:

Vin comes from your output voltage. Say it is 500mV and your regulator has a reference voltage of 1.25V. Vout goes to the feedback input of the regulator (where the tap on the voltage divider goes).

In general your output voltage will be Vout = Vref (\$\frac {R_G}{R_F+R_G})\$, so in this particular example we might pick Rg = 10K and thence calculate Rf = 15K.

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).