Electronic – TL431 as negative voltage reference

It will work, but it will not be very stiff.

(3.135V - 3V) / 560R gives 241 microamps. If the load is higher than that, the voltage drop across the 560 ohm resistor will lower the input voltage to below the setpoint you want.

Typically you should allow the 431 to eat 1 mA to keep all of the datasheet assumptions in check. You should go with a lower series resistor - 51R allows 2.65mA at minimum input. The 431 can safely eat a few milliamps so don't worry about power.

A quick analysis shows the lowest voltage = 2.980V, the highest 3.012V (nominal 2.996V) so we're looking at +/- 0.53%. Worst-case setpoints are when the two resistors are at opposite ends of the tolerance spectrum (one large, one small).

You state the you want to run 96 LEDs ...is that a string of LEDs in series?
If it is then you have other problems with a 20 mA constant current driver.

Working out the resistors for the TL431 is very simple, consider this simplification from Figure 39 of the datasheet.

The circuit of Figure 38 seems non-ideal for your application (if it's a string of LEDs you want to drive), and Figure 39 modified slightly may be better:

Here I've used LEDs with a Vf of 2.2 and a supply of 36 V, if it's different for your LEDS or supply, then you'd have to recalculate the values.

If you drive the LEDs from a low side constant current source than you get a couple of advantages...

1. The voltage above R1 is constant and unrelated or impacted by Vce for the transistor.
2. The TL431 Ik, which will vary to some extent (transistor Hfe and temperature dependent) is NOT added to the current for the LED string.
3. You can drive multiple strings with just one TL431 as shown below (now you might understand why I set the Ik at 5 mA in my scenario).

If you want to run 96 LEDs as a string it might be done as below:

Here with each 2n2222 Ib at 400 uA, there is still 2.2 mA current in Ik, well above the minimum requirements of 1 mA.
Each string of LEDs is now only 12 devices instead of 14, but the voltage (Vf) difference for the string simply appears across the 2N2222. So Vce would increase by 4.4 V. This does increase the power dissipation in the transistor, but it's well within it's ratings.

Note that you could put more LEDs in a string and reduce the number of strings, but you'd have to increase the Vce of the transistor. The 2N2222 is only 40 V Vce.
You could also use a more capable power FET instead of a transistor, but this increases the complexity of the design in terms of minimum voltage etc.

Hope this helps.