I'm using a TL431B (±0.5%) shunt regulator to get 3V ±1% (max.) from a 3.3V ±5% supply, so from 3.135V to 3.465V. The 3.3V supply is a switcher, so has a decent amount of high frequency (~3 MHz) noise on it.
This is related to my previous question;
Does the TL431 have a "dropout voltage" per se?
In fact the circuit configuration is very similar but I adjusted the values of the resistors, so I'll use the same diagram drawn in my TomCAD style:
560R 5%
+3.3V-+-/\/\/\---+----------+-------+-- 3V out
| | | |
--- 100n / | --- 4.7u
--- 16V \ 6.34k | --- 6.3V
| / 0.1% ___|__| |
--- | | / \ ---
- +--------/___\ -
| |
/ |
\ 31.6k |
/ 0.1% |
| |
--- ---
- -
Two questions:
-
Will it work?
-
What is the accuracy given the precision ±0.1% resistors and 2.495V ±0.5% reference? Is it the sum, so ±0.7%, or something complicated?
Best Answer
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).