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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:

  1. Will it work?

  2. 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?

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    \$\begingroup\$ I added a link to the TL431's datasheet. We're trying to make users aware of the importance of this, especially for less common parts, so that others don't have to go searching for it and that everybody is sure to be talking about the same thing. I realize this in an older post, so maybe you're already doing this now. \$\endgroup\$
    – stevenvh
    Aug 1, 2011 at 7:14

2 Answers 2

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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).

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  • \$\begingroup\$ Ah. The load is to an ADC's Vref so should be very high impedence (<1µA). I will reduce the resistor, thanks for pointing that out. Either 56R or 51R. \$\endgroup\$
    – Thomas O
    Oct 19, 2010 at 21:20
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    \$\begingroup\$ Thanks for your analysis of the tolerance - much better than expected! I didn't think they added together. In fact I remember asking this same question on AAC forums (all about circuits) and it went into really theoretical stuff. I'll try and dig up the post. \$\endgroup\$
    – Thomas O
    Oct 19, 2010 at 22:27
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Q1
Yes.

Q2
This deserves a longer answer.

\$V_{OUT} = V_{REF} \times (1 + \dfrac{R1}{R2}) + I_{REF} \times R1\$

To find the upper limit of \$V_{OUT}\$ you have to work with the appropriate limit for each of the factors. Appropriate means that for some it will be the lower limit, for others the higher limit. Here it's the lower limit for R2, the higher for all others. Plugging in the numbers gives us

\$V_{OUT} = (2.495 * 1.005) \times (1 + \dfrac{6340 * 1.001}{31600 * 0.999}) + 0.4 \times 10^{-6} \times (6340 * 1.001) = 3.0141V\$
\$= 3V + 0.47\%\$

Likewise for the lower limit for \$V_{OUT}\$ :

\$V_{OUT} = (2.495 * 0.995) \times (1 + \dfrac{6340 * 0.999}{31600 * 1.001}) = 2.9796V = 3V - 0.68\%\$

Why did I drop the term for \$I_{REF}\$ here? For \$I_{REF}\$ only a maximum value is given, no minimum, so thinking worst case it might as well be zero.
If you put on your wiseguy hat you could say "Right, that's the 0.7% I reckoned, I didn't need this complicated calculation for that." No you didn't because the tolerances are real low. For higher tolerances you'll find that simply adding isn't sufficient.

There's another important thing to note. Parameters in a datasheet are always specified under certain conditions. In this case it's a device current of 10mA. We have only a fraction of that, 240\$\mu\$A worst case. So the values may be a bit different.

Which brings us to our next class: philosophy!
Is it worth going through all this trouble and spend money on precision components if you're not quite sure of the outcome? How about the rest of the system? Did you do the same calculations for all other components? Are they all precision parts? Where's the weakest link?
In my freshman year in college I thought DMM showing 6 significant digits were cool. Now I know better. In most cases the last two digits are not reliable, for instance because they will drift as soon as the temperature rises half a degree. That reminds me: temperature controlled voltage references! What happens if the temperature in your product rises by 10°?
I remember an Analog Devices seminar I once attended, where the speaker showed a 5cm trace to an ADC input would introduce several LSBs of error!
I could go on, but the bottom line is: can I really achieve this theoretical precision, and, more importantly: do I really care?
Your product will probably perform as well if the reference is 3.01V, right? I guess I would rather focus on stability and noise. If you still think you need a precise voltage reference I wouldn't make if with a TL431. Google for precision voltage references. You'll come up with devices like Maxim's MAX6010 (one of the first hits for me), which has an initial accuracy of 0.2%. (You'll have to jack up your 3.3V a bit; it has a 200mV dropout.)
The Analog Devices AD780 is 1mV precise (or 0.033%), but requires at least 4V (maximum 36V, so you may have this available somewhere).

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