I want to get a stable 3V reference from a +3.3V ±5% supply (so it can go as low as 3.135V and as high as 3.465V.)

I know the TL431 is a shunt/zener regulator, so doesn't really have a dropout voltage, but there probably is some point above 3V at which it cannot maintain the 3V output. What I want to know is will a typical TL431 be able to handle this?

My circuit configuration is below:

+3.3V-+-/\/\/\---+----------+-------+-- 3V out
      |          |          |       |
     --- 100n    /          |      --- 4.7u
     --- 16V     \ 634R     |      --- 10V
      |          /       ___|__|    |
     ---         |       | / \     ---
      -          +--------/___\     -
                 |          |
                 /          |
                 \ 3.16k    |
                 /          |
                 |          |
                ---        ---
                 -          -

2 Answers 2


There is no dropout voltage as such. The reference voltage is 2.5V and it will run happily at 2.5V if you short circuit the ref pin to the cathode. The important spec to consider is the shunt current.

You must ensure that there is at least 1 mA biasing the device. (Based on the TI data sheet.) From your circuit, with 115R and a minimum voltage drop of 3.135 - 3 = 0.135 you will end up with a shunt current of 0.135/115 = 1.1 mA. However the voltage divider which has a total resistor of about 3.7 kR is going to use almost 1 mA itself, not leaving enough for the TL431. You also have to add in the current that you load will used, but if it is a high impendence - say into an op-amp - it shouldn't be a problem.

You can fix that by increasing the value of the voltage divider resistors. Say we multiply them by ten, to get 6.34k and 31.6k. The max input current to the ref pin is stated as 4 uA, which might start to give you some errors in the set point (depending on how accurate you expect the Vout to be). The worst case error will be about 4uA * 6.34k or 25 mV.

The last thing you need to consider will be the maxiumum shunt current. When the input voltage is at its max value the 115R resistor will have 0.465 V across it. This means the shunt current (including the voltage divider and load) will be 0.465 / 115 = 40 mA. Fortunately that is well within the 100 mA maximum of the 431.

Because we're still far below the maximum shunt current you could consider reducing the series resistor - too say 56R. That will help satify the minimum shunt current without changing the voltage divider and still stay under the maximum shunt current rating with the input voltage is at its maxiumum. At higher current you will also need to check the power dissipation. 100 mA * 3V gives a maximum power off 300mW. For some small smd packages that is getting a bit too high.

The basic problem trying to optimise the circuit as it is, comes from the fact that the voltage across the series resistor is too low and any variation in the supply voltage represents a big variation across the series resistor.

If you can't (as suggested in the comments) afford to waste the current you might need to consider a difference approach. Either a series regulator (will be tough with only 100mV dropout voltage), reduce your reference voltage or increase the supply voltage to the series resistor.

  • \$\begingroup\$ Hmm this config was recommended by TI's calculator... \$\endgroup\$
    – Thomas O
    Oct 15, 2010 at 9:12
  • 1
    \$\begingroup\$ I'd like to avoid wasting 100mA just in the TL431, is there any way to waste say only 10-20mA with the same output? The output 3V is high impedence, it is a Vref for an ADC. \$\endgroup\$
    – Thomas O
    Oct 15, 2010 at 10:50
  • 4
    \$\begingroup\$ Clint, 1up for this very good answer. Just one tiny thing: The power dissipation in the TL431 needs to be taken into the calculation just as much as the current into the 431's output ("Cathode"). \$\endgroup\$
    – zebonaut
    Oct 15, 2010 at 11:00

Also bear in mind that the TL431 has a maximum sinking current, beyond which the voltage will no longer be regulated (and the device could be damaged). With 3.3V and 115R you'll not be anywhere close to that.

It is also generally recommended to have a small capacitor from the cathode to the anode, for stability reasons. (431s can and do oscillate). The 'safe' range of capacitor depends on the setpoint and is usually specified in the datasheet. I'd recommend something from 1nF to 10nF, and check to make sure the output isn't a sawtooth.

Note that not all manufactures have the stability region curves, but it's been my experience that the curves in the National Instruments LM431 datasheet hold true for other vendors' 431 regulators (TI, ON Semi, Fairchild, etc. etc. and so on)


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