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I have used this circuit from the TLV431 datasheet and replaced the values according to what I need which is 2.48V. This output will be directed to an op-amp as a reference. My problem is I can't stabilize the output to 2.48V. It is increasing overtime to 2.49 to 2.5V with or without the load capacitor. The resistors used are 1% and my source voltage is 5V. I also tried different values for load caps like 10uF but the same increasing output. What could be the problem?

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I have changed my schematic to allow adjustments. thanks for all the help on my first schematic. Below is my new circuit. My values for R1 must be 1.492K and R2 must be 1.5K. I used series resistors to get the 1.4K and the pot(paralleled with a resistor) to get the 92R. Am i doing it right? Are my values now satisfies the current needs of my shunt? Based on my testing, this circuit now gives me a stable output but at some time, it randomly oscillates and then stabilize again. I dont know why that happens.

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  • \$\begingroup\$ What is your input voltage and actual resistors that you selected? \$\endgroup\$ – Michael Karas Jan 9 '15 at 13:53
  • \$\begingroup\$ I have used 5V as my input voltage and used 10k ohm resistors as illustrated on my post \$\endgroup\$ – Rhonald Rei Pahayac Jan 9 '15 at 14:05
  • \$\begingroup\$ Can you check the circuit with an oscilloscope to make sure that it is not oscillating? The TLV431 can oscillate if operated in an unstable region. Take a look at the stability boundary condition diagrams in the TI data sheet. \$\endgroup\$ – Michael Karas Jan 9 '15 at 14:21
  • \$\begingroup\$ i dont have an oscilloscope here as i am doing this at home. i have looked at it and it seems that the problem is im on the unstable region \$\endgroup\$ – Rhonald Rei Pahayac Jan 9 '15 at 15:20
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The problem is your capacitor. The IC is oscillating. See Figure 20 of the data sheet.

To make matters worse, you are running it starved; that is, with too little current. If you check the data sheet you'll see that it is spec'ed to run at 10 mA. At 10k for the source resistor, you're only allowing it .125 mA when you include the effects of the other 2 resistors, and that's right at the lower limit of the recommended operating conditions. And, as Figure 20 shows, at these low currents almost any capacitance less than several uF will cause oscillation.

Try replacing the top resistor with a 240 ohm unit, and get rid of the cap.

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  • \$\begingroup\$ i have seen that figure and replaced the caps with 10uF as it would be in the stable range with any operating range of current i will be using. i have tried replacing the top resistor with less than 500ohms and this seems to be the problem because of the low current. the output now seems to stabilize seconds after i connected it to the source. guess i would be careful now on choosing the values for my divider. thanks! \$\endgroup\$ – Rhonald Rei Pahayac Jan 9 '15 at 15:13
  • \$\begingroup\$ Have you tried getting rid of the cap altogether? I really don't see how you need it. \$\endgroup\$ – WhatRoughBeast Jan 9 '15 at 15:15
  • \$\begingroup\$ yes i removed it together with replacing the resistor value and that really helped. \$\endgroup\$ – Rhonald Rei Pahayac Jan 9 '15 at 15:18
  • \$\begingroup\$ I cant seem to find the right values for my divider, the output is unstable even if the top resistor was changed to about 260 ohm. I use a calculator from TI and it appears that the Idiv/Iref parameter is the one that determines the right resistor values. What is the more applicable value for that? The initial value stated there is on 200. \$\endgroup\$ – Rhonald Rei Pahayac Jan 10 '15 at 17:01
  • \$\begingroup\$ Link to the calculator, please. \$\endgroup\$ – WhatRoughBeast Jan 10 '15 at 17:13
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Your stabilization issue could be one of temperature changes. Does it settle down after minutes?

You may also want to consider decreasing the value of the top burden resistor to increase current through the TLV431.

Or you could be dumping too much current into the TLV431 such that it is self heating and causing a thermal shift of its characteristics.

BTW, what is your input voltage and actual resistors that you selected?

Edit: So the supply is 5V with three 10K resistors. Some things to consider. With these resistors you have 3.3V-2.48V=0.82V across the top burden resistor. This results in 0.082mA for the load, shunt and reference divider. Seems way to low.

You need to have the shunt current be ideally 10X the load current. You also want the current through the burden resistor to be ideally 10X the current through the reference divider. Try re-selecting the burden resistor to put 10mA through the shunt assuming that the load will be less than 1mA. A 100 ohm resistor may be a good starting point.

Also putting more current in the shunt will more quickly stabilize its internal temperature.

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  • \$\begingroup\$ i think it keeps on increasing as long as the circuit is connected to the source. I have left it connected for a couple of minutes and got a few millivolts of increase compared to the first reading. I used the values illustrated on my post. 3 10k ohm resistors and 5V input voltage. I also tried to replace the top resistor with 1k ohm to allow more current but still the same \$\endgroup\$ – Rhonald Rei Pahayac Jan 9 '15 at 14:01
  • \$\begingroup\$ Temperature stabilization of a circuit could take a while. So if it is drift due to temperature characteristics then you realize that thermal time constants are relatively slow. \$\endgroup\$ – Michael Karas Jan 9 '15 at 14:14
  • \$\begingroup\$ so changing resistor values would actually do the same right? \$\endgroup\$ – Rhonald Rei Pahayac Jan 9 '15 at 14:24
  • \$\begingroup\$ Do the same as what? Not clear what you are asking. \$\endgroup\$ – Michael Karas Jan 9 '15 at 14:29
  • \$\begingroup\$ i have tested replacing the top resistor with 100 ohms and it seems to have a stable output considering the right amount of current on the cathode now. i would now try to change the values of my divider resistors and check if the output would now stabilize. thanks! \$\endgroup\$ – Rhonald Rei Pahayac Jan 9 '15 at 15:06
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You don't need those resistors at all, just do this. Choose R as 1k5-2k2 for about 1 mA through the device, it should be enough. Keep the output capacitor, 10 uF would be even better.

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  • \$\begingroup\$ im really having a hard time getting the right values for the divider circuit. i want to have atleast 2.48V at my Vout and cant determine what values to choose that will keep my part operating and not oscillating \$\endgroup\$ – Rhonald Rei Pahayac Jan 9 '15 at 15:39
  • \$\begingroup\$ i use TLV431 which has a breakdown voltage of 1.25V, you are illustrating a TL431 which has a breakdown of 2.5V \$\endgroup\$ – Rhonald Rei Pahayac Jan 9 '15 at 15:40
  • \$\begingroup\$ Yeah nevermind that, my mistake. \$\endgroup\$ – downthewire Jan 9 '15 at 20:33
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What is the exact magnitude of change you are measuring, and over what time interval? Can you plot it over, say, 30 minutes.

I don't see anything obvious wrong with your circuit with the values shown (it indeed should be stable against oscillation and regulate well), but the TLV431 is not a very accurate voltage reference in instrumentation terms (low ppm/°C), and the worst-case specs are not that great. You're towards the lower end for regulation (128uA nominal when 80uA is required up to 85°C) which should minimize self-heating (which is good). Typical performance over room temperature to a bit above is really quite decent, so there is a bit of an anomaly if you think you're seeing mV of change.

If you are soldering to the part, then you can expect it to take some time to stabilize.

If you are not soldering to it, at it's apparently changing more than a few hundred \$\mu\$V, I would strongly suspect the battery in your multimeter is dying! That would cause the reading to rise over time with a perfectly stable input voltage.

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  • \$\begingroup\$ i'm also looking at the possibility that my battery is dying because of these strange readings and im also not soldering at it. i will have to try to replace my batteries and get readings again. also, the big value of my top resistor seems to also contribute to the problem because this gives a low current to the shunt that could cause it to oscillate. thanks! \$\endgroup\$ – Rhonald Rei Pahayac Jan 9 '15 at 15:16
  • \$\begingroup\$ I really don't think it's oscillating. You're well within the stable area on the 'tunnel of death' with more than about 0.5uF (stable down to the minimum cathode current of 80uA). \$\endgroup\$ – Spehro Pefhany Jan 9 '15 at 15:21
  • \$\begingroup\$ the recommended operating conditions said that a minimum current of 0.1mA should be on the cathode \$\endgroup\$ – Rhonald Rei Pahayac Jan 9 '15 at 15:27
  • \$\begingroup\$ @RhonaldReiPahayac Right and you've got 0.128mA nominal assuming you don't draw significant current from the reference (so a multimeter or an op-amp buffer is fine), well within the recommended range, and even guaranteed to work to 125°C ambient for the BQ variant) \$\endgroup\$ – Spehro Pefhany Jan 9 '15 at 15:46
  • \$\begingroup\$ I cant seem to find the right values for my divider, the output is unstable even if the top resistor was changed to about 260 ohm. I use a calculator from TI and it appears that the Idiv/Iref parameter is the one that determines the right resistor values. What is the more applicable value for that? The initial value stated there is on 200. \$\endgroup\$ – Rhonald Rei Pahayac Jan 10 '15 at 17:04

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