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I'm working on a project that uses an LM4040 precision shunt in order to obtain a 2.5 V reference voltage (in particular, I'm using the LM4040-N-25). My circuit is similar to the one shown below (from TI's website):

LM4040 circuit

In designing my project, I was basing my design on a prior project that used a 3.3 V input voltage Vs and an appropriately sized shunt resistor Rs. Unfortunately, I hooked Vs to a 5 V input and exceeded the maximum specified current for the device for some minutes (which is 15 mA, according to the datasheet). However, I only exceeded it by 1 mA, so I haven't let out any magic smoke.

What is the expected behavior of a precision shunt like this if it's failed? Currently, I am obtaining 4.3 V across my output and ground, indicating to me that the LM4040 is doing something, though I'm not getting the desired 2.5 V.

Thanks!

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  • \$\begingroup\$ Check the pin out diagram of the LM4040. Check it carefully. You may find a connection error. \$\endgroup\$ – Dwayne Reid May 3 '16 at 1:22
  • \$\begingroup\$ The LM4040 has snuffed it . \$\endgroup\$ – Autistic May 3 '16 at 1:28
  • \$\begingroup\$ As @Dwayne suggest, make sure the do-not-connect pin, if any (pin 3 on SOT-23 and pin 2 on SC-70, not present on TO-92) either is floating or goes to the anode /ground). \$\endgroup\$ – Spehro Pefhany May 3 '16 at 2:05
  • \$\begingroup\$ @DwayneReid -- Yup! This seemed to have fixed it... But the plot thickens: Get this... I ordered the part from Mouser, which said that the part in question was a TI component (whose TO-93 package is shown in top view on the datasheet). I connected it according to that datasheet linked above... So: looking carefully at the component, I noticed that it was a National Semiconductor part-- Googling turned up that their version of the LM3030 had a pinout that is different (looks similar on the datasheet, but it's a bottom view). The datasheet linked on the Mouser part pertains to the new TI part! \$\endgroup\$ – J. Murphy May 3 '16 at 6:54
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The 15mA listed is the recommended operating condition. The absolute maximum given is 20mA, which is still well above the supposed current that was applied to the device. This means that the device will likely not last as long as would have been expected otherwise, but (as an example) 90% of 20 years is 18 years, still a decent lifetime.

Nonetheless, don't let this keep you from swapping out the device if you suspect that it has been damaged and is directly responsible for any issues.

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  • \$\begingroup\$ Thanks-- I hadn't noticed that I was actually still under the absolute maximum for the component. \$\endgroup\$ – J. Murphy May 3 '16 at 7:01

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