I've always thought that the absolute maximum ratings on a part are the limits that thou shalt not violate. Period. End of story.

However, another engineer is making the case that it's okay to exceed the absolute max rating for the input voltage on a microcontroller I/O pin. Specifically, he wants to apply 5v, current limited to 30uA, to a micro with an absolute max voltage of 3.8v (Vdd + 0.3V <= 3.9V). The argument being the clamp diodes will take care of the excess voltage.

I couldn't find anything in the datasheet about the I/O hardware on the micro.

When is it okay to exceed the absolute maximum rating on a part?


User Guide

  • 14
    \$\begingroup\$ For a bench test, sure. For mass production, no. \$\endgroup\$
    – winny
    Commented Oct 18, 2016 at 12:23
  • 25
    \$\begingroup\$ When you don't mind destroying things. \$\endgroup\$
    – user16324
    Commented Oct 18, 2016 at 12:26
  • 7
    \$\begingroup\$ Probability of failure or infant mortality rises sharply, when exceeded. MTBF can go from decades of years to microseconds depending on which parameter and excess amount \$\endgroup\$ Commented Oct 18, 2016 at 12:28
  • 26
    \$\begingroup\$ It's physically impossible to apply '5v, current limited to 30uA'. You can apply 5v through a resistor which will limit the current to 30uA when the other end is at 3.8v, or 0v, or what ever you choose, or even a 30uA constant current source that has a voltage clamp so it doesn't exceed 5v. When the 30uA hits the I/O clamp pin, it will be limited by it. Try it, and measure the resulting pin voltage. \$\endgroup\$
    – Neil_UK
    Commented Oct 18, 2016 at 12:40
  • 6
    \$\begingroup\$ An utterly crucial point to note is that "absolute maximum ratings" are almost always given in a non operating state with guaranteed survival being what is at stake. "Recommended operating conditions are given for when OPERATING. Doing what your "engineer" associate recommends exceeds not ONE limit but two. || Frabjously tiny amounts of current in pin clamp body diodes will SOMETIMES result in bad or fatal results. Murphy controls the value of "sometimes". This answer of mine addresses this point. \$\endgroup\$
    – Russell McMahon
    Commented Oct 18, 2016 at 14:33

11 Answers 11


Its never safe to exceed the maximum ratings. Even operating at a point within the ratings can result in failures if for example the manufacturing process has drifted out of spec (I've had power transistors fail in a prototype run soak test, and the manufacturer admit to a fault).

The further from the 'safe' region you operate, the higher the chance of early failure. Maybe seconds, maybe months - generally the analysis won't exist. Rarely, (and sometimes more commonly as devices become more mature) a manufacturer might relax some of the maximum ratings - particularly ratings which relate to time limited stresses.

In the case you specify, you've identified that the absolute maximum ratings are probably an approximation. Its plausible that \$\mu A\$ currents with a high drive impedance can be accepted on the pins quite reliably without exceeding the breakdown voltages (and arguably you don't exceed the rating like this, since the pin will clamp). There is additionally the risk of latch-up if unexpected parts of the silicon are conducting with various voltage states.

Don't expect this to work in 100,000 parts which have a working lifetime of 10 years. If you can live with the occasional catastrophic failure, maybe the design is still reasonable. If its a debug port on a $5 product with a 6 month lifetime, it would be more reasonable.

  • 2
    \$\begingroup\$ It is very rarely condoned by the manufacturer to exceed absolute limits, in controlled situations. For instance, I've seen DRAM memory modules for which an absolute temperature limit of 100 degrees Celsius was stated, but had a derogation stating one could exceed this limit by 10 degrees if one halved the refresh interval. So talking with the manufacturer might let you push the envelope on one absolute limit while giving some ground on other parts of the envelope one cares less about or can work around. \$\endgroup\$ Commented Oct 20, 2016 at 3:13

Exceeding absolute maximum ratings is a bad idea.

In some very limited circumstances, carefully pushing something past the limits might be worth the risk. This might apply to one-off situations where you know, for example, that the temperature will always be below 25°C and you think you can get away with violating something else a bit as a result. It might also apply to McGyver-type situations where you either have nothing or something that might work.

It is not OK to exceed limits in a production design.

In your particular case, there are likely two limits, the maximum voltage on a pin and the maximum current into that pin. You are not really applying 5 V if that is limited to 30 µA. With only 30 µA thru the protection diode, it's possible that the maximum voltage is not actually exceeded. Read the datasheet carefully.

  • \$\begingroup\$ The abs max ratings for an AVR I/O pin are 0.5 V above Vcc or below GND and 40 mA current. I don't see any way for the user to shove/suck microamps and get the clamping diodes to let the voltage go 0.5 V outside the rails. \$\endgroup\$
    – Nick T
    Commented Oct 19, 2016 at 23:47

I once came across Atmel App Note AN2508 - AVR182: Zero Cross Detector (not TI, I know - still interesting) that condones such a construction... For zero-cross sensing on mains!

AVR circuit for mains sensing

To protect the device from voltages above VCC and below GND, the AVR has internal clamping diodes on the I/O pins (see Figure -1). The diodes are connected from the pins to VCC and GND and keep all input signals within the AVR’s operating voltage (see the figure below). Any voltage higher than VCC + 0.5V will be forced down to VCC + 0.5V (0.5V is the voltage drop over the diode) and any voltage below GND - 0.5V will be forced up to GND - 0.5V.


The series input resistor is a 1MΩ resistor. It is not recommended that the clamping diodes are conducting more than maximum 1mA, and 1MΩ will then allow a maximum voltage of approximately 1,000V.

So, apparently Atmel thinks it's okay to use the clamping diodes on their MCUs in this way, up to 1mA. (Although you can argue about the authority of App Notes)

Personally, I'm still not entirely sure what to think of it. On the one hand, if Atmel specifies it's ok to source/sink up to 1mA through the clamping diodes, then I see no problem if you steer well clear of that current (and 30µA would certainly qualify for that). Also, if used this way, you don't actually exceed the voltage specs; the diodes clamp it down, after all.

On the other, is it ok to use the clamping diodes like this? I never found anything about clamping diode current in the datasheets, so the only source for this is an App Note.

So you could try and find documentation from TI specifying the maximum current through the clamping diodes. Maybe they also have information in their datasheets or App Notes allowing or disallowing these usages.

But if you want to be safe, you're better off adding your own clamping diodes, preferably low-Vf ones, i.e. Schottkys. Or use a simple voltage divider. That way you won't have to worry if you're violating the specs or not.

Update, January 2023

When I came across the app note in this answer, I was actually making a hobby project where I ended up using this construct for mains zero-cross sensing. (For some more details, including a schematic, see this question; it's R8/R9).

The circuit connects 230VAC through 2MΩ directly to PB3 on an ATTiny85, putting about 58µA RMS / 163µA peak through the ESD diodes. I'm still not entirely sure how to feel about the whole thing; my motivation for using it was that the project was in part an exercise in minimalism; seeing how far I could reduce the circuit and still have it work well.

Whatever the feelings, over six years of extensive use later, the MCU is still working fine.

Make of that what you will ¯\_(ツ)_/¯

  • 1
    \$\begingroup\$ very interesting. I think they are going for absolute minimum parts count, otherwise yes i'd add all sorts of dividers and external clamps myself. \$\endgroup\$
    – KyranF
    Commented Oct 18, 2016 at 16:58
  • 2
    \$\begingroup\$ This is nasty even by Atmel app-note standards. Fun though. \$\endgroup\$
    – user98663
    Commented Oct 18, 2016 at 19:52
  • 2
    \$\begingroup\$ I like my chips fried. \$\endgroup\$
    – copper.hat
    Commented Oct 19, 2016 at 17:04
  • 3
    \$\begingroup\$ This circuit is absolutely fine. Obviously it's not isolated and you'll need to take the appropriate precautions, but it's the engineering way to do the zero-cross detector. A divider is fine, but if you reduce it too much you might not be as able to treat it as a digital edge signal (you want to have as little time spent between the max low voltage and min high voltage). Insist on clamping diodes and you might have just lost your company thousands of dollars for no reason. \$\endgroup\$
    – Nick T
    Commented Oct 19, 2016 at 23:32
  • 1
    \$\begingroup\$ Besides, re: "but if you reduce it too much you might not be as able to treat it as a digital edge signal". While not mentioned in the app-note, Atmega163 has an analog comparator on two of its pins, which should be able to handle this edge case (har har) in a more well-defined manner. \$\endgroup\$
    – nitro2k01
    Commented Oct 19, 2016 at 23:46

Regarding exceeding absolute maximum rating in general, i think the other answers have covered this (ie don't do it).

Regarding an I/O pin's absolute maximum voltage rating, it's a little more complex that it appears on the surface. In the (usual) case where the I/O have internal protection diodes to VCC and GND, you need to take into account two absolute maximum values: the absolute maximum voltage, and the absolute maximum injection current. If you don't exceed the absolute maximum voltages then you are fine. On the other hand, if your input is current limited to below the absolute maximum injection current (eg with a resistor), then the internal protection diodes will clamp the voltage the I/O sees and no absolute maxmimums will be exceeded (You should be able to test this on a prototype by connecting the current limited input and measuring the voltage - as long as the voltage and currents going into the pin are below their respective abs max values, you should be ok :) ). A excellent application note describing this is: http://www.nxp.com/assets/documents/data/en/application-notes/AN4731.pdf

Specifically for the device you listed, i was not able to find any values for the absolute maximum injection current.

In situations like this, where you're getting close to limits and/or you can't find the data you need, I would always recommend contacting the manufacturer directly and discussing the issue with one of their Application Engineers (don't be afraid of reaching out to manufacturers, they are usually super happy to help!)

  • 2
    \$\begingroup\$ If you respect the maximum injection current of an IO pin, you're basically guaranteed not to exceed its absolute maximum voltage rating. \$\endgroup\$ Commented Mar 23, 2017 at 16:21

Altough it might be true what the engineer is thinking, it certainly is not wise.

Clamp diodes are for unforeseen situations. They are NOT intended to compensate for ignorance and sloppy designs. By doing so all safety margins are gone. A little worse in tolerance by design, manufacturer, or what ever reason and the design fails. When a technician stumbels accros such a situation without knowing the background he or she can waste a lot of time to figure out what happens.

Therefore don't and stay within the specs.


Since it wasn't mentioned in other answers, exceeding the maximum ratings on one pin of a microcontroller can also result in the following:

  • If applied before the microcontroller powers up (even by microseconds) it can cause the micro to latch-up and fail catastrophically.

  • If applied while the micro is completely off or is shut down, that current will flow into it's power rails through the protection diodes, powering it or preventing it from powering off completely.

Dave Jones of EEBlog has a nice video demonstrating this behavior.


Ω The safer solution is put a TVS diode to clamp overvoltage, rather than depend on Device leakage effective series resistance. The series R will limit current and AS LONG as that current is safe, continuous, it should be ok. But IF capacitive coupling and ESD protection is compromised, a low Z clamp TVS clamp diode is best (3.6V TVS) to Vcc.

This Answer may use Ohm's Law with some reasonable estimates not precise values.

Probability of failure or infant mortality rises sharply, when ABSOLUTE MAX is exceeded.

MTBF can go from decades of years to microseconds,depending on which parameter and the excess amount.

  • Here is how interface current is limited and protected from ESD.

ESD clamp diodes, like all diodes, are rated for certain voltage drop, Vf at some rated current, If and are often in two stages with a series current limiting resistor in between to attenuate 3kV spikes to less than 0.5V or less than the Vgs threshold the CMOS. These ESD diodes are usually limited to 5mA DC current due to the small junction size to get a small reverse bias capacitance of 1pF for fast response of the interface and also fast response of the diode.

Let's assume the ESD rating protection from a standard discharge of 100pF is 1kV @5mA. All Diodes have an internal ESR which is inverse to it's W power rating.

We can estimate the voltage drop on the 1st diode and the voltage drop from the 5mA typical current limit for ESD diodes. If we estimate Vf=1V then we see it could be a 5mW diode (5mA*1V),which has an estimated ESR of 1/(5mW) = 200 Ohms.

But 1kV ESD over 200 Ohms would cause a 5V spike on the 1st diode.

Thus we need a 2nd diode with an estimated 10K in series. Now the ESD spike is 5V/10k= 0.5V which is just enough to be below the Vgs sub-threshold trigger level of CMOS gates.


simulate this circuit – Schematic created using CircuitLab


Is 30 uA small in this context?

How about computing the power dissipation in the clamp diode, divide through by the diode volume (i.e. look up the geometry size), and then see how quickly the silicon in the diode will heat up when this peak stress level is applied - What temperature would it reach? Will it melt?

These are simple reasonable calculations you can do to get a handle on the real loadings going on and explore them with your colleague. If you can cover the thermal effects, the voltage stress, the dV/dt from stray capacitance (1) and the like then you just might have a design.

But I suspect you will find that at least one issue will thwart the ambitions (maybe that's why they are abs max limits ;-).

(1) the stray capacitance of concern is the one spanning the current limit resistance, which will be discharged via that little protection diode, and may not have enough thermal capacity, especially as it is followed by a steady DC power load even if it survives.


With most Microchip PIC devices this will work and is also within the specification. The current limiter (30µA) works as an voltage divider.


Sometimes if it's OK that what you make breaks the first time it's used, you can care less about the rating. Assume you want to make a controller that drives an solenoid valve that releases a gas from a flask. It's going to be useless after the gas is released. In that case, you can drive the solenoid valve with only a transistor. When it's turned off, it will break, allowing current to pass between its collector and emitter. But it's OK because the device isn't needed anymore.


Perhaps is't not strictly electronics, but a pyro-igniter. A length of nichrome wire and a 12V car battery. The hobby rocketry people do this all the time to set off their motors.

A fuse is similar in that it's rated capacity is designed to be breached (in a safe manner).

  • \$\begingroup\$ However, if you exceed the rated voltage for a fuse, it might not interrupt the fault current. Fuses have maximum ratings, which are best observed, just like most other components. \$\endgroup\$ Commented Oct 23, 2016 at 13:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.