I'm using an analog input on a NANO board to measure the midpoint between two photo resistive (Cadmium) cells, with the two outside ends tied to GND and a source voltage. Basically a simple voltage divider. So it almost wouldn't matter what reference voltage I used in a case like that, as long as the AnalogReference() was correctly set corresponding to the reference voltage, AND the load did not pull down that voltage.

Well if I used the 5V for my source, and set AnalogReference(DEFAULT), then assuming a worst case load of 4K-ohm, the total load on the 5V regulator is still only 1.25mA. But even so, I figured I could minimize the system load even more by setting AnalogReference(INTERNAL), and using the VREF pin as a source for my voltage divider.

Well it seems to work fine, but I don't know that I've seen any actual examples of what I'm doing, which is using the AREF pin as a source output voltage for my voltage divider. Using vREF as a source, a 4K load would only draw 27.5 microamps with a 1.1V source. But I can't find any spec for the max current I should avoid exceeding, That has me worried maybe I'm not even supposed to be doing that!

So is it acceptable to use the VREF pin as an output for light loads, when using the INTERNAL AnalogReference?


EDIT: After receiving a "no" answer yesterday, I did some bench checking. I configured my NANO board for Internal, and connected a fully open 100K linear pot from AREF to Ground, with a DVM in parallel. The AREF terminal output was about 1.076, with or without the 100K load. Not quite the 1.1 specified but close enough. I then slowly rotated the pot to increase load. At just under 9K the voltage dropped a somewhat insignificant amount, to 1.075. From there on there were additional small decreases, down to about 1K, at which point i measured about 1.062V. Below 1K, there was a very steep decrease.

Note that this was a "quickie" test, done while powering the NANO from a USB port. However, repeating the test with 12V on the DV input did not alter the results significantly.

Even considering the USB source, from this test I conclude:

1) There does appear to be some buffering on the VREF output. It certainly did not behave as there was a simple case of a 1.1V source in series with a high impedance.

2) Up to a load of 10K, the VREF output seems at least as stable as the 5.0V regulator output.

3) Since a 1.062 drop from an initial no-load voltage of 1.076 equates to less than a 2% difference, I have to conclude that a load on VREF (to ground) greater than 1K (lets say 2K for margin) makes VREF probably useful in cases like i described, with a worst case load of 4K.

4) I repeated my test with a second NANO board. This time the initial no-load voltage was salightly higher, 1.084. However, the response to increasing load and the K point where the slope changed dramatically were unchanged.

Again, this is an arduino NANO board, and probably a clone. Can anyone else verify or disprove my results? I'm sure someone else will need to consider using VREF as I did, and it would be good to know.

2 Answers 2


No, it is not acceptable.

The datasheet states:

AVCC is connected to the ADC through a passive switch. The internal 1.1V reference is generated from the internal bandgap reference (VBG) through an internal amplifier. In either case, the external AREF pin is directly connected to the ADC, and the reference voltage can be made more immune to noise by connecting a capacitor between the AREF pin and ground. VREF can also be measured at the AREF pin with a high impedance voltmeter. Note that VREF is a high impedance source, and only a capacitive load should be connected in a system.

The 1.1V reference is only presented on the AREF pin for the purpose of adding an external bypass capacitor, or for measuring it using a very high impedance meter. Any load applied to the pin will cause the reference voltage to droop.

While it may work in certain circumstances it's certainly not recommended. If you want to use it externally you should buffer it through a high-impedance unity-gain voltage follower:


simulate this circuit – Schematic created using CircuitLab

  • Hmm! OK that's good information! I've been doing it wrong, though it "worked", I suppose it only did so because for a voltage divider it did not matter as much as it otherwise might have! I'll have to re-think my circuit a little bit though. Since I just need an ADC count representative of the voltage divider, it looks like I'd do better to simply use the DEFAULT and work with the 5V terminal for my reference!
    – Randy
    Commented Jul 14, 2019 at 23:55
  • Please see my edit. At least based on my bench test, it seems that using the AREF pin to power a low load voltage divider works, within the constraints I documented
    – Randy
    Commented Jul 15, 2019 at 16:26
  • It may work, but it's not what it's designed for. My car works when I drive it off a cliff. It's not designed for that though, and I doubt the long-term outcome would be good. If you want to use it externally you should buffer it through a high impedance unity-gain voltage follower.
    – Majenko
    Commented Jul 15, 2019 at 16:55
  • I note that you quoted the spec as saying AREF should only be measured with a high impedance voltmeter. But did YOU add the phrase "Note that VREF is a high impedance source, and only a capacitive load should be connected in a system."? Your car analogy is poor one. I've been doing what I came for confirmation on for a few years now with no trouble, and it was the lack of examples that made me second guess. I guarantee if you drive your car off a cliff and measure the results, they won't lead you to the reasonable question I raised.
    – Randy
    Commented Jul 15, 2019 at 17:03
  • Also I'll point out that a unity gain buffer done with a TL081 (not exactly a "rail-to rail" spec OPAMP) will also start to droop beyond a 1k load. But based on my bench test, I'd not be surprised if I found there already was such a buffer internal to the ATmega328. I don't have the internals, but based on my measurements, there must be some kind of buffer there already.
    – Randy
    Commented Jul 15, 2019 at 17:07

Now, having had the benefit of a support ticket and subsequent discussion with an engineer at TI on this subject, I can confidently offer a qualified YES to my own question. It turns out my case is probably the only one that would be within those qualifications.

  1. In the case where the RATIO between two hi-resistances is the ONLY thing needed from an A/D measurement, it is acceptable to use AREF and GND for the applied voltage, provided the load is low (see item 2). The main reason is that any droop in the AREF caused by the added load would not affect the ratio.
  2. The use of AREF for such a situation should still ensure limiting any load on the AREF pin to about 91uA. For the 1.1V (INTERNAL) source selection, that equates to about a 12K load. The number is based on a reverse calculation that would cause a 1 LSB difference in analog readings, IF other analog inputs are expected to be used for other non ratio-metric measurements.
  3. Using the AREF pin this way makes it even more important to add a capacitor from the pin to GND.
  4. Apparently there is no danger of harming the chip in any way, should the AREF pin become shorted.

I will comment that the engineer responding to my support ticket understood I was trying to avoid a PCB change, but recommended a different approach if I had occasion to make other board changes anyway. His recommendation was to use the EXTERNAL Analog reference option, using the available 3.3V pin for AREF voltage, and using a digital output to supply my voltage divider. The idea here is that I would still have the benefit of short circuit protection, and since I was concerned about power saving, I could switch the Digital output pin to a low state until readings needed to be taken, observing sensible delay for settling time.

A good suggestion indeed, but my reason for answering my own question here is because under the constraints given, using AREF as described in my OP is acceptable, will not affect the ratio measurement, and could not harm the chip.

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.