13
\$\begingroup\$

I have seen some development boards (for example. BL652 dev kit) for low power chips have battery power connected directly to the MCU without a regulator.

For the example case, the battery used is a 3V CR2032. The datasheet for the MCU defines the following parameters:

datasheet page 16.
Absolute Maximum Ratings            Min           Max
Voltage at VDD_nRF pin             -0.3           3.9

datasheet page 17.
Recommended Operating Parameters    Min    Typ    Max
VDD_nRF                             1.8    3.3    3.6

I'm interpreting this as "If your battery voltage drops to a value between 0-1.7 it isn't defined what will happen".

Why this worries me is because I've seen regulators having the Power Good pins and have found no explicit statements in the datasheet that the MCU from the example won't be damaged by the undervoltage.

How can I decide if a regulator is needed between a battery and a load, to guarantee there are no damages when the battery voltage starts dropping?

\$\endgroup\$
  • 4
    \$\begingroup\$ I'm very much an amateur, but my impression is that regulators do a couple of things. First, they constrain the supplied voltage to within a certain range. However, if the supply voltage 'goes away' they can't make it magically reappear. Loss of power, be it from a battery or from any other source is still loss of power. Secondly, they reduce any ripple to an acceptable amount. Batteries don't really have this issue. I don't think you're at any more risk running directly from a battery than you would be from something like a lab power supply. \$\endgroup\$ – MickeyfAgain_BeforeExitOfSO Nov 24 '16 at 22:30
30
\$\begingroup\$

If your battery voltage drops to a value between 0-1.7 it isn't defined what will happen

This is often true, but it won't, for sure, destroy anything. Because, if it was destructive, the min Vdd in "Absolute Maximum Ratings" would have been given as a positive value (which I have never seen in any datasheet, and I hope I'll never see that in my life - it wouldn't make sense).

So at this point, you are guaranteed the MCU won't be destroyed with undervoltage. However, it could still behave erratically (potentially damaging other external circuitry).

Now, in this kind of MCU, there is often a feature called "brown-out detection", or, sometimes, "undervoltage lockout". This is a feature that monitors the supply voltage and guarantees that the chip is held in reset state when the voltage is under a given level (sometimes programmable).

Good news: There is such a feature on the specific chip you're using. See chapter 5.1 in the datasheet you linked.

Therefore, you don't need to have a regulator with "power good" detection or an additional supply monitor circuit in your specific case.

Note that, if the MCU didn't have the brown-out detection included, there are tiny chips that just offer this feature (often combined with a timed power-on reset generator) without being voltage regulators.

| improve this answer | |
\$\endgroup\$
  • 2
    \$\begingroup\$ Additionally, external power supervisors can be used in the case the MCU doesn't have these features. \$\endgroup\$ – scld Nov 24 '16 at 14:35
  • 1
    \$\begingroup\$ For chips that don't detect this, putting an undervoltage protection between the battery and the device usually does the trick. They aren't complicated, expensive or power hungry. \$\endgroup\$ – Mast Nov 24 '16 at 17:03
  • \$\begingroup\$ Couldn't there be a latch-up with a lower supply voltage (that wouldn't happen with a higher one)? \$\endgroup\$ – Peter Mortensen Nov 24 '16 at 18:57
  • \$\begingroup\$ @PeterMortensen No, unless maybe on very unusual chips and for very specific cases (which would be clearly stated multiple times in the datasheet), or if there is a bug in the chip, there is no way it could experience latch-up because of undervoltage. It wouldn't make sense also because on power-on, it takes a bit of time for the supply to go from 0V to its nominal value (same on power off). You can't avoid it. If you risk latch-up each time your system powers on, it's bad. The worst that can possibly happen is erratic behavior, but this risk is eliminated by the brown-out detector. \$\endgroup\$ – dim Nov 24 '16 at 20:01
9
\$\begingroup\$

...between 0-1.7 it isn't defined what will happen

Actually below 1.8 V there is no guarantee what will happen.

Don't worry about damage these are the operating parameters. To prevent damage you must not exceed the Maximum ratings, which are not included in the linked sheet. If you know the chip(s) which are used you can look up their datasheets and see the Maximum ratings. I have yet to come across a chip which can suffer damage from a too low supply voltage.

You do want your product to "know" and respond when the battery is too low though. Add a battery detection circuit (or using the internal one) which will only release the reset when the battery voltage is high enough.

| improve this answer | |
\$\endgroup\$
  • \$\begingroup\$ One wonders what happens if Vdd rises above the 3.6v recommended spec, to the 3.9v absolute max spec. Data sheets rarely (if ever) say. My guess is that the manufacturer would say, "hey, we test up to 3.6v, it may still work above". \$\endgroup\$ – glen_geek Nov 24 '16 at 14:39
  • 3
    \$\begingroup\$ @glen_geek The issue is guaranteed lifetime. It's not impossible that an IC with the specs. you mention will work fine even at Vdd = 5 V. But it might only last one hour, one day, a week, a month or a year. The manufacturer will only guarantee a certain lifetime (for example 10 years continuous operation at 125 degrees Celsius) at 3.6 V. If the IC is always below 50 C then you can expect an even longer lifetime. At higher Vdd and temperatures, effects like hot carriers and electromigration slowly damage the IC internally. At the recommended conditions, these are not such an issue. \$\endgroup\$ – Bimpelrekkie Nov 24 '16 at 15:06
8
\$\begingroup\$

There is no guarantee your processor will not run amuck and scramble memory or provide unpleasant and possible damaging waveforms on the GPIO pins. It is guaranteed that the micro will not be damaged physically, but it could cause damage of a soft or, possibly, with bad design, a hard nature.

For example, if your battery powered micro is controlling the temperature in a terrarium via a MOSFET- acting as a remote thermostat and the micro runs amuck it could kill the reptiles if the battery ran down. An extreme example, and in reality there should be many safeguards against that happening. It's also rare that a battery powered micro can damage anything outside of itself. A more common example would be scrambling of battery backed RAM or of EEPROM.

To make sure that never happens you must inhibit the micro (hold it in reset) for any voltage that is below 1.80V. Since the circuit that does that won't be exact (there is always a tolerance on the threshold) you might pick 2.0V or 1.90V. +/-0.2 or 0.1V. Usually there is also some hysteresis so it might even be reset at 2.2V and out of reset at 1.9V. There is usually also a minimum reset pulse width for a proper reset to occur so that should also be guaranteed.

You will get most of the juice out of a CR2032 even at low temperature by cutting off about 2.4 or 2.5V so there is little reason to call it so close. enter image description here

| improve this answer | |
\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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