2
\$\begingroup\$

I'm building an optical wireless mouse based on nRF52833 and a 16340 cell, I need a sanity check on some considerations regarding the DC regulation.

The battery voltage goes from 4.2V when fully charged to 2.5V when fully discharged, so to power the nRF52 at 3.3V I need a buck/boost converter. I have selected TPS63031 for the job.

I'm running the nRF52 at 3.3V because according to this it draws a lot less current than it would at, for example, 1.8V or 2V. If I were to use those voltages I could just use a linear regulator instead of the buck boost.

During normal operation I estimate the MCU to be drawing between 1mA and 10mA, and when the MCU is sleeping something between 1uA and 10uA. If we look in the datasheet for efficiency figures we see that the 1-10mA range is fine, but 1-10uA has not been characterized, going as low as 100uA.

enter image description here

Questions:

  • Would the regulator output be stable (enough) when the load is less than 100uA? Judging by this very similar question with a very similar regulator I assume the answer is yes, but the output would have some extra ripple because of the power save mode that is also present on my selected regulator, although not as well explained in the datasheet.

  • Assuming the output is stable in the 1-10uA range, the efficiency of the regulator would be terrible, but would this even be a problem? Let's say the efficiency is 20% at 10uA, does this mean that to get those [email protected] the regulator would have to waste 80% of that, i.e. 50uA, for a total of 60uA plus the quiescent current of the regulator (35uA max)?. Is my understanding of efficiency in this context correct? If it is, I wouldn't call this current waste optimal, but certainly low enough to be negligible.

Edit: typos

\$\endgroup\$

1 Answer 1

1
\$\begingroup\$

Just because an efficiency figure is not given at loads under 100uA does not mean the device is not well behaved below that threshold.

The "Device quiescent current less than 50 μA" blurb, along with the typical spec of 25uA quiescent current means that at zero load, it will draw at most 50uA worst case, and 25uA typical.

The efficiency at low loads will be abysmal (eg, consider a 10uA load and 50uA quiescent/leak current, which would indicate a 16% efficiency at that load), but it's all relative; hardly any energy is used from the battery.

The fact that a quiescent current spec is given practically guarantees the device was designed to work even with no load, and its output (voltage) will be in spec.

For your 2nd question, "what is the meaning of efficiency in standby mode", the quiescent current spec allows you to compute shelf life of a battery charge while the device is not used. Divide the energy capacity of the battery by the total standby current. Whatever that figure is, it says how long the device will last off the charger. It's a user-experience concern. Do you want your customers to brag that their mouse is always fresh even after weeks on the shelf? To extend the standby time you'd need to implement a shutdown mode (perhaps shutdown after a few hours, requiring a button press to awaken), which might cost a few more pennies, so it's up to you to decide cost/benefit.

In your second question, I think you understood the meaning of quiescent current relatively well, but with some gaps, which I will clarify:

  1. The max quiescent current in your application, based on the TPS63031 datasheet you linked, is not 35uA but an unspecified figure, which is less than 50uA. The 35uA figure is given at VIN = 3.6V and Iout = 0mA, while your operating point is VIN = 2.5-4.2V and Iout = 10uA.

  2. The draw from VIN is not Iout + Iq, but Iin + Iq, where Iin = Iout * (Vout / Vin) / η. At η = 20% and Vin = 2.5V, Iin will be at most 10uA * (3.3V / 2.5V) / 20% + 50uA, or 116uA, and typically ~91uA drawn from the battery. The point is that Iq is measured at the input, while Iout is at the output of the device, so you need to convert Iout into the input domain before adding it with Iq.

This means 90%+ of energy is wasted, but it's relative. A typical 700mAH 16340 cell at end of life (80% of new capacity) will power the idle device for at least (700mAh * 80%) / 0.116mA = 4827h (more than 6 months). This seems reasonable for a consumer device.

BTW, you wouldn't use a linear regulator instead of a buck converter, because you'd be wasting most of the battery's energy as heat.

\$\endgroup\$
1
  • 2
    \$\begingroup\$ There are cases where a linear regulator can be more efficient. In this case, for instance, if you could find a linear regulator with a quiescent current of substantially less than 50 μA, and you expect the device to be in sleep mode most of the time. It's increasingly rare that this is practical, though, with better and better micropower switching regulators coming out. \$\endgroup\$
    – Hearth
    Commented Aug 31, 2023 at 2:55

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.