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I'm currently designing a circuit for a long term device that will consume 1-2mA at 3.3V constantly. The system is run with a default lithium polymer battery and I need to get as much time out of it as possible.

In the current state, I'm searching for the most efficient voltage regulator possible. The best one I could find was the TPS63805. It is a switching voltage regulator which means that it takes more space and costs more, which is luckily not a big problem for my project.

It seems to be that these kind of regulators are a bit more efficient on higher current consumptions. When only using 1mA, it is at about 90% efficiency, which is good but can it get better with another component or technique? (I'm already making sure to minimize the power loss through PCB design.)

Thanks for any help in advance!

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PS: Sorry for any grammar or spelling mistakes, I'm not a native English speaker.

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    \$\begingroup\$ Is the current pretty much steady or are there bursts of higher and long periods of lower current? There are regulators such as MAX20343 that are optimized for that situation. Needing buck-boost does limit the options. \$\endgroup\$
    – Spehro Pefhany
    Apr 13, 2021 at 21:32
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    \$\begingroup\$ Save power by running at 2.7V possible or 3.0V? Then a simple buck Reg may do. \$\endgroup\$
    – Tony Stewart EE75
    Apr 13, 2021 at 21:40
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    \$\begingroup\$ Krauseler, I think you should explain a little more about the 1-2 mA requirement. When you are niggling over 90% vs 100% efficiency, it draws my attention immediately away from the power supply regulation system over to "what in the heck are you really doing here?" Because it's far, far, far more likely that discussing the what will result in longer operational time than worrying over the last 10% of some power supply system. Discussing how to squeeze that last bit is just a waste of everyone's time when the real elephant in the room is standing over in that corner, undiscussed. \$\endgroup\$
    – jonk
    Apr 13, 2021 at 21:56
  • \$\begingroup\$ Exactly, which is why I questioned the need for 3.3V. What about UVLO to save the battery from self-destruction? What frequencies is your source sensitive to for EMI? What ripple V is acceptable? \$\endgroup\$
    – Tony Stewart EE75
    Apr 13, 2021 at 22:03
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    \$\begingroup\$ @Krauseler I wrote code for an instrument where I operated an ADC (16-bit) at 1 MHz. (C8051F061 device.) And I still spent most of the time asleep. The data, however, was captured for a variable time period (anywhere from 10 us to 50 ms per measurement) and then processed for offset subtraction and logarithmic decay curve fitting actions. In between each measurement, I'd take a burst of offset measurements, too, every single time. And with all of that, it still was "mostly asleep." Of course, you know your situation better. So I'm just giving you a nudge, is all. \$\endgroup\$
    – jonk
    Apr 14, 2021 at 0:06

2 Answers 2

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As Tony mentioned, you might consider running at a lower voltage, so that you can take advantage of an ultra efficient step down regulator, designed for energy harvesting applications. For example, the ADP5304.

If you can run at 2.5v instead, then your efficiency could be 90% - 95%, depending on your battery voltage:

ADP5304 effiency 2.5v

If you really need 3.3v out, then you'd only be able to run until your battery ran down to 3.3v. At that point, the ADP5304 switches to 100% duty mode. However, your efficiency will be better than 95%.

ADP5304 effiency 3.3v

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  • \$\begingroup\$ Thank you for the helpful answer! It might sound silly, but I only use 3.3V because it is the recommended voltage on all of the ICs I'm using. If it doesn't have a negative impact on the performance of microcontrollers & micro SD cards, there would be no problem at all to go with 2.5V or 3V because the range of all components are also between 1.8V and 3.6V. \$\endgroup\$
    – Krauseler
    Apr 13, 2021 at 23:09
  • \$\begingroup\$ @Krauseler - In that case, you might also want to run it at the lowest voltage that they can all accept. That should maximize the amount of run time you can extract from the battery. \$\endgroup\$
    – Rocketmagnet
    Apr 14, 2021 at 22:05
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The problem with most any switching regulator is that they don't give good efficiency at light load. They also have quiescent current that further reduces efficency.

A more reasonable approach is to specify your powered logic to support a voltage range compatible with your battery, say 3.0 to 3.6V. Then, use an ultra-low dropout LDO to take care of the range above that (up to about 4.2V for a fully-topped LiPo cell, typically 3.7V for most of the range.) For most of the discharge cycle the LDO will only be adding dropout loss (as low as 50mV for some, like the ST LDO40L.)

For your 1mA load, and 100mV overhead at 3.7V (regulator set for 3.6V):

  • total power: 3.7V * 1mA = 3.7mW
  • overhead loss = (0.10V * 1mA) = 0.1 mW
  • efficiency: (3.7-0.1)mW / 3.7mW * 100% = 97.3%

This gets even better as the battery voltage gets lower and the regulator is in dropout:

  • total power: 3.5V * 1mA = 3.5mW
  • overhead loss = (0.05V * 1mA) = 0.05 mW
  • efficiency: (3.5-0.05)mW / 3.5mW * 100% = 98.6%

MORE: You mentioned that your peak power is ~400mA. One strategy to deal with that is power the RF front end directly from the battery. Many of the ICs used for this purpose support a wider voltage range than digital.

BONUS: a fixed 3.6V regulator with about 2mV dropout at 1mA, and up to 150mA: Maxim MAX887_EUK36 https://www.mouser.com/datasheet/2/256/MAX8878-1389229.pdf

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  • \$\begingroup\$ I think there should be no problem going with 2.5V/3V. Ideally it would dropout just above 3V so there wouldn't be a need for an extra protection circuit. I'm not very familiar with LiPo batteries, but there is a little bit of buffer needed because a RTC is directly connected to the battery (consumes less then 1uA though). Your recommendation is great, thank you! \$\endgroup\$
    – Krauseler
    Apr 14, 2021 at 0:03
  • \$\begingroup\$ The idea of connecting the RF module directly is good. Shouldn't be a problem since it's not used for the majority of the time. I think a load switch might be helpful. \$\endgroup\$
    – Krauseler
    Apr 14, 2021 at 0:04

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