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I am powering a PIR sensor which has minimum 3.4V of VCC and 70µA supply current with parallel Li-ion batteries and I want to switch on a boost converter when battery voltage goes below 3.5V. But the voltage controller should consume the minimum possible current.

I searched for low power devices and designed the circuit below:

circuit

I would be pleased if I get suggestions and warnings about my design. My only concern so far is that it would be bad if Q1 and Q2 is conducting at the same time for a short time, and I can't guarantee they won't. I added the C1 capacitor to delay the startup of U5 but I don't know if it is enough.

Parts and Datasheets: TLV803EF29DBZR 74LVC2G14 TPS61322 IRLML6402 PIR Sensor

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  • \$\begingroup\$ When a boost converter isn't running, the input voltage is passed through to the output through the inductor. But you have a fully-integrated boost converter which will always run if it's got input voltage. Do consider using a different boost converter part, perhaps one that has an enable input and doesn't block the input voltage when disabled. \$\endgroup\$
    – Criticizing Israel not allowed
    Commented Sep 26, 2022 at 14:18
  • \$\begingroup\$ Q2 will be back-feeding circa 100 uA to your battery when the booster is activated (via the body diode). Have you considered putting your parallel cells in series and using a low power buck regulator? \$\endgroup\$
    – Andy aka
    Commented Sep 26, 2022 at 14:33
  • \$\begingroup\$ Yeah I noticed that body diode issue after I posted, everything can work under 3.7V including my MCU so I don't want to increase voltage and cause LDO on the sensor to waste power. I know boost converters will connect input and output when shut down but there will be a diode in between causing 0.6V drop, so shutdown wont last much. I got 2 options: 1: directly connect PIR to battery and waste some capacity. 2: An always on boost converter to supply with 3.6V but that means Vin will be greater than Vout for boost converter even it isn't shut down. How would the boost IC behave in that case? \$\endgroup\$
    – CaveScientist
    Commented Sep 26, 2022 at 14:39
  • \$\begingroup\$ I found the answer for my question: e2e.ti.com/support/power-management-group/power-management/f/…. There is still voltage drop insight the IC causing power waste. So which will last longer? Battery to PIR directly and waste capacity after 3.4V or always on Boost converter with constant voltage drop + ripple + quiescent. \$\endgroup\$
    – CaveScientist
    Commented Sep 26, 2022 at 14:44

1 Answer 1

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I am powering a PIR sensor which has minimum 3.4V of VCC and 70µA supply current with parallel Li-ion batteries

You can get a buck (not boost) regulator than consumes less than 20 μA and will run from series cells and, although your load current is 70 μA at 3.6 volts, when the two series cells are producing a voltage in the range 8.4 volts down to 5.4 volts, the current drawn from the cells will be less than 70 μA due to the buck converter's power efficiency. It's not like a linear regulator.

If the two series cells are at nominally 3.7 volts each, the voltage into the buck will be 7.4 volts and, the load current drawn by the sensor won't be 70 μA from the cells but more like 39 μA. Add to this the nominal current taken by the buck and that means less than 60 μA taken from your battery arrangement.

If you used something like the LTC3632, it has a nominal quiescent current of only 12 μA. Two series cells (instead of two parallel cells) allows the cells to discharge to their lower recommended limit thus giving you more life from your batteries and still enough voltage to regulate down to 3.6 volts: -

enter image description here

And, there are bucks with even lower quiescent current when run in burst mode such as the LT8618C.

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  • \$\begingroup\$ Good point of view. But this doubles my cell count. I can double my cell count now without any buck or boost and can still power until 3.4V with %100 efficiency. I think your solution is great for >=3.7 V applications. I will now test if the PIR sensor can really run with 3.4V when protection diode is shorted then decide to use buck converter or nothing at all. Because boost converter option doesn't really help much. \$\endgroup\$
    – CaveScientist
    Commented Sep 26, 2022 at 15:16
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    \$\begingroup\$ You said parallel cells and that means at least two cells. I said series cells and that means at least two cells. Why does it double your cell count? \$\endgroup\$
    – Andy aka
    Commented Sep 26, 2022 at 15:18
  • \$\begingroup\$ By using parallel cells I can certainly double the run time, but I waste %20 of the capacity. Can we make the analogy of a %80 efficient converter? You on the other hand can also approximately double the run time by drawing nearly half current from batteries but you are not %100 efficient either. The converter will use current itself and also conversion is %90 efficient at most. So we seem like we are going for a draw. The numbers I gave are assumptions that's why I am never sure. I just wouldn't bother adding extra part + shrinking battery life by discharging to 2.7V if it won't add much. \$\endgroup\$
    – CaveScientist
    Commented Sep 26, 2022 at 15:45
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    \$\begingroup\$ The important thing about it is that it clearly has no problems (even if the math says a draw) and, it is available and it doesn't leave you head-scratching about whether the boost will perform adequately and trying to figure out if you have enough hysteresis on your comparator so that you don't get into a state where it's switching back and forth. Less messing around I would say but, you can do the math and think about it @CaveScientist <-- also boost converters usually require a minimum load to get proper voltage regulation. \$\endgroup\$
    – Andy aka
    Commented Sep 26, 2022 at 15:52
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    \$\begingroup\$ Ofc your method is better than boost converter option, but I compared your way with "connecting the PIR sensor directly to battery". Then it is a draw for now. I'm testing minimum working voltage of PIR sensor with LM317, I'll share the results. \$\endgroup\$
    – CaveScientist
    Commented Sep 26, 2022 at 15:56

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