I'm planning to make a battery application with a PIR sensor. The sensor consumes 70 μA and will be supplied with 4-5 V. I have two Li-ion batteries in series which gives around 8-9 V in total.

I have 2 options:

1. Supply directly from the batteries (there is a battery protection module in between) and let the sensor's linear regulator obtain 3.3 V;

2. Put a buck converter in between and take the voltage down to 4-5 V before sending it to the sensor.

I was planning to use an LM2576 as a buck converter but I realized it draws 80 μA even in off mode so I'm sure it draws even more than that for its operation.

Let's assume 80 μA input for the IC itself and 100% efficiency from the converter to be optimistic. Input voltage will be 9 V and input current will be 70 μA * 3.3/9 = 26 μA so wasted power is {input power} - {necessary power} = 9*(80+26) μA - 3.3*70 μA = 723 μW. I don't know if the power calculation is relevant or not, but the necessary current is more than doubled with a buck converter anyway. On the other hand, a linear regulator will only waste (9-3.3) * 70 μA = 400 μW.

So is it true that linear voltage regulators can be more efficient than switch mode ones?

  • \$\begingroup\$ Please provide a link to the datasheet of your linear regulator. I think you have overlooked the regulator's quiescent current in your calculations. \$\endgroup\$ Aug 7, 2022 at 16:10
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    \$\begingroup\$ "and let the sensor's linear regulator obtain 3.3V". Please clarify. This linear regulator is in the sensor so it's quiescent current draw will be present in either scenario? Or does the switching regulator replace the linear regulator completely ? \$\endgroup\$ Aug 7, 2022 at 16:18
  • \$\begingroup\$ The linear regulator will always be there (I can desolder it from the sensor if battery span will greatly increase otherwise I dont want to touch it :D), I can add a buck converter if I want to beforehand or I can let the sensor take 9V. The regulator is, let me google... It is HT7533 \$\endgroup\$ Aug 7, 2022 at 16:54
  • \$\begingroup\$ The calculations say yes. Why do you doubt them? Supporting material: When is an LDO More Efficient Than a Switching Power Converter? \$\endgroup\$ Aug 7, 2022 at 17:28
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    \$\begingroup\$ If you want a really low power buck, then you need one that will skip / shut down at very low currents, and have very low Iq when it does. Take care with choosing linear regulators, some have surprisingly high Iq. The only solution is plenty of time, a spreadsheet, a good supply of coffee and data sheets. \$\endgroup\$
    – Neil_UK
    Aug 7, 2022 at 19:55

3 Answers 3


So is it true that linear voltage regulators can be more efficient than switch mode ones?

If you choose an inefficient-enough switcher and an efficient-enough linear regulator, sure.

I was planning to use LM2576 for buck converter but I realized it draws 80 microamps...

See above. The reason there's a bazzilion switching regulators on the market is roughly the same reason that there's a bazzilion different op-amps on the market: it's a hard problem to solve with one generic solution, so there needs to be lots of options.

That's an old part, designed at a time when the notion of selling a switching regulator than any circuit designer could just plop onto a board and gain success was revolutionary. For most consumers of that part, the quiescent current was the least of their concerns.

So -- find a different part, or go with your linear regulator scheme.

I did a bit of digging, and it looks like really low-quiescent current switchers are out there, but not common. It also looks like the selection is significantly better if you're looking at parts that only go up to 5V or so of input voltage -- so you may need to trim your system design to the parts available, and go with a single LiPo cell and either a buck/boost converter, or just run the thing down to 3.3V on the battery and call it "discharged". This will reduce your effective cell capacity, but you'll have longer cell life.

  • \$\begingroup\$ I've also searched and found some regulators with 1-2 μA even 60 nA quiescent currents but as you said they are uncommon. I don't seem to be able to buy them easily. I'll go with linear one, it will run the system more than a year even with the smallest battery if my calculations are correct. 1200mAh/0.1mA = 12000 hours = 500 days and I assumed current is 100 μA. \$\endgroup\$ Aug 7, 2022 at 19:22

That comes down to which regulator you can engineer to have lower quiescent current.

Both need some kind of reference voltage generation.

The switching circuit needs some kind of comparator. The linear regulator needs some kind of op amp.

I would argue that a hysteretic switching regulator would have the lowest quiescent current, but of course wouldn't be quite so precise.

An actual switcher would use a pretty fast oscillator which usually needs more quiescent current than a slow opamp for a linear regulator. You might even skip the opamp if the output impedance of the reference circuit is low enough, which might happen for ultra-leight loads.

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    \$\begingroup\$ While I have not attempted a comprehensive survey, in my experience, the lowest Iq linear regulators are always lower than the lowest Iq switchers, in practice. \$\endgroup\$
    – user57037
    Aug 7, 2022 at 18:13

In this context I would prefer to use the Lithium cells connected parallel and use a capacitive charge pump IC with low operating current (S-882140AMH, LTC1522, LM2775) to lift the cell voltage a little bit with high conversion efficiency). 30-40 µA operating current and efficiency > 82 % should be possible.

Because the cells are parallel, you have twice the runtime compared to the series circuit at the same operating (or GND) current. And you don't need a cell balancing circuit.

  • \$\begingroup\$ I've never heard of charge pumps before and I watch a video about it, it has a nice principle and looks simple. I've found that LM2776 has 0.1 mA quiescent current. Hmm seems nice. The linear regulator draws less current but wastes my second battery whereas I could use a charge pump and use that extra battery to double the capacity. Also it would eliminate the current drawn by the battery balancing module (it's around 10 μA I guess not a big deal). Seems like a nice deal. I'll think about it with a fresh mind. Thank you. \$\endgroup\$ Aug 8, 2022 at 2:02

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