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I want to make a wireless sensor node that will run at about 2.5v. I am trying to decide between running 3 alkaline batteries in series with a buck converter or in parallel with a boost converter. I guess if both converters have the same efficiency, let's say 88%, they would run just as long as each other?

The current required for this project should be <200mA

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    \$\begingroup\$ Why not get a buck/boost that works from 3V to 1.8V and just use two in series? It allows some pass through so you get better than 88% efficiency. Or what's the input range in your sensor? Would you even need a regulator if it takes 3V to 1.8V? \$\endgroup\$
    – Passerby
    Commented Apr 27, 2014 at 22:18
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    \$\begingroup\$ I would love to skip a regulator, but I'm using the ADC on the MSP430, which I believe requires 2.2v minimum. Can you elaborate a little on the buck/boost in series? Are you saying something like having the voltage boosted and then bucked will have higher efficiency? \$\endgroup\$ Commented Apr 28, 2014 at 1:12
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    \$\begingroup\$ Ah no. I meant a buck/boost, an ic that regulates both up and down as needed. And two in series, I meant two batteries in series for 3.0V - 1.8V full range. Which MSP430? \$\endgroup\$
    – Passerby
    Commented Apr 28, 2014 at 1:15
  • \$\begingroup\$ Also, I'm not locked into MSP430. It's just attractive because of the low component count and low power. I am very open to the STM32 line. \$\endgroup\$ Commented Apr 28, 2014 at 1:17
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    \$\begingroup\$ @Passerby Alright, so you are saying use two alkalines in series and a buck/boost combination chip so when the voltage drops below the desired level, it will begin boosting up. Right? Any recommendations for this type of IC? And as a note, I'm really liking the sound of the STM32L line now that I'm reading the datasheet. \$\endgroup\$ Commented Apr 28, 2014 at 2:18

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I make this sort of thing for a living. Presumably your sensor will spend most of its time asleep, only waking up to take a reading or send data periodically. In that case a linear regulator might be a better option.

For example, a TPS78225 would give you 2.5V @ 150mA. Keep in mind spikes in current demand can be met by caps and inductors. This device has a 0.5uA quiescent current, so if you device spends most of its life asleep you should be able to get down to the single digit uA range.

Rather than using 3 cells consider using either 4 (2x2) or perhaps a different chemistry (lithium will give you 3.6V per cell). Obviously you need to minimise the voltage drop for the LDO.

This method will likely be much more efficient than any switching system if your idle current is very low. You could even do a hybrid solution where you use an LDO for the microcontroller and switching regulator that is powered up only when needed.

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  • \$\begingroup\$ That's a very interesting idea! It makes sense to me. \$\endgroup\$ Commented Apr 28, 2014 at 13:12
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In general, and all other things being more-or-less equal, I2R losses are going to be the chief cost of efficiency in a power converter. Therefore, it would be a goal to keep currents as low as possible.

This would imply that the overall system efficiency will be better with a buck converter than with a boost converter. The average currents in the batteries, coil and the switching element will all be lower.

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With a booster you might still be able to get it to work down close to 0.5 volts. If the 3 in series all dropped to about 0 9 volts you'd probably be bust. So it's a booster for my choice but choose your chip carefully. Also, you need to consider that all three batteries need to be of equal charge or there'll be other losses. Not so in series.

Extra info on battery ESR that should be taken into account- the effective series resistance of a "typical" alkaline battery is about 0.3 ohms (see Energizer PDF file). For a series configuration, the ESR will be about 0.9 ohms and for 3 equally "charged" parallel batteries, the ESR is 0.1 ohms.

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  • \$\begingroup\$ If 3 in series were at .9v, doesn't that mean that the buck would get 2.7? Would it fail because many chips need adequate margin for vin over the vout or for some other reason? \$\endgroup\$ Commented Apr 27, 2014 at 16:36
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    \$\begingroup\$ yes, you need a bit of margin and maybe about 0.8333 is the theoretical limit. \$\endgroup\$
    – Andy aka
    Commented Apr 27, 2014 at 17:07
  • \$\begingroup\$ Can you recommend a potential chip? \$\endgroup\$ Commented Apr 27, 2014 at 17:17
  • \$\begingroup\$ What's your load current range? With that we can make a recommendation for both buck and boost options. Some buck converters have bypass or 100% duty cycle capability, so they could work all the way to nearly the theoretical limit, but Andy is right that you could make better use of the energy in the cells with a boost converter. \$\endgroup\$
    – John D
    Commented Apr 27, 2014 at 17:21
  • \$\begingroup\$ I edited the question. I need about 200mA, rounded up. \$\endgroup\$ Commented Apr 28, 2014 at 1:08
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"I guess if both converters have the same efficiency, let's say 88%, they would run just as long as each other?"

Correct, but efficiency varies with load, so they would have to have the same efficiency profile over your operating range. Also, some controllers/converters have light load efficiency modes that can make a big difference for battery life if your sensor doesn't draw much current or if it has a sleep or energy saving mode.

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