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I'm currently building a circuit used in a personal portable cooling system.

The fan is a 12v turbine fan (0.1A)

My plan is to use 2 9v batteries in series to give me 18v then use a buck-converter to step the voltage down to 12v.

My question is, if I have a switch between the turbine fan and the output of the buck converter, and turn the switch off, will the buck converter its self still slowly drain the batteries?

Or would it be best to have the power switch between the batteries and the buck converter?

Ideally due to the way the product will be implemented and used, it would be more user friendly to have the switch between the buck-converter and the turbine fan however if this would drain the batteries even when turned off, then I will need to have a little re-think.

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  • \$\begingroup\$ A 3S Li-ion battery will need little-to-no regulation and will last longer than two 9V batteries in the same or less space. Or you can go 4S LiFePO4 if you're worried about safety. \$\endgroup\$
    – Ignacio Vazquez-Abrams
    Feb 28, 2017 at 2:38
  • \$\begingroup\$ @IgnacioVazquez-Abrams Im glad you picked up on the safety factor - the device will be situated inside of a suit that is worn by the user and could be difficult to remove should there be any serious issues develop. Ive not heard of a 4S LiFePO4 before; i've had a quick google and it keeps bringing up circuit boards rather than batteries (apart from 1 that cost $100) - keeping costs down aswell as import/export regulations are something I keep being reminded have to be taken in to considerations, do you have any more info/advice on these? Many thanks. \$\endgroup\$
    – PaulF
    Feb 28, 2017 at 2:53
  • \$\begingroup\$ yes, any electrical device would drain the batery eventually, if powered - the buck converter still have its 12V on output, so it have to still oscilate inside, even when no current is asked for. A that internal work would take some energy from bateries, but less, than if it is fully loaded with the fan. So switch the bateries off save more power. If it is for some costume for ocassional use, than more user friendly switching would won over the price of bateries and the bateries can be removed later the use. If it is for everyday use, then switching bateries would be much better. \$\endgroup\$
    – gilhad
    Feb 28, 2017 at 3:34
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    \$\begingroup\$ The amount of current consumed by the buck converter when there is no output load is called the quiescent current. Check the datasheet. Some converters may be down in the uA range, but others may be several mA. So there is a pretty wide variation. \$\endgroup\$
    – user57037
    Feb 28, 2017 at 7:07
  • \$\begingroup\$ You can get 14500 LiFePO4 batteries (AA-size), put them in a 4S AA case, and then use a separate 4S charger to charge the device. \$\endgroup\$
    – Ignacio Vazquez-Abrams
    Feb 28, 2017 at 9:01

4 Answers 4

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Since the Buck converter will be running, the battery will get drained, if the switch is placed after the DC-DC output. you had mentioned placing the switch at the output is convenient, you can use the switch to shut down the buck converter also.

Most of DC-DC converter IC will have shutdown pins, which will put the device to low power mode in which it consume very low power, so you can combine this shutdown pin and the switch to reduce the power drain of the battery significantly

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  • \$\begingroup\$ This is a good idea. Then the switch doesn't actually switch the current. It is just a signal switch to the enable pin. \$\endgroup\$
    – user57037
    Feb 28, 2017 at 8:13
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As said in the comments, yes it will drain the battery eventually. However, there is a larger problem as switching regulators are not (usually) meant to be driven with no load. Sometimes they can actually break, as I have done to my own projects by accident. The safer option would be to put your switch between the batteries and the switching regulator, so you not only save some (albeit very small amounts of power as these things are rather efficient in terms of power draw) power, but protect your switching regulator from accidental damage.

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    \$\begingroup\$ I would not worry about it. In last ten years their feedback circuit is enough load. \$\endgroup\$
    – user76844
    Feb 28, 2017 at 5:02
  • \$\begingroup\$ With off the shelf parts showing a simple wire connecting the feedback pin to the output and nothing else I'm a little skeptical of that, I would assume it would depend on whatever parts you're using in terms of designing the thing from scratch. \$\endgroup\$
    – bit0fun
    Feb 28, 2017 at 15:25
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Although to save the most power, you would want to use a switch before the DC DC converter, there are lots of them that are very very low quiescent(standby) current! Since you are using two 9V batteries, which both have approximately 500mAh capacity, lets say you use a converter that draws 5 microamps of quiescent current. 1000mAh/0.005mA ~ 200,000 hours before it drains both batteries completely! Now this is only a very rough estimate, but you can clearly see that it's most likely your load would drain the batteries much faster.

(here is one app note for a very low quiescent current converter http://cds.linear.com/docs/en/design-note/dn142f.pdf )

Another thing you might want to consider is a buck-boost converter. If you are just using a buck converter, then when your input voltage drops to 12Volts, you will not be able to use the rest of the energy in the batteries. With a buck-boost, it can continue to maintain 12Volts even after and fully drain the batteries.

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    \$\begingroup\$ The two batteries are in series, so the capacity can't be added. \$\endgroup\$
    – JRE
    Feb 28, 2017 at 8:46
  • \$\begingroup\$ @JRE Well, yes you're correct if we're considering the battery at 18 volts then it is the same mAh. If we are considering it at 9 volts then it doubles.. \$\endgroup\$
    – Elixir
    Feb 28, 2017 at 10:14
  • \$\begingroup\$ Well,the question was about using a buck converter to get 12V out of 18V so it would be about two 9V batteries in series. \$\endgroup\$
    – JRE
    Feb 28, 2017 at 11:23
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Very simple, if anything connected had an impedance / resistance it'll draw power. If your switch is off (provided not an electronic switch), there will be no power flow. Electronic circuits will draw leakage current and hence leakage power.

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