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Well as you know many hardware modules work with 3.3V. Our product has a PCB with a few components which most of them require 3.3-3.6V to work properly. A little bit below that, they will not perform well anymore.

Till now our approach was to use a lithium polymer chargeable cell of 3.6V which works great with a regulator. But we want to go back to coin cell battery.

Most of the coin cell batteries are 3V. The only one that has a 3.6V is LIR2450 but its a chargeable battery with very low capacity.

I see two options:

  1. A charged coin cell battery is usually more than 3V. Can we make that assumption and just put it in our products? Is it "not professional"?

  2. A step up converter from 3 to 3.3V. Does it take lots of current? Are there modules that just do that?

How generally engineers approach this issue that all cell batteries are 3V and most of the hardware components are 3.3V?

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  • \$\begingroup\$ I've looked a bit a CR2032 cells. Brand-new from the manufacturing packaging I've seen anywhere from 3.18V to 3.35V but after just a few weeks of 2 micro-amp load, they all drop to around 3.1V. I wouldn't depend on such a cell to directly power a circuit that requires 3.3V. \$\endgroup\$
    – DoxyLover
    Oct 29, 2015 at 19:18
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    \$\begingroup\$ I have honestly not seen a Joule thief (your currently accepted answer) used in any professional products. If you expect a more informed/professional answer, we'd need to know what kind of current you expect/need [and for how long]. As DoxyLover says, coin cells don't have much capacity. Typically coin-cell powered stuff runs on 2.5V rails or similar. It's possible however to use a 3.3V [output] charge pump (i.e. a type of boost converter) like ti.com/lit/ds/symlink/tps60212.pdf \$\endgroup\$ Oct 29, 2015 at 19:29
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    \$\begingroup\$ Regarding current you can hope to draw from a coin cell, TI has a nice app note: ti.com/lit/wp/swra349/swra349.pdf \$\endgroup\$ Oct 29, 2015 at 20:01
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    \$\begingroup\$ @RespawnedFluff Well , we have somw wifi module that wakes up once in while and draw 200ma for about 50-100ms. In some cases rarely, it draws 100mah for 30 seconds \$\endgroup\$
    – Curnelious
    Oct 29, 2015 at 20:12
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    \$\begingroup\$ The latter is a lot, i.e. 3 Coulombs. To buffer that you'd need a 1F [super]capacitor at 3V. This is just a rough calculation, ignoring any inefficiencies in voltage conversion etc.; the voltage of the [super]capacitor will drop as it discharges which in turn affects the efficiency of any regulator you'd use. \$\endgroup\$ Oct 29, 2015 at 20:24

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You may want to look at the concept of a Joule thief. It is a "self-oscillating voltage booster". It is relatively simple to build and it will allow you to use almost all the power in the battery, even when its voltage drops.

From https://en.wikipedia.org/wiki/File:Joule_thief_schematic_de.svg

Image courtesy of Wikipedia

Note As others have pointed out in my comment this solution is not going to work well for what you are doing.

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  • \$\begingroup\$ very interesting, but its not really answer me for other stuff such as- can you draw lets say 200mA from a coin cell for 1 second pick ? the circuit idea is nice but the battery after all has to fit. \$\endgroup\$
    – Curnelious
    Oct 29, 2015 at 18:54
  • \$\begingroup\$ I don't know how much current you can draw from the coin cell through this converter. If your load is transient you can store energy in capacitors for sudden bursts. \$\endgroup\$
    – HighInBC
    Oct 29, 2015 at 18:57
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    \$\begingroup\$ You can't power a constant load with a joule thief - at least, not without adding enough extra circuitry you may as well use a regular boost converter. The LED's forward voltage drop is an integral part of what makes the joule thief 'tick'. \$\endgroup\$ Oct 29, 2015 at 19:32
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    \$\begingroup\$ @Curnelious If you need to draw 200mA, a coin cell is a poor choice. They have very high internal resistances that will result in a lot of voltage drop if you try and draw that much. \$\endgroup\$ Oct 29, 2015 at 19:33
  • \$\begingroup\$ @Curnelious: Now that you've mentioned the current you want, I agree with Nick that a coin cell is indeed not a bright prospect. We've actually had a similar question (about almost the same current) a while back: electronics.stackexchange.com/questions/59303/… You should have put the current in the question; I almost missed it here. \$\endgroup\$ Oct 29, 2015 at 19:54

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