two 18650 and low current for many month

[Johannsen]

Hi,

for a outdoor beam brake I need a battery solution and thinking about using two 18650 cells in series and a LDO MCP1307 3.3V. The current is around 0.8mA. Using a battery calculator that should run for over 10 month.
Is this solution possible with the 18650 cells? I'm also looking for a discharge protection circuit with low power consumption. I'm using a 3.3 voltage regulator to keep the intesity of the IR LED at a constant level. MCU is a 328P

[Rerouter]

a 328p is quite the power hungry thing for that level of time frame, but should be possible

You will probably need to look at everything in your circuit if you want a reliable time frame, e.g. making sure your micros clock is slow enough to run for your entire discharge range, Also you will need to probably include the leakage of any capacitors in this total, some leak worse than others, and in previous projects a poorly chosen cap has taken up 10-50% of my power budget

Things like what are the idle currents of even the battery discharge protection switch.

[Johannsen]

Actually the 328P uses very little energy in my device. They whole circuit uses as I wrote around 0.8mA..so this problem I solved. The issue with the batteries and a protection circuit however not

[Rerouter]

sorry forgot about if the ADC was properly switched off, OK, so he draws nothing for a pin interrupt wake up and act system.

For your application, your device is well within a normal lithium discharge curve, so unless you need the higher voltage for something else, you could use them both in parallel and loose the LDO, leaving a small led buck regulator (possibly buck-boost if you need an odd voltage led)

That should leave you with about 250 days of runtime, (I am assuming IR leds, and system cuts out around 2V)

Looking back at the original math, your 10 months seems wrong for an LDO regulator, as the mAh of series batteries do not add, only there voltage.

[GeorgeOfTheJungle]

two 18650 cells in series and a LDO MCP1307 3.3V. The current is around 0.8mA. Using a battery calculator that should run for over 10 month.

How do you arrive at that figure?

With an LDO and 3Ah cells I get 5 months: 3e3mAh/0.8mA= 3750 hours, /24/30= 5.2 months.

With a switching regulator (say 90% efficient), you'd get:

a) Cells energy: 2*3.7V*3Ah= 22.2 Wh
b) Device power: 0.8e-3A*3.3V= 2.64e-3 W
c) efficiency 0.9
runtime= c*a/b= 7568 hours, /24/30= 10.5 months.

[Johannsen]

I used digikey's calculator https://www.digikey.com/en/resources/conversion-calculators/conversion-calculator-battery-life
and since I have two batteries I assumed 6000mA in total capacity (7.4V*4A=22,2W)
So this calculation is only correct with a switching regulator.

>you could use them both in parallel and loose the LDO

Yes that would be propapbly the best way but I want to keep the LEDs at a constant brigthness till the end and every now and then I want to measure the batteries voltage so I need some reference

Thats the LDO
http://ww1.microchip.com/downloads/en/DeviceDoc/20005122B.pdf

would that be a better solution?
http://www.ti.com/lit/ds/symlink/lmz21701.pdf

[aix]

With an LDO and 3Ah cells I get 5 months: 3e3mAh/0.8mA= 3750 hours, /24/30= 5.2 months.

But didn't the OP say they had two cells?

Never mind me.

[Siwastaja]

It's insane to connect two li-ion cells in series to power 3.3V circuits with a linear reg. One of the cells is doing absolutely nothing expect provide extra energy to be turned into heat.

Assuming you can afford (cost or size) exactly two 18650 cells, put the cells in parallel instead and pick the absolute lowest-dropout LDO (or figure out if you can run it without an LDO, directly off the cell) so you could run the cells down to 3.4V which is basically near 0% state-of-charge at very low currents.

[Johannsen]

Two cells because of the size. I need a fixed voltage for Aref and keep the brightness of the LED at a constant level.
>is doing absolutely nothing expect provide extra energy to be turned into heat

It provide additional capacity (20.2W) as mentioned below and the heat dissipation is marginal compared to the gained capacity?

[Rerouter]

For the LED look into a constant current driver to keep it consistant. Many are built for li-ion based torches

For your battery voltage. You have an internal bandgap reference in the micro. Yes it consumes a little power when switched on. But you should not need to measure more than once a day. So can keep it switched off when not in use.

[GeorgeOfTheJungle]

It provide additional capacity (20.2W) as mentioned below and the heat dissipation is marginal compared to the gained capacity?

Only if you connect them in parallel, or in series and you're using a switching regulator. Look at this figure:



There's not much to gain by going below 3.4 volts => Siwastaja is right. There are ultra low dropout voltage LDOs, down to 50mV, for example:

https://www.st.com/en/power-management/ultra-low-dropout-ldo-regulators.html

[Johannsen]

Thanks for your advice. I will have a look at the suggested regulators.
I'm not quite sure if I want to use the Li-ion batteries anymore. They would perfectly fit into my planned housing and have a great capacity but yesterday I tried to test how long my old power bank would work with the circuit..when I tried to recharge it and use it it exploded.. and those 18650 cells are not even protected. The device should also be maintained by other persons and I'm worried that they could insert it the wrong way or other things.

[viperidae]

What's the overall service life of this thing? You could use 3 AA cells and replace them every 9 months instead of recharge.

It would work out cheaper in the long run. Much cheaper cells, no charging, no protection required.

[Johannsen]

Hey,
at least 6 month would be perfect.

>Much cheaper cells

I'm not sure. rechargable batteries like eneloop only have 1.2V and good lithium batteries are also expensive and can't be recharged.

>Don't forget about self-discharge and temperature-based capacity loss.
I checked a data sheet of a Panasonic 18650 and that looked ok for that time span.

Somehow my 328P freezes when I go below something around 3 volts. My low-power library disables the BOD in sleep mode..I'm not sure if it's because of that or because of the pulsed current by the IR LED. When I remove the LED it's more stable and I even had to add a capacitory 470µ at the 328P's VCC..otherwise it restarted all the time. Never had that before 

[I wanted a rude username]

Are you trying to run it at 16 MHz or something? That would be outside spec at 3 V. For battery applications it is worth running at the slowest acceptable frequency ... especially if you use the low-power 128 kHz oscillator and/or the 32 kHz asynchronous oscillator on Timer 2.

Ikea ALKALISK AAs have about 2.5 Ah and are very cost effective (AUD 50¢ each in Australia).

But I would not accept a 0.8 mA current consumption ... it's probably possible to get this way down, e.g. just by PWMing the LED using whichever timer you are using for sleep.

[mariush]

If the volume is not an issue, you can get C or D type batteries that go up to 18-20 Ah in capacity :

https://www.digikey.com/products/en/battery-products/batteries-non-rechargeable-primary/90?FV=-8%7C90%2C32%7C323994%2C32%7C336742&quantity=0&ColumnSort=-33&page=1&pageSize=25

For that low amount of current you could even use a single battery and use a voltage doubler to boost 1.5v to ~ 2.8v..3v
With two of those you could potentially get everything working straight from battery.
 
Put 3 of those in series and you get 3 x 1.35v..1.65v

Maybe be worth thinking about adding a small solar cell and a capacitor that would charge from the solar cell and at least give you power during the day, while at night the battery can take over.

[Siwastaja]

Don't forget about self-discharge

Self-discharge for commercially available li-ion cells is around 2-6%/year (based on what I measured, depending on brand) so almost meaningless - and this is with a full cell. Self-discharge drops near zero (or maybe 1%/year tops) already at 50% state-of-charge.

[Siwastaja]

It provide additional capacity (20.2W) as mentioned below and the heat dissipation is marginal compared to the gained capacity?

Be careful with the physical entities and their units. W is unit of power, which is speed of spending energy. Wh or Joule are units of energy; Ah (or Coulomb) are units of charge. With batteries, colloquial term "capacity" as-is almost always means charge (Ah).

Two 10Wh cells in series is indeed 20Wh of energy, but that energy is delivered at doubled average output voltage of approx. 7.2V. What a linear regulator does, it burns the excess 3.9V as a heat, to give 3.3V out. The runtime can be directly calculated from the current consumption, and charge capacity Ah - not Wh.

With the cells in parallel instead, the Ah doubles, not the voltage. The regulator only burns 0.3V average as heat, and the runtime is doubled, because Ah is doubled.

A switch-mode regulator would be theoretically capable of transforming the voltage while preserving energy, but for very small loads, it can be tricky to get good efficiency.

[Siwastaja]

Only if you connect them in parallel, or in series and you're using a switching regulator. Look at this figure:
There's not much to gain by going below 3.4 volts => Siwastaja is right.

And, if you further approximate what a, say, 0.01*C curve would look like, I bet you would be able to get the full 100% capacity.

This is because at very low state-of-charges, the equivalent DC resistance of the cell shoots up, causing a significant voltage drop at higher loads. Even if the load isn't that high, say "only" 0.5*C, it becomes significant at the end of discharge. But with extremely low discharge rates, the voltage approaches the open-circuit voltage curve, and you can measure a fully discharged cell at rest to see it will be a bit over 3.4V actually, even if you discharged it down to 2.5V under (significant) load!