Problem with Lithium 18650 battery that are not stable at 3,7V

Hi. I'm facing a problem with the 18650 Lithium battery I'm using. The voltage they are delivering is decreasing too quicly vs what I want to do and what I have calculated.

Basically I am using boards ProMini with SX1278 Loard, with sensors that are consuming around 1494,97A iver 1h, that is equivalent to 0,42mAh.
I'm using 3,7V Lithium 18650 battery, with a capacity of 3200mAh.
It means that I should be able to keep my battery for around 10 months (I have validated my calculation).

The problem I'm facing is that the voltage delivered by the battery doesn't stay at 3,7V but decrease too quickly.
After 1 month, it is at 1,9V... that is a problem as it is not enough for the board with Lora to start. And so, the board doesn't send the measures on Lora.

Do you have any ideas on a solution ?

Did you face the same pb ? Does it make sense what I see or do you think that the battery I'm using are of poor quality ?
Should I use other technology for the battery ?
Or shall I use 2xbattery 3,7V Lithium in sequence/series so that I start at the beginning with a higher voltage that should reach 1,9V after 2 months ? (that is not enogh for me).

if you actually mean 1494.97mA over 1 hour then a 3200mAh battery would only last 2.14 hours

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That is a problem as you damaged your battery letting it discharge so low. Over discharged battery is not safe.

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The battery is dead at 1.9V. Your math is wrong matter how many times you checked it.

Is the 0.42mAhr figure the average from the actual measured current consumption of the board ?

Have you actually measured the actual capacity of the battery ?

See link below for some tips, 14 months from a 155mAhr battery;

https://stuartsprojects.github.io/2020/02/20/just-how-long-can-a-sensor-battery-last.html

How did you validate your calculations? Because it seems there are some significant errors.

Assuming that "1494,97A" was a typographical error, and that you meant 1495 mA, you seem to have done the following:

(1495 mA)/(1 h) × (1 h/3600 s) = 0.42 mA/s

3200 mAh / (0.42 mA/s) / (720 h/month) = 10.6 month-seconds

So your numbers may be arithmetically correct, but there are serious errors in the units of measurement (and in the formulas used).

If your circuit draws approximately 1.5 A of current, then you will need a battery capacity of about 1 kAh (kilo-ampere hour) for every month that you want the circuit to be powered (i.e., 10 kAh for 10 months). So with your 3200-mAh batteries, you would need to connect over three-thousand such batteries in parallel to get a 10-month running time.

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Hello. Yes you are right. My measures are in mA.

Hello. Sorry but you are wrong. See what srnet and grb wrote : he is right.

Please continue reading the rest of my comment, and show the exact calculations that you performed if they are different from the ones I show above.

For the record, the contents of my comment above are in complete agreement with the response you received from @jim-p:

(3200 mAh) / (1495 mA) = 2.14 h

So I don't understand why you believe that @srnet and I are "right", but that @jim-p's comment is "wrong". This could possibly indicate that you did not understand what I wrote.

Hi. I was wrong when I wrote 1 494A over 1h. It is in mA. And so it consumes the equivalent of 0,42AmAh (0,42×60×60) = 1500mA.
And so, with a battery of 3200mAh, it should long around 10 long. With just 1 battery.

Hi. When it consumes in total 1500mA, it is equivalent To 0,42mA delivered every seconds over 1h. So 0,42mAh.
And the capacities or the batteries are specified in mAh.

Srnet le right.

My batteries should be ok for 10 montage (3200mAh/ 0,42mAh) = 7 705h = 321 days.

Do u agree ?

There is something wrong in your current measurement, or in your units of measurement.

If the circuit draws 1500 mA continuously, then, over the course of an hour, the amount of charge (capacity) consumed from the battery will be (1500 mA)×(1 h) = 1500 mAh, not 0.42 mAh.

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I asked the right questions I believe.

But I have not seen answers to the questions.

Hi. No : it consumes in total 1500mA over 1h.
That is different than consuming 0,42mAh.

1500 / 60 / 60 = 0,42mAh. = delivering continuously 0,42mA per second over 1H = 1500mA.

We are not going to get anywhere until you get the units right
Did you mean 1.494mA?
Or did you mean 1494mA which is 1.494A?

if you actually mean 1494.97mA over 1 hour then a 3200mAh battery would only last 2.14 hours
3200mAh/1494mA = 2.14 hours.

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You are editing your post so many times that it is difficult to follow. As of right now, you've written:

If your sensors are only using power intermittently, then the measure that you need is how many mAh are consumed by your sensors every second, not how many mA are drawn.

Let's suppose that you have made another mistake in the units of measurement, and you meant that your sensors consume 0.42 mAh every second (not 0.42 mA). This means that your 3200-mAh battery (in theory) would last for

(3200 mAh) / (0.42 mAh/s) / (60×60 s/h) = 2.12 hours

However, until you start using correct units of measurement in a consistent way, it is very difficult to understand what you have in front of you.

In the above expression, there are units of mAh in both the numerator and then denominator, so the units will cancel. The ratio you are calculating does not give a result in "hours", it gives a result that is dimensionless.

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I Will read carefully.
Yes 1500mA over 1h is a measure I did:
The board is meanly in a sleep mode over the h. It consumes 0,007mA (real time measure I did) in sleep mode.
3 times per h it wakes up and it consumes 10mA over 7s each time (so over 21s in 1h) and 105mA over 4s (so 105mA in 1h).
Si my calculation : it consumes 10 x 21 + 105 x 12 + 0,007 X (60×60 - 21 -7) = 1500mA over 1h

1500mA for 1h makes 1500mAh
0.42mA for 1h makes 0.42mAh
Quite a difference...

I assume you mean it consumes 1500mA/Secs per hour.

And what was the actual measured capacity of the battery ?

I may be wrong (always possible) but when you consumes 0,42mAh and you have a 3200mAh battery, how do you calculate the duration or the battery ?