Why do i get better battery life than my calculation?

[ElectricPower]

I've been doing electronics for a while now, but I can not get my calculations to agree with what happens in practice regarding battery life.

I have made a temperature gauge (wireless) that sends temperature to a server that I can read again in an app. This is controlled with a Wemos D1 Mini (ESP8266). It is programmed to send data once a minute and the process takes approx. 5 seconds before going into "deep sleep" mode again.

I have measured that it uses 200uA (micro amperes) in deep sleep and 80mA in the five seconds it is alive per. minute. I use 1 pc. 18650 battery with 2000mA capacity. (brand new battery taken from a newly purchased but cheap battery bank)

Can anyone help me to calculate approx. how long will such a battery last with the consumption figures I have mentioned above?

The reason I ask is because the battery has lasted as long now as I have assumed it should last, but when I measure the battery it still has 4.1V in it ...

Either I have calculated completely wrong, or I have a multimeter that is lying. Thelast, however, is very unlikely ....

[kaz911]

What did your own calculation spit out ?

If it is 80mA for 5 - that is then 400mA(seconds) - is that not about 0.11 mAh? <- sorry just me being absolutely idiotic.

Just "reaching" here - have not had time calculate it.

It runs for 80mA for 1/12th of the time. So 6.66mA "all the time"

[NickEdison]

It is important to realize that the capacity in your case is likely expressed in mAh; meaning how much mA it can deliver for 1 hour.

The average current drawn each minute:
200[uA] * 55/60 + 80[mA] * 5/60 =  0.2[mA] * 55/60 + 80[mA] * 5/60 = 6.85 [mA]

The average current each minute is the same as that per hour, hence your consumption remains 6.85 [mA] (average for 1 hour). The battery life can now be estimated as follows:
2000[mAh]/6.85 [mA] = 291.97... [h] (291.97.../24 gives an expected life of approximately 12 days)

The main parameters that impact the consumption above is the time it is alive and what it consumes then:

• Maybe it is only alive for 3 seconds or less
• Not sure what else happens before it goes into "deep sleep" mode, but maybe it is in another state where it is consuming less current

[ElectricPower]

Thanks for good explanation

My calculation was almost the same, but since we now know the average consumption (6.85mAh) and the battery should last around 12 days, i can tell that the battery has now last 14 days, and when i measure the voltage on the battery i get 4.11V.

I do not understand why. I'm absolutely sure that my measurement is correct, but it is a chance that the units on time is +- 1 or 2 sec.

I thought the battery should be almost empty now, but 4.11V is just 0.6ish volt usage....

[RoGeorge]

Are you sure about the 5 seconds activity to transmit data?

That, seems a huge time to send a temperature.  A single data packet would be enough, so more like 0.5s, maybe way less than that.

Wire a small series resistor in series with the battery (1 ohm or smaller), and capture the voltage drop on that resistor during a full minute.  If you don't have a digital oscilloscope to capture the voltage drop, maybe use a soundcard to record the voltage, then look at the results with a program like Audacity.

[BeBuLamar]

I would expect you to get better battery life than the rated 2000mAh because your discharge rate is too low. Most Li-Ion battery would have higher mAh rating if discharged at lower rate.

[ElectricPower]

I would expect you to get better battery life than the rated 2000mAh because your discharge rate is too low. Most Li-Ion battery would have higher mAh rating if discharged at lower rate.

That can be an good explanation yes, if it's true.

[ledtester]

I have measured that it uses 200uA (micro amperes) in deep sleep and 80mA in the five seconds it is alive per. minute. I use 1 pc. 18650 battery with 2000mA capacity. (brand new battery taken from a newly purchased but cheap battery bank)

Battery capacity depends on the current draw and the 2000mA rating is usually measured when drawing 100's of milliamps (like 500mA or 1A). A battery will last longer if you draw less current.

To reduce transmission time you might look at this video:

An Efficient WiFi Button For Home Automation Using ESP-NOW | ~68 Years From 2xAA Batteries
https://youtu.be/iEkGRI8_txE

They got the transmission time down to 280ms.

Edit: Here's an example showing how the discharge rate affects the battery capacity measurement:

https://www.powerstream.com/9V-Alkaline-tests.htm

In the table note how the capacity measurement depends on the discharge rate -- e.g. 310 mAH @ 100 mA discharge rate vs. 90 mAH @ 1A discharge rate. This example uses alkaline batteries, but it is generally true for all batteries.

[richard.cs]

For what it's worth it's possible to measure and report temperature way quicker than that, and to achieve lower sleep currents. For example at work we have a sensor measuring and transmitting every 30 seconds with a 7 year life on an AA. I've not looked at all at the components you're using but in general I would expect you to be able to get the sleep current down to a couple of microamps, and the measure and TX to a few tens of milliseconds. 80 mA during TX sounds kinda high too. Getting to the lowest power sleep modes often takes some careful datasheet reading and quite a bit of care.

Of course if 12 days meets your requirements that's cool, well done for making something that works. :-) If you'd like to improve it or shrink the battery then there are probably options to do so.

When working out battery lifetimes I prefer to work entirely in charge units (Coulmbs or mAh) but some people like to calculate it as an average current which is equivalent.

[ElectricPower]

I see.

So all my calculation regards to battery life do not give me any good data at all. That was new for me. Measured voltage to day (17 days in use) and the voltage shows 3.98V.

[Siwastaja]

Many wrong answers, real reason is simple: multimeter was never supposed to measure spiky current correctly. There's a reason why it says DC.

For measuring sleep time leakage (which is pretty much DC), it's OK, but that 80mA peak could be actually anything. It's a complex waveform, you need the integral (or average).

You really need a scope, and a device such as Dave's µCurrent, or your own similar shunt amplifier circuit. You may be just barely be able to it with just scope, with relatively high-value shunt resistor.

As others have said, 5 seconds just to send temperature is massive waste of time. Even if the current actually isn't 80mA, it's still something quite high, so fix this and it'll last even longer. You should be able to do this in 50ms easily.

[bdunham7]

Many wrong answers, real reason is simple: multimeter was never supposed to measure spiky current correctly. There's a reason why it says DC.

For measuring sleep time leakage (which is pretty much DC), it's OK, but that 80mA peak could be actually anything. It's a complex waveform, you need the integral (or average).

You really need a scope, and a device such as Dave's µCurrent, or your own similar shunt amplifier circuit. You may be just barely be able to it with just scope, with relatively high-value shunt resistor.

As others have said, 5 seconds just to send temperature is massive waste of time. Even if the current actually isn't 80mA, it's still something quite high, so fix this and it'll last even longer. You should be able to do this in 50ms easily.

And among those wrong answers--Peukert's effect doesn't apply to Li-Ion cells.

Given his results, I'd guess that his device isn't actually staying on for 5 seconds.  I've actually done something like this to look at battery draw where I used a 1-ohm resistor as a shunt (which is probably what his DMM has anyway) and then measured across it with two different channels of a DSO so that I could set them at different scales to see the baseline and the high points.  This is probably marginal at 200uA depending on the scope, but should capture the power-on interval fairly well and allow him to piece together a pretty good idea of total draw without using an SMU.

[bdunham7]

So all my calculation regards to battery life do not give me any good data at all. That was new for me. Measured voltage to day (17 days in use) and the voltage shows 3.98V.

As they say, Garbage IN, garbage OUT.  Your measurements of the current draw are pretty clearly higher than actual, likely by a huge amount.  So yes, your DMM is 'lying'.  What exactly are you using?

[ElectricPower]

So all my calculation regards to battery life do not give me any good data at all. That was new for me. Measured voltage to day (17 days in use) and the voltage shows 3.98V.

As they say, Garbage IN, garbage OUT.  Your measurements of the current draw are pretty clearly higher than actual, likely by a huge amount.  So yes, your DMM is 'lying'.  What exactly are you using?

I will rather say in this case, "garbage in, refinement out"

I measure input voltage (DC) and current with Brymen 869S. I know how to read, and when i measure the current it says between 60-90mA. In about 5 seconds.

[NickEdison]

multimeter was never supposed to measure spiky current correctly. There's a reason why it says DC.

For measuring sleep time leakage (which is pretty much DC), it's OK, but that 80mA peak could be actually anything. It's a complex waveform, you need the integral (or average).

You really need a scope, and a device such as Dave's µCurrent, or your own similar shunt amplifier circuit. You may be just barely be able to it with just scope, with relatively high-value shunt resistor.

This is a good point and to do accurate measurement good hardware is definitely required. However if you would like to get a better estimate of the behavior while transmitting data, you could measure the current over a shunt with a 2nd micro controller. This will not be accurate (although you might be able to improve it significantly with an amplifier circuit) but might give you a better idea of what is really happening.

If you are interested in how to, let us know and either me, or someone else will be able to help you with that.

As they say, Garbage IN, garbage OUT.  Your measurements of the current draw are pretty clearly higher than actual, likely by a huge amount.  So yes, your DMM is 'lying'.  What exactly are you using?

Although possible I think it's more likely that a DMM is not the right tool for this task because of what Siwastaja stated.

[bdunham7]

I measure input voltage (DC) and current with Brymen 869S. I know how to read, and when i measure the current it says between 60-90mA. In about 5 seconds.

60mA for 5s per minute is 5mA average.  Your battery should be stone dead in 400 hours from that alone. 

Do you have a DSO?  If so, use a 1-ohm resistor as a shunt as I posted above and have a close look at that power on cycle..  If not, try these:

Look at the bargraph display--does it show a steady(ish) reading for five seconds?

Use the AC+DC current mode, see if anything changes.

Put a capacitor, maybe 10uF, across the multimeter.

[Siwastaja]

And among those wrong answers--Peukert's effect doesn't apply to Li-Ion cells.

Exactly, although really high currents accompanied with not shifting low-voltage cutoff limit result in effective loss of capacity, which can be gained back by lowering LVC limit or CV discharging until current drops to datasheet conditions. In any case, this could account for maybe 10% capacity loss and is completely insignificant in such low-power gadget.

Capacitor over the multimeter is a good low-cost, simple idea, while it won't make the readings accurate, it will significantly reduce the measurement error and give the right ballpark.

[james_s]

I think the question has been pretty well answered, but only 12 days on a battery for something that is meant to be used continuously sounds grossly inadequate. For a simple wireless temperature measurement each minute you ought to be able to get *years* out of a battery. The ESP8266 is probably not the best choice here. No matter what you use, you will want to avoid linear voltage regulators, especially the cheap ones that are used on most of the various modules from China. The quiescent draw can easily be more than what the load is drawing for micropower stuff.

[Siwastaja]

Linear regulators are OK if you use low-Iq types. Low-Iq yet high efficiency switchers are a relatively new thing. That's OK too.

The key here is to minimize that 5 seconds into something sensible. It could go down to 5ms with proper software design. 50ms should be very easy to achieve regardless of what bloat libraries are used or not.

[ElectricPower]

I using an HT7333 regulator with an 1000uF and an 100nF capacitor.

[james_s]

In the small number of battery powered devices I've made I didn't use a regulator at all, I used a 3V lithium battery powering the microcontroller directly. Low quiescent regulators designed for battery powered devices do exist, but all of the cheap Arduino-like devices I've seen have cheap regulators with a high quiescent draw.

[ElectricPower]

16 days now, and still going.

Voltage is 3.88V

[james_s]

I think you will find that the voltage will drop rather suddenly at the end. Lithium ion batteries have a very flat discharge curve with a sudden steep slope when they are depleted.

[ElectricPower]

I think you will find that the voltage will drop rather suddenly at the end. Lithium ion batteries have a very flat discharge curve with a sudden steep slope when they are depleted.

Maybe. Time will show

[SteveyG]

How are you measuring battery SoC? Battery voltage on Lithium batteries is not a reliable measure. What load are you applying whilst trying to infer SoC?