# How to calculate consumed energy effects on battery voltage?

I have 2 3,6V batteries. I want to send a SMS when voltage drops down to 3,4V. I ve chosen a voltage detector which has a 3,4V threshold. What I am trying to figure out is how long does it take the battery voltage drop down to 3,4V from 3,6V... So what are the elements to consider while calculating the time of this 0,2V consumption? What do I need to know? I checked the battery datasheet and there is this figure indicates what I need to know. But if there is not a datasheet, how can I calculate the time? What values sould I know?

Here is the battery datasheet: http://www.farnell.com/datasheets/301017.pdf

P.S I ve search the entire questions and answers but I could not find anything but voltage drop coversations. Sorry if this is asked before.

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If you do not know the discharge characteristic of the battery you have a few options to estimate the time for a particular voltage drop.

1 - Perform an experiment.

For small drops, put a resistor equivalent to your circuit load across the battery. Measure the time for the drop to occur. Most batteries have a reasonable manufacturing tolerance so this value would be pretty close to the average.

2 - Estimate the time based on the battery chemistry.

For this method you need the effective energy capacity of the battery (mAh) at your particular current draw, and a function describing the voltage drop over time. These two values can be estimated by looking at the data sheets of similar parts

e.g. Alkaline batteries:

Typical 1.5V alkaline cells start of at 1.6V, drop quickly to 1.5V, then fairly linearly to 1.1V, and finally drop off past 1.0V quickly after that. You a very rough approximation would be that the energy remaining in the battery is a linear function of voltage difference.

Let d be the voltage drop, t be the time to drop, a be the current draw, and e be the battery energy.

Then drop / 0.6V * energy / current = time

i.e. 0.2V / 0.6V * 2800mAh / 10mA = 93h

e.g. Lithium batteries:

Lithium batteries tend to hold their voltage as their energy is used. So any drop probably happens as the last 5% is consumed. Thus time for the drop is

95% * energy / current = time

i.e. 0.95 * 2400mAh / 10mA = 228h

3 - Gut feeling

This is by far the most common method. It involves ingesting cups of tea and/or coffee to get the creative cogs turning, and making up a figure that is palatable to all involved. Be aware though that this cognitive dissonance must be passed on to the client in order to transform this imaginary figure into one that can be relied on.

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