# Why does voltage not appear in the battery life calculation?

It does not appear in your equation because this equation assumes you're using the battery at its output voltage during the whole usage without conversion.

This is not the case here, because you're using a step-down converter. So, to build the correct equation you:

- get
*Vavgbat*: the average voltage of the battery during the whole discharge cycle: the discharge graph of the battery datasheet shows it's around 3.6V for the low currents, such as the one you use. - get
*Iavgbat*: the current you'll draw from the battery, in average, during the whole cycle. It is**not**the current you use at the output of the DC-DC converter (that is where you missed something, I think). If we say the converter output current is*Iout*, then*Iavgbat*= (*Iout***Vout*/*Vavgbat*) /*efficiency*(*efficiency*being the efficiency of the DC-DC converter, usually around 80-90%, check the datasheet). - then you apply the forumla you mentioned:
*time*=*capacity*/*Iavgbat*.

So you have: $$time = \frac{capacity}{Iout \frac{Vout}{Vavgbat}} efficiency$$

Now, you see the output voltage in the formula.

So, if *capacity* = 3.4Ah, *Iout* = 400µA and *efficiency* = 85%, we have:

- time = 8670 hours (about one year) for a 3V output
- time = 14450 hours (more than a year and a half) for a 1.8V output

**One more thing**: given the large times resulting, I think you have to account for the batteries self discharge (or leakage current), which may be significant. Unfortunately, I didn't see it mentioned in the batteries datasheet.

**Additional detail**: Where does the *Iavgbat* = (*Iout* * *Vout* / *Vavgbat*) / *efficiency* formula comes from ?

It comes from the fact a DC-DC converter, unlike a linear regulator, is able to output (almost) as much power as it draws from its input. So *Pin* = *Pout* / *efficiency*. If we say *Pin* = *Vavgbat* * *Iavgbat* and *Pout* = *Vout* * *Iout*, we can obtain the above formula.

On the opposite, with a linear regulator, the voltage is dropped without any consequence on the input/output current. So *Iavgbat* would be equal to *Iout* (not accounting for the quiescent current), which was your initial (inaccurate) assumption.

My question is, why does voltage not feature anywhere in my battery life calculation?

Because your calculation is missing one aspect.

You can use two kinds of voltage regulators:

- linear or
- step-down switch mode.

Now, with linear, the Energy per Charge (= Physical definition of Voltage) that is "too" much is just converted to heat (and subsequently, lost).

So, the current going into the linear regulator is pretty much the same as the current used on the regulated output. The power going into the regulator is higher than coming out of it – because the current is the same, but the voltage is lower.

With switch-mode converters, the Energy from the "input" side is stored, typically in a magnetic field inside a coil (but for your low powers, the cheap and small switched capacitance voltage regulator ICs might make sense, too, where the energy is stored within an electrical field only).

Then, only as much voltage is "generated" from the stored energy as required.

That means that the *power* going in the regulator is the same as the *power* coming out (aside from non-100% efficiency), which implies that if you, for example, half the voltage in your regulator, your regulator only draws half the current it supplies!

Now, the question is, if all your modules support a wide input voltage range, it means they all have integrated supply regulators. Now, if these are linear, you're probably right to use a switch mode step-down converter to boost efficiency. If these modules contain switch-mode power supplies, you shouldn't be using your own regulator – it's very likely that the cascade of regulators will be less efficient than the integrated one alone.