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

I have a PCB with a few radio modules on it. In its various states it currently uses anywhere between 100 µA and 100 mA. I can calculate how much time it spends in each of its states over a given year.

The radio modules on my PCB all have a wide acceptable input voltage range. My main processor and Bluetooth Low Energy module, for example, accepts anything from 1.8V to 3.6V. Right now I'm running it at 3.0V, using a step down DC-DC converter.

The battery is a 18650 lithium-ion (datasheet).

When fully charged, it provides about 4.3V. I'll step it down to 3.0V. The battery has a capacity of 3400 mAh.

Assuming the average current I'm drawing from it is 400 µA. My calculation for battery life is simply:

time (h) = capacity (Ah) / current (A)

3.4 Ah / 400 µA = about one year

Now, I know that in order to reduce my power consumption, I should run my circuit at the lowest possible voltage, so I'm considering changing my DC-DC converter and running my main processor and BLE module at 1.8V instead of 3.0V.

My question is: why isn't the voltage featured anywhere in my battery life calculation?

• dim's answer looks good, what I can add is that you probably don't want to deplete your battery below a certain level. Depth of discharge during each cycle could be important to consider if your device is to work without battery change for long periods of time (we are talking year). May 30 '16 at 8:07

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.

• at very low currents, converter efficiency might be appalling, even if it's 85% at higher current. It needs to be measured at the target current for any given converter. Dec 30 '17 at 14:14
• @Neil_UK You're right, the 400µA mentioned by OP is rather low, and standard switching converters have bad efficiency at these levels. But if the requirement of high efficiency at low load is correctly considered, finding appropriate converters isn't very hard: just google "micropower buck"... Besides, this wasn't really the point of my answer.
– dim
Dec 30 '17 at 14:42

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.

Regardless of the converter efficiency (or assuming 100%), the battery voltage is being used to calculate the capacity (mah) of the battery. More correctly, the usable voltage drop, 1.4v (4.2v - 2.8v).

In your particular use, your voltage drop is only 1.2v (4.2 - 3.0) and the actual efficiency might be 90%, both tend to reduce the length of time. However, your average current is only 400uA, which tends to increase the length of time, so your answer of about one year seems right.