3
\$\begingroup\$

I know this is going to be a broad question, but I'm looking for a broad answer that can get me a ballpark figure. When building a board with microcontroller, what's a good way to estimate how long a battery will last?

For example, I'm building a circuit with Texas Instrument's CC 3200. One of the requirement is that the microcontroller has to be always connected to the AP, so I'm assuming I need to look at the current consumption under MCU / NWP active profile. For this chip, it widely varies (I think? I'm not too profound on what the test condition mean in page 31 of the above datasheet) from 59 - 278 mA for MCU / NWP on.

Let's say I use a 9v battery to power this on. A 9v battery typically has around 500mAh capacity rating. So does the estimation go something like

Min: 500mAh / 59mA Max: 500mAh / 278mA

Battery will last between 8.47 hrs to 1.8 hrs?

\$\endgroup\$
8
  • 1
    \$\begingroup\$ Doing measurements would be a good start \$\endgroup\$
    – PlasmaHH
    Commented Feb 2, 2016 at 23:09
  • \$\begingroup\$ @PlasmaHH Well let's assume then you're just designing the circuit and you don't have the parts ordered. Before ordering the parts, you want to get a ballpark figure of how long the battery may last. \$\endgroup\$
    – Xiagua
    Commented Feb 2, 2016 at 23:10
  • \$\begingroup\$ Then you do calculations of an average current draw over a reasonable long period of time. \$\endgroup\$
    – PlasmaHH
    Commented Feb 2, 2016 at 23:11
  • 1
    \$\begingroup\$ Look at a 9V battery's actual capacity at different charge rates, it'll be in the datasheet. At 278mA you'll be lucky to get an hour. farnell.com/datasheets/2006032.pdf \$\endgroup\$
    – user16324
    Commented Feb 2, 2016 at 23:19
  • 2
    \$\begingroup\$ Note that this assumes a direct battery current/processor current identity - which occurs with a linear regulator. If you use a switching buck regulator, 100 mA at 9 volts will give about 220 mA at 3.3 volts, assuming 80% efficiency in the converter. \$\endgroup\$ Commented Feb 2, 2016 at 23:23

1 Answer 1

3
\$\begingroup\$

You are not a million miles away with your estimate. BUT you need to think about what your device is going to be doing to make a real estimate. Use your calculations of:

500mAh / 59mA = 8.47h if your device is only going to be used as a receiver.

Or use the higher value of:

500mAh / 278mA = 1.8h if the device is going to be used as a transmitter.

But you may want to take some other possibilities into your calculations, if you dont need to device actually doing anything for some of the time, you can use the hibernate mode on the device. The current consumption in this mode is 4uA. So if your device lived in this hibernation mode for the whole of its life cycle:

500mAh / 0.004mA = 125,000h or 14 years...

but in general the low power deep sleep is 250uA so:

500mAh / 0.25mA = 2000h so a measly 8.3 days.


One other consideration is the battery life itself, when you look at datasheets, the battery capacity changes with the discharge current. For instance, in this datasheet, at 300mA discharge, the battery has just under 400mAh of capacity, a reduction of 20% from your calculations! but at 25mA, the battery could last over 600mAh, an increase of over 20% from your calculations.


One final thing to consider is that your device doesn't run at 9v, it runs at 3. This means that you need to do some trickery with the voltage. If you want to use regulator, you will instantly reduce your battery life by 60% due to current losses. Another way is to look into buck converters, where you could get a decent efficiency, you might only lose 20% of the battery life. The final consideration would be to change the batteries; if you were to use AA batteries, not only would you increase battery life (much higher mAh rating), you wouldn't get the same losses related to regulation because the voltage would already be lower.

Also, have you thought about using multiple chips instead of a single chip solution? may be easier to quantise current draw for each part. Just a thought

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.