measuring current and power needed

Using Arduino General Electronics
1

Hi guys,

I am new in arduino and currently I am trying to use multimeter and how to measure the power needed by the arduino

My simple circuit on the breadboard with arduino Uno R3 is that two 10kohm resistor in series connect to 5V pin and ground pin.

Using my multimeter I measured that one of the 10k ohm resistor needs 2.5V, thus Current = 2.5V / 10k = 0.25mA right?

I am going to use my Xiaomi power bank (10400mAh / 3.6V ).So it is correct that 10400mA / 0.25mA = 41600 hours , that means can use for 1733 days?

Thanks in advance for answering my doubt ::slight_smile: ::slight_smile:

2
  1. Yes
  2. ?
    How do you plan to connect your power bank? 3.6V is not 5V. If 3.6V is applied to 2x10kOhm resistors the voltage on 10k resistor will be 1.8V and the current will be 0.18mA...
3

put 100 ohm in series with your uno supply line and measure the voltage drop across the 100 ohms.
The current drawn will then be measured as V100/100 amps

This will show what the uno requires

Your 10000 mAh battery pack will self discharge within 1 year so irrespective of what the Uno requires do not expect several years of continuous use.

Apart from that manufacturers claims of capacity are wildly inaccurate

4

Why not use your multimeter directly to measure the current - if you haven't got a multimeter
with current ranges you don't have to spend much money to get one and they are very useful.

I'd be a tad sceptical of the capacity of any power bank, some companies just lie about the capacity,
and capacity drops markedly with use for any realistic battery chemistry, its a fact of life.

If your power bank is boost converting to 5V, the capacity in Ah will drop since energy cannot be
created from nothing - an output current of 100mA will be pullling more like 150mA from the
3.6V cell.

5

Budvar10:

  1. Yes
  2. ?
    How do you plan to connect your power bank? 3.6V is not 5V. If 3.6V is applied to 2x10kOhm resistors the voltage on 10k resistor will be 1.8V and the current will be 0.18mA...

But the breadboard is connected to 5V pin no?
So the whole circuit power supply should be 5V?

6

It was a good attempt, but you missed some things.

The big one is that if your Arduino is connected to the power bank too, it will draw some current, and quite a lot of it too. One of the moderators Nick Gammon has written an article (linked in the "Useful links" sticky) about this, and found that an Uno draws about 50 mA in normal use, no matter what sketch is uploaded. That brings the estimated operating time down to 208 hrs.

The next thing was mentioned by MarkT. 10,400 mAh is very likely the sum of the batteries' capacities. Even a 100% efficient boost converter will derate that capacity by 3.6/5, giving a 150 hr estimated operating time.

This estimate needs to be further reduced to account for inefficiency in the boost conversion and the quiescent current of the conversion circuit. There's too much uncertainty about that to calculate it from first principles though, and probably won't make that much of a difference anyway, so lets just leave 150 hrs. as a pie-in-the-sky ideal maximum.

So we went from almost 5 years estimate to less than a week. Successfully running an Arduino project from a battery is much more difficult than just finding a power source capable of producing 5V. As you can see from this calculation, with a normal sketch your battery life is horrible even with a pretty chunky bank. If you want to get anything resembling good battery life, you need to get off a development board and start using ATmega chip barebones. That way you can use the substantial power reduction and sleeping features of the chip to potentially reduce the current consumption by hundreds of times.

These things may require substantial rewrites of your code though. If you are using large amounts of delay(), it will be very difficult to properly work in the different sleep modes and get a large amount of power saving.