I am pairing an Arduino Uno along with both a HC-05 bluetooth module and a sensor circuit.

Show below is the circuit conditions as well as the circuit. enter image description here

As far as powering up the system, I am planning to use a 9 Volt battery(550 ampere) to connect to the Arduino which draws in 45 mA and the sensor circuit which takes in a total of 28 mA. I calculated the total life of one of the batteries to be 7.53 hours by (550/(28+45)). Can someone help me with the total power consumed by the system?

UPDATE: I read on the sensor data sheet that there are two parameters which heater power as well as sensor consumption. Heater Power consumers approximately 140 mW and the sensor consumes approximately 10 mW. I believe the Arduino will consume 405 mW because of (9 volts * 45 mA) and the bluetooth module will consume 26.5 mW because (3.3 volts * 8). Therefore, the total power consumption is 571.5 mW. Is this correct??? Also will the 9 volt battery be able to support the total power?

  • 550 ampere - What does it mean? Short circut current? Must be some kind of car battery – KIIV Dec 4 at 7:52

Your calculations are not totally right. I did not check the data you provided, so I'm just using the numbers you put.

The arduino draws up 45mA @5V. The sensor has a power of 150 mW and is powered at 5V, which means a current of 30mA. The bluetooth module gets 8mA (are you really sure? it seems too low) @3.3V.

Now, the battery gives you 9V. So you need to convert them to the proper (5V and 3.3V) voltages required by the devices.

You have mainly two options:

  • you use a linear regulator
  • you use a switching regulator

(Please note that if you use the regulators already present on the UNO board - the 1117 and the LP2985 - you will use the first approach)

In the linear case, the regulator will drop the voltage by dissipating the excess power through heat. In this case, you will need to sum the currents:

Total current = 45mA + 30mA + 8mA = 83mA. Please note that if using a 9V battery, the 5V regulator will need to dissipate (9-5) * 83mA = 332mW of power (or slightly less if the 3.3V regulator directly interfaces to the 9V battery).

Now, according to the first datasheet I stumbled upon on google, this kind of battery have a capacity rated at 10mA, but at 100mA it loses about 40% of it. In fact, batteries tend to last less when the current increases (if you double the current you will get less than half of the time).

Considering so a mere 30% of loss at 83mA, your battery will last slightly above 4 hours and a half (550mAh*70%/83mA)

Now, if you use a switching regulator to get 5V from the 9V (and then a linear one to get the 3.3V), you will have to generate 83mA at 5V. This means that, considering an efficiency of 85%, your battery will have to provide roughly 54mA (83mA*5V/9V/85%).

According to the datasheet of the battery, this lower current reduces the degradation of the capacity to about 25%. The total uptime is consequently 7 hours and a half (550mAh*75%/54mA).

If you put a second switching regulator to get the 3.3V, you will have - 75mA at 5V, which with a 85% efficiency will become 49mA - 8mA at 3.3V, and with a lower efficiency due to the very low current (let's say 75%) become 4.4mA

Total current from battery: 53.4mA (which can be considered equal to the case with just one switching converter.

Just one last note: please consider that a heated sensor like the one you have is not engineered for battery-powered devices, since a lot of power becomes heat. When you want to develop a real battery-powered sensor, you will need to go with a lower-power solution for both the microcontroller (and the board) and the sensor

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

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