Hello-


Thanks for taking the time to read my question. I have a few questions I need answering regarding a project I am working on.


First some background information about my project:


-It is powered by a rechargeable 9V 600mAH battery

-The components I have within my circuit are; two vibrating motors, a voltage regulator using two resistors (10K and 1K resistors), a Bluetooth HC-05, an Arduino Uno board, and a H-bridge (L293D)

-the load of the current is 304.1mA


QUESTION 1: Is the correct formula for battery life [mAH of battery/load of current in mA]?


Using this formula my battery life would be around 2 hours. Am I correct with my calculation?


QUESTION 2: Does voltage of the battery influence the life of the battery? Or is it solely dependent on the battery capacity and load of the current?


QUESTION 3: What are the best ways to prolong the battery life? I need my battery to last at least 2-3 days if possible before needing a recharge. Is it possible somehow to put the circuit in sleep mode when not in use?


Thank you for answering and let me know if any part of my post confused you I tried to explain the best I can.


Fobio Design-

 

1. Yes, that's the correct formula, but at that high a current draw the life will likely be less than 2 hours due to losses in the battery.
2. Voltage has nothing to do with it. It's strictly a function of battery capacity and current load.
3. Whether you can put the circuit into sleep mode or not depends upon the nature of the application and programming of the Arduino board.
If you are regulating the voltage down to 5V from 9V with a linear regulator, then you can increase battery life significantly by using a switching regulator, since they have higher efficiency and the current from the battery will be less than the load current.

 

I have a book from Radio Shack which contains graphs of duration compared to load. Radio Shack doesn't exist in most places, but the internet surely has capacity curves provided by battery manufacturers.

 

The useful amp hours varies considerably based on how much current is drawn. Also, battery life (time between replacement) varies based on how deeply the battery is discharged. The greater the level of discharge the sooner the battery will fail. The optimal battery life verses discharge is about 20% of charge.
Here is a typical chart:

So, lets work some numbers.

.3 amps X 3 days X 24 Hours = 21.6 amp hours.
20% discharge - At 20% discharge this would imply a 100 amp hour battery.
50% discharge - At 50% discharge this would imply a 50 amp hour battery.

Please note: The above numbers are based on deep cycle batteries. Automotive batteries are not deep cycle batteries and are not designed to provide current over long periods.

 

1. Yes, that's the correct formula, but at that high a current draw the life will likely be less than 2 hours due to losses in the battery.
2. Voltage has nothing to do with it. It's strictly a function of battery capacity and current load.
3. Whether you can put the circuit into sleep mode or not depends upon the nature of the application and programming of the Arduino board.
If you are regulating the voltage down to 5V from 9V with a linear regulator, then you can increase battery life significantly by using a switching regulator, since they have higher efficiency and the current from the battery will be less than the load current.

I'll give you a little more detail because I am fairly new to electronic circuits so my knowledge is limited at the moment. The purpose of my project is to create a vibrating pillow. The two motors will be placed inside the pillow I have created and they will be controlled via your smartphone (I made an app as well), hence the use for the bluetooth module. One of the functions of the pillow is to be able to set an alarm via the app and then in the morning the pillow will vibrate and wake you up. If my battery life is short then it cannot work as an alarm because the current load will drain the battery. My current load of 304.1mA is when the motors are vibrating (which is selected by the user by the app). When the motors are turned off my current load is 126.1 mA (which gives gives me a battery life of somewhere around 4 hours, which is still not enough). This is the reason I asked about the standby mode or sleep mode that would be capable of decreasing the current load when the pillow isn't being used at all. I hope my explanation makes sense to you. Also, one last bit of information, all off the components I listed in my initial post are in parallel. I know if I increased the overall resistance in the circuit there would be a lower current draw but in series circuits the current is the same across all components. Thanks in advance for your help!

 

- Assuming that the motors run for less than a minute, makes their load almost insignificant.
- .13A X 3 X 24 = 9.4 Amp hours. 20% = 50 AH. 50% = 20 AH

You are now in the range of emergency lighting batteries. Also a 100 AH automotive battery might work.

 

- .13A X 3 X 24 = 9.4 Amp hours. 20% = 50 AH. 50% = 20 AH

May you please further elaborate on that equation. Also, I need the battery to be fairly small. A car battery is way too big for my project. Thank you!

 

You want your battery to be rechargeable, right? I'd look at 2 or 3 lithium ion cells (like the ones in laptop battery packs) in series. That battery chemistry has a superior energy density in terms of weight and volume, although they are more expensive for a given capacity. Another option would be a pack of 6 AAs. You'd probably want to take them out for recharging in a regular commercial charger. Either of these options would give a lot more juice than a puny 9V.

I'm wondering why the "quiet" current is so high. Could you maybe reduce that by putting the BT to sleep most of the time? I'm just speculating about where the power is going. Do you know?

 

You want your battery to be rechargeable, right? I'd look at 2 or 3 lithium ion cells (like the ones in laptop battery packs) in series. That battery chemistry has a superior energy density in terms of weight and volume, although they are more expensive for a given capacity. Another option would be a pack of 6 AAs. You'd probably want to take them out for recharging in a regular commercial charger. Either of these options would give a lot more juice than a puny 9V.

I'm wondering why the "quiet" current is so high. Could you maybe reduce that by putting the BT to sleep most of the time? I'm just speculating about where the power is going. Do you know?

I am going to list my components and how much current they draw in the circuit that I have configured. The two motors, voltage regulator, bluetooth module, H-Bridge, and Arduino are all in parallel on a breadboard. I am looking to eventually create a PCB as well to insert in the pillow.

1) two motors: 178mA
2) voltage reader (which consists of a 10k and 1k resistor in a series with a node between them that has a wire which is plugged into the A0 pin of the Arduino: 0.68mA
3) bluetooth module HC-05: 40.3mA
4) H-bridge driver L293D: 27.5mA
5) Arduino board: 57.02mA

 

May you please further elaborate on that equation. Also, I need the battery to be fairly small. A car battery is way too big for my project. Thank you!

.13A X 3 X 24 = 9.4 Amp hours. 20% = 50 AH. 50% = 20 AH

.13 amps is your current draw. You will draw that current for 3 days. There is 24 hours in a day. Thus the power drawn is .13 X 3 X 24 = 9.4 amp hours.

What is the amp hour rating of a battery when 9.4 amp hours is 20% of its capacity. answer: 50 amp hours.

Understand??

 

I am going to list my components and how much current they draw in the circuit that I have configured.

You need a sleep mode. That's burning a lot of power all night while it's not doing anything. It's like having 20 LEDs on all night (eg. 10 parallel strings of 2 LEDs in series).