battery operated devices

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hi freaks,

 

I just recently completed my dinky project which is used to read rfid tags and than i would transmit it to my pc to keep track of things. 

 

But my RFID reader became very static because of power supply which i give it through adapter, so now I'm planing to make it handheld, so my questions are:

 

1. How should I calculate how much time it will stay on, on a 9V battery?

2. And how should I calculate it, if I'm given Ampere Hours or just amperes current dissipation rating?

 

Novice freak.

This topic has a solution.
Last Edited: Thu. Dec 15, 2016 - 05:13 AM
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"Dare to be naïve." - Buckminster Fuller

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Measure the current draw of your device at max load - meaning while reading a card.

 

Hopefully this is in milliamps.  Then simply divide the milliamp/hours by the measured value to get a pretty close idea of how long the battery will last.  Although I dont think a 9v battery is a good choice.  I would use AA cells and a boost regulator, but that's the subject of another thread.

 

JIm

 

Edit:  Mr. Chapman I see provided a better, and more accurate solution.

I would rather attempt something great and fail, than attempt nothing and succeed - Fortune Cookie

 

"The critical shortage here is not stuff, but time." - Johan Ekdahl

 

"If you want a career with a known path - become an undertaker. Dead people don't sue!" - Kartman

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Last Edited: Fri. Dec 2, 2016 - 01:13 PM
This reply has been marked as the solution. 
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9V batteries are one of the worst for energy density but they can generally supply about 500mAh. That means it could supply 500mA for 1 hour. 100mA for 5 hour, 10mA for 50 hours, 1 mA for 500 hours or whatever. So if you could get your device down to 1mA then 500 hours which is about 21 days. If you could get your app down to 100uA (say) then it could be ten times that - so 210 days.

 

Obviously to achieve this you make the micro sleep most of the time. In low sleep modes some models of AVR are down at 1uA levels or even better. You have the micro wake up infrequently, do a quick bit of activity at full speed then get back to sleep as soon as possible. Let's say it spends 95% of the time asleep at 1uA and 5% of the time awake and using 3mA then the average is (3000 * .05) + (1 * (1 - 0.05)) = 150uA + 0.95uA = 150.95uA. Clearly that short time awake is the real killer. But if it is 500mAh (that is 500,000uAh) and it averages 151uA (roughly) then it is 3,311 hours = 138 days.

 

So you are going to need to make measurements of the current consumption when awake, the current consumption when asleep and how long it spends in each state.

 

The battery calculator link is interesting (especially that 0.7 factor!) but it could do with being extended to take into account both awake/sleep current and relative lengths of time.

 

 

Last Edited: Fri. Dec 2, 2016 - 01:21 PM
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Now I wear FREAKS badge on my shoulder,
such a great community.

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I'm not getting why there is 0.7 in the equation of battery calculator link, is it for rms value.

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rockingSharma wrote:
I'm not getting why there is 0.7 in the equation of battery

See the note below the calculation:

*The factor of 0.7 makes allowances for external factors which can affect battery life.

David (aka frog_jr)

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Go with AA, not 9 volt.

 

Run your processor fast, but sleep it as much as possible.

 

Keep everything else turned off as much as you can.

The largest known prime number: 282589933-1

It's easy to stop breaking the 10th commandment! Break the 8th instead. 

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I have use a NXP reader chip and it use about 200 mA 3.3V when reading, but depending how you handle it, it could be only for about 1/10 sec, if you push a button.

 

If it's waiting for the ID tag, there are some modes that use way less power it "see" if there is a tag near by, if so it make a real read. 

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frog_jr wrote:
See the note below the calculation: *The factor of 0.7 makes allowances for external factors which can affect battery life.
So what it's really saying is that if you were hopeful a PP3 might supply 500mAh that in reality the capacity is really only like 350mAh !

 

Why PP3/9V anyway? Is this about the physical size? Does it really need to be 9V?

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Possibly also logistics (damn!  forgot to pack the battery ... off to the convenience store)

If it does have to be 9V then likely will expend a number of batteries; could consider secondary (rechargeable) instead of primary batteries.

A series of Low Self Discharge (LSD) NiMH :

Maha Energy

Maha Energy

Imedion

http://www.mahaenergy.com/imedion/

 

"Dare to be naïve." - Buckminster Fuller

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Battery University

BU-502: Discharging at High and Low Temperatures

http://batteryuniversity.com/learn/article/discharging_at_high_and_low_temperatures

...

A battery that provides 100 percent capacity at 27°C (80°F) will typically deliver only 50 percent at –18°C (0°F).

...

Some of the industrial nickel electro-chemistry cells do have significantly better capacity than 50% at -18C.

Obtaining capacity at -40C is difficult.

 

"Dare to be naïve." - Buckminster Fuller

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Concur

Can get split rails from one AA or AAA via a boost converter with a simple add-on for the negative rail, or, boost into a rail splitter.

Sometimes don't need much negative voltage and/or current (is the op amp truly rail-to-rail in?)

Battery supplied circuits have less noise though with a relatively high frequency SMPS and some filtering a SMPS can be acceptable.

There are "low" noise SMPS; IIRC that's by limiting the slew rate.

 

 

"Dare to be naïve." - Buckminster Fuller

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jgmdesign wrote:
Edit:  Mr. Chapman I see provided a better, and more accurate solution.
My post was cursory ... the following is not :

Battery University

BU-503: How to Calculate Battery Runtime

http://batteryuniversity.com/learn/article/bu_503_how_to_calculate_battery_runtime

...

(linearity, discharge Q = charge Q)

This is not possible because of intrinsic losses and the coulombic efficiency is always less than 100 percent.

...

"Dare to be naïve." - Buckminster Fuller

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Yes, but the formula on Digikeys website follows Peukerts law.

 

Cursory or not its accurate.

 

JIm

I would rather attempt something great and fail, than attempt nothing and succeed - Fortune Cookie

 

"The critical shortage here is not stuff, but time." - Johan Ekdahl

 

"If you want a career with a known path - become an undertaker. Dead people don't sue!" - Kartman

"Why is there a "Highway to Hell" and only a "Stairway to Heaven"? A prediction of the expected traffic load?"  - Lee "theusch"

 

Speak sweetly. It makes your words easier to digest when at a later date you have to eat them ;-)  - Source Unknown

Please Read: Code-of-Conduct

Atmel Studio6.2/AS7, DipTrace, Quartus, MPLAB, RSLogix user

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jgmdesign wrote:
Cursory or not its accurate

It might be accurate for led acid batteries but not really applicable to a random 9V cell of unknown chemistry.

 

This reminds me of school where I had to use complicated formulaes to do some calculations on transistors to "compensate" for Hfe + 1 becasue the base current also flows through the Emitter.

My argument that this was a stupid idea because hfe varies widely because of production paramaters was nullified by the statement that calculation would be to simple otherwise, we had to learn something at school and i would not pass the test if I did not use the qouted formulae correctly.

I was not a popular guy at school.

 

Back then we only had books with tables to compare transistors to each other. Nowadays every transistor has an easy accesable datasheets with a nice graph which shows how Hfe changes with temperature and collector current.

 

To get back to OP.

No AVR runs on 9V (Exept some unobtainable HV variants). and with a linear voltage regulator you would waste most of the energy.

With the low average currents you want the idle current through the ground pin of your voltage regulator could well define your battery life.

Most AVR's run happily on 2 AA's  and these usually do something between 2 and 3 Ah.

An AVR which runs officially down to 1V8 would even run reliably on 2 almost empty AA's (Zinc -carbon, NiMh, Alkaline, whatever).

 

Getting the most out of a battery is an art in itself and is an combination of the right hardware with optimised software.

Search & research "Jeenode" / "Jeelabs". That guy has written some blogs about power consumption & Joule counting on low power battery powered uC's.

 

 

 

Doing magic with a USD 7 Logic Analyser: https://www.avrfreaks.net/comment/2421756#comment-2421756

Bunch of old projects with AVR's: http://www.hoevendesign.com

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Somehow, nobody understands linear regulators. (I exaggerate) Perhaps they don't understand electricity?

 

You have a 3v system running on a 9v battery drawing 5mA. So you device uses 3 * .005 = 15mW of power. You might think, "Cool," but look what your regular is doing. It's dissipating 6 volts times 5mA or 30mW. That's twice the energy your device is using just going to heat things up.

The largest known prime number: 282589933-1

It's easy to stop breaking the 10th commandment! Break the 8th instead. 

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Torby wrote:
nobody understands linear regulators
I might agree with that if it instead referenced "noobody"cheeky

David (aka frog_jr)

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Does providing a PWM supply lines to microcontroller will help or it will trigger brown out detector​?frown

 

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Crowd Supply

snapVCC

by Mahesh Venkitachalam

https://www.crowdsupply.com/electronut-labs/snapvcc

...

The 3.3/5 V super-efficient regulator that snaps right on your 9V battery.

...

 

"Dare to be naïve." - Buckminster Fuller

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Of course, there still isn't as much energy in a 9v as there is in 2 AA cells.

The largest known prime number: 282589933-1

It's easy to stop breaking the 10th commandment! Break the 8th instead. 

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A good 9V id 1.1 AH equ to 9.9wH

A good AA is 3 AH equ to 4.5wH => 2 AA equ 9wH

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That depends.  If you compare apples to apples, 9V loses.

 

Typical AA capacity for an alkaline is 2700 mAh.  At nominal voltage, that's 4.05 Wh, or 8.1 Wh for 2xAA.  Typical PP3 capacity for an alkaline is 565 mAh.  At nominal voltage, that's 5.085 Wh.

 

The difference lies in the fact that the 'best' AA chemistry is Li–FeS2 with a nominal cell voltage of about 1.5 V, similar to alkaline, but with a capacity only about 11% better than alkaline:  3000 mAh, or 4.5 Wh, or 9 Wh for 2xAA.  Where Li-FeS2 greatly outperforms alkaline is in high-current applications.

 

Meanwhile, the 'best' PP3 chemistries can be any one of the several ~3 V chemistries, which can yield as much as 1200 mAh, 112% more than an alkaline PP3, or 10.8 Wh.

"Experience is what enables you to recognise a mistake the second time you make it."

"Good judgement comes from experience.  Experience comes from bad judgement."

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Last Edited: Mon. Dec 12, 2016 - 04:06 PM