Picking the right battery

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Sambuchi
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Picking the right battery

Post by Sambuchi » Wed Oct 08, 2008 5:41 am

I was planning to make a widget that would run on batteries.
I've never had to select one before.. thought it would be fun.

I have a device that would draw 1.2W at 5 volts.

So here are some silly questions.
--mAh--
Is that the max current rating the battery can handle? If so how does the hour come into play. I'm sure it has to deal with the life of the charged cycle but how?

Putting 2 of the same batteries in series would increase the voltage
and maintain the current rating?

All of this boils down to me picking a battery that can handle a load and will operate for a given time. I'm just not seeing it clearly right now.

Well... let me know if I'm making sense. thanks

Tony

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philba
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Post by philba » Wed Oct 08, 2008 7:10 am

the Amp Hour (or mAmp Hour) ratings refer to the amount of current delivered over time. So an idealized 1000 mA hr rating means that you can draw 1000 mA for one hour. For a rough determination of the life time of a battery, divide the capacity by the average current draw.

The reality is some what different, though. The given rating is usually specified at a certain current level. Higher current levels cause a pretty severe derating. That 1000 mAh battery probably would only run for 40 minutes at 1000 mA. Temperature and age also factor into the actual mAHs delivered. In addition, you can't squeeze all the rated capacity out of a battery - as it discharges, the voltage drops and at some point the voltage is too low to use.

In general, yes, you can put the batteries in series to get the same current capacity. However, the weaker battery will determine the usable life.

As a rule of thumb, I assume I'll get 80% of the capacity. If I get more, it's a bonus.

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jwax
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Post by jwax » Wed Oct 08, 2008 2:25 pm

Your 1.2W @ 5 volt device will draw 1.2/5=0.24 amp. If your 5 volt battery was, for example, 2.4 amp/hr, it would drive your load for 10 hours ideally. BUT, as philba said, probably only 80% of that. (8 hours)
Keep in mind these are generalizations, since we know battery ratings to begin with are always suspect. Age, temperature, discharge rate and history of (mis)use affect the final results.
Also, if your 5 volt device really requires 5 volts, you need to go with probably 6 volts worth of batteries- four (4) D cells for example.
John

Bigglez
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Re: Picking the right battery

Post by Bigglez » Wed Oct 08, 2008 3:30 pm

Sambuchi wrote: I have a device that would draw 1.2W at 5 volts.

So here are some silly questions.
--mAh--
Is that the max current rating the battery can handle? If so how does the hour come into play. I'm sure it has to deal with the life of the charged cycle but how?

Putting 2 of the same batteries in series would increase the voltage
and maintain the current rating?
Some good answers from others, I'd suggest reading a
battery maker's spec sheet to get a handle on what
happens when you change the load current (capacity
is non-linear) and other factors such as temperature
and aging. A PIX is worth 1kB words/bytes, here:

non-rechargeable battery

rechargeable battery

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philba
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Post by philba » Wed Oct 08, 2008 3:56 pm

At 240 mA, I bet you don't see anything close to the rated capacity.

Follow on questions:
- what battery(ies) are you planning on using.
- since you said 5V I would guess you are using a regulator. Linear or switcher?

If you are planning on a linear regulator, don't!

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Sambuchi
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Post by Sambuchi » Thu Oct 09, 2008 5:20 am

Thanks for the great input!

Let me give a little more information.

I will be using a AA size battery.
So lets say I buy a real nice Lithium 1.5V 3000mAH ($3.75 each)

Now I will be needing to drive a Display, Micro and and Sensors.

For the display I chose this part from digikey... let me first get this off my chest!
http://search.digikey.com/scripts/DkSea ... -1171-5-ND

This is from the data sheet.

TOTAL Package Power Dissipation at TA = 25°C[2]
4 Character =1.2 W

but if you look further down the data sheet you will find this

Peak Pixel Current:___________________________ IPIXEL VLED = 5.5 V
__________________________________________ All pixels ON
HCMS-29XX (Other Colors)___15.9mA___________Average value per

Soooo... this means that with one pixel on at 5.5V it will draw 15.9mA.
P=IxV=5.5x.015=.08W per pixel

Each char is 5x7
4(5x7)=140 total pixels
140x.08=11.2W

I'm not sure how the above statement (4 Character =1.2 W) is true.


ok... now that I got that out of the way...

This design will require a boost circuit. I went with a TPIC74100.
1.5 V->5V @ 120mA
I plan on putting a 200-400 uF at the load.

The ucontroller may be a PIC.

I am planning that I will NOT be able to display too may pixels at any
great length of time. Hopefully the charge caps can help me there.

Well... so far I feel good about this.

I look forward to hearing what you have to say.

Bigglez
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Post by Bigglez » Thu Oct 09, 2008 6:27 am

Sambuchi wrote:For the display I chose this part from digikey... let me first get this off my chest!
http://search.digikey.com/scripts/DkSea ... -1171-5-ND

This is from the data sheet.

TOTAL Package Power Dissipation at TA = 25°C[2]
4 Character =1.2 W

but if you look further down the data sheet you will find this

Peak Pixel Current:___________________________ IPIXEL VLED = 5.5 V
__________________________________________ All pixels ON
HCMS-29XX (Other Colors)___15.9mA___________Average value per

Soooo... this means that with one pixel on at 5.5V it will draw 15.9mA.
P=IxV=5.5x.015=.08W per pixel

Each char is 5x7
4(5x7)=140 total pixels
140x.08=11.2W

I'm not sure how the above statement (4 Character =1.2 W) is true.
There is an error in your math. The display is MUX'd
and the power is defined by eq4 page 14:
ILED(AVG) = N * IPIXEL * 1/8 * (oscillator cycles)/64

Where:
PD = total power dissipation
N = number of pixels on (maximum 4 char * 5 * 7 = 140)
IPIXEL = peak pixel current.
Duty Factor = 1/8 * Osccyc/64
Osc cyc = number of ON oscillator cycles per row

140*0.0159*0.125*(28/64) = 122mA
Where osc cycles = 4 digits * seven pixels = 28

0.122 * 5 = 608mW

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