Good news: project works. Bad news: eats batteries like popcorn.

firestorm.v1

Hello again, I wanted to thank you all for straightening out my programming problemearlier. Ends up that there was a bad solder joint in my cable I made that was preventing me from using the INS and OUTS correctly that were causing my LCD displays not to work. I have made massive improvements in my LCD + RFID scanner however I now have a new problem. I am eating batteries like mad!

I have the RFID reader running off of Vdd at the top of the mini breadboard while the LCD is behind it's own +5v regulator and is attached to Vin. The project works properly as expected however the battery only has about an hour's lifespan before the LCD will only turn on when data's been sent to it then will ultimately not work at all. It appears that there's not enough power to run everything, but I don't think I'm running anything too powerful.

I checked the wiring as I know that running the LCD off of the on-board regulator will closely push it's limits if not break it but I'm not sure about the RFID reader.

Am I simply running too much off of a 9V battery? Should I put the RFID reader on the same +5v regulator as the LCD? (The regulator is a 7805, rated for 1A and according to the datasheet for my LCD, it's current consumption is 363mA)

As always, thanks for your help and your recommendations.

achilles03

If your LCD is using 300+ mA constantly, then that's what's draining the battery. Most 9V batteries have about 500mAh on a good day (600 if they're drained slowly):

data.energizer.com/PDFs/522.pdf

Does the LCD have a backlight? That can be an extra drain. Read the datasheet and see what functions (if any) are the biggest current hogs for the LCD and then see what you can do to minimize current drain.

Also, from the documentation, it looks like whenever the RFID reader is enabled, it's searching for nearby tags. That means it's using more power than it would be if it was idle. Putting the reader on a duty cycle would help (say .5 sec on, .5 sec off). Unless you expect things to be whizzing by really fast, the reader shouldn't miss anything.

Hope that helps,
Dave

allanlane5

Yes, 300 mA is a "large drain" if run continuously. 1.5 volt AA batteries can have 1500 mA-hours of capacity, but that'll only get you 5 hours of operation (300 mA-H * 5 == 1500 mA-H).

Is there some reason you can't power this off a wall-wart? And why is your LCD pulling so much current, is it back-lit? And is this project supposed to be powered continuously, or would it be enough to press a button, read an RFID tag, make a display, then turn the thing off till next time?

SRLM

The RFID reader uses 90 mA when running (red light on), and only 10 mA in standby mode. www.parallax.com/dl/docs/prod/audiovis/RFID-Reader-v1.2.pdf

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Duffer

Along with the suggestion about a duty cycle for the RFID, it might be worth investigating whether you 1) need the LCD on all the time or just when the RFID reader is triggered and 2) whether you need the LCD's backlight on every time the LCD is on (it may be readable in daylight and only needs to be lit at night; controlled by a photocell). Turning on the LCD only when the RFID reader is triggered and leaving·it on for 5 seconds (RFID success or failure) would increase your battery·life by at least 12 times. I agree with the others regarding 9V cells.·If space is a consideration,·using 4 AAA (or 5 rechargable AAA) cells would give you·longer lasting power than a 9V.

Duffer

erco

A 7805 acts like a variable resistor. It just dissipates energy as heat to maintain 5V. That's just wasting power. The more current that goes through it, the more heat is given off. Wasted power.
The more voltage drop it has to regulate, the·more heat is given off. Wasted power.

So a 9V battery is a poor choice to power a 5V circuit. High voltage and low current capacity.

You could probably power everything off 3 alkaline AAs or AAAs (4.5 v) for a long time. Better yet, 4 NiMH AAs or AAAs (~5V). This completely eliminates the regulator and is·much more·efficient.

BTW, a 9-volt battery is made of six tiny AAAA cells spot-welded together. See http://www.youtube.com/watch?v=b28uZ427VUM·for useful info how to·open the case,·cut them apart and use them individually. Typically a good 9V alkaline can be had for $3 and yield 6 AAAAs, whereas IF you can find AAAAs individually, you will pay about a buck apiece.




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·"If you build it, they will come."

firestorm.v1

Good evening everyone. I wanted to tahnk you all for your helpful suggestions. I was completely unaware exactly how much power a battery could provide or how much power that the LCD and the RFID reader could use.

The LCD is indeed backlit using an LED array. According to the datasheet for the LCD, the LCD *itself* only draws about 2 or 3mA. If this device makes production, I might have the BS2 raise an IO pin high and turn on the LCD for a brief amount of time, then turn it off while the LCD itself remains displaying (like a lot of watches and touchscreen devices do).

This is more of an incremental project, I'm not sure what I'm going to do with the finalized device. It's more of a way for me to get used to using the BS2's IO pins and the RFID reader/LCD

I think the suggestion to go to a wall wart is definitely my next goal however I am concerned about the "noise" with most unregulated wall warts out there. Will the regulators guard against any reasonable amount of noise/brownouts?

Along the same note should I go with a +5V source instead of a +9v replacement without damaging the regulators? My thinking is this would be fine given that there's less power to regulate against, but this is a very short tiptoe into this realm and I am still very new and don't have my footing. [noparse]:)[/noparse]


Thank you all as always. I appreciate your suggestions.

SRLM

For wall warts, you can test to see what the output is under no load. The cheaper ones will output a higher voltage than the label says at no load, and load will drop that. The nicer ones will output the correct voltage from no load up to the maximum load. I have some warts that I scavenged from an old palm pilot docking station, and they output 5 v directly and it's well regulated.

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allanlane5

A linear regulator has a minimum input voltage that it requires, so that it can put out a 'regulated' output voltage. The difference is called the "drop-out voltage". For a 7805, this is 2 volts. So, if you're going to get 5 volts out, you need 7 volts DC in.

Now, most "unregulated" wall-wart adapters have simply a transformer inside them with a couple of diodes and a capacitor to make a full-wave rectifier. As such, their output tends to be higher than rated at no-load. With a linear regulator, this tends not to be a problem, as long as the voltage doesn't fall below the (output + 'drop-out voltage').

Now, there DO exist "low-drop-out" regulators, like the LM3940, which can regulate with 0.5 volts difference between input and output.

I believe the BOE board has a plain 7805 on it, so a 7.5 volt unregulated wall-wart works fine with this. A 9-volt wall-wart will work also, though the 7805 will get a little warm. As has been said, a linear regulator works by being a variable resistor, so it 'drops' the difference voltage to insure the output voltage stays at 5 volts. This means there's some power (P = IV) being disappated by the 7805 as heat. The larger the voltage difference, the warmer the 7805 gets. But this is not a problem for what you're trying to do -- even 500 mA is pretty small current for a wall-wart with 7805.