Electronics HELP!

[Bender]

Hi all

I have a 7.4v 4400mA lion battery (from a portable dvd player) that I need to power a 5v 550mA led light set (usb book light for a mini marquee)
I think all I need is the right resistor, but I know nothing about electronics
so I have no idea how to figure out what resistor I need, or even if that is the right thing
any help would be much appreciated
Thanks,

Bender

[Ed_McCarron]

Depends on the current draw of the light.  I'd go with a regulator capable of handling the 550 ma.  In fact, I'd question if 550ma is actually the current draw - thats over half an amp - it'd make for one pretty damn bright LED.

The 7805 is a 1A regulator - heat sink it to be safe and the rest should be a cakewalk.  A bit inefficient, but hey, they're like a buck.

If you want efficient, try one of these:

http://www.dimensionengineering.com/DE-SW050.htm

It'll work down to V_out + 1.3v, so 6.3v.

[BobA]

Did you get that 550 ma off the power adapter for your book light?   Seems like an odd figure even for a power brick. 

[Bender]

Thanks for the info guys

I got that info off a 5v plug in adapter that I used to test the light out with (standard USB charger)
there are actually like 18 led's in the fixture it was made to plug into a usb port, so whatever the USB standard mA are is what it should be
EDIT: seems that USB should be 500 mA so lets go with that

you have to remember I'm a total noob at the electrical stuff, so if you could give me a little more info on how to set up a regulator that would rock
I'm trying to keep this project super cheap so I think I'd go with the 7805 thingy

[MonMotha]

If these are super bright white LEDs, a set of 18 (likely 6 strings of 3 in series or 9 strings of 2 in series if they operate off 5V) is likely to draw on the order of 200-300mA, not the full 500 allowed on a USB port after enumeration.  Still, designing for half an amp isn't a huge deal.

Pack voltage spec on Li-Ion batteries is usually at full charge.  It'll go down by ~10% before it "falls off a cliff" when the pack is almost fully depleted.  So figure you're range here will be about 6.5-7.5V.  You want 5V out.  This isn't really enough to justify a switch mode regulator.  Just use a 7805.  Your maximum power dissipation will be (7.4V-5V)*0.5A=1.2W, and you'll be delivering 5V*0.5A=2.5W to the load for an efficiency of ~68%.  A cheap buck regulator could attain 80-85% efficiency, so we're not doing too bad with our linear solution, though a high-efficiency design can top 95% at the cost of significant circuit complexity.

Typical drop-out voltage on a 7805 is 2V@1A, but lower at 0.5A out, so we need at least ~6.75V in.  That's probably down into the range of near full depletion on the pack.

Free air thermal resistance on a TO-220 LM7805 is 65C/W, and we're dissipating 1.2W (max), so we could expect a worst case temp rise of 65*1.2=78C.  That's a bit hot (~100C in a typical room ambient), so a heat sink would be recommended.  Note that the regulator is spec'd to 125C, so if you don't mind it being too hot to touch during operation, a heat sink is not strictly required.

You'll be pulling a constant 500mA from the pack irrespective of pack voltage, so expect 4400/500=8.8 hours of battery life if you run it all the way down.  In practice, with an older pack and not running it all the way down (hard on the battery), you'll probably get 7-7.5 hours.


If you go switch mode, things change.  Much less power is lost in the regulator, and your current draw is non-constant (it will vary as the pack voltage drops).  Might be worth it if you want to play around.  In a commercial product, I would probably go SMPS here, but on something I'm hacking up to play with, I'd just grab a 7805 from my parts bin.


Make SURE your Li-Ion pack has an internal safety board.  DO NOT use it otherwise as there is a MAJOR fire hazard!  If it does catch on fire, whatever you do, don't put water on it!  Even with an internal safety board, take care not to short circuit the output.  Currents in excess of 200A can easily flow before the safety board cuts things off, which could cause injury due to heating or sparking.

Also, how do you plan to recharge this thing?

[Bender]

Again thanks for all the info
It's starting to make sense to me now
I'm actually using the display from the DVD player and the whole PCB and I was just going to get the power from where the battery connects to the board(that way I get power when it's plugged in too) I took out the dvd drive ect. so I was hoping to get about 2 hours out of the thing with the video and marquee on
as for recharging I'm just planing on using the power-cable that came with the DVD player

[MonMotha]

Should be OK.  Most battery packs like that have some sort of internal safety (though it is often non-resettable).  Make sure you at least turn off or, even better, disconnect the light from the battery when charging.  If you don't, the charger won't charge the battery right.  It shouldn't create a hazard, but it will take a very long time to charge the battery and could damage its capacity.

There's probably 5V available on the DVD player PCB somewhere if you can find it.  Of course, how much current you can pull from it is anybody's guess.

[Bender]

ok, so the the 7805 looks good, now how do I set up the circuit?
I see there are three pins on there what hooks to what?
and How do I get it to drop from 1A to 500mA?

[Ed_McCarron]

Don't worry about the 1A/500ma part; current isn't like voltage -- having too much won't blow things up.  

As for the 7805...

Three pins - left to right with heat sink down.


V_in   Gnd   V_out



Don't forget the bypass caps -- odds are it'll work without them, but its good practice to use them to prevent stray oscillation.

[Bender]

GREAT!

I can do that!
sorry for the silly questions, but what kind of caps should I use for a bypass cap and what does it do?

[Ed_McCarron]

GREAT!

I can do that!
sorry for the silly questions, but what kind of caps should I use for a bypass cap and what does it do?

Electrolytics, negative side to the ground pin of the 7805.  Values as in the pic.  Voltages - whatever you have, as long as its over whatever voltages you expect to see.  They'll shunt anything thats not DC to ground, keeping the 7805 from going into oscillation and eating itself.

[Bender]

Thanks Everyone for the help! (and patience with the noob questions)

I'm gonna try it out tonight!

[Bender]

Here it is my first little circuit, everything seems to be working now


Thanks Everyone and especially Ed for the diagram couldn't have done it without that

one more quick thing, do I also connect the ground? or just the heat sink, seems to work both ways but the light is a touch brighter if I connect the ground, weird

I gotta take a course in Basic electronics or get a good book, I wish I understood all this better

[MonMotha]

Ground on the regulator needs to be connected to whatever's the "other half" of the power supply.  In this application, that should be the negative side of the battery.  I do believe the tab on the regulator (connected to the heatsink) is internally connected to the ground pin, so either works as a connection point, but use of the pin itself is recommended since your connection from the tab to the heatsink is probably not electrically good (anodized heatsink, etc.).

[Bender]

MonMotha, thanks, that's the way I have it set up now and everything seems to be running smoothly
I'm going to test out how long it'll run on the battery and make sure nothing explodes
then run it a few hours plugged in to check it that way, before I put all this in a tiny cab

[MonMotha]

If you have this hooked directly to the battery, then it will interfere with the battery charging operation if you try to use it while the battery wants to charge.  This is because the charger has to limit the input current to the battery and does so by monitoring said current, but the current your LED setup draws (which is substantial) will be included in that measurement even though it doesn't go to the battery.  This will, at minimum, cause the battery to charge more slowly than normal, and it could in fact result in the battery not charging at all (and perhaps even continuing to discharge) if the LED setup draws more current than the programmed battery charge current.

You should either turn off or disconnect (better) the LED setup while charging the battery, or you'll need to find the battery rail on the other side of the charger and tap in there, so the charger won't include your LED current in the charge current measurements.

Also, ensure that you never let the battery pack become fully depleted.  The charger circuit is often responsible for cutting power before the battery goes completely kaput, but since you're on the battery directly, that won't happen.  Fully discharging a Li-Ion battery tends to damage its capacity pretty badly.  Moving to the system power side of the charger would fix this, too, but finding that net may not be the easiest task.

[Bender]

Again, thanks for the heads up, I can see what you mean, I was wondering how the battery would know when to shut off and assumes the the LED screen had the largest draw and it was made to shut down for that, so I thought I'd be ok
With everything on it does seem to still charge the battery(and stop charging when it's full) I don't know it it's any slower or not
also the light is hooked to a switch that shuts it off when the dvd player is off
Right now I'm tapped into the contacts on the PCB where the battery connects, but not directly to the battery
Seems like the dvd drive would draw a similar amount of power as the LED's put that's an uneducated guess