Where is this HEAT coming from???

Dr. Yo · Post by **Dr. Yo** » Sun Nov 12, 2006 2:53 pm

Figures. The first time I try lighting a model in five years and I get
a mystery. Its two wide-angle red LEDs, 1.7v, hooked up to a pair
of AAA batteries in a Radio Shack bettery box, with a simple two
position micro switch. Lights up just fine, but after a few minutes in the
battery box, the batteries are almost too hot to touch! I'm baffled. I
think I have everything hooked up properly....The batteries are RS
akalines, if that matters. Any help/insight would be appreciated.

Styrofoam_Guy · Post by **Styrofoam_Guy** » Sun Nov 12, 2006 6:02 pm

It sounds like you have a short. Any chance you can draw your wiring diagram?

macfrank · Post by **macfrank** » Sun Nov 12, 2006 8:37 pm

Craig, almost certainly your LEDs are drawing as much current as they can from the battery. You need a current limiting resistor.

If you haven't already, use the LED calculator to find a combination of LEDs and resistors that'll match your circuit.

jwrjr · Post by **jwrjr** » Sun Nov 12, 2006 11:59 pm

I believe that macfrank is likely right. You need that resistor (I use one per led unless they are in series).

Sparky · Post by **Sparky** » Mon Nov 13, 2006 1:24 am

You have them hooked in sereis right? if so then you have this:

1.7*2 is 3.4 volts,

2 AAAs, are 2* 1.5 3 volts.

I this case you are already limiting the voltage the LEDs will get exposed to.

If you add a resistor you will be under powering the already under powered LEDs. A properly operating anything will only draw the current it needs (steady state). A limitng resistor is there so that if the LED starts to go crazy (say burnout or over heat) the increased current draw will result in more of the power supply voltage being dropped across the resistor. This starves the voltage left for the LED, the effect is that the LED never has to dispate more Power (current times voltage) than it was rated for, the resistor will not regulate current.

If anything the sereis LEDs are drawing more current than the batteries can supply. Try double AAs or a double C pack. Some of the ultra brights will draw 150 mA, more if you are starving them for voltage.

One other thing to watch, this happened to us this last weekend. The battery pack we had started getting real hot. The back 2 springs in the battery pack are held in place by some type of pop rivet, this means that those connections are exposed in the back. wires and brass rod are protruding into this houseing (one as a support spine for the model, the other as the support stand, and the blinky board is in there too). One of these must've been shorting the battery pack. I had to remove the over heated battery and put some electrical tape over the battery pack ends. The heat problem went away.

Dr. Yo · Post by **Dr. Yo** » Mon Nov 13, 2006 11:11 am

I'm inclined to suspect the battery pack myself-I remember getting a
sort of mild shock from it wit hbatteries installed when it wasn't hooked up to anything. I'll take a closer look at it tonight ( unfortuntely, it is
superglued to the bottom of the base which may limit my options...) and
see what I can figure out.

I did think about a resistor, but hardly ever use them, and honestly I
don't think thats the problem here. If I understand the term 'series'
correctly, maybe. Both the positive leads are soldered to one post of
the switch, along with the positive lead from the battery pack. The negative lead from the BP is soldered to the center post of the switch and
the negative leads from the LEDs are soldered to the other end post.

Thanks all, for the advice and condsideration.

macfrank · Post by **macfrank** » Mon Nov 13, 2006 12:12 pm

Dr. Yo wrote:
I did think about a resistor, but hardly ever use them, and honestly I
don't think thats the problem here. If I understand the term 'series'
correctly, maybe. Both the positive leads are soldered to one post of
the switch, along with the positive lead from the battery pack. The negative lead from the BP is soldered to the center post of the switch and
the negative leads from the LEDs are soldered to the other end post.

Thanks all, for the advice and condsideration.

It sounds like you have them tied in parallel, which was what I was guessing.
You'll need a resistor if this is the case.
I don't understand why you have the switch wired as you've described - it sounds like you could short out the battery pack if you flip the switch in the wrong direction.

For the circuit you've described, the ideal cercuit would be:
BP+ to switch center post.
A switch post to two 68 ohm resistors
The other ends of the resistor each go to ONE LED each.
The - end of each LED would be tied directly to the - end of the battery pack.

If you don't want to use a resistor, you'll have to wire them in series:

BP+ to switch center post.
A switch post to the + end of one LED
The - end of this LED goes to the + end of the next LED
The - end of the last LED would be tied directly to the - end of the battery pack.

If you wire them this way, you have to make sure that the BP never goes above 3.5 volts or so (ie, don't hook up a 5V power supply or 9V battery)

Frank

Sparky · Post by **Sparky** » Mon Nov 13, 2006 2:32 pm

macfrank is right, you have them in parallel.

If you draw the diagram of how they are hooked up you can more easily see this. If the leads from the two items (in this case the LEDs) are hooked up so that all the positives conenct to one buss and all the negatives to another they are in parallel.

If they are daisy chained the way the battery back hooks up it's batteries (plus to minus or end to end) then they are in series.

In parallel you also have to conisder that the current draw will be additive, that is if each LED needs 20 mA of power, the power supply needs to provide 40mA. Each LED in the parallel loop needs its own 20 mA.

For how you have the switch hooked up, we need to see a drawing. It sounds fishy, but if you're using a double pole single throw then it makes sense. You can switch both the positive and the negative with one lever, the two sets of posts are not electrically connected to each other.

BTW it's not nesseary to do this, double pole switches are usually twice as big, since they have to have 2 switches interanlly, isolated from each other.

All you need to do is brake the connection to one of the battery terminals.

So say, the positive lead of the circuit goes right to one of the switch's pole, then the other pole goes to the battery's poistive lead.
The negative lead of the circuit goes to the neagative lead of the battery.

heres some wiki diagrams on series and parallel circuits:

http://en.wikipedia.org/wiki/Image:Seri ... rcuits.png

Sparky · Post by **Sparky** » Mon Nov 13, 2006 2:51 pm

Is this what your switch looks like?
http://myweb.tiscali.co.uk/nuukspot/dec ... dpdt.l.jpg

Say without the E and F terminals.

Dr. Yo · Post by **Dr. Yo** » Mon Nov 13, 2006 4:59 pm

Sparky wrote:Is this what your switch looks like?
http://myweb.tiscali.co.uk/nuukspot/dec ... dpdt.l.jpg

Say without the E and F terminals.

Not quite-Mine is a simple slide switch, with half the number of contacts-
three, instead of six. I think I've got a little better handle on the problem.
It'll probably be Tuesday eve before I can attempt a fix, but I'll keep
y'all posted.

(I'm sorry I can't show you fellows a diagram, or pictures of the set-up,
believe me, I wish I could, but the home computer is getting increasingly
balky. Again, thanks for all the help, all of you. )

macfrank · Post by **macfrank** » Mon Nov 13, 2006 5:42 pm

Like Sparky said, you only want to break the power at one point.

BTW, it's always a good idea to place a switch on the positive side of the circuit, so that when open (off) there's no path at all to the positive side of the supply (the power).

Placing the switch on the negative side will work, too but since grounded connections and points are more common than powered ones, it's possible to accidentally power up a circuit by grounding the unswitched side.

Frank

Slide · Post by **Slide** » Tue Nov 14, 2006 2:01 pm

and now it's my turn to confuse everyone:

Electrical theory works backwards from what actually happens.

power [electrons] flow out of the - end of a battery, but for simplicity/logic's sake we say power goes + to -... when really it goes - to +

so it really dosn't matter where he puts the switch, as long as it breaks the chain completely... but you are correct: it is good Practial/Theoretical Practice to place the switch Before the active elements of a circuit.

just thought i'd help

jwrjr · Post by **jwrjr** » Tue Nov 14, 2006 2:46 pm

In a low voltage circuit (like a lighted model) it is not so important where you put the switch. In a higher voltage circuit (like house wiring) putting the switch in the neutral line could get you killed. If you are working with enough voltage to hurt you, you ALWAYS put the switch in the 'hot' wire. It is a good habit to get into to put the switch in the 'hot' wire all of the time.

Sparky · Post by **Sparky** » Tue Nov 14, 2006 3:12 pm

If you have regulators or filters, they work by shunting energy to ground, or negative side. It's bad practice to break the ground conenction since this disables the regulators and filter's abilties to shunt power away from the circuit they drive.

If you ever add votlage regulators or blinky circuits to your model lighting you can expect weird results and damaged components breaking ground and positive connections together. Lots of voltage bounces occur when flipping a manual switch. Since the double pole system is really two switches inside, there's not much chance that they will both make contact at the same time.

Dr. Yo · Post by **Dr. Yo** » Tue Nov 21, 2006 10:45 am

Well, for what its worth, I beat the heat. I tried the wiring plan that
Frank suggested, but that only lit one of the LEDs-so I went back to
paralell, but I think I reversed the connection-its now got the negative
battery lead to the switch, or something. Anyway, it doesn't burn my fingers after being on for three minutes, and it lights up. My photographer
friend took some pictures of that an a number of other projests this past weekend, so hopefully I'll be able to show y'all the reults before too long.

Steven Marshall · Post by **Steven Marshall** » Sat Dec 30, 2006 8:47 pm

I apologize for reviving a month old thread, but there seems to be some serious misunderstanding of how LED's work. So I'm going to get a little pedantic here.

LED's like all diodes are like a voltage controlled switch; once the voltage across it it is above a certain level (it's forward voltage) it will conduct just about any amount of current you supply it. So, if you put a LED with a forward voltage of 2.5V across a 3V battery, it will draw all the current the battery can supply just as if you put a wire across the battery. That is why Dr. Yo's batteries were getting hot, he was essentially shorting them out. When the voltage is less than the forward voltage it will not conduct.

The purpose of the resistor in series is to limit the current by defining an "operating point". For the previous example of a 2.5V LED and a 3V battery if you want to limit the current to 20 mA you would need a resistor that would give you .5V with 20mA flowing through it. Using Ohm's Law where voltage across the resistor equals the resistance times the current (V=I * R)
.5 = .020 * R
R = .5/.02 = 25 ohms

Why is it important to use a resistor? If your (no resistor) LED is across a small battery that can't put out much more than 20mA then you'll probably be okay, but now use a 3V wall wart that can put out say 100mA and your LED will go up in a puff of smoke. With the resistor in series, it doesn't matter what the current rating of the supply is, the circuit will only draw 20mA regardless.