Question - 9v Battery Heating in simplest circuit

Huh?

If it's off, then there will be no current drain from the battery and thus the battery can't get hot.

Show us what you've done. You should be using only two terminals on the switch, no matter how many terminals there are.

|--battery+------switch------resistor-----LED---| | | |_______________________________________________|

The switch type won't matter one bit, it's an "AC switch" because that's how it's rated voltage wise, and you are nowhere near the voltage limit of the switch.

Michael

=A0 =A0 =A0 =A0 =A0 =A0 =A0 |

The switch only works when connected to +V and GND. I tried it just in- line on +V and it doesn't work. So, it's more like this:

| --battery+ --------switch------[optional resistor]-------LED-------|

+V | | | | |___________________ |____________________________ |_______| GND

=A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 |

=A0 =A0 =A0 =A0 =A0 =A0 =A0| =A0 =A0 =A0 =A0 =A0 =A0|

Well, THAT

=A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 |

=A0 =A0 =A0 =A0 =A0 =A0 =A0| =A0 =A0 =A0 =A0 =A0 =A0|

Well, THAT didn't format very well! Battery, switch and LED are tied to GND.

The switch should not be connected between +V and GND. When it is connected this way, the switch is providing a dead short to the battery when the switch is closed. This results in a very high current coming from the battery through the switch. You can put a resistor in series with the switch and LED to limit the current. See below.

Your circuit should have V+ connected to one side of the switch. The other side of the switch should be connected to one end of a resistor. THe other end of the resistor should be connected to the LED. The other side of the LED should connect to GND. When the switch is closed, there will be current flow through the switch to the resistor and then to the LED. The resistor should be chosen to limit the LED to the normal operating current for the LED. A typical red LED will have about 1.5 volts across it when lit. A typical operating current is

20 milliamps. (If you have the specifications for the LED use the values that you are given.) The voltage across the resistor will be 9 volts minus the 1.5 volt drop across the LED. This would mean that the resistor should be (9 - 1.5) / 0.020 = 375 ohms. If you not have a 375 ohm resistor (it is not a common value), then use the next higer value that you do have.

A final note: There are several different ways to put this circuit together. You can change the order of the switch, resistor and LED without having an effect on the results. The description above is probably the most common.

,

Thanks VERY much! That was a very thorough and informative answer.

So, if I take the above at face value... in order to switch off the LED, you're shorting the battery's positive terminal to its negative terminal, right? I think we've found your problem.

You are obviously shorting out the switch (in parallel) - that's BAD. Your switch must go in series with the LED and resistor.

IT had nothing to do with the AC or DC rating of the switch.

Dave.

This is a problem so basic that I am even getting a "Troll-O-Meter" reading. However, for better chance to good for the world including electronics newbies, I will go along.

If the switch is across +V amd GND, it will short the battery rather than interrupt the current through the LED. I do suspect that is known to most who know enough to mention the specific abbreviations of +V and GND. That is why I am getting a "Troll-O-Meter" reading.

If you are not aware that a resistor (that you did not mention) is necessary for most LEDs with 9V batteries to an extent to have been mentioned by you, you should be - lack of mention of resistor contributes to my "Troll-O-Meter" reading.

If you need to hurry out and grab a resistor, get 330 ohm 1/2 watt or

470 ohm in either 1/2 watt or 1/4 watt. But adding a resistor won't solve the switch problem - the switch needs to interrupt the current supplied to the LED for "off" rather than short the power supply.

- Hope I am not doing too much troll feeding, - Don Klipstein ( snipped-for-privacy@misty.com)

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Well, don't know how to disprove I'm a troll. I'm an old retired guy with no experience in electronics or electricity, never taken a class, and just dove into the deep end of the pool and started swimming. I have no mentors, no one knowledgeable to talk to, so I'm dependent on what I learn on the internet, a couple of books and by following and posting to forums. My gulf of misunderstanding is wide an deep.

My mental model was completely wrong for how a switch works. Yeah, I know it's among the most basic of concepts, but what can I say... I try to work things out by myself, analyze, try something else and that usually clarifies things for me. I don't like to bother people in forums with trivial questions, I try to find things out myself first. However in this case, I got so far off on the wrong foot that that style of debugging wasn't going to work for me.

Anyhow, Thanks to everyone who responded, it was really good info.

I stuck a 9 volt battery in my pocket a while back, and it managed to face down into some coins and get shorted. It got HOT! Weird sensation.

John

Just last night I helped someone measure the voltages of a bag full of assorted batteries, and she had no idea that it could cause problems if the terminals shorted out. The 9 volt batteries are the worst offenders, with two closely spaced bare terminals, and it could be tempting for someone to stick two of them together by mating the male to female.

I have heard of fairly serious burns from Lithium batteries in cell phones that people have had in their pocket, and it is suggested to carry them in an external case. I think NiMH batteries are also capable of dangerous overheating and possibly fire if shorted.

I replaced two AA cells in a flexible head LED flashlight with a couple of freshly charged NiMH batteries, and I noticed a small arc when I screwed the end cap on. The LED flashed a little and went out. Then I noticed one end of the metal canister getting hot. I took it apart and found that the connection to the flexible stalk was a spring inside a plastic insulator, but the spring was distorted and somehow shorted to the metal case, and partially melted the plastic part. I was able to fix it, but I'm keeping the batteries out just in case it shorts out again. AA cells have as much as 2500 mAH of energy, and can probably generate enough power to cause a fire.

Paul

yeah, Baked Beans 420g = 1620 Kj NiMH AA, 2500mAh = 3000 Kj

Bye. Jasen

NiMH (and NiCd) batteries _are_ capable of dangerous overheating if shorted. They won't catch on fire themselves AFAIK, but they'll certainly burn the insulation right off of the wire that's shorting them out!

Lithium batteries _will_ (once again, AFIAK) burn directly -- lithium burns in air, and they're pretty hot by the time they burst open.

When model airplane enthusiasts started using lithium polymer batteries instead of NiCd to power airplanes, we found out this distinction -- with NiCd you may scorch the inside of a fuselage; with LiPo batteries you may catch your house on fire.

--

Tim Wescott
Wescott Design Services
http://www.wescottdesign.com

Do you need to implement control loops in software?
"Applied Control Theory for Embedded Systems" gives you just what it says.
See details at http://www.wescottdesign.com/actfes/actfes.html

On Mon, 20 Oct 2008 18:33:29 -0700, sys_spud wrote: ...

The switch shouldn't be connected to ground at all. You're shorting out the battery. Like Michael said, you should use only TWO terminals on the switch - the one labeled "C" (common) and the one labeled "NO" (normally open).

To do ascii diagrams, youi need to set your newsreader for fixed font. I think this is possible even on google.

What you want is this:

Switch c no +-----------------o/ o-------[R]------+ | | [battery] [LED] | | +-------------------------------------+

Good Luck! Rich

Obviously not.

Carbohydrates are about 5 food calories per gram, so a pound of beans is about 2250 food calories.

1 food cal = 1 kcal, and times 4.186 joules per cal the beans are 9.2 MJ, thats megajoules.

The AA bat is about 10 kilojoules total. Thats 10 kJ Besides, most AA cells are under 2 Ah.

These are trivial calculations, you should check them yourself.

bw a =E9crit :

s=20

2 MJ,=20

And every time I read your post I get 1000 times more! check your units please. POM

yeah, open mouth, insert foot, tell the world, I got the units wrong too, and after being corrected only a few days ago.

closer to 470, baked beans are mostly water.

the figures are straight from the can, Nutritional energy is measured in SI units here (the 470 is extrapolated from a different brand of beans advertised on the web as 105cal/100g)

New ones like that are hard to find. Within reach I had a Sanyo 2.5, Energiser 2.5 and (the oldest one) a Lenmar 2.3,

yeah, I should have. 10.8kJ is right.

that puts things in perspective NiMH 10.8kJ/31g = 0.35kJ/g Beans 1620kJ/420g = 3.8 kJ/g

Beans is over 10 times better (more if dried) but NiMH is self contained, beans need oxygen.

Bye. Jasen

And every time I read your post I get 1000 times more! check your units please. POM

energy = volts x amps x time

Given a AA cell rated at 2 amphours then

watthours = 1.2 volts x 2 amps x 1 hour (actually should be 0.2 amps for 10 hours)

1.2 x 2 x 1 = 2.4 watthours

One watthour = 3600 joules so

2.4 times 3600 = 8640 joules