Battery Blew Up!

Hi,

I was working with my breadboard and I started to smell something bad, which I thought was the heater coming on in the house. It does that some times.

Then 9V blew up on my circuit.

I was trying to run power to each of my breadboards, and I added a switch to turn the circuit on and off so that I wouldn’t have to keep disconnecting the battery.

The switch was off, and the LED was off, so there shouldn’t have been any power running through the circuit but the battery still blew up. It took 30 mins for the battery to cool down enough so I could dispose of it.

I am going to start over with some beginning books to get more knowledge before I play with any more circuits because it freaked me out.

I uploaded a few pictures of my circuit, which I still have. Can someone tell me what I did wrong? I would like to know so I don’t make the mistake again. I would like to know how I screwed up.

Hi adele.

Your 2nd. picture shows the switch.
Switches like that often have 4 pins (for some structural strength), which are connected 2 x 2.
There might also be 3 actual connections and 4 pins where the 2 common pins are conneted, and the switch connects either A or B to that.
The picture shows you have all wires to the switch in one row of the breadboard (which would be your 2nd mistake here), so they are always connected.
That means the switch doesn't even do anything.

Far worse (the 1st. mistake) is that you have connected the red and the black line coming from the battery to the switch.
That will blow up a battery allright.

Switches like these are only to connected to one wire from a battery, and the red (for positive) one is preferred.

Solution:
Remove all black wires to the switch.
Move the switch 4 positions to the left, bridging the gap in that breadboard.
Use another color wire (orange ?) between the other side of the switch to the 2nd red power rail on top.
Move all other red wires that you want to switch to the 2nd. red power rail (else the switch doesn't do any switching).

Hi,

This switch has 6 pins. It did work when I pressed the button, the light came on. I will look at the data sheet again for the switch to make sure I know how to hook it up correctly. The "in" always went to the same pin, but depending on which pin I used for the "out" it would set the position of the switch. One way, the switch would be on when it was up, and the other it would be on when it was down (it is a latching switch). There was also another pin that would make the switch on regardless of the position (up or down).

So you think the switch is what made the battery short out?

If you wire a switch across a battery you will cause it to supply all the current it can. If that is a lot then yes it will blow up every time. I have difficulty believing that you do not know this. You need to do a lot more reading about circuits before doing any more practical work.

Grumpy_Mike:
If you wire a switch across a battery you will cause it to supply all the current it can. If that is a lot then yes it will blow up every time. I have difficulty believing that you do not know this. You need to do a lot more reading about circuits before doing any more practical work.

I only know what the Arduino book has showed me so far. The Arduino Projects Book that came with the starter kit shows on page 26 a switch where one end is connected to power and the other end is connected to an LED and ground.

So I don’t understand what the difference is between the book’s example and what I did.

The book doesn’t show that both ends of the switch are connected to power. It shows power going into one side of the switch and ground on the other. So I was just copying that.

I am going to start back over with the book from the beginning and work through all of the projects again. I also have another projects book that I bought so I can start to understand it better.

I just saw such a switch a few moments ago, during my daily visit in an online shop.
Having another peek at the 2nd. picture, you seem to have the red wire to a top pin, and the black wires to a center pin, the lower pin is then left unconnected.

The problem stays that you connected the black and the red wire, i do not know if that was while the switch was pressed or released.
The LED would have been on while the switch was open, and went off when you closed (shorted) it.

Let's assume you have learned today :cold_sweat:

I see you have just replied.
The book shows a switch with one end to the battery and the other to the LED, which then is connected to the battery's black wire (i hope there is a resistor somewhere in there too).
That is not what you did.
You connected both wires from the battery to the switch.

So I think I understand what I did wrong. By connecting the "in" on the switch to red and "out" to black, I basically shorted the battery in the same way that I would have if I connected a wire directly between the + and - of the battery.

So my problem is that I didn't have any components between the + and - of the battery. The switch basically acts like a jumper wire, not really a component.

So I could have connected one end of the switch to the + of the battery and the other to the - but I needed to have some components like resistors, LEDs, etc... in the circuit, so something was using the energy.

If I'm understanding correctly.

Edit: I understand that I can't short the + and - of a battery together, but I didn't realize that's what I was doing with this circuit. Which is why I asked here for help.

Yes that is correct.
The clue is in the name 'circuit' it is a path that the current follows from the + to the -, unless there is something to limit the current then you end up making a heater out of the battery.
The relationship between the voltage and the current is given by ohms law, it is voltage divided by resistance equals current. All components have resistance. Your switch, which is a component, has a very small resistance, almost zero. That allows a very large current to flow.

I knew that you couldn't short a battery out, but sometimes when I start hooking up the wires, I lose track of what is going on. It's still hard for me to understand how everything works when it's connected, since I can't actually see the electricity moving.

This is still a good lesson learned that will help me understand how to (not) do things in the future. When I look at the mess of wires and components, I can forget that there is a specific path that the circuit is following.

I want to think of a circuit as a point-to-point connection, always between the positive and negative. Then I see circuits with several components connecting to ground and several connections to positive, and I can't understand how it's working. It seems like it's no longer a circuit.

In my head, I think that only one component should ever be connected to positive, and one to ground, so it forms an actual "circuit", but the schematics don't look like that.

Hopefully, when I go back through the books, it will start to make sense.

Thanks

Then I see circuits with several components connecting to ground and several connections to positive, and I can't understand how it's working. It seems like it's no longer a circuit.

It is in fact several circuits all connected at the same time. This is known as connections in parallel. When the components are all in a single line from + to - this is known as "in series". Any none trivial circuit will have many current paths in series and in parallel.

Learn to think "schematic" and then translate that schematic into the "mess of wires" while you make it. Do not be tempted to do this the other way round because that eventually will get you nowhere.

Here are some links to help you start to think schematic:-
Reading a schematic
http://blog.makezine.com/archive/2011/01/reading-circuit-diagrams.html
Colin's Lab video on reading a schematic

Grumpy_Mike:
The relationship between the voltage and the current is given by ohms law, it is voltage divided by resistance equals resistance.

Sorry Mike, 2 out of 3 aint bad as they say… :wink:

He meant voltage divided by resistance equals current.

Graham

Here is a little something that is easy to remember to help with Ohms law, the triangle, if you cover V, you are are left with IR, V=IR. If you cover I you are left with V/R I=V/R. If you cover R, you are left with V/I R=V/I. Where V=Volts, I= Amps, R= Ohms.

Ohms.jpg

Hi, what sort of battery was it?
Tom.... :slight_smile:

TomGeorge:
Hi, what sort of battery was it?

Well we know for sure it wasn’t one of these:-

PP3.jpg

ghlawrence2000:
Sorry Mike, 2 out of 3 aint bad as they say… :wink:

He meant voltage divided by resistance equals current.

Yeah, I figured that was a typing mistake. I knew what he meant. The book explained that on the first few pages.

Grumpy_Mike:
It is in fact several circuits all connected at the same time. This is known as connections in parallel. When the components are all in a single line from + to - this is known as “in series”. Any none trivial circuit will have many current paths in series and in parallel.

Thanks!

That helps a lot. The book explains series and parallel, but they use a very basic example. There are so many connections going on in larger circuits that I had no idea this is what was happening. It was hard for me to pick out that they were running in parallel. It just looks like a huge mess to me.

These little details help a lot. It’s kind of like when I first started learning to program. I understood some of the concepts, but when I saw a huge “mess of code” I couldn’t make heads or tails of it. Now I can look at it and tell that they are classes, with functions, etc…

TomGeorge:
Hi, what sort of battery was it?
Tom… :slight_smile:

I attached an image of the battery I was using, both before, and after the explosion. Let’s just say, it was a lesson learned.

b1.jpg

HI, waaa???
Sorry, but I hope you mean a duracell, 9V with stud terminals like the picture posted like Grumpy_Mike posted.
Was it rechargeable.

Or was it a group of cells in series or parallel.
You see the problem I have is that you should have at least some damage to the protoboard, the leads and the switch if you cooked a battery, that is a lot of energy.

Did you make up a pack of cells, if so can you post a picture of a circuit diagram, hand drawn will do.
Sorry for the questions, but I fix electronic stuff, you see “its my job”.

Tom… :slight_smile:

TomGeorge:
HI, waaa???
Sorry, but I hope you mean a duracell, 9V with stud terminals like the picture posted like Grumpy_Mike posted.
Was it rechargeable.

Or was it a group of cells in series or parallel.
You see the problem I have is that you should have at least some damage to the protoboard, the leads and the switch if you cooked a battery, that is a lot of energy.

Did you make up a pack of cells, if so can you post a picture of a circuit diagram, hand drawn will do.
Sorry for the questions, but I fix electronic stuff, you see “its my job”.

Tom… :slight_smile:

It was a 9V battery. The only thing I have remaining of the circuit is in the pictures I posted earlier. I have taken it apart.

The breadboard and the components are fine; however, I have not tested the switch since it happened. Afterword the battery was really hot. I couldn’t even touch it to remove it from the connector. I had to use some ESD gloves. After it cooled I threw it in the recycle dumpster outside.

I don’t know how to draw a schematic yet, but you should be able to see all the connections in the pictures I posted, if you would like to try and draw one yourself.

I don’t know if it was rechargeable. It was an old battery. I got a picture from Google, if you really need it.

images.jpg

Rechargeable batteries should have a relieve valve, to prevent it from actually blowing up.
Overheating will generate gasses, which will come out one way or the other.
A relieve valve will offer a way out, but of course the battery will be damaged after that.
If there isn't a way out available, an alternative will be forcefully created.
A standard battery doesn't have a valve (too expensive to create).
The release of the gasses doesn't need to be a speedy process.
With a lesser current (but still too high) it might take a while before the battery blows.
Any component has some resistance.
That of the battery will be the highest, compared to wires, breadboards and switches.
So it isn't impossible for other parts of the circuit to come out relatively undamaged compared to that battery, they should be inspected though.
I'm not sure whether the heating up of that battery will create a higher or a lower resistance.
If it goes down, an avalanche effect will occur and the destruction will be speedy.
If it goes up, the battery will live it's (by then miserable) life a bit longer.

The reason for rechargeable batteries to have this valve, is the charging process.
A too high charging current will also create gasses.

Some rapid charge methods also incorporate discharging too.
It will rapid-charge for a while, in such way that small gas bubbles are formed.
After a set time, a very short but high discharge pulse will be created, which is supposed to revert the gasses.
I remember some Elektor project (about 20 years ago) where this was explained, and played around a bit with this ICS1700N based project.
That process is explained in datasheets for that chip that can be found online.

adele:
I knew that you couldn’t short a battery out, but sometimes when I start hooking up the wires, I lose track of what is going on.
<…>

One of the great fascinations with Arduino is that every project is an experiment… You will look back on this a one of your fond memories. With over 50 years of electronic experience, I still manage to smoke a project from time to time :grinning: In my growing up days, I had a year where I was fascinated by high voltage experiments: Jacob ladders and Tesla coils. When these experiments go wrong, bad things happen! My pilot friend says that any landing that everyone can walk away from is a “good” landing.

Ray