Shake it like a Tic-Tac!

Step 7 Circuit:

This is the circuit:

Comments

[anfegori91]

Why my joule thief is flashing the led? (~30 seconds off, ~10 seconds on)

[moep]

this diagram is actually really well drawn, for those who still don't understand because it is not a circuit diagram it looks like so...
I may have messed up direction of LED in diagram so just make sure you put it in the correct way :)

Circuit Diagram

[gopher72]

i thought it was hard to read at first, but after looking at the one you had, it was a whole lot easier to understand.

[Xyndicator]

The diodes on the left side are incorrect. The solenoid (wire coil with the magnet inside) will produce AC and the bridge rectifier is used to change AC to DC. In your case, the diodes on the left side will cause your battery to "recharge" in the opposite direction, which will damage your battery.

[seaplusplus]

can i use 2two 10 ohm resisters isntead of a 22 ohm?

[BOOJAN]

yes you can...

[Xyndicator]

Here's a cleaner schematic for the TicTac flashlight. The diodes that are being used here have a black body and a silver/gray stripe. The gray stripe indicates the positive end of the diode. A good rule of thumb when putting together a bridge rectifier is that you want the gray stripes to point towards positive side of the battery or polarized capacitor. The diode symbol is an arrow with a line on the tip. So, the arrow tip is "pointing" towards the positive side of the battery and the line represents the gray stripe on the diode. If you're planning on using a capacitor to replace the battery use a polarized capacitor. Polarized capacitors have a stripe as well and it represents the negative end. Make sure you don't reverse the capacitor's polarity. You can always use an unpolarized capacitor but they tend to be bigger. I hope some of this info is helpful to you guys. If you can't see the picture, right-click on it and save it to your computer.

[Otherman]

A few people have mumbled about using a capacitor in this project rather than a battery, simply because it's easier to get. I was told by some guy at a battery store that something like a 1000 uf capacitor should be able to power an L.E.D. for a couple of minutes. I don't remember all my stuff about analog electronic and I'm too lazy to dig up my notes to do the math so I have two questions. First, would a capacitor (of 1000 uf or higher) be something worth trying in this circuit? And secondly, if it is worth trying, how should it be hooked up? I have a feeling that just hooking up the capacitor straight to the DC input and having the switch cause it to discharge wouldn't work (of course with the correct resistors to drop the voltage), but I don't really know.

[ilostmydrink]

I am beginning this project today with a cap instead of the battery. I'll post how it goes. It should turn out to be similar, since a battery is in essence a capacitor.

[Mike060187]

can sum1 draw a better diagram for me plz? no offence but i cant seams to work out where most of the things r ment to be.. thanks great instructable tho :P

[Mike060187]

nevermind... i got it :P

[super_genius]

how are we supposed to know what side of the battery to put what part on? like I know that the 2 of the 4 diodes are connected to the same side, either positive or negative, and the resistor obviously goes on the other side, so somebody just tell me what the diodes are attached to, and I'll be done with it all. also.. you might need to put up a more powerful resistor ... cuz even with a 22 ohm... even with a 50 ohm if somebody is charging it for too long, it can blow the led. I will probably use a ... hmmm.... 125 ohm, (5 banded) just for safety purposes, cuz LED's aren't cheap!.

[aiden120000]

they are cheap

[babylonfive]

Heya Super.

Here's why 22 ohms is great, and won't blow anything. Don't forget that the diode has it's own 'forward voltage' that does not count toward the voltage difference falling on the resistor (which sets the current). The way to figure the resistance is: r = e / i The e = e(source) - e(diode drop). This is 3.6V for the battery and 3.2V or slightly more for the diode. Therefore, only around .5V falls over the resistor. So, .5V/.020A = 20ohms!

A subscript here is that the voltage over the diode actually increases with current; check the datasheet for the diode to see what I mean. because of this, each new current level set by the resistor causes a slightly different voltage level on the LED. Again, check your datasheet to be sure of the final level. Some LEDs drop about 3.6V @ their max current and those LEDs don't even need resistors at all with this sort of lithium battery! They automatically accept the current value associated on the curve with that forward voltage.

Also, as to the 'greater voltages' concern, the battery will keep any higher voltages from being reached. It would actually continue to accept the pulses of current as recharging current and the maximum voltage would not appreciably rise until the battery were completely full or beyond. Check a lithium battery book for the charge/discharge voltage curves - you'll find them quite flat.

So, summary: shake away, and you won't reach dizzying voltages as long as the battery is there... a capacitor might be a different story! And, 22 ohms is the right value because the resistor only limits the voltage it is exposed to... only about .5V or less.

Regards,
David

[super_genius]

oh wait. nevermind

[barri_kid]

ummm... Why would u need 4 diodes. The only thing diodes do is make sure the current goes in 1 direction. There is no need to have 4 in this circuit, Also the 22 ohm resistor is kinda, Very small. Did u get the instructions on a site?

[Daddio_UK]

Its a bridge rectifier, it converts ac voltage to dc voltage, single diode would transfer only 1/2 as much power to the batteries

[Genevamech]

The polarity of the energy pulse coming from the coil changes depending on the direction the magnets are moving. With two diodes you would have a half-wave rectifier. That means you will only get a charging pulse when the magnets travel in one direction, and nothing when they travel in the other direction. Half of your shaking effort is wasted. With four diodes, you have a full wave rectifier and you get two charging pulses per shake. Refer to the diagram on the page xenobiologista linked to.

[barri_kid]

Ah, thanks, also, do u know how to read diodes? Like u know how they look like mini AAA batteries, would the smaller lighter colored part be the + end, and the longer darker end be the - end?

[Slushba132]

I think usually there is a white or silver stripe called a cathode on the outside of the diode and that tells the positive end...Im not sure but I think so

[watermotors4]

Yes, and no. There is usually a stripe on one side indicating the cathode (opposite side being the anode), but the cathode should be connected to the NEGATIVE and the anode to the positive.

[super_genius]

alot of times, im not to sure with these kind of diodes, but they can be used like resistors... it doesn't matter all to much which direction they are facing or not.

[ilostmydrink]

Diodes are definitely direction dependent..... even the LED (Light Emitting Diode) has one leg longer than the other to indicates that directions of diodes are important.

If you're using a diode that is orange with a black stripe, the black stripe is negative. Most common LED's negative leg is longer than the positive leg. Other diodes will be marked.

If you are unsure about the polarity of your diode, look up the specifications sheet. Good places to do this (as well as price the materials) are Digikey and Mouser.

[Slushba132]

ok...know I decided to use a bridge but I still cant figure it out... The capacitor will charge with a batteries and the coil produces about .075 volts but together they just dont work

[technofreak]

what type of battery should we use? would triple a work?

[Slushba132]

Whats the resistor for?

[watermotors4]

The resistor limits the flow of current from the battery so that you do not blow up the LED.

[electronicfreak22]

I'm right in the middle of winding the wire on the pen but the circut isn't very clear to me (and i can't read circutry yet) but could you amke like a list of what connects to what?

[downgrade]

This may or may not help. However just remember all the lines are wires (or leads), and positive will always connect to the next negative (thus making it a loop, all the positives should be pointing the same direction, not physically just if you look at in track view). If you start from the positive lead of the battery, you will connect the resistor, then the negative of the led to the resistor, then the switch, then the negative of the battery to the switch. then the coil between the diodes... but that part can fit in after the diodes as well.

[CowboyGA]

Electronicfreak22, let me know if you successfully finished the project. I'm in the process of trying to set it up using a capacitor and I'm questioning this.

[super_genius]

if a capacitor doesn't work for you he first time, just try switching the polarities. peace.

[Roflolommo]

im not to good at reading circuitry things either but it looks like the - part of the battery connects to one lead on the switch the other switch lead connects to the positive on the led the - connects to the 22 ohm resistor which connects to the plus which connects to 2 diods connecting to the ends of the coil connecting to two more diods then the - part of the battery

[xenobiologista]

The diodes are supposed to be like the diagram on this page: http://en.wikipedia.org/wiki/Diode_bridge

I think the circle with (-) in it is supposed to be the negative terminal of the battery too, he just didn't label it "battery".

[Tierra]

Ah is this the circuit that changes AC current into DC current? We were looking at them in physics this year.