16 replies to this topic

[Noodleownz]

(Sorry for the bad pictures)
Well I soldered my battery snap wires to my transistor then soldered my yellow LED. My only problem is where do I solder my on/off switch. As shown in the picture it has 4 prongs, and you push the red button for on and off. But I have no clue where this goes and what I solder it to. If anyone can help me that would be great! Thanks.

[KingBouyah]

You should be able to solder it anywhere in line. I'd put it on the black wire. Just snip it and solder one snipped end to a terminal on the switch, and the other snipped end to the other. Push buttons shouldn't have polarity like an LED would.

By the way, that looks like a basic push button where you have to hold it down for the light to stay on. It's hard to tell, and I know switches come in all shapes and sizes, but I just wanted to make sure you've covered all your bases before you solder anything.

[Noodleownz]

You should be able to solder it anywhere in line. I'd put it on the black wire. Just snip it and solder one snipped end to a terminal on the switch, and the other snipped end to the other. Push buttons shouldn't have polarity like an LED would.

By the way, that looks like a basic push button where you have to hold it down for the light to stay on. It's hard to tell, and I know switches come in all shapes and sizes, but I just wanted to make sure you've covered all your bases before you solder anything.

Thanks for the help though I'm still confused. So your telling me to cut the black wire and solder one end to any part of the prongs, and the other end back to the other black wire? Or, since I got 5 snap ons in a pack cant I just cut one full black wire from one?

EDIT:Also wheres the terminal out of the four prongs? On the back of the packaging it gives a diagram but none of them say terminal. DC IN is for the top and bottom, and load is for the left and right

Edited by Noodleownz, 07 September 2009 - 07:11 PM.

[KingBouyah]

By "snap-ons" you mean the battery connections? I'd get some other wire and use the snap-ons for other projects. I would expect that you'd want some extra wire anyway so you can put the LED, Battery, and switch in convenient locations and not be limited to the length of wire that your snap ons have with them.

Essentially your circuit should look like this.

Switch
	(-)-----------|   |-------|------(-)
Battery						  R	   LED
	(+)----------------------|------(+)

If that makes sense.

[Noodleownz]

By "snap-ons" you mean the battery connections? I'd get some other wire and use the snap-ons for other projects. I would expect that you'd want some extra wire anyway so you can put the LED, Battery, and switch in convenient locations and not be limited to the length of wire that your snap ons have with them.

Essentially your circuit should look like this.

Switch
	(-)-----------|   |-------|------(-)
Battery						  R	   LED
	(+)----------------------|------(+)

If that makes sense.

Makes sense. But when I cut my black wire, where does it go on the switch? On the load ends? Or DC IN ends?

[flashflint]

I am pretty sure that resistor should not be connected to both wires. I am not sure which, but not both.

[1337]

Switch	   Resistor
	(-)-----------|   |--------{_}-----------(-)
 Battery											 LED
	(+)-------------------------------------(+)

This is the correct way to wire it, right?

[nerfdude123]

Switch	   Resistor
	(-)-----------|   |--------{_}-----------(-)
 Battery											 LED
	(+)-------------------------------------(+)

This is the correct way to wire it, right?

Yes that is correct.

[Noodleownz]

After I cut the wire where do the 2 black parts go!?!??! On the Load prongs, or DC IN prongs? I need to know this

[nerfdude123]

After I cut the wire where do the 2 black parts go!?!??! On the Load prongs, or DC IN prongs? I need to know this

Well after rereading that it has four prongs it should go like this


(-)----------(Dc in prong)(Load prong)------(-)
BATTERY................................................LED
(+)---------(DC in prong)(load prong)------(+)

Or so I think. The load is the LED so the load prongs should be connected to it. The DC in is where the power goes.

Again this may not be correct. If it is not try other methods of by a simplier switch.

Edited by nerfdude123, 07 September 2009 - 08:52 PM.

[Noodleownz]

After I cut the wire where do the 2 black parts go!?!??! On the Load prongs, or DC IN prongs? I need to know this

Well after rereading that it has four prongs it should go like this


(-)----------(Dc in prong)(Load prong)------(-)
BATTERY LED
(+)---------(DC in prong)(load prong)------(+)

Or so I think. The load is the LED so the load prongs should be connected to it. The DC in is where the power goes.

So your saying cut the red wire also. And your saying 2 red wires go to the load prongs, and the 2 black wires go to the DC in prongs?
OR
Are you saying half the black goes to DC in and the other half goes to load. And vice versa with the red?

Edited by Noodleownz, 07 September 2009 - 08:52 PM.

[Blacksunshine]

Switch	   Resistor
	(-)-----------|   |--------{_}-----------(-)
 Battery											 LED
	(+)-------------------------------------(+)

This is the correct way to wire it, right?

Yes that is correct.

No its not correct. the resistor should be on the positive side to limit the power that is being delivered to the LED. on the negative side it does nothing except slow the amount of current that leaves the LED increasing its ability to burst into flames.
Also the switch being on the negative side also will only hinder the removing power from the LED. switch all that shit around.

Edited by Blacksunshine, 07 September 2009 - 08:53 PM.

[nerfdude123]

Switch	   Resistor
	(-)-----------|   |--------{_}-----------(-)
 Battery											 LED
	(+)-------------------------------------(+)

This is the correct way to wire it, right?

Yes that is correct.

No its not correct. the resistor should be on the positive side to limit the power that is being delivered to the LED. on the negative side it does nothing except slow the amount of current that leaves the LED increasing its ability to burst into flames.
Also the switch being on the negative side also will only hinder the removing power from the LED. switch all that shit around.

Ahh, yes you are correct. I didn't see that it was negative on top. Resistor goes on postitve.

Also can you take a better picture of the switch? The first one is blurry.

Edited by nerfdude123, 07 September 2009 - 08:57 PM.

[Blacksunshine]

That switch looks like it might be a momentary switch. If it is it won't work either. you need an on/off switch.

[Hipponater]

Switch	   Resistor
	(-)-----------|   |--------{_}-----------(-)
 Battery											 LED
	(+)-------------------------------------(+)

This is the correct way to wire it, right?

Yes that is correct.

No its not correct. the resistor should be on the positive side to limit the power that is being delivered to the LED. on the negative side it does nothing except slow the amount of current that leaves the LED increasing its ability to burst into flames.
Also the switch being on the negative side also will only hinder the removing power from the LED. switch all that shit around.

No.

It doesn't matter where the resistor is, as long as it's in there. It doesn't need to be before the LED, it doesn't have to be after. The resistor doesn't "slow" current leaving the LED if it's behind it, it "slows" the whole damn thing.


Noodle, it doesn't matter what you cut or what color it is, so long as the +/- terminals of the LED are correct with the battery. Make it looks like this, as was posted earlier:

Switch	   Resistor
	(-)-----------|   |--------{_}-----------(-)
 Battery											 LED
	(+)-------------------------------------(+)

If your switch has more than 2 prongs, you didn't get the right one.

Edit: fixed spacing on the code thing

Edited by Hipponater, 08 September 2009 - 12:24 AM.

[KingBouyah]

Sorry, to have led you astray. Since you already had the resistor in parallel with the LED, I thought you knew what you were doing, so I left it like that.

Bob's right, Hipponater knows what he's talking about.

[Longbow]

Hipponater is completely correct here. The amount of current going into the LED will always be the same as the amount of current coming out.

I have been seeing questions about LED's and general circuit design questions, so I am going to make an electronics questions and answers thread. I'll link it once it is done.

EDIT:
Here is the link to the guide. I have finished the first section, and will be continuing to add more as time goes.
Electronics Guide

Edited by Longbow, 08 September 2009 - 07:02 PM.