No. In series, If one blows, be it the 1st, 2nd or 3rd LED, then all the others shutoff b/c the connection is broken. Only one LED will blow, but the whole device will shut off. This should work. In series,
voltage is cumulative. The train example was bad b/c it was AC, you're right. Here's another example, a boombox radio using 6 D size batteries. The voltage that the boombox is using is not 1.5 volts, but 1.5 x 6 = 9 volts. The batteries hooked up in series produce the 9 volts for the boombox.


EDIT: And just to clarify, hooking the 3 LEDs in series is hooking the 1st LED positive to the 12 V source, the negative of the 1st LED goes to the positive of the 2nd LED, the negative of the 2nd LED goes to the positive of the 3rd LED, the negative of the 3rd LED goes to ground.

DW

 

hmm but that would mean if i down the voltage to 3 volts... and then hooked up the leds in series, the last led would get 0 volts basically?

or if the leds were hooked up in parallel then the last led would receive 0 volts?

sorry, just not sure about this part.

For me, when i hooked up 12 volts to 4 leds a while back, I hooked them up as:

(+)-----:------:------:-----:
........O......O......O.....O
(-)-----:------:------:-----:

the "O" is the led. the ":" is the led metal connector.
the "-" is the wire.

was the above setup the reason why the leds blew?

 

multiplexor, yes it's the autodynamic LEDs ... just the single LEDs with the inverted cone shape for light dispersion. I decided against using the 4-LED because the dimmer switch is rated for only around 3A (I think) and I don't know what current those extra-bright 4-LEDs draw. Plus, the 4-LEDs don't have that inverted cone so there's a possibility it might focus the light very narrowly on the meter.

and yes, that's the reason your setup blew. You were running 12V across each LED so they blew.


dwapenyi is correct, you can run 3 LEDs in series (anode connected to the cathode of the other diode to make a chain of LEDs)

If the acceptable voltage drop across each LED is 3.7V, running 3 in series will let you drop a total of 12.1V, hence no need for ressitor but you're looking at a freakingly bright light!

dwapenyi nailed it down to a tee : you have to hook them up in a certain way to make sure you have current flow through the circuit. And even though you have 12V at one end of the first LED, you'll only drop 3.7V.

It's not the voltage that kills the LEDs, it's the current. Of course, the current goes up as the voltage goes up so ...

 

hmm cool... i'm going to give that a shot tonight. I've got waaaay too many leds lying around my house... hehe

thanks

 

The given the resistance of the conponents there will be a decrease in voltage and current. You not solving the problem by arranging the three LEDs in series. THe first LED will see the full 12V and the full amount of current (depending on the internal resistance of the diode). After the first diode, there is the voltage drop to 8.3V. This is measured between the cathode side of the first LED and the ground. After the second diode the potential will drop to 4.6V. This would be measured at the cathode end of the second LED. In effect, you are exceeding the voltage rating of all three LEDs. Yes it is true that if you add up the total voltage drop of the three LEDs you will arrive a low enough voltage, but this voltage would be measured AFTER the three diodes...you'd have three LEDs, the first one being the brightest (because it see the full 12V) and the two others progressively dimmer...assuming they haven't failed them all from the excessive current.

 

...I may have mislead the discussion...the 3.7V you see on the packaging does not denote the voltage drop of the LED. It is the maximum voltage or potential that can be held across it.

Typical voltage drops for diodes are between 0.6V and 1V. For LEDs in some cases, when at temperature (resistance increases as temp goes up - thus a higher voltage drop) can drop voltage by up to 2V.

 

Thank you for you support. Please drive thru . . . .

DW

 

hmm what i had always thought was that no matter how they are hooked up, that first led will get those 12 volts.. in series, it would blow the first led, and then the others would die out cause no power would go to them.

hmm I know leds pretty well, but this has me confused a bit

 

You know, I assumed a simplistic voltage drop across the diodes like a resistor. Given the values you suggested, I would measure the voltage on each one and just wire up more LEDs so they sum up to 12 volts or so. Since they are in series, they would all experience the same current. The voltage *might* vary between each one, like you said,if they are so in-consistant voltagewise.

DW

 

You're right. That why it is critical to have that resistor in front of the LED anode

In this case, that is exactly what would happen.

 

You can hook up 3 3.7v led's in a series circuit. I have 4 sets of them in my car right now so I know it works. Each LED does drop around 3.8 volts. It doesn't work that the first recieves 12 then the others get whats left, it's divided according to the amount of resistance each load has. Your forgetting that the path to ground has to loads (leds) on its path. If a series circuit worked in the way you guy's described it would be useless. Every load will drop a given amount of voltage until you have 0 volts at the end of the circuit. You can check this by measuring voltage drop across each LED in a series circuit.
BTW what I meant before was exactly what some said. If you hook up a 3.7v LED to a 12v circuit it will probably blow.

 

 

And the lesson is now over

 

That's exactly what happens. Voltage drops, current remains the same. Voltage would only be constant through all the LEDs if they were wired in parallel, where current is now divided.

 

The LEDs will not appear dimmer as you go along the circuit from hot to ground. Current remains the same. But you WILL exceed the voltage rating on the LED.

And, for us scholars, the lessons are never over.

 

I believe the 12 volts from the cig lighter carries 1 amp current. Could this be right?

Hmm... so the current would remain the same, but the voltage would drop. Is that what your saying?

So putting a resistor before all the leds, would drop the current and voltage to a level that could power all the leds?

 

I don't know the answer to that. I might be able to find it in the FSM.

Yes.

Only the voltage would drop. But, I repeat, wire the LEDs up in parallel. Just make sure you sue bright enough LEDs...somewhere in the 2000-5000mcd range.

 

hmm I don't trust the whole, hook it up without a resistor thing... doesn't feel right. I'm not in EE, but I've been playing with electronics for a long time. I feel much more comfortable dropping the power using a resistor and working with that.

But that's just me. If it works for others, go for it hehe I feel safer putting a resistor before the device i'm going to connect.

 

The 1 amp you refer to is how much the cig lighter puts out. The LED , or any electrical load, for that matter, does not care. They take what they need. If the LEDs need 2A, and your circuit supplies only 1A, then fuses will be blowing regularly. The LEDs will be fine, but your circuit will not.

In this case, the circuit can supply up to 1A, the LEDs, altogether will take maybe 0.2A.

DW

 

Hooking up the LEDs in parallel while using a resistor does have its advantages. The resistor can help protect the load, and also, in parallel, if one LED dies, the others continue to work.

In series, if one LED dies, they all go down.

Still, given that, I would hook them up in series b/c LEDs are not sophisticated delicate electrical devices that need that kind of protection, and it's only 4 or 5 LEDs, so if one dies, it's easy to figure out which one kicked it. If it was an array of thouasnds of LEDs, then a combination parallel series circuit would have o be set up.

DW