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LEDs Instant Burn Out - Page 2

Quote:
 Originally Posted by liljoejoe54 Current flows from negative to positive so the resistor would have to be on the negative side in order to reduce the voltage before it goes through the LED. If its on the positive side, the full 12V will go through the LED and then go through the resistor on the way out.
Incorrect.

While in metallic objects (e.g. copper wire) the charge carriers (electrons) flow from lower to higher potential, current is defined to flow from in the direction of positive charges (protons) regardless of the charge carrier e.g. from higher to lower potential. In other words, the charge can flow from positive to negative or negative to positive based on the conductor; but the current is always from positive to negative.

That aside, it does not matter in what order the components (LEDs and resistors) in a series string are wired, as all that's important is the voltage potential across that component. For example, if you wire a light across the +12V and +5V lines, that light only sees that 7V potential across it.

Now technicalities aside, to address the concerns of the OP:
A key point to remember about LEDs is that they are polarity sensitive, if they are connected with reverse polarity they will in general just not work but if exposed to too high a reverse voltage can also be fried. Typically the maximum reverse voltage is 5-10 (or more) times higher than the rated forward voltage, so in general they will not be damaged from simply wiring them in reverse polarity to the correct supply voltage.

Most LEDs have a supply voltage somewhere within the range of 1.8V and 3.2V (2.1V is probably the most common) and have a forward current of 20-30 mA; these values vary between different LEDs and should be looked up on their datasheets. With supply voltages higher than the rated forward voltage of the LED (e.g. the 12V rail of a PSU), you need a current limiting resistor (inducing a voltage drop across the resistor) in series with the LED such that the voltage potential across the LED is equal to the rated forward voltage.

The necessary resistor sizes can be computed using Ohm's law (V = I * R) and the power law (P = V * I).
Say for example you want to use a 12V supply to light 2 LEDs rated for 2.1V @ 20mA connected in series (they are also in series with a current limiting resistor).

So we need an induced voltage drop across the resistor of Vr = 12V - 2 * (2.1V) = 7.8V
So the desired resistance is Rd = Vr / I = 7.8V / 0.02A = 390 Ohm.
Next we round that up to the next nearest standard resistor value which coincidentally 390 Ohm is standard, so we use an actual resistance Ra = 39 Ohm.
Now we determine the necessary power rating of the resistor.
The power consumed by the resistor is: P = V * I = I^2 * Ra = (0.02A)^2 * 39 Ohm = 0.156W
So the appropriate resistor to use would be a 390 Ohm 1/4 Watt resistor.

BTW, resistors and voltages add in series; currents add in parallel; the voltage drop across parallel items is constant, and the current in a series string is constant.
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Quote:
 Originally Posted by radodrill Incorrect. While in metallic objects (e.g. copper wire) the charge carriers (electrons) flow from lower to higher potential, current is defined to flow from in the direction of positive charges (protons) regardless of the charge carrier e.g. from higher to lower potential. In other words, the charge can flow from positive to negative or negative to positive based on the conductor; but the current is always from positive to negative.

So then it can be different with other electronics? When I swap out the incandescent bulbs in my dash panel components on my car and replace them with LEDs, I have to put the resistor on the negative side of the LED or else it will pop as soon as I put power to it. In school I was taught that the current flows towards the +.
Quote:
 Originally Posted by liljoejoe54 So then it can be different with other electronics? When I swap out the incandescent bulbs in my dash panel components on my car and replace them with LEDs, I have to put the resistor on the negative side of the LED or else it will pop as soon as I put power to it. In school I was taught that the current flows towards the +.
As I said before, the direction the actual charge flows can be to +, to -, or even in both directions depending on the charge carriers within the conductor; in metallic solids the electrons (negative charge carriers) carry the charge to +, while in electrolytes (e.g. the acid in an lead-acid battery) charges flow in both directions.
By convention, Current flow is defined to be in the same direction that positive charges flow regardless of the direction the actual charge is flowing (or net charge flow when bi-directional flow is present).

I know for testing LEDs I have on some occasions had the resistor on the - side while other times I've had it on the + side and both methods work just fine. The common practice is to withe the resistor on the + side of the LED, although I have seen some vendors that have had the resistor on the - side.

As for your car issue, I'm not sure why that would be the case, because the LED only sees the voltage potential across it not the actual voltage at each node. The only thing I can think of is that the resistor you used may have been slightly undersized (e.g. by rounding down to the nearest standard resistance) an it just happened to work fine one way but not the other.

That said, I'm not trying to create an argument, just providing information from personal experience and what I've learned in EE (electrical engineering) and physics courses; though sometimes there are anomalies to theory and common practice, so I guess your car just happens to be one of them.
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Quote:
 Originally Posted by radodrill That said, I'm not trying to create an argument, just providing information from personal experience and what I've learned in EE (electrical engineering) and physics courses; though sometimes there are anomalies to theory and common practice, so I guess your car just happens to be one of them.
Nobody is arguing, i was just asking an honest question.
Quote:
 Originally Posted by liljoejoe54 Nobody is arguing, i was just asking an honest question.
A lot of people get very defensive when someone else has a differing opinion, it's always nice to see people like you who are rational and open minded

You were indeed right that in metallic solids the actual charge flows to +, but by convention the current flow (for graphical representation and calculations) is from +
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Quote:
 Originally Posted by radodrill A lot of people get very defensive when someone else has a differing opinion, it's always nice to see people like you who are rational and open minded You were indeed right that in metallic solids the actual charge flows to +, but by convention the current flow (for graphical representation and calculations) is from +

Lol If im wrong I have no problem if people correct me, thats how I learn. It would be bad to keep giving out wrong information.
Thanks for the info guys. So these should work then, right? It's just strange that all of them burnt out. Could it be a problem with my power supply?
Quote:
 Originally Posted by tw33k Thanks for the info guys. So these should work then, right? It's just strange that all of them burnt out. Could it be a problem with my power supply?
most likely the manufacturer either used the wrong resistor or intended it for the 5V rail and inserted the pins into the 12V position of the molex connector.
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i would say they are probably 5v leds which you have to connect to a 5v lead.
not enough info on the site to say really...
love how the voltage isn't marked on them.
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