LED Problems

[cerberus]

Ok. What am I doing wrong I wire up 4 led's in series and they work fine with a 470ohm resistor and either a 9v or2aaa batteries. soon as I add a fifth led I get nothing everyone say's wiring is relatively simple but I am pulling out what hair I have trying to figure this out. I've watched the youtube tutorials I've read almost every post on this site and I'm just not getting it. Please Help because either I'm lower than a newbie or I'm just that dumb and I hope it isn't the latter. All the led's are the same color (white) with the flat tops.

[tetsujin]

Ok. What am I doing wrong I wire up 4 led's in series and they work fine with a 470ohm resistor and either a 9v or2aaa batteries. soon as I add a fifth led I get nothing everyone say's wiring is relatively simple but I am pulling out what hair I have trying to figure this out. I've watched the youtube tutorials I've read almost every post on this site and I'm just not getting it. Please Help because either I'm lower than a newbie or I'm just that dumb and I hope it isn't the latter. All the led's are the same color (white) with the flat tops.

When operating, an LED has a certain amount of "voltage drop" - for a white LED this is usually around 3V. If you wire up the LEDs in series, the voltage drops accumulate - and eventually you reach a point where the minimum voltage drop for the series circuit is greater than your voltage source.

It is helpful to have a good understanding of the nature of voltage and current. Voltage is a measurement of potential; how much force there is to push current through a part of the circuit. It is always measured at one point, relative to another point. In that regard you could think of it like altitude: A mountain may be 1000 feet above the surrounding countryside, but the countryside itself may be a couple thousand feet above sea level... But if you were rolling a cart down a hill, the relevant measurement would be altitude of where you're starting versus where you wind up... That would determine about how much energy (speed) you'd gain.

So if you have a 9V battery, then you'd expect one terminal to have potential that's 9V higher than the other terminal. But as you follow the path of the current through the circuit, that potential drops (you've used energy lighting the LED, so you have less electro-motive force (voltage) to work with.) Each white LED cuts it down by about 3V, if that LED is operating... So after a few LEDs you're out of power.

Of course, it's slightly more complicated than that. A LED doesn't -just- operate at 3V - its behavior is governed by a voltage/current curve that gives you a bit of leeway with regard to different voltage levels. But if you have less voltage across the LED, you'll have less current through it (because the force pushing current through the device is less) - so it may be possible to run the LEDs at 2V, but get a seriously diminished current flow. This could explain how you can get four of them in series, and the fifth causes a failure...

There's also an important law called Kirchoff's Voltage Law, which basically explains how voltage drops across different terminals relate to one another... For instance, if V(a,b) is the voltage at (a) with respect to (b)...

• (V(a,x) - V(b,x) = V(a,b)) - If you measure a and b with respect to some other point, and subtract them, you get the voltage of a with respect to b. (Measure two hills with respect to sea level and you can tell which one's taller and by how much)
• (V(a, b) + V(b,c) + V(c,a) = 0) - Basically, if you take a series of voltage drops that form a loop and add them together, you get zero. If V(a,b) is your battery, it might be 9V - then V(b,c) and V(c,a) are negative numbers that add up to 9V. Again, the altitude analogy works pretty well here...

Current is the rate at which charge (electrons, generally) flows through the circuit. It is governed by a rule called Kirchoff's Current Law - which basically states that the amount of current flowing into a single point in the circuit equals the current flowing out. You can think of this like a closed system of pipes carrying water: there are various ways the water can move around, but the total amount of water doesn't change, which means if some water leaves a place, other water must come in to take its place. In the case of a simple loop circuit, the rate of current flow at any point will be the same.

That last bit is important because it explains how you can relate the behavior of different devices in the circuit together: the current through each LED, and the resistor, in a simple single-loop circuit, will be the same. So if there's 5mA of current flowing through the circuit, the voltage drop of each LED will correspond to the 5mA point on that device's voltage/current curve... And through the resistor, voltage drop will be determined by Ohm's law (V=IR, where I is current and R is resistance... Double the current means double the voltage, and vice versa) - finding out the behavior of your circuit means finding the solution that satisfies Kirchoff's Current and Voltage laws as well as the operating characteristics of the individual devices.

[ChrisQ]

So going by the problems demonstrated here and then explained by tetsujin, is it better to just always run LEDs in parallel? Is there a scenario in model building when series is better?

[paraclete1]

So going by the problems demonstrated here and then explained by tetsujin, is it better to just always run LEDs in parallel? Is there a scenario in model building when series is better?

I perfer to run paraellel so if one fails (rare, but happens), you don't lose all of them.

As for the op, other things to check is the LED to make sure it works, turn it around to make sure polarity is correct, is resistor to large now that you've add another LED. I've wired 5 LED's before and ran them constantly for 6 months on a 9V battery. Actually about a year, but they faded and flickered as the blinking LED switched on, so it is doable. Good Luck.

[MillenniumFalsehood]

I run parallel for a similar, but different reason: if one fails, it's FAR easier to find the troublemaker if they're parallel. Anyone remember hunting for that one bulb in the Christmas tree light strand that was burnt out? There is a dark stain on our living room ceiling where the accumulated swear words generated by this has permanently darkened the paint . . . It's almost as big as the one above the TV stand where we had a SNES which ran many a game of Mariokart . . .

Like they've all said, parallel comsumes far less voltage than series. It's also easier to design circuits for, because you can get away with hooking up triplets of white/blue LEDs and sextets of red/green/yellow LEDs to the same 9v battery (with an appropriate resistor, of course) ad infinitum.

Well, almost ad infinitum. After a certain number of sets of LEDs, amperage begins to come into play.

[paraclete1]

I remember those lights, start at the end with a new bulb, and if that didn't work, remove the next one and put back in the first one you took out and just keep moving the bulbs up the line until there was light. Thankfully, I don't ever recall a time when it didn't work. They wouldn't dare burn out two at a time...

[Ant]

Remember though - in parallel, each LED needs its own resistor!

Ant

[MillenniumFalsehood]

I do that, but I've never really understood why. Can you or someone explain why you should use a resistor for every lamp as opposed to a single one leading to multiple sets of LEDs?

[ChrisQ]

I will attempt to answer your question to see if I have a proper understanding.

If the lamps are all connected one after the other, and one dies, the connection is broken to all that follow. But if they each have a dedicated path to the power source, they are not affected by the other LEDs burning out.

If they are in series, they are all behind a single resistor and so are already receiving the proper current. But if they are in parallel, they each need their own resistor.

[Ant]

Here's an analogy - not a very good one, but hopefully you will get the idea...

Think of the current in a circuit as cards driving down a road. The battery can supply a large number of cars coming out one after the other as quickly as possible.

The LED is a small bridge in the road. If the cars come along too quickly, the bridge ends up overloaded and collapses due to the weight of all the cars (too much current).

The resisitor is like a toll booth in the road. It provides an obstrution which slows down the rate at which the cars pass through and then onto the bridge, so the brige remains safe. The ideal situation is that as many cars are crossing the bridge as possible whilst staying safe.

If you have more than one bridge one after the other, you still only need one tollbooth, because the rate of cars over each bridge will be the same.

Now consider if you have several bridges in parallel. With one tollbooth the cars are free to use any bridge, so in order to get the best utilisation of all bridges, the toll booth needs to let through more cars faster. BUT once through, all the cars are then free to pile over one bridge, overloading it. Disaster

So the solution is to use one tollbooth per bridge. This limits the flow across any one bridge, but allows all bridges to be used safely.

This also means that you can cope with different sized bridges at the same time - if you had a big one and a small one, the big one would have a larger tollbooth. If these were one after the other, the flow would have to be limited to the capability of the smallest bridge and the larger one would be under used.

In terms of LEDs, different colours have different specs which equates to the "size" of the bridge. Red, green & yellow typically require 2v whereas blue & white may need 3-4v. You cannot mix these in series and have them operate at the same brightness, because one resistor does not 'fit all'. In parallel, each can have a resisitor tailored to its specific needs.

( note the tollbooth analogy is not entirely correct because it would have to be before the bridge. However with LEDs, the resisitor can be before or after the LED, it does not matter)

Hope this helps!

[ChrisQ]

So the effect is not cumulative? What if I soldered a bunch of resistors in a row and then sent the current through, would it come out the same as if I used only one resistor?

To use an analogy, if I fire a bullet at a board, it will be slowed down but still pass through. When it hits a second board it will slow down further, and so on until finally it is stopped. Is it the same with resistors?

[paraclete1]

I think if the resistor is large enough to handle all the LED's you can get away with one. I wired 5 LED's with one resistor to light the fiber optics in a Wall Map of the places I've been in the world. It's ran pretty much constantly since late 2004 early 2005 and each optic is as bright today as they were then.

Plus a search on Parallel LED circuits shows circuits both ways. But the advantage of using a separate resistor for each LED will prevent all lights from going out if one of the resistors fails.

[naoto]

Here's an analogy - not a very good one, but hopefully you will get the idea...

Think of the current in a circuit as cards driving down a road. The battery can supply a large number of cars coming out one after the other as quickly as possible.

The LED is a small bridge in the road. If the cars come along too quickly, the bridge ends up overloaded and collapses due to the weight of all the cars (too much current).

The resisitor is like a toll booth in the road. It provides an obstrution which slows down the rate at which the cars pass through and then onto the bridge, so the brige remains safe. The ideal situation is that as many cars are crossing the bridge as possible whilst staying safe.

If you have more than one bridge one after the other, you still only need one tollbooth, because the rate of cars over each bridge will be the same.

Now consider if you have several bridges in parallel. With one tollbooth the cars are free to use any bridge, so in order to get the best utilisation of all bridges, the toll booth needs to let through more cars faster. BUT once through, all the cars are then free to pile over one bridge, overloading it. Disaster

So the solution is to use one tollbooth per bridge. This limits the flow across any one bridge, but allows all bridges to be used safely.

This also means that you can cope with different sized bridges at the same time - if you had a big one and a small one, the big one would have a larger tollbooth. If these were one after the other, the flow would have to be limited to the capability of the smallest bridge and the larger one would be under used.

In terms of LEDs, different colours have different specs which equates to the "size" of the bridge. Red, green & yellow typically require 2v whereas blue & white may need 3-4v. You cannot mix these in series and have them operate at the same brightness, because one resistor does not 'fit all'. In parallel, each can have a resisitor tailored to its specific needs.

( note the tollbooth analogy is not entirely correct because it would have to be before the bridge. However with LEDs, the resisitor can be before or after the LED, it does not matter)

Hope this helps!

Let's assume that the driver of each car tends to follow the rule of "path of least resistance" (i.e. chooses whichever bridge that seems to have traffic flowing freely). Let's say there's a minor mishap on one of the bridge, causing the traffic to slow -- this results in more cars trying to use the remaining bridges -- increasing the chance of a collapse. Let's say an overloaded bridge collapses. The cars approaching the bridges divert to the remaining bridges -- increasing the load, thereby causing another collapse...

[tetsujin]

Let's say an overloaded bridge collapses. The cars approaching the bridges divert to the remaining bridges -- increasing the load, thereby causing another collapse...

Actually, from what I've heard, when an LED burns out it tends to become an easier current path, not a harder one.

Let's take this back out of the realm of analogy, deal with voltage and current again...

Suppose you have two points in your circuit, we'll call them "A" and "B". You have power hooked up so that if you make a connection from A to B you'll complete the circuit and current will flow.

The simplest variation of the circuit would be to simply connect A and B with an LED:
A -- LED -- B
Given the right voltage between A and B, this is perfectly fine. The problem is, the LED is very sensitive to minor changes in voltage. A small change in voltage can double the current through the LED. So unless you can be reasonably sure that voltage V(A,B) will be what you want, this usually isn't a good idea.

We stabilize the circuit by adding a resistor:
A -- LED -- resistor -- B
The resistor wastes power, but the upshot is that it makes the circuit less sensitive to minor voltage changes. Resistors are governed by Ohm's Law (V=IR) - which means that doubling the current would require doubling the voltage. According to Kirchoff's Current Law we know the current through LED and resistor must be the same, so as long as we're OK with the waste, we get a circuit with more stable behavior.

If you add a second LED in series with the first one:
A -- LED -- LED -- resistor -- B
the result is that the voltage V(A,B) has to be greater in order for the LEDs to operate, but otherwise it behaves very similarly to the first circuit. Again, due to KCL, the current through each LED must be the same...

If you add a second LED in parallel to the first one:
A -- (LEDx2) -- resistor -- B
This can be unreliable. The reason is because according to Kirchoff's Current Law, the current that flows through the resistor is divided between the two LEDs - but to determine how the current is divided, you have to look at the voltage across the LEDs. Each LED will get the same voltage - but the characteristics of each LED determine how much current each one passes. With just a minor difference between the two LEDs (either different types of LEDs or two "identical" LEDs with minor differences as a result of wear or manufacture or temperature) - this voltage-to-current relationship can shift, causing one of the two LEDs to take the lion's share of the power.

If you run two LED-resistor sets in parallel:
A -- (LED -- resistor)*2 -- B
This is different from the previous example because the voltage drop across each LED is not necessarily the same. On each branch, the voltage drops across LED and resistor must add up to the same total (V(A,B)) - but the voltage across the two LEDs may differ and the voltage across the two resistors may differ.
As a result, the resistors can individually stabilize the voltage/current relationship of the LEDs, making the circuit less sensitive to unexpected variations in the LEDs' behavior.

Now, for some of the other questions:
"Is there ever any advantage to running LEDs in series?"
The simple answer is usually that running LEDs in series is more efficient. To explain this I have to understand the concept of "power". In electronics, power is measured in Watts. The power dissipated in an electrical component is described like this:
P=IV
That is, power equals the current through the device, times the voltage across it.
Now, the power dissipated by a single LED passing a certain amount of current is going to be the same no matter how the circuit is laid out... Because you're targeting a certain amount of current that you want to go through the LED, and that results in a certain known voltage across its terminals - which in turn means a certain known amount of power being used by each LED - as long as it really is operating at that level of current you wanted.

But the resistors are using power, too. So if you have this:
A -- LED -- LED -- LED -- resistor -- B
the amount of power the resistor is using is (I(LED) * (V(A,B) - V(LED)*3) - that is, the current you're passing through the LED times whatever voltage is left after the three LEDs have eaten their share.

If you had this:
A -- (LED -- resistor)*3 -- B
Assuming the voltage between A and B is the same, then each resistor has a greater voltage across it (and therefore is dissipating more power) - and there's three of them.

If you had this:
A -- (LED)*3 -- resistor -- B
It works out the same as the previous example: the voltage across the resistor is (V(A,B)) minus just one LED's worth of voltage - but the current through the resistor is the sum of the currents through the three LEDs - so again, a fair bit of power wasted.

Wasted power means faster battery drain, or the need for a bigger power supply, and it means more heat generated inside your model. So it's worth being careful about it.

[Madman Lighting]

Actually I'm impressed by Tetsujin's explanations, good and detailed and accurate.

The reality is more complicated of course but I dont think anyone here wants to know about V-I curves for diodes.

The simple explanation is the battery acts as a voltage source, it provides a fixed voltage (assume a perfect battery) but unlimited current. The LED acts like another voltage source, but going the opposite way, and smaller in value.

Lets say you have a 9V battery and two white LEDs, each of which has a 3V forward voltage drop so the total LED voltage is 6 volts. If you put those LEDs across the battery directly there would be nothing to limit the current and nothing to "absorb" the voltage difference (9V - 6V = 3V). This would cause the LEDs to over current and blow up.

What do you do? You add in a resistor to limit the current and absorb that 3V drop. Lets say you want to run 15 milliamps in your LEDs. Do a little math with Ohms' Law. 3 Volts divided by 15 milliamps equals 200 ohms. So you put a 200 ohm resistor in series with the two LEDs and now all is well. You are supplying enough current to light the LEDs BUT are below their absolute max so they wont blow up.

There are some pretty good articles over in Wikipedia about basic electronics that are a good place to start. Lookup resistors, LEDs, voltage and current sources and read about series and parallel combinations of things.

Reality is more complex and I use an active current source for my products but the result is the same. LEDs light up and last a long time.