Resistance of LED's

The resistance of a led is dependent on the current through it... its a diode!

In the specifications there will be a figure for the forward voltage of the led at a certain current... What you need to do is find what current gives the best light from the led, use the forward voltage which in your case sounds like 3 volts and then calculate the resistor needed to drop 240 volts (actually 237 volts) at that current...

Say you wanted 10 mA of current, the resistor would need to be 240/10 kohm = 24 kohm.

Now you must throw the led you were using away as you have put it across the 240 volts ac and it has suffered reverse breakdown about 50 times every second you had it lit! So it won't be in the best of health...

Put a new led with a diode in parallel with the resistor in series of say 24 k or 50k which ever gives the best illumination...

Now the resistor will have 10 mA flowing through it so with 240 volts across it it going to have to dissipate the waste power... which is....

I x V = 10 x 240 mW = 2.4 Watts!! So use a big one that's rated for high voltage use above 400 volts...

A very wasteful way to power a led! John.

"Now you must throw the led you were using away as you have put it across the 240 volts ac and it has suffered reverse breakdown about 50 times every second you had it lit! So it won't be in the best of health..." I understood that Electroman created a resistive divider using 50K and 1K resistors, so only about 1/50 of the line voltage was applied to the diode. I don't believe I'd throw it away.DickL

led's like transistor base emitter diodes, have a very low breakdown voltage in the order of 5 to 7 volts...

It may still work but for reliability I would definitely chuck it out, after all how expensive is a led and how expensive would it be to replace a dud led?

John.

Horace: "Resistance" is the measure of a linear characteristic of a device. Ohm's Law, E=IR, defines the relationship of resistance, current and voltage associated with a resistive device. Diodes are not linear. First, they conduct in only one direction and block current flow in the other. Second, as you begin to apply voltage in the "forward" (conducting) direction, there is very litle current flow until, for a silicon diode for example, the applied voltage reaches about 0.6 volts. Third, once conducting, the diode maintains a relatively constant voltage drop. There is some increase with current flow, but it isn't linear. The diode can be damaged by excessive voltage in the reverse direction or excessive current in the forward direction unless limiting circuitry is used. Your circuit apparently provided adequate limiting for the characteristics of your diode. I hope this helps. DickL

Dick just as you disagree with my post, I disagree with yours...

Resistance is NOT a measure of a linear characteristic of a device!!

Resistance is the resistance to flow of electrons in a device with a potential difference across it... The device might well have a very nonlinear characteristic!

John.

Why not use a power blocking diode in series with a power resister in series with the LED. Put this little circuit in parallel in one leg of your ac circuit. It will not glow as bright because it is on only half the time but it will allow things to also be off half the time. Some simple Ohms law calculations and you can have the whole circuit put together. If you really have your heart set on a bright LED my next best suggestion is to us an inductive pick up on one leg of the circuit and run it though a step down transformer and rectify it and get a dc current and voltage and then the LED will be on the whole time thru the entire 360 degrees of the ac circuit.

Oh dear Guest.... that sounds messy...

If I had this problem I'd do one of two things:

1) throw the led away and use a neon indicator!

2) Use a cheap little ic that you put 240 volts in and get 3 to 5 volts out at a low current... I have some of these chips and they work extremely well, even with NO capacitor dropping.

John.

Only if he has is heart set on a diode that is on for the whole 360 degrees of the ac cycle. Remember nothing is free and you want to keep the loading of one leg of the ac circuit as low as possible to keep it from being unbalanced. Anything you put into the circuit can unbalance it so you want to keep it negligible. A neon indicator is fine but you must make sure you agitate the gas enough to get it to conduct and European 50 Hz or American 60 Hz sometimes is not fast enough to agitate the gas into conduction. Guest #13 has a pretty good little circuit and your chip is also good but remember at some place you must dissipate the excess power somewhere even for a chip.

This is a great place to trade ideas!

Charles

Thanks all for various replies.

The current consumed by the LED is limited by the 50k resistor - so it cannot be more than 4.8 mA, But I am still a bit puzzled about what 'constant' characteristics of the LED to use in calculations. I appreciate the amps volts and ohms relationship might not be linear - but you have to start somewhere.

In my case I thought I would measure the resistance of the LED - and I came unstuck.

To re-phrase the question, why does my LED - after all it is a diode - not show a low resistance in one direction when tested with the meter.

Is it because an LED has some sort of ionisation characteristic that requires 'striking' to get a current flow. Which then comes back to 'enough' voltage.

Therefore is it my test meter??

I just wondered.

As has been said the forward voltage of a silicon diode is around 0.4 to 0.6 volts depending on current through it...

A led has a much higher forward voltage and depends on its construction as well as the colour of the led...

A red led usually has a forward voltage of about 1.8 to 2.2 volts, a blue or 'white' led will have a much higher voltage drop up to 4 volts...

So its simply that your ohmmeter hasn't enough volts to turn on the led, or rather the resistance of the forward biased led is too high for your low voltage ohmmeter...

John.

use a 1/4w carbon film resister(47k ohm) in series with a IN4007 Silicon diode and put it in series with your led. it will light and work for 2yrs. no issues.

237*237/47000=1.2watts or, if you use a series diode so that current flows 1/2 cycle then average of 0.6 watts of heat in the 0.25 watt resistor is not a no issues scenario.

Hello

Why use a resistor for the voltage drop, it will bee warm and consume energy. Instead you can use a capacitor, 0.1 - 0.2 µF in serial and put a 1N4148-diode anti parallell to the LED. It will do the work. The capacitor voltage must bee rated 600V for 230V power

Anders

Hi,

please look at this link. www.turbokeu.com/myprojects/acled.htm

I have been using simular circuits before without problems.

br

"Why use a resistor for the voltage drop, it will bee warm and consume energy. Instead you can use a capacitor, 0.1 - 0.2 µF in serial and put a 1N4148-diode anti parallell to the LED. It will do the work. The capacitor voltage must bee rated 600V for 230V power"

This is not a good method of controlling the current through a resistor. If you turn the power on at the peak of the mains voltage the current through the diode will be enormous and will most likely destroy the diode. At best it will degrade the junction and result in premature failure of the diode.

Another thing to keep in mind is that with a 240 V system the peak voltage that you get is 340 V, 240 is the RMS voltage. So the circuit needs to be able to cope with a forward and reverse bias voltage of 340 V not 240 V.

As Electroman said the forward bias voltage drop across a red LED is about 1.8-2.2 volts while the forward drop across a normal silicon diode is about 600 mV.

The simplest circuit would be to use a resistor power diode and LED in series across the mains. To calculate the value of the resistor you need to know the maximum forward bias current that the diode can take. I would then design the circuit to only drive the diode to 75% of this current. The formula for the resistor would then be

R = (340 – Vf_LED – Vf_Diode) / (75% x I_max)

Where

I_max = Maximum forward bias current the LED can take

Vf_LED = Forward bias voltage drop of the LED

Vf_Diode = Forward bias voltage drop of the power diode

So if we had a red LED that could take a 20 ma forward bias current and had a forward bias voltage drop of 2 volts we would get

R = (340 – 2 – 0.6) / (75% x 20 mA) = 22.5 KΩ

You also need to check the power that the resistor is going to need to dissipate so

Power_Resistor = ½ (V_RMS – Vf_LED – Vf_Diode) x (75% x I_max ) = 1.78 watts

That means you cant use a ¼ or ½ watt resistor you will need to use a 5 w resistor and that is going to add considerably to the cost. The ½ in front of the calculation is due to the fact that current is flowing through the circuit for only ½ of the time or while the diodes are forward biased.

The fact that you will need to use a 5 watt resistor is one of the reasons that it is unusual to find LED used as indicators for mains voltages. Another factor is the intolerance that LED have to the reverse bias voltages that mains voltages produce across the junction. I would therefore ask if it were possible to install the indicator led later on in the circuit after the voltage had been reduced to a more workable level? If not then the type of regulator that Electroman referred to would be the next best option. Failing that you can use all this information to calculate the value of the resistor needed to build a circuit along the lines you initially envisioned. Keep in mind though you will also need to install a power diode to stop the led from being damaged by the reverse bias that the mains produces.

Hello Masu

Thanks for the formulas. They are what I would use - if I knew which numbers were applicable.

In my case I have used a potentiometer type circuit as a voltage divider so that without the LED connected, the the max volts across the 1k resistor will be 4.7 v (6.65v peak). Is this Ok. It has not 'popped' yet. So is it an LED on borrowed time'

However, I would still like to be able to calculate the voltage and current.

It seems to me, from your example, that you take the rated current (20ma) and the rated voltage (2v) to calculate the resistor value. Except you have reduced the current to 15ma (75% of 20ma) - but kept the same voltage.

The effect of this in the overall circuit (for the purpose of calculation) would be the same as using a resistor that has a notional resistance of 133 Ω (ie. 2v / 15ma).

Except the current is 'chosen' thus using 10 ma I would get a resistance of 200 Ω.

Which is odd, because I cant imagine an electrical device having an 'assigned' resistance. It must have a resistance related to a voltage or current or some other intrinsic characteristic - and that is back to my original question of what parameter to choose for use in calculations.

In the meantime I have bought a digital multimeter that has a diode measuring setting. This sends a current through the diode that gives a reading of volts drop across it. My red LED does not give a reading - ie. the digital reading is the same with the probes open circuit or with the LED connected - which suggests high resistance (no current) - and this time there is an internal 9v battery.

So I am still puzzled.

From another angle, can it be said that the there is no current in the forward direction (high resistance) ) until the bias voltage is reached when the current flow becomes a linear function of supply voltage (constant low resistance?) - or what.

Regards.

The LED is a semiconductor, it is a diode. It does not conduct until the voltage across it reaches a threshold voltage, typically ~2v for an LED or 0.6 volts for a simple diode. Once that voltage is reached, current flows and with continued increase in current, the voltage will rise somewhat depending on the size and composition (things that effect the effective internal resistance). You might do better to think of it as a voltage clamp that kicks in at ~2 volts and that is in series with a relatively low resistance. The LED is a current mode device. The light output is a function of the current flowing in the device. The voltage across it is only a consequence of its operation. You can, at a given current and voltage, calculate what its effective resistance is but it is not generally a useful parameter in the application of LED's. You won't find resistance specified on an LED data sheet. You will find a curve that plots the forward voltage as a function of current. Pick a spot on that curve, divide the voltage by the current and you will get the effective resistance but there is no need to be concerned with this unless you are operating from a voltage very close to the forward voltage of the LED.

Well described by almost everybody. !!!!!

Almost 24K resistor + a Diode in series to block reverse current [400vPIV rating]

Hi Horace40, I think I know where you are getting confused.

For the explanation you will need to refer to the diagram on the right. This is a diagram of the voltage (horizontal axis) and current (vertical axis) for a diode (red line) and resistor (blue line).

Lets look at the blue line first. This line represents the characteristics of a 2Ω resistor and we can see that the current through the resistor is always twice the voltage across the resistor. The line on the graph is straight and we say that the resistor has a linear characteristic.

Lets now look at the red line. This line represents the current passing through a resistor against the voltage across the resistor. Now starting at the origin we can see that as we increase the voltage across the diode in the positive direction the current remains zero until we reach just under +2 volts. If we no increase the voltage just a little more the current suddenly rises and shoots off the top of the graph. The voltage that the diode starts to conduct in the forward direction is call the Forward Bias Voltage and is usually represented by V_f.. You can clearly see that no matter how much current is passing through the diode the voltage remains pretty close to V_f.

Ok lets now go back to the origin and look at what is going on when we increase the voltage but this time in the reverse direction. As you can see as the negative voltage increases there is only a small reverse current until we reach about -4 volts. Suddenly at about -4 V the current stars flowing in reverse direction and no matter how great the current becomes the voltage stays at -4 V. This is called the reverse breakdown voltage and is usually represented by V_r.

As you can see the line on the graph is not a straight line and we refer to the diode as having non-linear characteristics.

A couple of thing to remember about diodes is that they normally they DO NOT TOLERATE REVERSE CURRENTS WELL so it is normally good to try and avoid a diode from breaking down in the reverse direction..

Here is a summary of what we now know

• While forward biased the current through a diode will be negligible wile the voltage is less than V_f
• Once the diode switches on in the forward direction the voltage will not increase appreciably above V_f with an increase in current.
• While reverse biased the current flow through the resistor will be negligible until the voltage reaches V_r.
• Once Vr has been reached while reverse biased the voltage will not increase greatly from V_r with an increase in current.
• Most diodes don not tolerate current flowing through the junction in the reverse direction so applying a negative voltage greater than the reverse breakdown voltage V_r should be avoided.

Now the values of V_f and Vr are highly variable and depend on the type of diode being used. For example a MSB64DA-1 red LED has a V_f of around 2 V a V_r of 5 V and a maximum forward current of 20 mA.

Compare this to a 1N3673A rectifier diode has a V_f of 1.35, a V_r 1,000 V and maximum forward current of 12 A

Lets now look at our circuit and work out what we need

When both diodes are forward biased the led will have 2.0 V across it, the power diode will have 1.35 V and we wish to limit the current to 20 ma. Therefore the resistor we need is

R = (240 – 2.0 -1.35) / (75% x 20 mA) = 15,777 ≈ 16 KΩ

and the power the resistor needs to dissipate is

P_R = ½ (240-2.0-1.35) x 15 mA = 1.8 W

So you need a 16 KΩ 5 W resistor to limit the current through the diodes and the power diode need to be able to tolerate a reverse bias of 340 V to protect the LED from the reverse bias voltage.

"In my case I have used a potentiometer type circuit as a voltage divider so that without the LED connected, the the max volts across the 1k resistor will be 4.7 v (6.65v peak). Is this Ok. It has not 'popped' yet. So is it an LED on borrowed time' "

What is happening with you circuit is that you are adjusting the voltage to 4.7 V but as soon as you install the LED it forward biases and consequently limits the voltage to its forward bias voltage of 2.0 V. If you now adjust the pot as you would to increase the voltage the current through the LED will increase and it will glow brighter but the voltage across it will still remain at around the 2.0 V. Be careful doing this as you can see from the above calculations that the pot will be dissipating something like 2 watts which is normally above what they are rated for.

I hope that has helped and that I havn't totally confused you but if you are sill having problems please don't hesitate to ask. I have no problems spending a week explaining something to somebody you that is willing to and wants to learn.

Nicely put Masu.... what software do you use to draw those pictures and how do you put them on here?

John.

Hi John,

I use a really expensive drafting package called DeltaCad, you can download a demo copy and purchase a license from their web site for US$40. Its only a 2 D drafting package but if you know your way around a drawing board it will take about an hour to do the tutorial and become reasonably proficient with it. I have been using it for over 10 years now and draw all my technical drawings with it. I even use it to do PCB design.

To use it for CR4 you need to introduce an intermediate step. First you creat the drawint then cut the part you wish to use and paste it into draw saving it in a format that can be used at CR4. Following that you click on the picture button on the CR4 editor and follow the instructions.

Thanks Masu. Again.

I can see how work out my resistor circuits now. But interestingly enough Gustav180 suggested a capacitor. I will give some thought to that as well.

ps. I use Visio for my desktop drawings.

If you use a capacitor be sure to have a parallel diode across the LED in the opposite direction of the LED. It is required to reverse charge the capacitor. Also you may want to use some resistance in order to provide some protection from power line spikes that the capacitor will couple into the LED at a very low impedance.

There is another problem associated with using a capacitor to limit the current through the diode. The current through the capacitor is proportional to the rate of change of voltage. Now if you turn the circuit on while the voltage is at or near the peak of the cycle you will be applying a 340 V step to the circuit. The result is that the capacitor sees a waveform like the blue line shown on the right. This sudden increase in voltage will cause very high currents through the diode that may damage and probably will degrade the junction of the diode. You also still have the reverse bias problem.

Thanks rcapper, Masu and everyone else for helping me. That's enough for the moment. I will let my LED run until it pops.

I note what has been said about the capacitor circuits. If I use one I will make sure that I turn it on when the mains is at zero volts.

Thanks.

Most LED's will tolerate a pulse current of two amps or more. So if you place a 180 ohm resistor in series with the device it would limit the current to less than 2 amps with a 340 volt transient and with a correctly sized capacitor the resistor wattage could be kept low since average current will be low.

Your demo in intersting but I don' t know how you derive the formula to find the resistane and the power.

Your demo in intersting but I don' t know how you derive the formula to find the resistane and the power.

Resistance is the relationship between the voltage across a component and the cirrent through that component. Put mathematically

V = IR

So if we had a resistance of 1 Ω at o.5 volts the current would be .5 A and at 4 V it would be 4 A. This is a linear relationship and a resistor is called a linear device.

With a diode on the other hand the current that flows through the diode is no linear. At 0.5 V the current it zero and if you tried to put 4 V across the diode a diode shown in my graph in post #19 the current would be so great that the diode would burn out. This is because. Because the diodes resistance is dependant on the voltage across it you can't really give it a value.

Now to calculate the power that a diode is consuming you need to multiply the voltage across it by the current through it. If the voltage across the diode is less than the forward bias voltage then the power is zero because the current is zero. If the diode is forward biased then the voltage will remain at something pretty close to the forward bias voltage and the current will need to be controlled by an external device like a resistor. In this case the power going to the diode will be the forward bias voltage drop multiplied by the current.

To sum up diodes are a non linear device so they don't really have a resistance as such and the power they dissipate is either close to zero or the voltage drop across the diode multiplied by the current through it.

To Masu - and all others who helped.

I used the simplest resistor potential divider volt drop circuit. My LED's are working OK. But the resistors I used had to be selected on the basis of the characteristics of the LED - ie, desired current and voltage - which gives a notional resistance of the LED to introduce into the calculations.

But what is the underlying explanation for the non linear characteristics. Is it to do with the forward voltage required to energise (whatever glows - gas?) and once excited it becomes linear with rising volts for a while until it gets too hot - and then burns out. Or what?

And what happens in reverse flow - what breaks down when the voltage is too high.

Just wondered.

Let me see if I can keep this simple and brief without screwing it up. The LED is a semiconductor. That means there is a junction between two substrate materials that are doped with different impurities. One impurity causes an excess of electrons in the atoms of the substrate, the other a deficiency. Consequently, at the junction of these materials atoms try to share electrons and do to the degree possible until as stable state is reached at which point a barrier threshold is created equal to a potential that is particular to the material substrate and the impurities. It is this barrier threshold voltage that must be overcome before current will flow in the junction. I'm sure if you look up semiconductor junction on Wikipedia you will find a better description. Actually, it is not as thorough as I thought it would be but here it is:

http://en.wikipedia.org/wiki/P-N_junction

Probably if you look around you can find some additional info.

Oh yeah, the light comes from electrons in the material that first jump to a higher energy shell and then collapse back to their original shell. Then to return to their original energy state the must give up energy that is radiated as a photon. The frequency at which this occurs determines the color of the emission.

"The frequency at which this occurs determines the color of the emission." No, the energy difference between states determines the color of the emission. The frequency determines the brightness.

The theory behind how a diode works is very complicated and a full description requires a good knowledge of chemistry, physics and mathematics. None the less here is a somewhat better description of the concept of a diode that may give you a basic understanding.

Well done Masu, an EXCELLENT answer.

Wow, lol. Um ok I just need to know one thing. What resistance dose an LED create?? I'm connecting about 300 into a single circuit. It isn't ideal and the most efficient way. But I need to save space. And I need to know how to efficiently power all of them. They are basic 3mm LED's but I don't remember the voltage they require. Because I bought them last year and already hooked them up. I'm just at the powering stage and don't wana burn any out. If someone would please respond but to my email. If someone would please do that, please respond to ipad36@yahoo.com. All replies will be greatly appreciated and credit will be given if asked in the overall product.

G'day gals, guys & gurus,

Light emitting diodes LED don't have a resistance as such but rather a voltage drop that remains relatively constant regardless of the current flowing through them.

There are two parameters that you need to take into account when designing a circuit that drives LEDs.

• Forward Voltage Drop V_f: This is the voltage drop you will see across the LED whenever there is current flossing through it.
• Maximum Forward Current I_f: The is the maximum safe working current that can flow through a resistor without damaging it.

I've been building a model of the Starship Enterprise that has over 200 LEDs spread throughout the model so let's look at one of the white LEDs that have the following characteristics.

• Diameter Ø: 3.00 mm
• Forward Voltage V_F: 3.50 V
• Forward Current I_F: 30 mA

So what we have is a LED that will have a voltage drop of 3.5o V across it whenever there is current flowing through it regardless of how much current is flowing through it and this is where things start to get confusing. For example if I were to connect a 5.0 Vdc across the resistor there would be nothing to limit the current flowing through the LED and is would burn out almost immediately. In fact it would not matter what supply voltage you used if it is above the VF of the LED then it will overload the LED and burn it out.

What we need is some way of controlling the current that is flowing through the resistor so that is remains no greater than the forward bias current or IF which in this case is 30 mA.

What we need to do is add what is referred to as a current limiting resistor as shown in the circuit on the right. Here the resistor R will limit the current flowing through the LED and prevent it from burning out.

So how do we calculate the value of the current limiting resistor?

If you recalled earlier I stated that the LED had a VF of 3.5 V while the supply voltage is 5.00 Vdc. As a result the voltage drop across the resistor in our circuit will be

• V_R = 5.00 V – V_F
• V_R = 5.00 V – 3.50 V
• V_R = 1.50 V

We now wish to limit the current flowing through the LED and as a consequence through the resistor to no more than the forward bias current I_F of the led which in our case is 30 mA. As a result we can now use Ohm's Law to calculate the value of the resistor as follows:

• R = V_R / I_R
• R = 1.50 V / 30 mA
• R = 50 Ω

So, we can now safely connect our 5 Vdc supply voltage to the LED Resistor circuit and be assured that the current flowing through the LED will not exceed the 30 mA limit.

Now this is for a single LED but it doesn't matter how many LEDs you are utilizing there must be some sort of current limiting device in the circuit that prevents the LED from burning out.

Another example would have say 10 LEDs connected in series with a single current limiting resistor. If we were utilizing the same LEDs then the total voltage drop across the ten LEDs would be 35 V. If we were to then supply this with say 40 Vdc the voltage across the resistor would be 5 V and since the current is still 30 mA the value of the resistor would be:

• R = V_R / I_R
• R = 5.00 V / 30 mA
• R = 166.66 Ω
• R ≈ 180 Ω

So in this case we would have ten LEDs and one 180 Ω resistor in series all supplied from a single 40 Vdc supply.

Unfortunately you need to know the forward bias voltage V_F and maximum current I_F for your LEDs which you stated you do not have. So what can you do.

You could measure the characteristics with something like a component tester or guestimate the values.

This is a guess but if you are using high intensity LEDs then the forward current IF is likely to be around 30 ma.

The voltage is a little more complex but is generally tied pretty closely to the colour of the LED with the following being a good place to start:

• White 3.0 to 3.5 V
• Red 2.0 to 2.5 V
• Green 3.2 to 3.5 V
• Blue 3.5 to 4.0 V
• Yellow 2.1 to 2.5 V

I hope that has helped and if you need anything else please either post it in this thread or use this link to send me a Personal Message.

Regards, masu

Masu,

you deserve a praise for all the time and effort you put in your postings. They are very informative.

I just bought a set of LED Christmas lights (100 count) that was on sale. I didn't want to brake it down but was wondering if there is any current limiting component involved or if the all LED's are simply connected in series, and plugged into wall (120V AC). The way they're interconnected is so every two LED's are connected in reverse to each other, and then they're connected in series to the next two, and so on.
So, technically there is 50m LED's connected in series for both sides of AC cycle.
Is 50 LEDs connected in series enough to make up its own current limiting resistance, so it can be powered by "raw" 120V AC (here in the US)?

Or, in other words, how would you calculate the total number of standard white LEDs that can be connected in series and powered by 120V AC, without any other components such as resistors or capacitors, but so they operate at their nominal values (i.e. 3.3 V, 0.066W)?

Thanks, Henry.

The image got lost somewhere. Sorry...

Here's a new one

(See it here)

The forward bias voltage drop or V_f across a LED is primarily dependent on the colour and ranges from about 1.2 V for the red ones up to 3.5 V for the blue and white ones.

Given that in the circuit you would have 50 LEDs operating in series you need to add up all the V_f values for each of the 50 LEDs which I would hazard to say comes pretty close to 120 V. However, you would still need a single current limiting resistor to control the current which at a guess is probably hidden in the plug.

I have seen people drive LEDs with the exact V_f requited and no current limiting resistor, but if the supply voltage increases by even a small amount you run the risk of blowing up the LEDs while a similar drop means they don't work at all.

If I were designing a circuit like the one you described I would have set it up so that the total V_f for the LEDs came in at around 110 V then installed a single 1 kΩ resistor to limit the current to 10 mA. That way if the voltage goes up to say 125 V you will only end up with 15 mA through the LEDs which while higher than desirable should prevent them from blowing up. Conversely if the voltage drops to say 115 V the current will drop to 5 mA giving you a reduced intensity rather than total shutdown.

Does that help?

Yes, that helps. I was just intrigued by the idea of powering a long series of LEDs with AC voltage considering the fact that LEDs don't have passive resistance (like resistors) and their susceptibility to currents even a little out of their safe range. I couldn't resist my curiosity so I broke open the AC plug of the lights set just to see how it was designed and here's what I found.

(See it here.)

The components are very small and the resistors are not more than 1/8W in size (almost the size of a surface mount resistor). Could you shed some light on the purpose of that circuit there? Is it designed to compensate and/or protect from the variations in AC level, like you mentioned?

Thanks,

Henry.

The 1N4004 diodes are designed for use in power supplies and in particular in bridge rectifiers where they have to cope with high reverse bias voltages of up to 400 V.

It's an interesting circuit and my first thought was that they were using the 1N4004 diodes to protect the LEDs from overly high reverse bias voltages (abbreviation V_r) but that doesn't make sense because the V_r across each of the LEDs is limited to the V_f of the parallel LED.

The reason that they are using both resistors and capacitors to limit the current is to get rid of as much of the flicker as possible. When the AC voltage is low or in other words has just crossed over zero Volts the rate of change of voltage is the greatest and the current through the capacitors the highest. On the other hand near the peak voltage the rate of change of the voltage is quiet small and thus the current through the cap is reduced. What this means is that the voltage across the diodes is smoothed out and thus the flicker reduced.

However, to get a cap that has a sufficient value and voltage rating you have to go to electrolytic capacitors which have a nasty habit of blowing up if you connect them the wrong way. That's why the 1N4004 diodes are they, they prevent the capacitors from being reversed biased and emitting little puffs of smoke.

Finally they have doubled everything up and paralleled all the LEDs with reversed ones so that it works on both sides of the AC waveform.

At least that's what I think they are doing.

PS: You are obviously intrigued by machines and electronics so may I suggest signing up with CR4 and participating in other discussions, I think both you and CR4 could benefit.