**Important**: Attempt only if proficient in working with AC Mains

A simple yet effective circuit for a single LED on AC Mains....

**Signing Up**

## Step 1: Parts & Assembly

**Parts**:

1. An LED - 5mm or 10 mm of any colour

2. A diode, preferably 1N 4007

3. A resistor of 1 watt rating, of value anywhere from 20 kilo ohms upto 100 kilo ohms.

4. A two pin male plug (hollow)

Lower resistor values will give more brightness and higher values will prolong LED life. So choose the tradeoff, depending on requirement.

Lower wattage of resistor like 1/2 watt or lower

**will not do and may burn**since they are meant for 6v DC circuits, not for 220v AC Mains.

**Assembly:**

1. Connect black anode of diode to negative of the led.

2. connect the resistor to the positive of the LED.

3. Connect the free ends of the diode and resistor to the male pins.

Done. See the attached pic for clarity.

Another circuit with the diode connected "across" the LED is also attached. A bulb base adaptor is used instead of a male pin.

It should work on 110v AC as well as 220v AC Mains systems.

**DC Feature:**

It will also work on any battery!!

Once again, read "

**Important**" first line, in case you missed it....

how this circuit will perform if space is no bar. Details of diagram is given below:

1)R1= 220 Ω

2)R2= 1MΩ

3)C1= 470nF,400V, Polyester capacitor

4)C2= 10uF, 16V-50V, Electrolytic capacitor

5)D1= LED (any colour)

6)BR1= Bridge rectifier made using the diode

IN4007Comparing between this & your circuit, which will give better performance?

How would you configure this circuit for 120v ? I would also like to add a photo resister into this circuit to light up LED in the dark. Any suggestions?

stonesouls, This is the Automatic LED Night Light circuit for you.

The 5k resistor provides negative feed-back to 2N3906 giving a snap-on effect for the LED. Without it, the LED will glow under slightly dark light. See the effect by connecting and disconnecting this resistor.

You can increase the value of C1 to provide as much current as is needed by the LED connected. For example, using 2.2uF will give 100mA current which will light up a 1W LED brightly.

Remember during daytime, the zener diode will take up all the available current, you have to adjust the rating using W=VI based on what you connected. For instance, with 100mA current, use 1W 6.2V zener for 2 LED, use 0.5W 3.3V zener for 1 LED. Otherwise, 1/4W rating is good.

stonesouls, you can used the circuit on 120V without any change, and the circuit will supply 20mA current.

Remember the circuit will also supply 20mA current once plugged into mains. Adding a photo resistor together with a switching transistor or SCR can automatically turn on the LED at night, but may consume more power in daytime than when it is at night. Of course, it also involves a totally different design of the circuit.

I think putting C2 there is risky. Without it the circuit will work fine (LED will flicker at 100Hz on a 50Hz mains and 120Hz on a 60Hz mains with a very short time off in each cycle). If the LED fails, C2 will explode catastrophically.

If space is not a constraint, better circuits are always available. See "Dipankar" circuits on I'bles.

I have made two - one is without diode bridge (LED Bulb - Part 1) and other is with diode bridge (LED Bulb - Part 2).

In India, we have 220 V AC Mains and have therefore used 220 nF, 400 V rating cap.

No polyester cap was used in AC mains for smoothing....for obvious reasons!

reg

k

Ultra-bright LEDs need driving voltages of either 3 or 6 volts depending on what you have around or buy. Using a 3V LED, a resistor and a 5 Volt DC source and 20mA of current a similar circuit will be using only 100mW (0.10 W) - P=I*V. The method described in this intructable which uses a resistor pushes that up to almost 500mW (0.50 W) - thanks KrisBlueNZ. In this case the resistor reduces the current through the LED and produces heat to do so. What happens if we use an AC 250 Volt rated capacitor instead? Will this reduce the current without having to dissipate power as heat & thus bring the power used back down to 100mW? (You ask, why bother? First, the resistor wastes the power by producing heat as some have noted, second, if we get back to 100mW we're now using only one fifth the power, in other words using the resistor uses five times more power than needed! Third, getting rid of the heat reduces the fire hazard.)

While we NEED to use a resistor when using DC voltage, a resistor is NOT needed to limit the current when using AC voltage. Perhaps KrisBlueNZ can help us here. I've built this circuit for AC mains use, but I have not done the calculations and since the capacitor does NOT get hot, and the LED is nice and bright depending on the Capacitor's value, I assume all power used is that used by the LED and so we're down to 100mW or so. I was looking for the short piece I wrote for Nuts & Volts years ago so I could tell you the capacitor's value, it was rather small but still for a cap handling 250 Volts it might be as physically large as the resistor being used. I managed to fit the circuit into a 120 Volt AC un-polarized replacement plug and called it my "traveler's night light."

The circuit I ended up with was an LED, Cap, and a 1N4003 (240 PRV) or so diode on 120 AC Mains. The diode goes between the LED leads and is reverse connected, anode-to-cathode - cathode-to-anode and the cap is attached to the LED anode, if I remember correctly. In this case the diode helped protect the LED. It might also increase the total power used, I'm not sure about that. Even if we don't save power we're not dissipating power as heat in a resistor which has to make this a safer circuit to use. I believe I've now been using this in four of my AC light switches, that control outside lights which cannot be seen from inside the house. We're able to know when an outside light is left on without having to go outside and look now. These have been in use for about 10 years and before I included the reversed diode the light switch controlling the garage light died twice in two years, none have died since.

KrisBlueNZ - Can you do the calculations and tell us if this really will use less power and the value of the cap needed? (0.20 mfd? less? more?)

Hi PhilKE3FL,

Since there are no answers to the calculation question you asked, here it is.

Assuming 120V AC, 60 Hz, 0.47uF capacitor, and ignoring the voltage drops in both the diode and LED as they are insignificant comparing to the AC mains, the impedance of the capacitor, Zc = 1 / (2 * pi * f * C) = 5.644 kohm.

Next, calculate the average current that the circuit can supply. Here, Irms is not used because both the diode and the LED are non-linear components and P = VI will be used for power dissipation. Linear components like resistors will use Irms to calculate power when using P = I^2 * R.

So average current, I(avg) = ACV(avg) / Zc = [(2 * ACV(peak)] / pi / Zc = 19.14mA

In the first half-cycle of AC, assuming the LED will conduct and produce a potential difference of 2.2V; P(led) = VI / 2 = (2.2)(19.14) / 2 mW = 21.06mW.

During the other half-cycle, the diode is forward-biased with 0.7V voltage drop; therefore P(diode) = (0.7)(19.14) / 2 mW = 6.70mW.

Total power dissipation of your circuit; P(total) = P(diode) + P(led) = 27.76mW, which is significantly lower than the circuit with the current-limiting resistor connected. This difference is the result of efficiency between capacitive and resistive components.

All great information. JohnL8, based on the information that you have provided can I use a Metallized Polyester Film CBB Capacitor rated at 225k 250V 2.2uf for a 10mm LED rated at 3.6v @ 100ma on a 120v main? I will also use a 1N5408 Diode rated 3.0 amp. The diode is probably overkill, but rather be safe than sorry.

stonesouls, Sorry for the delay, I just notice your question now.

You can do in the way you said. The LED will run at 100mA in one-half cycle, and won't operate as brightly as under full-wave 100mA. Any diode rated 200V or higher is good as current has been limited to 100mA.

Be careful as you are dealing with mains voltage.

JohnL8,

Thanks for the calculations! I've used 0.10, 0.22, 0.33 and 0.47uf and the larger the capacitor's value the brighter the LED.

0.47uf ~ 21mW + 7mW for ~ 28mW (19.14mA for 0.47uf)

0.33uf ~ 13mW + 7mW for ~ 20mW (13.44mA for 0.33uf)

0.22uf ~ 9mW + 7mW for ~ 16mW ( 8.96mA for 0.22uf)

0.10uf ~ 4mW + 7mW for ~ 11mW ( 4.07mA for 0.10uf)

much less than using that big resistor and much safer too.

Alternately, the diode can be removed and a capacitor can be attached to one lead of the LED and a 1 kilo ohm resistor, rated for atleast one watt can go with the other lead. (Any lead, since the other end of the cap and resistor are directly on AC Mains).

In the above pic, i have used a Male socket pin. There is no place inside for a capacitor. So i have used the circuit, as above - fully aware that heat is produced by the resistor. But since the resistor is rated for one watt, it does not give way and safety is addressed.

There are many "robust" circuits on the internet, however they all require space and will not fit into something this small. This is the first safe circuit (i couldn't find any other, atleast!).

reg

ketan

Anything from 104 (0.1 uf) to 103 (1 uf) works but I found using 474 (0.47 uf) put 2.2 Volts across the 3V ultra-bright LED. I then used a 1N914 (signal diode?) across the LED and also a 470 Ohm 1/4-watt resistor in series with the LED. The resistor and diode were put in to protect the LED and since doing this none of the LEDs have died as of Sept 24th 2006 so that's 7+ years so far. This circuit was published in the Sept 2009 issue of Nuts & Volts magazine.

Lets illuminate the world!

--------------------------------------------------------

"May the good belong to all the people in the world.

May the rulers go by the path of justice.

May the best of men and their source always prove to be a blessing.

May all the world rejoice in happiness.

May rain come in time and plentifulness be on Earth.

May this world be free from suffering and the noble ones be free from fears"

---- Vedic blessing

hello sir, how to connect only 1watt or 5watt led on 220 volt ac. plz help me

You can try a mobile charger. Dont forget a heat sink, since its high wattage LED.

reg

For more LEDs in series, just remove the 1N4007 diode.

Just keep a 3 watt rated 22k resistor on any side.

Diagram for LEDs in series!? Haha!

I strongly suggest you work with 3 volt DC battery first, not 220 Volt AC Mains!

Hope it will be taken in the right spirit! ;-)

I made it this project very nice.

1. GENERAL

This design uses a half-wave rectifier to eliminate the negative half-cycles of the mains voltage, and a series resistor to limit the current. The brightness of the LED, and the power dissipated in the resistor, can both be calculated if you know the mains voltage and the value of the resistor. The formulas, written the way you would enter them on a calculator, are:

2. CURRENT AND POWER FORMULAS

I = V / R / 2.22

P = V * V / R / 4000

where:

I = average LED current, in milliamps;

V = mains voltage, in volts RMS (in most countries, either about 115 or about 230);

R = resistance of the resistor, in kilohms.

3. RESISTANCE FORMULAS

These formulas can be rearranged so you can calculate the right resistance for a given current or power dissipation:

R = V / I / 2.22

R = V * V / P / 4000

4. EXAMPLE CALCULATIONS

Using mains voltage = 115V and R = 6.8 kilohms:

I = 115 / 6.8 / 2.22 = 7.6 milliamps;

P = 115 * 115 / 6.8 / 4000 = 0.49 watts.

Using mains voltage = 230V and R = 27 kilohms:

I = 230 / 27 / 2.22 = 3.8 milliamps;

P = 230 * 230 / 27 / 4000 = 0.49 watts.

5. COMMENTS

(a) LEDs are typically operated at 10~20 mA (milliamps) so in both of these examples, you will not get particularly high brightness out of them. If possible, use LEDs with high efficiency, i.e. high light output for relatively low operating current.

(b) The resistor should be rated for at least twice the expected power dissipation. In these examples, use a resistor rated for at least one watt.

(c) You may have noticed that the power dissipation in the resistor is the same in both examples, but the 230V example has only half as much LED current as the 115V example. This is correct. It's a consequence of one of the laws of physics, and ye cannae change those (Cap'n)!

(d) Heating in the resistor is proportional to the power dissipated multiplied by the "thermal resistance" to the surrounding air. The thermal resistance, and therefore the temperature, can be reduced by (a) using a larger resistor with more surface area; (b) thermally coupling the resistor to something with more mass - for example, using a rectangular resistor and squeezing it between the two sides of the plug; (c) there may be other tricks I haven't thought of.

(e) The polarity of the diode and the LED must match. Connect the anode of the diode (that's the end opposite the stripe) to the cathode of the LED (that's the shorter lead).

(f) The version of this circuit with the diode connected across the LED, instead of in series with it, will work but the resistor will dissipate twice as much power as in the main circuit (the one with everything connected in series). The calculations are for the main circuit.

(g) There are various ways to improve the performance of this circuit but none of them are as compact as this design.

(h) Yes, the LED will flicker 60 or 50 times per second (depending on your mains frequency). This is not normally noticeable, but you will notice it if you turn your head quickly from side to side while watching it. You will also see this effect with a clock that has an LED display. When you look straight at it, you probably won't notice the flickering.

(j) Many blue and white LEDs are ESD-sensitive; that means they can be damaged by electrostatic charge from your body during handling. Try to get the manufacturer's data sheet for the device you're using, to see whether it's ESD-sensitive. If so, keep the leads shorted together with a paper clip or conductive foam until the circuit is fully assembled.

(k) If the LED is ESD-sensitive, it can actually be damaged by ESD after the circuit has been assembled and mounted in a mains plug. To protect the LED, connect a low-value capacitor across it before you remove the shorting clip or foam. For example, a 0.1 uF (microfarad), 25~50V ceramic capacitor. To save space you can use an SMT (surface-mount technology) MLCC (multi-layer ceramic capacitor) such as these ones from Digikey:

http://www.digikey.com/product-detail/en/C1206C104K5RACTU/399-1249-1-ND/411524 ("1206" size code)

http://www.digikey.com/product-detail/en/C2012X7R1H104K085AA/445-7534-1-ND/2733606 ("0805" size code - smaller).

(m) Connecting this circuit to your mains supply may be illegal, depending on your country's regulations.

(n) Every point in this circuit must be considered live and dangerous. You must do everything possible to prevent any person from coming into contact with any point in the circuit. Some examples I suggest: (a) replace the screw that holds the socket together with a nylon screw and nut; (b) recess the base of the LED into the plug and/or surround it with glue to prevent any possible access to the LED wires in the area between the LED base and the plug.

Ignoring the voltage drops of the diode and LED, both are insignificant compared to the 230V AC mains, the RMS of the half-wave rectified pulsating DC voltage is ACVpeak / 2 = 162.64V.

When the resistor chosen is of 27k ohm, I = V / R = 162.64 / 27k = 6.02mA. Power dissipation of the 27k resistor, W = I^2 (R) = (0.00602) (0.00602) (27k) = 0.9796W.

I am afraid you may use the wrong formula to calculate the RMS voltage of the half-wave rectified supply. Full-wave DC(avg) is directly proportional to half-wave DC(avg), as the latter is half of a sine wave and is exactly half of the full-wave DC. When calculating RMS, each value has to be squared first, rendering the two not directly proportional any more.

.

.. the RMSof the half-wave rectified pulsating DC voltage is ACVpeak / 2 =

162.64V.

No. That would be true for a square wave, but it is not true for a sine wave. This should be clear from the following diagram, which shows an AC square wave and a half-wave-rectified square wave.

squarewaves.png

The square wave is unique in that it completely "fills up" the half-cycle-width area. The mean and RMS voltages are therefore half of the peak voltage.

A half-wave-rectified sine wave does not fill up the whole area, i.e. the area under the curve for each half-cycle is less than 100% of the area available, so the mean and RMS values are obviously significantly less than half of the peak value. Neither the mean voltage, nor the RMS voltage, of a half-wave-rectified sinewave is half the peak voltage. The numbers can be calculated using the formulas given in my earlier post.

When the resistor chosen is of 27k ohm, I = V / R =162.64 / 27k = 6.02mA. Power dissipation of the 27k resistor, W = I^2

(R) = (0.00602) (0.00602) (27k) = 0.9796W.

This formula is wrong because of your earlier incorrect assumption.

Full-wave DC(avg) is directly proportional tohalf-wave DC(avg), as the latter is half of a sine wave and is exactly

half of the full-wave DC.

That is true, but not relevant to this case. We are not comparing half-wave- and full-wave-rectified voltages; we are calculating the mean and the RMS voltages of a half-wave-rectified sinewave.

I did mention in point (g) that there are ways to improve the design. Using full-wave rectification is one of these.

When using Ohm's Law, V = IR, the value used for V should be the RMS value of the applied voltage for any wave forms including DC.

From a book I read, Vrms of half-wave rectified sine wave = Vpeak / 2.

The reasoning: for Full-wave Sine Wave, Vrms = Vpeak / SQRT(2)

Squaring both side, (Vrms)^2 = (Vpeak)^2 / 2

Half-wave rectified sine wave is exactly one-half of the full-wave sine wave, therefore, for Half-wave Sine Wave, (Vrms)^2 = [(Vpeak)^2 / 2] / 2

= (Vpeak)^2 / 4

Taking square root on both sides, Vrms = Vpeak / 2 for half-wave.

It sounds like you have under-calculated the power consumed. That explained for my last reply to you.

For full-wave square wave, Vrms = Vavg = Vpeak.

For half-wave rectified square wave, Vavg = Vpeak / 2,

However, Vrms = Vpeak / SQRT(2).

For a 10V full-wave square wave, the Vrms of its half-wave rectified form will be 7.071V, not 5V.

When using Ohm's Law, V = IR, the valueused for V should be the RMS value of the applied voltage for any wave

forms including DC.

Not necessarily. If you apply an AC voltage to a resistive (ohmic) load, and use I = V / R to calculate the current through that load, you can use the RMS voltage to calculate the RMS current, or the peak voltage to calculate the peak current, if that's what you want to know. You can even use the peak-to-peak voltage if you want to calculate the peak-to-peak current.

When you're calculating power dissipation though, it's only meaningful to use the RMS voltage. This is because for a resistive load, power is proportional to the *square* of the voltage (from P = V^2 / R), or the *square* of the current (from P = I^2 R), so when the absolute value of the voltage is not constant, you need to apply RMS calculation to the waveform (for every possible point in the cycle, square the voltage, then take the mean of those squares, then calculate the square root of that) to calculate the mean power it represents into a resistive load.

From a book I read, Vrms of half-wave rectified sine wave = Vpeak / 2.The reasoning: for Full-wave Sine Wave, Vrms = Vpeak / SQRT(2)Squaring both side, (Vrms)^2 = (Vpeak)^2 / 2Half-waverectified sine wave is exactly one-half of the full-wave sine wave,

therefore, for Half-wave Sine Wave, (Vrms)^2 = [(Vpeak)^2 / 2] / 2= (Vpeak)^2 / 4Taking square root on both sides, Vrms = Vpeak / 2 for half-wave.Hmm. "I'm beginning to think you may be right", as one of my father's co-workers used to say to him when it was finally obvious beyond all doubt!

Here's an explanation by example, that makes sense to me.

Take a full-wave-rectified sine wave with a peak voltage of 10V with a resistive load of 1 ohm. The RMS voltage is 7.07V and the load will dissipate 7.07W.

Now remove every second half-cycle, so it's the same peak voltage but only half-wave rectified. Now the load resistor will dissipate 7.07W for half the time, and 0W for the other half, so its dissipation is 3.535W, half what it was.

Because power dissipation in a fixed resistance is proportional to V^2 or I^2, a 2:1 ratio in power dissipation is a voltage (or current) ratio of sqrt(2):1, not 2:1.

The ratio between Vpeak and Vrms for full-wave is sqrt(2):1. Halving the power reduces the Vrms by another sqrt(2):1 ratio. The two ratios combined give 2:1, so for half-wave, Vrms = Vpeak / 2. As you said.

Comparing a full-wave-rectified sinewave with a half-wave-rectified sinewave, half-wave has half the power (into a fixed resistive load), and half the average (mean) voltage, but not half the RMS voltage! And the same is true for a squarewave, as you showed in the last part of your post.

This seems pretty counterintuitive to me, but I'm trying to get my head round it. In the meantime I plead no contest on your calculations relating to power dissipation in the resistors for this project. Using the values I calculated, they should be rated at more than 1W.

Thanks for the corrections!

Hi Kris, can you tell me why you have the 2.22 in your formulas as opposed to just R=V/I only please? Is it because of the ac?

thanks

Yes.

The voltage of an AC sinewave can be measured in several different ways. Normally, the stated value is the RMS (root mean square) voltage.

For more information, see http://www.rfcafe.com/references/electrical/sinew...

In the USA, the AC mains voltage is around 115V. This is the RMS voltage. (It produces the same heating effect as a DC voltage of 115V DC into a given load.)

Using this as an example, several other voltages can be calculated from an RMS voltage of 115V.

The peak voltage is equal to RMS multiplied by the square root of 2, which is 1.414. So it will be 162.6V. The peak-to-peak voltage will be twice this, i.e. 325.3V.

The average (mean) voltage after full-wave rectification is 63.662% of the peak voltage. That will be 103.5V. With half-wave rectification, only half of the waveform is used, so the average voltage will be half that, or 51.8V.

Since all these voltages are in fixed proportions, you can calculate a fixed divisor value that will convert an RMS voltage directly to an average (using half-wave rectification) voltage. This divisor is the square root of 1/2 (about 0.707) divided by half of the 0.63662 from the average formula, which works out to about 2.22.

Therefore you can take an RMS voltage, divide it by 2.22, and you will get the average voltage that will be produced when the voltage is half-wave rectified.

These figures aren't exact because the AC mains voltage varies somewhat, and there are voltage drops in the diode and the LED, but they're close enough.

Thanks for the very thorough explanation; you should be a lecturer (if you are not already)!

When using Ohm's Law, V = IR, the value used for V should be the RMS value of the applied voltage for any wave forms including DC.

From a book I read, Vrms of half-wave rectified sine wave = Vpeak / 2.

The reasoning: for Full-wave Sine Wave, Vrms = Vpeak / SQRT(2)

Squaring both side, (Vrms)^2 = (Vpeak)^2 / 2

Half-wave rectified sine wave is exactly one-half of the full-wave sine wave, therefore, for Half-wave Sine Wave, (Vrms)^2 = [(Vpeak)^2 / 2] / 2

= (Vpeak)^2 / 4

Taking square root on both sides, Vrms = Vpeak / 2 for half-wave.

For full-wave square wave, Vrms = Vavg = Vpeak.

For half-wave rectified square wave, Vavg = Vpeak / 2,

However, Vrms = Vpeak / SQRT(2).

For a 10V full-wave square wave, the Vrms of its half-wave rectified form will be 7.071V, not 5V.

Ignoring the voltage drops of the diode and LED, both are insignificant compared to the 220V AC mains, the RMS of the half-wave rectified pulsating DC voltage is ACVpeak / 2 = 155.59V. Therefore I = V / R = 155.59 / 22k = 7.07mA. Power dissipation of the 22k resistor, W = I^2 (R) = (0.00707) (0.00707) (20k) = 1.100W. I am afraid 1W rating for the 22k resistor is too low.

Variant on this idea ...

http://www.marcspages.co.uk/tech/6103.htm

Now need to try & figure out which is best option

Here, since i have used a male pin, there is practically "no space" for a capacitor.

The idea was to make something small "and" at the same time simple enough.

Concept:A single diode (black positive anode part of diode to the negative of the LED) works as a half wave rectifier.

The 22 kilo ohm resistor limits the current (Rated for at least one watt, since it has to reduce 220 volt AC Mains current).

Ignoring the voltage drops of the diode and LED, both are insignificant compared to the 220V AC mains, the RMS of the half-wave rectified pulsating DC voltage is ACVpeak / 2 = 155.59V. Therefore I = V / R = 155.59 / 22k = 7.07mA. Power dissipation of the 22k resistor, W = I^2 (R) = (0.00707) (0.00707) (20k) = 1.100W. I am afraid 1W rating for the 22k resistor is too low.

That circuit, as suggested, has a full wave bridge rectifier and a capacitor also - which is a robust circuit, useful when you plan many LEDs are in a LED bulb light. See my I'bles LED light part 1 and Part 2.

Part 1 has a different arrangement of LEDs, whereby i eliminate the need for the 4-diode bridge rectifier.

In Part 2, i have used a bridge rectifier to remove a slight flicker (not generally visible to the naked eye). The LED arrangement is different here - 3 LEDs are in series and such triplets are then in parallel, so if one LED fails, it can be identified and replaced - making the bulb EVERLASTING!!

reg

k

I kept forgetting to turn off the coffee maker, which has no indicator that it's on.

I used this circuit to add a LED to the coffee maker, it stays on while the machine is on.

I used a 32K ohm resistor and a 1N4007 diode, and a white 10mm LED.

It works great! Many thanks.

Innovative adaption! Congrats for out of the box thinking!

Hope the LED outlasts the coffee maker! ;-DDD

reg

k