Recently I came across a high quality transformer selling for under $1.00. The reason they were so inexpensive was the fact that their output was AC only, while most consumer products required well filtered DC.

This Instructable is put together with the goal of getting AC-transformers working with LEDs without diodes and capacitors. I will show enough maths here so the concept is applicable to most other AC-only transformers.

Interestingly, many Black&Decker; Dust-Buster transformers are AC only, and they are well suited for conversion, since many only use 1/2 of the output (half-wave rectification) only.

Step 1: Working the Numbers

The subject transformer was made for many AT&T; cordless phones, it is rated for 110v/60Hz and has a 10VAC 500mA output.

First, we have to be aware that the 10V rating is known as the RMS voltage, and is the effective average power of the sine-wave. The maximum voltage, which we will subject our LEDs to, is about 1.4 times higher.

We can demonstrate this by hooking up our transformer and taking some measurements.

The second image shows 10.8 VAC, which the unloaded output of the transformer. So we should expect a peak voltage of 1.4 x Vrms or 15.3v

Next we add a simple diode with a smoothing capacitor and measure the voltage across it: 14.5VDC.

This number is about .8v less than our calculations because the diode has a voltage-loss across it of .8V

This is one reason we try to avoid diodes because each one inherently loses (as heat) a bit of power - .8v is 25% of the power for a 3.2v LED.

So, we will be using 15.3 volt as the basis our calculations.
could you put 34 or 68 leds across 120VAC
You will need <em>more</em> than 34 (or 68) LEDs to run on 120vac without current limiting.<br> <br> For AC, always start by calculating PEAK voltage, which is 170v here, then divide that by 3.3v to get the number of LEDs to safely drop the voltage, which is 52. You will need another 52 to cover flow in the opposite direction unless a full wave bridge is used.<br>
Ah, but the actual peak only goes to half of that, because it is measured in peak to peak voltage, my friend. :D Same would go for rectified ac. it's about 1.4*1/2(RMS)
<p>@spark light is just plain wrong when he says the actual peak only goes to half of 170V! The calculations above convert AC to Peak, not AC to Peak-to-Peak, so 120 VRMS * SQRT(2) = 169.7 V PEAK, approximately 170 V Peak for one half of a sine wave cycle. The other half cycle will have the same magnitude of 170V Peak, but in the opposite direction. Therefore, the peak-to-peak voltage is 170 * 2 = 340 V Peak-to-Peak.</p>
ah i see
&nbsp;Hi, great posts about AC led lighting!!<br /> But, &nbsp;how about heating?
Hey, I have a question: what is the value (farads) of the smoothing capacitor you´re using?
It's a 0.1uF capacitor. Since there is no load on the circuit (other than the high resistance of the multimeter), just about any value will work.
I'm trying to understand something here. Normally one has to use a current-limiting resistor with LEDs since they don't create appreciable load... Why isn't this necessary here? It just seems to me like you'd be pushing something like 500ma instead of 20ma through these suckers. Obviously I'm wrong since this works for you, but I'd like to understand WHY it works :-)
It's a fairly popular misconception, but LEDs DO have internal resistance, typically about 17ohms for 25mA units. The practice of adding the limiting R is mostly because LEDs are usually based on battery power, and battery voltage can be quite different from the stated value. For example, 1.5v alkalines can go from 1.8volt when new all the way down to 0, and if the designers are lazy (or uninformed), the stick the resistor in, just in case.<br/><br/>To see how this is true, let's take the common formula for a limiting resistor:<br/><br/>R = (V-Vf) / Iled <br/><br/>Basically it says, take the { (Supply voltage minus the LED's Voltage) and divide it by (the LED's current) } Now ask yourself, what happens when V is the same as Vf? Then (V - Vf) equals zero! And 0 divided by any non-zero number is ALWAYS zero! Which means you don't NEED a resistor!<br/><br/>I've also added a couple of safety features here:<br/>(1) I am using a relatively low Vf of 3.3v, which means the LED is never over-driven. And<br/>(2) Remember we are using AC here, which means that we only reach the peak current for a small percentage of the time. The rest of the time, the LED is simply &quot;coasting&quot; along.<br/><br/>
There are other reasons as well though. Everything you said is right. But if an LED is in parallel with another device, they sometimes add a resistor because the current drop across the LED changes as the other device pulls different currents. An initial pull drops the current/voltage to the LED, but when it stops both increase. AC power for example, when a compressor kicks on the lights dim, and when it stops they brighten. It's very brief but visible. In DC applications as well, you want to beware that a device doesn't dump a voltage spike across the led and destroy it. Voltage spikes happen on AC and DC and go above the normal supply voltage level, so it could go above your LED's voltage. Make sense?
I think we are talking absolutes here - there will never be a circuit that can protect against all possible mishaps. How many appliances can survive a direct lightning strike? The key is to be sensible about it. If it makes you feel better, then feel free to add resistors, MOVs and a fuse if and when you decide to build one! However, I DO know that in actual use these circuits have all continued to work as designed. The light from this article is the walkway light outside the furnace room - with fan blowers and pumps cycling on and off, and nothing has 'zapped' yet. Other facts to keep in mind: (1) The transformer has a huge capacity to absorb spikes, (2) LEDs can withstand temperatures up to 250C briefly and (3) LEDs are routinely used in automobiles where the voltage goes from 10v to 16v, with surges as high as 75-volts, not to mention electrostatic charges. LEDs these days are nowhere as fragile as earlier in their development history, and a lot of the precautions we take go back to the days when we expect 10%-15% of any lot to fail within weeks.
Good points. I'm impressed with your articles as well as how much you know on the topic in general. Out of curiosity, what causes 75 volt surges in automobiles?
When the engine is cranking over, especially with a weaker battery, a surge of 75v can occur for up to half a second. Look up "load dump" for more information.
Huh... I have some super-cheap LEDs that start to work at around 2v so I threw three of them in series and attached to a 6.5v DC power source and it worked fine just as you say. :-) But now I'm confused about something else. I measured current at 16.5 milliamps. 6.5 volts and .0165 amps means the resistance in the circuit has to be nearly 400 ohms. Resistance in series would just be summed, yes? That'd be something like 130 ohms per LED, not 17... I don't think it's the power source limiting current...
I see you actually already answered my question in another response. never mind! Basically I was underdriving them so resistance was in effect greater. :-)
hmm, im wondering. what is the &quot;turn-on&quot; voltage of the LEDs you are using here? most diodes have about a .7v &quot;Von,&quot; meaning if you put .7v across them they wont light at all, between .7 and 3.3v they are partially on, and at 3.3v+ they are fully on.<br/><br/>If you've got a bunch of LEDs in a row, say, 5, thats about 5*.7 = 3.5v of your 10V sine wave when NONE of your LEDs are on.<br/><br/>might there be a way to recapture that lost portion of the voltage swing? because youre losing it for both the positive and negative directions!<br/>
You're confused between diodes made of Silicon and Light-Emitting Diodes made from (among other materials) Gallium Arsenide. They each have very different conducting regions.
the wikipedia article on LEDs shows an Id vs. Vd curve very similar to the silicon diode curve I am familiar with, complete with a Von and a turning on region... perhaps LED's Von are closer to the 3v than to .7, but the idea still holds, even more so. I dont mean to be a pain, just confused.
The danger with basing beliefs with a generalistic analysis is that they leave too much to guesswork. The following is a composite of the Current vs Voltage for (1) the 25mA white LED from CREE, while the second is from the datasheet of the 1N4001 diode. While the LED shows that it will "consume" 20mA at 3.4v and 80mA at 4.4v, it regresses to 0 somewhere around 3.0v. Meaning that there is NO significant current flow below the threshold, Vth. The 1N4001, in contrast, will do 20mA just above 0.6v and 200mA by 0.8v. It should be clear that there is a tremendous world of difference between these two devices - despite their sharing the term "diode" in their name.
The diode charts:
So are you saying that LEDs have no von? The data sheet for the LED doesn't show it dropping to zero. It just gives no info. I would think all diodes have a von region. They work by creating an electrically insulating gap and a specific voltage is needed to overcome that gap even when operating in the right direction. The von is a small current bleed if I'm not mistaken when that gap is just bridging. I could be wrong on this, I'm just guessing. Only way to reduce waste would be to regulate the voltage above 3 or whatever min is, but then you're leaving it on and wasting more voltage anyways. No point, and to convert it to square wave uses more power than is lost. I mostly ask out of curiosity.
If you look carefully, the chart for the 1N4001 doesn't go to 0 either. The key is that we are talking about effective resistance in the hundreds if not thousands of ohms, and can be ignored for our purposes. What you call Von for the LED is what I term Vth - the point you start seeing a definite glow in the LED - which happens around 2.9v, or therabouts, but only for these LEDs. The turn-on voltage for semi-conductors do depend on the electron-gap and are fixed by the chemicals in their composition: Silicon is around 0.6v, Germanium is 0.25v, but neither can produce any usable light. For red LEDs they only need 2v to work, while blue (and hence, white) LEDs needs 3+volts to work. That's a rule of nature - it's not something we can "make" happen.
but what i dont understand is if the current is constant through a circuit how canthe led "use current" and why does this matter when choosing a resistor
The basic calculation of current through a LED for a given voltage can be estimated to be:<br/>Iled = (Vsupply - Vth) / Rled where Vth is the point at which it starts to glow (the <em>threshold</em>), and Rled is about 17-ohms for 25mA LEDs.<br/><br/>Let us say our white LED starts glowing at 2.9v, and that will be our Vth. Then at 3.3v, our current will be: (3.3-2.9) /17 = 23.5mA.<br/><br/>That means, that, without ANY external resistors, if we put 3.3volt across this LED, it will &quot;consume&quot; 23.5mA of current.<br/><br/>If you work out the current for this LED at other voltages (try 3.2 and 3.4 first), you will see that the current varies quite a bit for very slight voltage changes.<br/><br/>That is the primary reason resistors are added LED circuits - insurance, at the cost of lowered efficiency.<br/>
so where did you get 17ohms from?
The datasheet is your friend!<br/><br/>This is a chart for a 25mA white LED from the CREE website. Most manufacturers have a similar chart with almost identical figures.<br/><br/>We can see that the LED &quot;uses&quot; 20mA at 3.4volt, and 80mA at 4.4volt. Therefore, the current increased 60mA (0.060amps) for a 1-volt increase.<br/><br/>The <em>dynamic</em> resistance of this LED, therefore, is ( R = V / I ) or 1 / .06, or 16.667 ohms, which I've rounded to 17.<br/>
ahh understood thanks
what do you mean leds can be any current? if that is so can you explain why the current matter with working out a resistor value for leds <br/>vsupply - vled / led current = resistor value needed. <br/>also why if they run on any current are the operating currents specified for leds like my blue leds run at 20ma <br/>thanks <br/>
If you consult the data sheets of LEDs, there is a voltage (Vf) when it will consume the designed-for current. Generally, for White, Blue and UV LEDs, this is around 3.3v. In the resistor formula you quoted, when Vsupply and Vled are the same, then R is zero, at whatever current the LED is designed for - and that is why we can leave it out of our calculations. Only when you are using external parts (resistors, in theis case to waste extra power) do you need to look at overall current consumption. White LEDs are basically Blue LEDs with a coating of Yellow phosphor over it, so the two should have very similar Vf of about 3.3volts.
Any ideas for just one LED? (Usually component count is a killer). Like, say you've got a heater you want to tell when it's on or something. I've heard like MOSFET's can pick up current induced off an AC line. Is there some quick and simple way to sap off just enough to drive an LED?
MOSFETs can be used to sense current, yes, but it cannot, by itself, generate the power to light a LED.<br/><br/>You can find your circuit in the newly posted <a rel="nofollow" href="https://www.instructables.com/id/Using_AC_with_LEDs_Part_2_and_make_this_handy_/">&quot;Using AC with LEDs, part 2&quot;</a>.<br/>
This Instructable is put together with the goal of getting AC-transformers working with LEDs without diodes and capacitors. its funny to me that LEDs are diodes. and you said without diodes. Also. Flashing a LED can save half as much energy being used--Thats good.
I think he meant without extra diodes to convert AC to DC.
Point taken. How 'bout "without extraneous diodes?"
The flicker effect is not bad, anyway. I read once that LED power indicator lights are designed to cycle on and off 50 or so times a second to extend the bulb's life. But, it is so rapid that the human eye does not notice it. What you have done has probably been done before by someone, but it is still a clever idea. I have a bicycle and I have a generator for AC. The voltage is too high for most LED circuits, but I could utilize your idea very well. Thanks.

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