Using AC With LEDs (Part 1)





Introduction: Using AC With LEDs (Part 1)

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.

Step 2: Getting Light

We know that most white and blue (and UV) LEDs range between 3 and 3.6 volts. So by dividing our PEAK voltage by an average LED voltage, we get an idea of the number of LEDs our transformer can support:

15.3 / 3.3 = 4.6, which we round up to 5, giving about 3.1v per light. But remember, that AC has an identical NEGATIVE cycle! Which means we can add a mirror circuit that work on alternate phases.

The advantage of using voltages to start our calculations is that, as long as we stay with similar LEDs, and stay within its operating voltages, the current will stay within safe limits.

So, by adjusting the number of LEDs in use, we can handle most AC transformer outputs.

Now a quick check of the voltage shows that it is still at 10.8VAC. Our LEDs are only using a miniscule portion (4%) of the 500mA capacity of the transformer that...

We can multiply the light output up to 15 times just by adding chains of 10-LEDs arranged the same way across the supply! Imagine running 150 LEDs in an vast array off one tiny transformer. Pure simple direct drive all the way.

Step 3: The Pitfalls

One safeguard is that we have limited the drive to our LEDs to a very safe level - it will only reach its rated peak once per cycle. In fact it will be off completely when the opposing chain is lit. So we can expect extreme longevity from this arrangement.

The fact that each chain is off for half the time means there will be some flicker, which you can see in the photos below, taken with a high shutter speed.

By alternating on and off rows, the effect is minimized, and is no worse than using fluorescent lighting.

Step 4: Some Variations.

Sometimes, you cannot get the right number of 3.5v LEDs for what you need. Then you can 'cheat' by substituting an amber LED in each chain - they operate around 2.4 volts, so that allows you to fudge your numbers a bit.

And about those Dust-busters - if you applied our method to their wall-warts WHILE the unit is charging, you may well find that one chain of LEDs never lights - this is because they only use half their circuit to charge the unit. Think of using the OTHER half of the cycle for LEDs as free power.

You can also adapt this method for DC supplies - but make sure you always measure the actual output first! Commercial units are notoriously bad for making up numbers.

Step 5: Recapping

So, to find out what a transformer can support:

Measure its output:
- If it is AC, use the V-AC scale on your multimeter, and multiply the results by 1.4 to get V-peak
- If it is DC, use the V-DC scale read out V-peak.

The number of white (or blue) LEDs it can support is:
- Vpeak / 3.3 and round up to the next integer. (E.g 4.2 is 5)

(Use V-peak / 2 for Red, Orange and Yellow LEDs)

That is the number of LEDs you can put in a series to operate off the transformer safely.

For AC circuits, you will need to duplicate another chain in the opposite polarity.

LEDs can be any current, as long as they are all the same, and the transformer has the current (A or mA) to support it.

Note: AC transformers can also have a VA rating instead of amps - just divide that number by the volts to get amps.

- end of Part 1 -
(Continued here)



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    What about using a 24vAC transformer for 12vDC LED strips? I ask this because I mistook an 18.6vAC transformer for a similar looking 12vDC I had been using and hooked it up to a short strip (1/2 meter) and it lit up fine, just a bit dim. I checked the voltage (DC) and it wavered and wouldn't read until I switched it to AC on meter. I think I was getting about half the AC going through the strip in a DC way (9.3v) and it explained the dimness. What's going on?

    2 replies

    Unless it's a misprint or a 240v transformer working on 110v, I can't explain it.

    If you are sure your readings are right, you can short out the 100-ohm (or so) resister over every third LED on the strip, it will now run at 10.5v, closer to the output of the xformer, and give you 85+% of full brightness

    I'll do that.

    No writing still visible, encased in black plastic so don't now if current limited, etc. So I guess I should get a 12vAC transformer with known specs to do my experimenting? Definitely. Thanks for replying; it was a curious accident that then led me here.

    Great! I'm gonna use all these old door bell transformers (18v AC).


    You will need more than 34 (or 68) LEDs to run on 120vac without current limiting.

    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.

    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)

    @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.

     Hi, great posts about AC led lighting!!
    But,  how about heating?

    Hey, I have a question: what is the value (farads) of the smoothing capacitor you´re using?

    1 reply

    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 :-)

    7 replies

    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.

    To see how this is true, let's take the common formula for a limiting resistor:

    R = (V-Vf) / Iled

    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!

    I've also added a couple of safety features here:
    (1) I am using a relatively low Vf of 3.3v, which means the LED is never over-driven. And
    (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 "coasting" along.

    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. :-)