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.
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.
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.
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.
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.
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.
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.
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.
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.
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.
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)
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)