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.
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Signing UpStep 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.
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.
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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.
But, how about heating?
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.
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.
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!
Iled = (Vsupply - Vth) / Rled where Vth is the point at which it starts to glow (the threshold), and Rled is about 17-ohms for 25mA LEDs.
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.
That means, that, without ANY external resistors, if we put 3.3volt across this LED, it will "consume" 23.5mA of current.
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.
That is the primary reason resistors are added LED circuits - insurance, at the cost of lowered efficiency.
This is a chart for a 25mA white LED from the CREE website. Most manufacturers have a similar chart with almost identical figures.
We can see that the LED "uses" 20mA at 3.4volt, and 80mA at 4.4volt. Therefore, the current increased 60mA (0.060amps) for a 1-volt increase.
The dynamic resistance of this LED, therefore, is ( R = V / I ) or 1 / .06, or 16.667 ohms, which I've rounded to 17.
vsupply - vled / led current = resistor value needed.
also why if they run on any current are the operating currents specified for leds like my blue leds run at 20ma
thanks
You can find your circuit in the newly posted "Using AC with LEDs, part 2".