# AC transformer for LED

How do I select the correct transformer (voltage) to drive a series of LED's.

The voltage for a series of ordinary LED's are simple:

Example: Using 10 pcs. @ 1.6 Volt requires 10 x 1.6 Volt = 16 Volt (and at a voltage source above, using voltage divider/regulator).

But transformers, have:

1. a non-load voltage factor, and;
2. a the peak voltage on the AC sinus.

So take a 16 Volt transformer, then the sinus peak voltage is 16/0.707 = 22.3 Volt

Note: it is presumed the AC is already rectified to DC-sinus with diodes (bridge).

Problem is:

1. If I go for the 16 Volt (the transformer RMS) in the LED calc, then the LED'S are overloaded from their 16 Volt max. to the sinus peak at 22.3 Volt. Considering the fraction of time they are, may still not be a healthy solution.
2. If I go for the 22.3 Volt (and use 14 diodes x 1.6V = 22.4V), then they are underpowered most of the time (only full at the sinus 22.3 Volt peak). A capacitor could help, but considered as the "bulky" solution.
3. The I could use one of those famous L78xx/LM317 regulators. But this poses another problem: To get the most from the DC sinus, I need the lowest sinus part AKA not the small top to get most efficiency. That means regulating a higher voltage down (chopping off the small sinus top). A capacitor could also help. Either both burns off energy as heat in the regulator.

Are there any suggestions (here answers) for the best compromise to make such LED power source?

• I'm not sure why you think adding a capacitor is the bulky solution if you're already planning to use a transformer. Have you looked into using a buck converter? They can be wildly efficient. – Sean Boddy Feb 21 '15 at 8:05
• You shouldn't look at the peak You need to divide the RMS multiplier by the average of the rectified sinus. This way you get a factor of 1.11 (as far as i remember). Multiply the RMS with this to get the ideally smoothed output from a 1ph full wave rectifier. Try to look it up somewhere, as you should check the factor. – WalyKu Feb 21 '15 at 11:09
• I guess you're just complicating things. A typical red LED would need 1.8V forward voltage to turn on and can handle up to a max of 2.2V. Assuming a basic transformer with bridge rectifiers, you need a transformer that has (1.8 + 0.7)Vrms output (1.8 for the LED, 0.7 from the rectifier drop). Remember, an LED is a diode: what can destroy it is not forward voltage per se but rather forward current and reverse voltage. If you're using VRMS in calculations, the current values produced by Ohm's Law equations is the DC forward current and can be compared to the datasheet. – shimofuri Feb 21 '15 at 11:53
• @Sean : It's not power LED's and as far I know buck's are for constant current. – Gearlos Feb 21 '15 at 14:38
• @Shimofuri: RMS is the equvialent DC voltage for for watt calc. and not the peak voltage. – Gearlos Feb 21 '15 at 14:47

Since the output frequency of a full-wave rectifier will be twice its input frequency, making flicker a non-issue, the easy way is to use a full-wave bridge with no smoothing (since its output frequency will be either 100 or 120 Hz ) and to connect the LEDs in series with an appropriate ballast.

For example, in the schematic, following, 120 volt 60 Hz mains are stepped down to about 12 volts RMS and used to drive a single LED through a 420 ohm resistor.

The transformer puts out about 17 volts, peak, and there are two diode drops across the bridge, so that leaves about 16 volts peak, which is about 11 volts RMS, across D5 and R1.

D5 drops about 2.5 volts RMS with 20 mA through it, which leaves 8.5 volts RMS across the resistor.

So, to get the value of the resistor, we can say: $$R = \frac{E}{I} = \frac{8.5V}{0.02A}=425\Omega$$

For more LEDs, you'd change the transformer's secondary voltage and possibly the bridge diodes and adjust the ballast to suck up whatever excess voltage was there that the LEDs didn't need.

Just for grins, Here's the LTspice .asc file you can run to play with the circuit if you want to.

I would go with solution 3 and use LM317. It is very simple to use, but it is very inefficient.

Use the datasheet to determine correct values. I have tried to calculate them for you, but you will need to check them again.

simulate this circuit – Schematic created using CircuitLab

The value of Cadj can be found in the datasheet. There it is explained more.

If something is wrong with the calculations for transformer out, please edit.

The voltage output of the transformer needs to be 19+2/0.707= 30V

• Thank You so much, but I have used the LM317 many times and I thought there could be a smarter way to use an ordinary trafo cheap and efficient. – Gearlos Feb 22 '15 at 1:01

A combination of #1 and #2:

simulate this circuit – Schematic created using CircuitLab

• This is a bad design in that you have not provided any current limit for ZD1. It will most likely die very shortly after applying power. – Dwayne Reid Feb 21 '15 at 6:15
• I agree @DwayneReid – Gearlos Feb 21 '15 at 6:31
• @DwayneReid, just click edit – Jon Feb 21 '15 at 10:41
• @Jon: this is the 2nd time that I've seen you post a schematic that is seriously flawed. You then ask someone else to fix it rather than doing so yourself. It's okay to post an answer that may not be correct but you should specifically mention that it is possible that your answer is NOT correct if you aren't sure. You should also then FIX your answer when a mistake is pointed out. It's also okay to mention that you don't know how to fix your mistake. People like me will jump in and help. What's NOT okay is to submit a wrong answer and then not fix it – Dwayne Reid Feb 21 '15 at 20:13
• @Dwayne: The 200 Ohm resistor limits current through Z1. If You think it will get overloaded then please tell us why ;) – Gearlos Feb 22 '15 at 0:58