AC transformer for LED

led

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

Note: I have seen other Questions about this, but not like this here.

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?

Best Answer

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.

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