3
\$\begingroup\$

I am trying to create an LED pool light that works off a 12VAC input. Upon disassembling a similar pool light (Intellibrite), I found that they appear to be using rectifiers and perhaps a few small inductors, as well as many large capacitors.

I am trying to do something similar.

INPUT:

12VAC, up to 20 Amps

OUTPUT:

5-30VDC (voltage doesn't matter as the LED driver I'm using can use any voltage between 3-30VDC) Capable of supplying at least 60 Watts of power (IV >= 60)

What is the best way of accomplishing this? My goal here is efficiency as well, as well as a simple BOM (less than maybe 20 components).

I considered using full-wave bridge rectifiers, and simply putting HUGE caps on the output. However, I am not sure if this is a good solution, considering I need 10Amp output, and the capacitor values might need to be humongous.

Any IC that makes this easier is a bonus.

The LED Driver that I am considering is the LT3496 (same one I see used in the Intellibrite). Also, if there is a better candidate for an LED driver, feel free to sugggest.

EDITS:

  1. Realized the LED driver can support up to 30VDC input, not up to 12VDC.
  2. Corrected the requirement from 10Amps of current to Wattage requirement.
\$\endgroup\$
8
  • 1
    \$\begingroup\$ Why do the capacitors need to be huge? Ripple may not be a major concern if you have LED drivers already. The bigger problem is that 12VAC is going to give you about 17VDC (minus diode drop in the rectifier).You probably want some sort of proprietary buck converter, but I don't know anything about them really, but someone else will. \$\endgroup\$
    – Ian Bland
    Commented Aug 16, 2017 at 0:23
  • \$\begingroup\$ Can we have a part number or datasheet link for your LED driver IC or module? \$\endgroup\$ Commented Aug 16, 2017 at 1:16
  • \$\begingroup\$ Where does the 12V ac comes from? \$\endgroup\$
    – Jason Han
    Commented Aug 16, 2017 at 2:43
  • \$\begingroup\$ The 12V ac comes from the pool's power control panel that contains a 110VAC to 12VAC transformer. The wire runs from the control panel through a conduit to the pool niche (hole for the light). \$\endgroup\$
    – Adam B
    Commented Aug 16, 2017 at 3:25
  • \$\begingroup\$ @IanBland, if the capacitors aren't large enough (in terms of capacitance), then the voltage drop between the peak of the waves would be too much. We are potentially drawing 10 amps, so the capacitor would be need to be pretty big to hold enough charge during the 8ms "dip". \$\endgroup\$
    – Adam B
    Commented Aug 16, 2017 at 3:30

2 Answers 2

4
\$\begingroup\$

For converting mains power to DC, "Full wave rectifier + filtering caps" is the only realistic option. 60W / 12V = 5A, which is easily achievable with reasonable size components. Your LED driver can handle a wide input voltage range and regulates LED current, so an unregulated supply should be OK.

For best efficiency you should use high current Schottky diodes and low ESR filter capacitors. The only question is what ratings and values are required? Incorporating all relevant factors makes the calculations a bit tedious, so I simulated the following circuit in LTspice:-

schematic

simulate this circuit – Schematic created using CircuitLab

Peak AC input voltage to the rectifier was 16.6V (blue line). The diodes dropped 0.5V each so the peak DC output voltage was 15.6V (green line). Peak diode current was 24.6A (red line). Between peaks the capacitors held the DC voltage up, but it dropped as the load current discharged them. With a total of 9400uF (2 x 4700uF 25V) the minimum 'trough' voltage was 12.6V, corresponding to 63W minimum output power.

The average mains input power was 81.1W (ignoring transformer core loss which wasn't simulated) and the average output power was 70.6W, for an AC-DC conversion efficiency of 87%.

enter image description here

\$\endgroup\$
3
  • \$\begingroup\$ Excellent information and answer. It turns out after closer examination of the Intellibrite PCB, I found 4 3.3mF capacitors, and the inductors are used in the LED driver part of the circuit, not the AC-DC power conversion. If the peak diode current was 24.6A, does that mean I have to find diodes that can support this high of a peak current? Most DO214 type (bigger SMD diodes) are around 1A (though I think that's average, not peak). What about using a rectifier that includes all 4 diodes (i.e. a rectifier chip)...I'm assuming such a thing exists. \$\endgroup\$
    – Adam B
    Commented Aug 16, 2017 at 19:45
  • \$\begingroup\$ Make sure you read the datasheets for the diode and capacitors you're selecting. Yes, the diodes will have a "peak repetitive surge current rating" and the capacitors will have a "max ripple current" spec at 120 Hz since this is a very common application. Make sure your design stays well below those else your components may not last long. \$\endgroup\$ Commented Aug 16, 2017 at 20:40
  • \$\begingroup\$ Thanks, just saw that he listed a part number for the rectifier/diode. As a future reference, I see that FERD30M45C might also be a good choice, with better performance (less Vf, and less power dissipation). \$\endgroup\$
    – Adam B
    Commented Aug 16, 2017 at 21:54
0
\$\begingroup\$

"Full wave rectifier + filtering caps" concept may not be applicable in your case due to high current requirement. Note that the rms current drawn from AC source will be approximately twice the DC output current. Are you sure that the transformer is able to deliver that high current? Also, if you require a good regulation at the output, you may need to put a few big caps in parallel (size issue). Another thing to consider is that the output voltage of the rectifier will be about 17VDC at best and this is higher than your requirement.

Do you have to use 12Vac input? If not, a good alternative way is buying (or making) a 110-to-12V - 10A converter.

\$\endgroup\$
1
  • \$\begingroup\$ The transformer is rated for 300W. Also I realized the LED driver I'm using can support up to 20V input. Made corrections to the questions. Thanks! \$\endgroup\$
    – Adam B
    Commented Aug 16, 2017 at 17:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.