covert incandescent landscape light to LED

Discussion in 'The Projects Forum' started by tp724, Aug 26, 2016.

1. tp724 Thread Starter New Member

Jul 27, 2015
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Hi All,

Can someone review the project notes below and let me know if my calcs are correct, and let me know if i need a smoothing capacitor and if I do, provide the calcs and process to calculate the value of the cap. Thank you.

Project: Convert a string of 3 incandescent landscape lights to LEDs. The lights are supplied by a Malibu timer/transformer. The transfer input is 120VAC 60HZ and the output is rated output 12V 60HZ. The output was measured at 12.89VAC 60HZ.

The plan is to use a full bridge rectifier and a smoothing capacitor to supply the LEDs with stable DC.

As I already have SB5150 diodes (150V, 5A, .85FC) I plan to use them for the rectification portion of the circuit.

In order to complete the project, I’d like to find the following:

1. The rectified DC voltage, then I can find…
2. The available DC voltage after the rectifier circuit, then I can determine…
3. How many LED’s I can place in series or parallel, and then…
4. The placement and values of resisters to limit the amount of current available to the LEDs.
5. The I’d like to calculate the size of the capacitor needed to adequately smooth out the DC after the rectifier circuit (if needed)

I think one find the rectified DC voltage peak first by multiplying the AC input by 1.414, which equals 16.968.

When the current passes through the rectifier circuit it passes through only two diodes at a time. Each diode has a forward voltage of .85V, so the voltage drop across two diodes would be 2 x .85 or 1.7V. The voltage after the full bridge rectifier would be 16.968 – 1.7 or 15.268.

If I were to use the 3W LEDs (DC Forward Voltage:3.6V~3.8V Forward Current: 700mA) and place them in series, then the total voltage drop across all the LEDs would be 3.8 x 3 = 11.4V, and the total current would be 700mA.

Each LED dissipates 2.66W (3.8FC x .7A), or 2660mW so the total dissipation for all three LEDs is 7.98W or 7980mW (3 x 2660mW).

The current remaining after the LEDs is 15.268 – 11.4 or 3.686V.

V = I x R so R = V / I. R = 3.686 / .700A or 5.525. So I need a resistor with a value of approx. 5.5 ohms. And the resistor will need to dissipate about 2.7W. (3.868 x 700mA), so I will need a 3 watt resistor or higher. I have an 8 ohm power resistor rated to 25Watts which I may try.

Can someone let me know if the above is accurate and if not, point out where I went wrong?

No, do I need a cap to smooth the output of the rectifier circuit? If so, can someone explain the formulas and process to calculate the value of the capacitor.

Thank you,

tp724

2. ronv AAC Fanatic!

Nov 12, 2008
3,292
1,255
Looks okay to me, but your calculation assumes a large filter cap to store the peak voltage. Without it you should use the RMS voltage for the power. The problem is you can then only use 2 LEDs in series. Maybe that is okay?
If not, here is a nice write up on the filter caps.
http://www.electronics-tutorials.ws/diode/diode_6.html

Aug 27, 2015
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