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12V Incand Desk Lamp - Retrofit with LED

Discussion in 'LEDs and Optoelectronics' started by mike271828, Jan 16, 2013.

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  1. mike271828


    Jan 16, 2013

    I have a desk lamp that I want to retrofit with an LED. The 'stock' halogen bulb is too inefficient- it gets really hot and burns out frequently. I happen to have some really bright white 60mA LEDs to use instead.

    The lamp has a transformer coil inside it. I took 2 pictures of inside the lamp still unmodified (2nd and 3rd pictures below). The voltage on the load side is about 14 volts AC (measured with a DMM). Its rated "12V." The bulb looks like a car headlight (1st picture below).

    In order to run an LED, I need to convert the AC signal to DC. So I figured I'd make a bridge rectifier out of 4 diodes and use a filter capacitor to smooth out the bumpy DC.

    I made a cheap diagram in MS Paint to denote the modification I plan to do (the 4th picture below). I plan to cut the output lines coming out of the transformer coil (that originally go to the bulb) and solder the home-made module there. The home-made module I'm adding is drawn in blue. It entails the bridge rectifier, the filter capacitor, and the resistor.

    My question: I need to know what filter capacitor to use. I don't know much about capacitors. I'm not even sure if I'm supposed to put it in parallel like the diagram shows. Can you give me advice on which capacitor to use and whether I should put it in series or parallel.

    Also, is there more to this than I plan? Do I need to do/add more stuff, or is this okay?

    Specs: I tried to provide as much info as possible, so I listed the specs of everything. Let me know if you need more to answer the questions.

    - The bulb is a 12V incand/filament bulb. The lamp hood says, "Use 12V/20W JC Type T or Smaller <bulb picture> G4."

    - The LED: Forward Voltage is 3.2V, and forward current is 60mA. I will use a 1-Watt 180-Ohm resistor.

    - Bridge Rectifier: I will use 4 1N4000 series diodes. These are rated at 1A continuously.

    - The measured output voltage on the load side is about 13.8VAC. I measured this by sticking the DMM probes into the socket holes where the bulb plugs into.

    Stock Bulb:

    Transformer Coil

    I hope I provided enough info. If not, let me know, and I'll do my best to provide what you need to answer the questions.

  2. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    Jan 21, 2010
    There are LED drivers designed specifically to run from 12VAC and which will delver a specified current without the heat generated by a resistor.

    Here is an example.

    Either get one for the correct current, or change a single sense resistor to set the current.
  3. mike271828


    Jan 16, 2013
    Thanks for the reply

    I don't want fixed current though. I'd rather leave it so I can swap out the LED some day if I wanted to and just change the resistor accordingly. If I was to do it the way I originally planned, what would be a good capacitor to use?
  4. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

    Nov 28, 2011
    If you're running the LED at 60 mA, a 1000 uF smoothing capacitor (rated at 25V) will be fine. This will be an electrolytic capacitor.

    The amount of smoothing determines how much the DC voltage "dips" in between mains voltage cycles. When AC mains is full-wave-rectified and smoothed, the resulting DC is never perfectly smooth. On the peak of every mains half-cycle, the smoothing capacitor is charged up, and it then gradually discharges somewhat during the gap between half-cycles.

    The amount by which the smoothing capacitor discharges between half-cycles is determined by (a) the mains frequency, or more exactly, the time period between half-cycles; (b) the capacitance of the smoothing capacitor; and (c) the amount of current you are drawing from the smoothing capacitor.

    The formula is dV / dT = I / C where
    dV is the change in voltage across the capacitor, in volts;
    dT is the time period between half-cycles, meaured in seconds;
    I is the current drawn from the capacitor, in amps;
    C is the capacitance, measured in farads.

    This formula rearranges to dV = dT I / C
    Plugging in the known values gives dV = 0.00833 * 0.06 / 0.001
    = 0.5V.

    Your peak DC voltage will be about 15V so with 1000 uF of smoothing and a 60 mA load, your DC rail will remain between about 14.5V and 15.0V. The ripple is only 3.33% peak to peak which is negligible. In fact you could decrease the smoothing to 470 uF or even less.

    If you omit the smoothing capacitor altogether, the circuit will still work (your LED supply is still unipolar, although it is not steady), and the flicker at 120 Hz will not be noticeable when everything is static, but it will be noticeable when you move, turn your head, move the lamp, or blink. So SOME smoothing is definitely desirable.
  5. mike271828


    Jan 16, 2013
    Got it, thanks very much to both of you.

    KrisBlueNZ, the differential equation works for me. Now that I understand how ripple span is a function of capacitance, I can work on my own with it.

    Steve, I definitely like the idea of the driver too but for other projects though- such as in the car (sometimes I add interior LEDs just for project fun). I hear cars' supplies are real spiky. A constant current driver, similar to the one you linked to, might help with LEDs that frequently burn out in the car.
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