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I need to design a constant current driver for a filament LED bulb. The input is 230VAC and the output should be 120VDC, 20mA (constant current) for 4 filaments arranged in series-parallel structure. Can somebody please help me to find a simple transistor-based solution or some small linear circuit with low BOM cost.

LED filament structure

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  • \$\begingroup\$ That looks like an incandescent bulb not an LED! A transformer will get you from 230 to 120. Vac. \$\endgroup\$
    – George Herold
    Apr 8, 2016 at 16:13
  • \$\begingroup\$ Not constant current, but I would consider two resistors and a bridge rectifier (optionally a capacitor if you want to reduce flicker). You will have to get rid of several watts . \$\endgroup\$
    – Spehro Pefhany
    Apr 8, 2016 at 16:16
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    \$\begingroup\$ @GeorgeHerold I've seen these things- they mimic old-style bulbs for use in chi-chi coffee shops and such like- but they are orange LEDs in the color that SED's JL likes. Very Bohemian. Perhaps the OP should add a few more words to the question lest others be 'led' astray.. \$\endgroup\$
    – Spehro Pefhany
    Apr 8, 2016 at 16:17
  • \$\begingroup\$ @GeorgeHerold No, there is a series string of LEDs inside each "filament". It's a particular design. Note the thickness and the yellow color of the phosphor on the outside. \$\endgroup\$
    – Kevin Reid
    Apr 8, 2016 at 16:20
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    \$\begingroup\$ @GeorgeHerold: You also need to specify whether you want 120 V DC or 20 mA. You set one and let the lamp determine the other. \$\endgroup\$
    – Transistor
    Apr 8, 2016 at 16:29

2 Answers 2

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You could consider a circuit like this one (one for each string, plus a bridge/capacitor to make +300VDC):

schematic

simulate this circuit – Schematic created using CircuitLab

M1 will have to be rated appropriately and have an appropriate heatsink, perhaps of a type typically used in off-line switchers.

Or see if two resistors (one per string) plus the bridge + capacitor is good enough, because that will be cheaper and more reliable while using the same amount of power.

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You can use a capacitor power supply where most of the voltage is dropped in a cap on the AC input side of the bridge rectifier.The losses can be very low .The cap should be a X rated type at least 250VAc .The DC side of the bridge rectifier should have a filter cap across it to prevent annoying flicker .Place a surge limiting resistor of say 1 Kohm 5W in the AC side to protect things at turn on .The Cap supply is nothing new and it is described elsewhere.In fact some mains LED string drivers use them .

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  • \$\begingroup\$ The "Capacitive Dropper" you are mentioning is as far I have understood a cheap, and easily modifiable, driver. (See my question here electronics.stackexchange.com/questions/227721/… ) You say that the dropper capacitor should be rated at least 250V, but since a 230V AC supply peaks at over 300V, I always use 400V rated caps - even for the filtering capacitor in case an LED goes "open circuit" (then the voltage will rise over it as well). I use 22µF-68µF smoothing (depending on available space) 400V (always) caps. \$\endgroup\$
    – 10100111001
    Jun 16, 2016 at 7:57
  • \$\begingroup\$ Also it should be mentioned though, that a capacitive dropper is referenced to the mains (not isolated like a transformer based solution) so every exposed conducting surface in connection with the driver should be considered being Live at mains voltage. \$\endgroup\$
    – 10100111001
    Jun 16, 2016 at 7:59

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