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I'd like to modify a color changing candle (example) circuit so it can be turned on without lighting the actual candle. It uses a photodiode and a optical fiber alongside the wick to cycle the RGB LEDs as long as it's burning.

I've found out I need to supply about 30 uA with a external power source (another 3V battery with a 100k resistor) to the PD leads to substitute its function in the circuit. Sourcing current from the two CR2032s to the PD doesn't work.

Is there any easier modification without the need of a external power source?

color changing candle circuit top enter image description here enter image description here enter image description here

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    \$\begingroup\$ You'll need to post a schematic, please use the tool and edit your question. (ctrl-m while editing) \$\endgroup\$
    – Voltage Spike ♦
    Commented Dec 27, 2017 at 19:11
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    \$\begingroup\$ It all depends on how the photodiode is hooked up. It could be as simple as replacing it with a resistor, but might be totally different too. SHOW THE SCHEMATIC. \$\endgroup\$
    – Olin Lathrop
    Commented Dec 27, 2017 at 19:16
  • \$\begingroup\$ It really depends on what is hiding under than black blob or epoxy. \$\endgroup\$
    – Trevor_G
    Commented Dec 27, 2017 at 19:27
  • \$\begingroup\$ Sorry guys, I don't a have a schematic. This is a commercial product and all I have is in the photos. \$\endgroup\$
    – Branislav Blesak
    Commented Dec 27, 2017 at 19:30
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    \$\begingroup\$ Hey, that worked! Why haven't I thought of this stupidly simple solution... Thank you Trevor. \$\endgroup\$
    – Branislav Blesak
    Commented Dec 27, 2017 at 22:23

2 Answers 2

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As suggested by Trevor, just short the PD leads.

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You want to defeat the flame sensing with some cheater control. You may increase value of 100k by 10 or 100 or trigger by aiming a remote control it.

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PD with cathode to 3V and 100k from Anode to 0V will raise Anode voltage with optical current ~0.5mA /mW.

By your comment 30uA , this is simply the optical current generated to drive 100k to 3 V. No need for more Voltage.

The PD needs some IR illumination to obtain ~60uW of light out of some source.

But since the PD has a daylight black epoxy blocking filter, it responds best to IR infrared light or heat.

Changing the 100k to1M makes it more sensitive and removing 100k makes most sensitive. But moving 100k from 0V to 3V biases the input to be on always

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  • \$\begingroup\$ Unless it's using the candle flicker to play with the color changing. Who knows what's under that blob. \$\endgroup\$
    – Trevor_G
    Commented Dec 27, 2017 at 19:44

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