Yes. I too think the resistor is for voltage drop. Parts list
specifies "an LED" so.....

My sense is that voltage stored in the cap would naturally taper off
but kind of moot if I used an LED(s).

I was dying for a way to make use of the Su circuit but as mentioned
would prefer to trigger light manually. That said, if using a largish
cap (e.g. 4000 uF), I wonder how long it would take to charge up
enough to power an LED festoon for 5 seconds or so? I'd think
measuring current draw of the LED would help determine things.

On 7/31/17, Alistair Bell <albell@shaw.ca> wrote:

> I saw your circuit, looks like the resistor (470ohms) used to drop voltage
> to foraged voltage required by led. And the cap is there to provide the
> power when main power removed. It's the draining of the cap and resulting
> voltage drop profile that I don't understand. Maybe it will have the led
> bright for a while then rapidly dim and off, or maybe it will start to dim
> immediately and then when forward voltage is insufficient the led winks
> out.
>
> Ab
>
>> On Jul 31, 2017, at 11:59 AM, Neil N <musomuso@GMAIL.COM> wrote:

>> Link to image of schematic on my google photos:
>>
>> https://photos.app.goo.gl/9taHoZwFDyv37H1l1

--
Neil n

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