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AdamWelchUK avatar AdamWelchUK commented on July 21, 2024 1

Hello all. I guess this is mentioned because of my mistake. I forgot to factor in the voltage drop of the LED into my calculation and suggested R2 was too low in my video. So firstly Stuart, sorry about that, I hope you will forgive me.

As you know I've put a 510ohm in for R2, which as you say Colin gives about 2mA. Now, although that original decision is based on incorrect maths I'm going to stick with 510 ohm. I've attached two images which I took with fixed settings on my camera a few seconds apart. One is running at 2mA the other 20mA. It's hard to reproduce light levels on camera, but it should give you a rough idea.

img_7301
img_7300

To add to the discussion I'll argue that the value of R2 will entirely depend on each users individual situation. A system placed in a dim environment rather than a sunny one is one factor, but the amount of time the LED is likely to be lit is another. My LEDs were very cheap and I'd rather not have to replace them, running them at a low current should help that considerably. I've also swapped the suggested blue LED for red - with a lower forward voltage. Red makes more sense to me as an indicator to suggest something is wrong (high) or we're burning off the energy.
Now I'll also argue that R2 may be chosen with consideration of R1. If you are aiming for a discharge current of 200mA for example (say with a 20ohm R1) then an additional 20mA is an additional 10%, but if you are going for a ~1amp discharge current (eg. R1 is 4.3 ohm), then 20mA extra is almost nothing.

Cheers

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CuriousTimo avatar CuriousTimo commented on July 21, 2024

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chickey avatar chickey commented on July 21, 2024

With 510ohm, wouldn't that give just over 2ma which seems really low.

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stuartpittaway avatar stuartpittaway commented on July 21, 2024

Thanks for the post @AdamWelchUK closing this issue now as you explain the reasons why very well :-)

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