Originally Posted by crutschow
Thus at the instant the capacitor starts charging its potential is at ground while the LED has a 2V potential

No...the LED and capacitor are in parallel, so they will always have the same potential difference across their terminals. At time=0, the potential difference is 0V across the capacitor and 0V across the LED.
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True. I worded that poorly. I meant that the LED requires a 2V potential to conduct, thus at the start all the charge goes to the capacitor.

 

I meant that the LED requires a 2V potential to conduct

That's closer, but still a little loose. Diode forward voltage is a really fuzzy number.

For example, can you tell what the forward voltage is for these four differently-colored LEDs from the same manufacturer?

The answer, which you can see is somewhat arbitrary, is to read along the 20mA line near the top of the graph (except for red which uses 25mA). It's where the solid lines turn into dotted lines. That's just how this particular manufacturer defines forward voltage.

So for the OP's circuit, the LED actually starts conducting almost right away--it just conducts very very very little. But it does conduct a non-negligable amount of current even before the rated forward voltage. It's not like a switch--it's a smooth transition.

But just because it's conducting doesn't mean it's giving off a visible amount of light. The LED will appear to be "off" until the current is sufficient.

It does leave open the question of what makes a particular electron turn right or left when it gets to the junction where it could go toward the capacitor or the LED. But that's more of a physics question than an electronics question. The electronics answer is to use Kirkhoff's laws or equivalent. The physics answer would probably just derail this thread beyond the OP's intended purpose, but look up "Energy transfer in electrical circuits" for some interesting explanations.

 

That the diode starts to conduct every so slightly right from the beginning, that there is no single magical voltage at which things suddenly turn on, that the current steadily transitions from the capacitor to the LED at the same time that its magnitude is decaying, and other similar things were pointed out at the very beginning, and throughout the thread. Given the level of the problem, it is perfectly reasonable to use an idealized model for the basic discussion and analysis.

 

That's closer, but still a little loose. Diode forward voltage is a really fuzzy number.

For example, can you tell what the forward voltage is for these four differently-colored LEDs from the same manufacturer?

The answer, which you can see is somewhat arbitrary, is to read along the 20mA line near the top of the graph (except for red which uses 25mA). It's where the solid lines turn into dotted lines. That's just how this particular manufacturer defines forward voltage.

So for the OP's circuit, the LED actually starts conducting almost right away--it just conducts very very very little. But it does conduct a non-negligable amount of current even before the rated forward voltage. It's not like a switch--it's a smooth transition.

But just because it's conducting doesn't mean it's giving off a visible amount of light. The LED will appear to be "off" until the current is sufficient.

It does leave open the question of what makes a particular electron turn right or left when it gets to the junction where it could go toward the capacitor or the LED. But that's more of a physics question than an electronics question. The electronics answer is to use Kirkhoff's laws or equivalent. The physics answer would probably just derail this thread beyond the OP's intended purpose, but look up "Energy transfer in electrical circuits" for some interesting explanations.

That's all certainly correct, but I though it would just muddy the waters bringing in that level of detail based upon the fundamental level of the original question.

 

Hi!

I'm having a really busy week at work till this sunday, but some new thoughts emerged as I followed this discussion in my breaks. I'll reply soon, thanks for helping.

Be right back.