Simple AC Powered LED

carbonzit said:

Ah, so. So when those folks were saying "The currents through the capacitor in each direction must be equal", what they really meant is that in order for the LED to light in the forward direction, some current must flow through the capacitor in the reverse direction in order to discharge the capacitor: would you agree?

And I suppose that if you look at the current over time (di/dt), then the currents would have to be equal.

Just an editorial note: in case you don't see what I'm driving at, it's not sufficient to make an assertion ("the currents must be equal") without some explanation, even if it makes perfect sense to you. (Which is also a roundabout way of thanking The Electrician and Diver300 for their patience.)

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I think that we have come to an agreement.

You have a slight mathematical error. The current over time is ∫i dt, which has to be zero on a capacitor (give or take leakage current) while di/dt is the rate of change of current, which could be anything.

carbonzit said:

But on negative half-cycles, doesn't the stored charge counteract the line voltage, as shown on the right? Doesn't this lessen the reverse voltage across the LED?

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Diver300 said:

No!

It increases the voltage on the LED, just like putting two AA cells in series, + of one connected to - on the other, gives twice the voltage.

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Just like the classic voltage doubler.

Aaaargh! I can never wrap my head around things that involve polarity. It's like I've got a massive mental block on that simple, simple concept.

Are there any tutorials out there that help explain this, so I know which way is which when looking at a circuit? I'm not talking about how to connect stuff, I know that; I mean this whole business of charges and polarity, particularly concerning capacitors. I always seem to get it wrong.

It is wrong to think of an LED as being a diode that conducts current when forward biased and blocks current (or leaks a small current) when reverse biased because an LED has a very low reverse breakdown voltage of only 5V to 20V when it has avalanche breakdown and conducts as much current as is available.

carbonzit said:

Ah, so. So when those folks were saying "The currents through the capacitor in each direction must be equal", what they really meant is that in order for the LED to light in the forward direction, some current must flow through the capacitor in the reverse direction in order to discharge the capacitor: would you agree

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Another way of saying this is to say that only AC current can flow in a capacitor. There can't be any unidirectional current except briefly (or very small leakage).

And the only way there can be reverse current is if the LED breaks down in the reverse direction; without reverse current, there wouldn't be forward current except for the first cycle or so which would charge up the capacitor. You should try the little experiment I suggested, Put a 1N4004 diode in series with the LED so that forward current can flow through the LED, but there can be no reverse breakdown because the diode's breakdown voltage is high enough to prevent reverse current flow (except for leakage). The LED then won't light up; try it.