Quote:
Originally Posted by TechniSol
The biggest problems I see with your reasoning is the idea that the cap can charge instantaneously -it can't
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Oh I agree
But if not, then you have a voltage drop across PWM (either pin directly or MOSFET like you say) and the entire power saving idea of PWM is ruined :/
In the extreme case, you might as well replace PWM with a variable resistor and you'll end up with roughly the same voltage drop, only constant instead of pulsing.
Like I said initially: it's all doable, but there is a bunch of "but"s.
Quote:
Originally Posted by TechniSol
Tiny cap by the way, just to smooth out spikes on either end of the PWM cycle waveform
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I suppose I would like to respectfully disagree: smoothing out the edges will not produce any realistic change to the visible pulsing effect. I would say people don't see the edges (which are in harmonics of switching frequency) but they see the switching frequency itself. Smoothing out edges doesn't reduce any energy from switching frequency, and doesn't add any energy to the DC component.
This is why I'd approximate the practical filtering effect by peak-to-peak measurement: if a design doesn't reduce p-p, it doesn't smoothen anything.
The initial talk of classic RC filter was better in this regard: it was leaving DC component intact, but was sinking all higher frequencies into a resistor.
[edit]btw,
Quote:
Originally Posted by TechniSol
continuously passing full rail voltage to your LEDs. Could be kind of embarrassing.
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I would say it's a bad way to think about this: a diode can never have more voltage across it than its LEDvdrop. If you try to connect it to a voltage source higher than that, it will either burn or the voltage source will burn. It's like a short circuit.
The resistor is there to solve it: like you said, current running through a LED is I = (E - LEDvdrop) / R (for E >= LEDvdrop).
Without R (R=0) this formula resolves to:
- infinity for E > LEDvdrop (ie something burns)
- undefined for E = LEDvdrop
- zero otherwise.
So, this resistor is there to decide what the current is (at each duty cycle), not to protect the diode under some duty cycles but not others.
[this uses this simple approximation of a diode having constant voltage drop. The truth is exponential, so not quite like it, but not far either]
[also I would like to apologise for off-topic-ness, but geeky discussions are always fun ;p]