capacitorcapacitor chargingcommutationdischarge
I was going through the working of class D commutation and the article said:
As soon as the capacitor completely discharges, its polarities will be reversed but due to the presence of diode the reverse discharge is not possible.
Why does the polarity of the capacitor reverse as soon as it completely discharges? Is it because it immediately ( at the speed of light) starts to charge?
Moreover, what does reverse discharge mean?
You had the right idea with a voltage divider, but as you say resistors will always dissipate power. The answer is therefore a capacitive voltage divider. It works just like a resistive voltage divider except that the impedance goes down with frequency. Your frequency is well known and controlled, so a capacitive voltage divider is quite appropriate. Adjust the capacitors so that the peak output voltage is the 200V you want to charge the output cap to, and put a diode to this output cap like you would from any other 200V AC source.
One thing to watch out for though. The power line will occasionally have large voltage spikes. These will make it thru the capacitive divider proportionally just like the intended voltage. This could possibly overcharge the output cap on rare occasions. It might be a good idea to put some kind of clamp circuit accross the cap that turns on when the voltage is a little above the normal value. This could save a few random mysterious field failures years after installation.
Here is what I was talking about in the comment:
You have 220V AC sine coming in, so the peak voltage is 220V * sqrt(2) = 311, and the peak to peak voltage is twice that or 622V. C1 and C2 must be sized to reduce that p-p to 200V, so C2 must have 2.11 times the capacitance of C1.
Think about the circuit with only C1 and C2. The node between these two now is at 200V p-p. Diode D1 pushes the DC bias on the capacitors so that the negative peak is 0V or more, and D2 pushes it so that the positive peak is 200V or less. When C3 is at less than 200V, then the circuit acts like a charge pump and each cycle will add a little charge onto C3. How much depends on the absolute value of the capacitors and the cycle frequency. Once C3 reaches 200V the AC peaks just hit the limits where the diodes conduct but don't actually transfer any current. In other words, once C3 is fully charged, there is no more power drain from the AC line.
Note that while there is no power drain when C3 is fully charged, there is current. However, this current is 90 degrees out of phase with the voltage, which is how no power is tranferred. Technically this presents a poor power factor. However, in general the AC power grid is more inductive than capacitive and power companies actually maintain banks for capacitors to try to even this out. In most cases you're actually helping the power company by offsetting some local inductive load. Also from what you say this is apparently quite low power. A few mA reactive current either way isn't a issue.
That article is crap, and you should probably forget you ever saw it. Besides the fact that it's totally wrong, as you've discovered, it's full of half-truths that cultivate an incorrect understanding of how electrical phenomena actually work.
The plate on the capacitor that attaches to the negative terminal of the battery accepts electrons that the battery is producing.
wrong. Batteries don't produce electrons. In fact, nothing in your circuit produces electrons. As far as physicists have been able to demonstrate, charge is never created nor destroyed. A physicist can tell you how to make an electron by assembling more fundamental particles, but unless your circuit includes a particle accelerator or operates inside a star, there won't be any relevant electron creation or destruction in your circuit. Batteries pump electrons. They don't produce them.
In the first few paragraphs, that article uses the word charge in several senses. An experienced engineer can distinguish the senses by context, but the novice is more likely to confuse them. Go read Bill Beaty to immunize yourself against this misconception.
The bulb will get progressively dimmer and finally go out once the capacitor reaches its capacity.
Not exactly true. The bulb will go out once the voltage across the capacitor is equal to the battery voltage. When this happens, there can be no voltage across the bulb, and thus no current, thus no light. I don't know what they mean by capacity in this sense, but it seems to me that they are phrasing it this way to avoid explaining how capacitors actually work. You might define a capacitor's "capacity" any number of ways, but they haven't defined it at all. This kind of half-thinking isn't going to help you understand once you figure out that the explanation is incomplete.
A capacitor's storage potential, or capacitance, is measured in units called farads.
If you take "capacity" and "storage potential" as synonymous, as anyone would naturally do, this is wrong and self-contradictory. If this were true, then we could re-write the previous statement as
The bulb will get progressively dimmer and finally go out once the capacitor reaches its capacitance
...which is totally bogus. Ordinary capacitors don't change capacitance under normal operation conditions. Again, the problem here is they haven't fully defined the underlying concepts. If capacitance is a capacitor's storage potential, then what is it storing?.
A 1-farad capacitor can store one coulomb (coo-lomb) of charge at 1 volt.
Here, they try to define the stuff that the capacitor is storing, but it hardly makes sense. The problem is if you use words like "capacity", this implies there is some sort of "full" or "at capacity". But, there is no concept for an ideal capacitor. If you push 1C of charge through a 1F capacitor, the capacitor will have a voltage of 1V. For 2C, you get 2V, and 1000C gets you 1000V. You can push 1000C through a 0.5F capacitor also, but it will then be at 2000V. An ideal capacitor is never full, and you can push charge through it forever and it will never be "full". That's not a capacity, it's the ratio of charge to voltage, which is evident in the definition of farad (a farad is a coulomb per volt):
$$ F = \frac{C}{V} $$
Real capacitors, incidentally, have a maximum voltage, which if exceeded will damage or destroy the capacitor. So in this sense, a capacitor can be "full" and can have a "capacity". But, this is not how this article is using the word.
I could probably write an entire article on the problems with that article, but hopefully you get the idea. Please wipe your memory of what that article has said, and seek a more sound explanation.
Best Answer
Here's the circuit in your link:-
Initially SCR2 is triggered to charge the capacitor through the load. Once the capacitor has charged up to the supply voltage SCR2 will turn off when current drops below its holding current.
If SCR1 is then triggered to power the load, the capacitor will discharge through the diode and inductor (which is now connected to V+ through SCR1). During this time the inductor produces an opposing voltage that is proportional to the rate of current change (principle of inductance).
The inductor and capacitor form a tuned circuit, so current rises until the capacitor is completely discharged, then starts to drop. Now the current change is negative so the inductor produces opposite voltage, charging the capacitor up in the opposite direction. In a tuned circuit this cycle would normally continue producing a sine wave, but the diode stops the capacitor from discharging again.
Now that the capacitor has 'negative' voltage on it, the '-' side is more positive than the supply. If SCR2 is again triggered it discharges the capacitor into the load, bypassing SCR1 which now turns off because it has negative voltage across it.
To illustrate the waveforms I simulated part of the circuit in LTspice. This graph shows the voltage across the capacitor, so it goes from positive (charged from V+ to V-) to negative ('reverse' charged by the inductor). In the actual circuit it generated a positive voltage of ~9V above the 10V supply.