Ok, so I'm reading up on Inductors, and to convert ac to dc with an inductor it is as simple as putting it in line with the ac, according to Wikipedia anyway.
I understand the theory, while the positive part of the sign wave charges the inductor, it allows current to travel through the circuit on the other side of the inductor, but when the AC changes direction, the inductor resists this and basically "stops" the flow on the other side. So, sign wave goes in with peaks ranging from 1 to -1 and a square (ish) wave comes out ranging from 1 to 0. I imagine that with the right rating of inductor you could even have sign wave of 1 to -1 one go in and get a relatively flat .5 dc line coming out (half of wave lost to charging, then discharges enough to "overwhelm" ac and continue producing output).
All of this is me guessing. Can someone please explain more about this? Or link me to a thread where this has been explained? Why use 4 diodes instead of an inductor? Is it just because you'd have to have a very specific inductor that might be hard to find, (one perfectly rated for 120 v 60 hz ac) as opposed to more "generic" diodes? or does the switching diodes produce a cleaner signal than the square wave of the inductor.
Also, does voltage drop on the other side of the inductor as it is charging? It would basically act as a resistor in that situation, right? It's used in step up (boost) circuits, for it's discharge property, but couldn't it also be used in a similar way to step down voltage, as in, the voltage while it's charging is used, and then it's switched to bleed off circuit to discharge the extra current. So, two of these circuits working together could reduce voltage, switching between each other with the added benefit of storing energy instead of converting to heat as in a resistor.
Thanks for any help. I also understand I'm basically asking two questions here. Sorry about that.
-Kris
I understand the theory, while the positive part of the sign wave charges the inductor, it allows current to travel through the circuit on the other side of the inductor, but when the AC changes direction, the inductor resists this and basically "stops" the flow on the other side. So, sign wave goes in with peaks ranging from 1 to -1 and a square (ish) wave comes out ranging from 1 to 0. I imagine that with the right rating of inductor you could even have sign wave of 1 to -1 one go in and get a relatively flat .5 dc line coming out (half of wave lost to charging, then discharges enough to "overwhelm" ac and continue producing output).
All of this is me guessing. Can someone please explain more about this? Or link me to a thread where this has been explained? Why use 4 diodes instead of an inductor? Is it just because you'd have to have a very specific inductor that might be hard to find, (one perfectly rated for 120 v 60 hz ac) as opposed to more "generic" diodes? or does the switching diodes produce a cleaner signal than the square wave of the inductor.
Also, does voltage drop on the other side of the inductor as it is charging? It would basically act as a resistor in that situation, right? It's used in step up (boost) circuits, for it's discharge property, but couldn't it also be used in a similar way to step down voltage, as in, the voltage while it's charging is used, and then it's switched to bleed off circuit to discharge the extra current. So, two of these circuits working together could reduce voltage, switching between each other with the added benefit of storing energy instead of converting to heat as in a resistor.
Thanks for any help. I also understand I'm basically asking two questions here. Sorry about that.
-Kris