Here, let me try! Hopefully I can unravel the mostly semantic difficulties you guys're having in this discussion.
Disclaimer: NEVER play around with house current like I'm going to describe here unless you know what you are doing around electricity and take appropriate safety precautions!!! I don't intend any of this discussion to be practical advice! You have been WARNED! If we start with an AC 60 Herts, 110 volt sinusoidal source, there are several things we can say about it. First, the source voltage will rise from 0 to positive something, back through 0, and to the same amount negative, before completing one cycle of the sine wave. The voltage at the peaks is not 110, but a higher number so that, when you average the amount of voltage across the entire cycle, the number works out to 110, which is the amount of useful voltage available to drive current through a load over time. That's where the ubiquitous factor of 1.4 comes in. It's approximately the ratio of average voltage to peak voltage for half of a cycle. For our example it gives you that 154 volts you heard about. So if we look at the voltage of our source on an oscilloscope, you'll see the familiar sine wave. However, if you check the calibration of the vertical axis, you'll discover that the voltage swings between about 154 volts and -154 volts and back again, following that graceful sine curve. This pulls electrons back and forth in the wire, each half cycle undoing the "work" of the last in terms of moving electrons, but with both half cycles yielding power to the load -- which averages out to 110V times whatever average Amps flow which equals power in watts. If you want to talk about it that way, you can say that there is a "Peak to Peak" voltage of 154 times 2 or 308 V p-p. That's not a particularly valuable number, since there's never 308 volts between any two spots at any given time. It's just the dimension of the total height of the sine wave on the oscilloscope tube. Now, to try to make DC from this. Put a diode in one leg of our source and you can imagine what happens. Look at the output on the oscilloscope and you'll see the voltage spends half it's time on the flat line that our formerly graceful sine wave used to twine itself about. Half it's soothing curve has been cut off as if with scissors! The voltage starts from 0, goes through half a sine wave with a peak voltage of 154 volts, then goes back to 0 and stays there, dead, for the remaining half cycle. If you think about it, obviously only half of the power is available and, in fact, if you calulate or measure the *average* voltage over time, it's been cut in half to 55 volts. It's now a 110v peak to peak signal, if you're honest about it, but now it's pulsating direct current, since electrons are moving in only one direction, and that, only half the time. The surviving half cycle performs just as much work as each half cycle of our original AC signal, but it's pulling the cart by itself. (Ain't anthropomorphism fun?) <ahem> This is called half-wave rectification. The next step is to give our struggling half-wave back it's partner, but get them pulling together in the same direction because we happen to want DC instead of AC. Put what's called a "full wave rectifier" or "bridge rectifier" across the original voltage source, and you'll neatly get both halves of our sine wave back, but now they're *both* going in the same direction. Once again, the 154 volt peak occurs, twice per cycle, but now both times of the same polarity. Just as much energy is available to do actual work as our original sine wave, but instead of the electrons travelling back and forth in the wire, they're driven in surges in one direction only. This is full-wave rectification. So if you look at the oscilloscope display, you'll see the bump, bump, bump, bump, of both half cycles right after another, instead of the bump, pause, bump, pause, of half wave rectified DC. Okay, you probably know that much already. Here's where we start playing with capacitors. Capacitors are basically two sheets of metal separated by some kind of good insulator. Jam electrons into one side, or plate, and they electrostatically repell electrons out the other side. Think of a water filled pressure vessel with a rubber diaphragm in the middle. Pressurize one side with a garden hose and some of the water in the other half is forced out. If you think about it, with clever use of valves, you can actually use this capacitor/pressure vessel thingie to *store* some pressure -- or electrical charge. And that's what we use it for in our circuit. If you put a capacitor accross the output of *EITHER* of our two rectified voltage sources, you'll get exactly the same thing. Each time the voltage swings up to it's peak of 154V, the capacitor will charge up to that same voltage. However, one more detail rears its ugly head here. Without any kind of *LOAD* attached, there's nothing to allow the pressure to bleed away and that capacitor will just stay charged to the full 154V. Now, our little CS generators are basically sipping so little power that they are for all intents and purposes, no load! So that's why it's likely that you'd see close to 154V DC on the output of either a half-wave or a full-wave rectified source. If you wanted to hook up a *BUNCH* of jars in parallel and series to make a *LOT* of CS this way, you could possibly be pulling enough current at some point to pull down that capacitor a little bit between the peaks when it gets recharged. How much so depends on the size of the capacitor. A really small capacitor might actually get pulled down to some lower voltage before being recharged by the next voltage peak, so your average voltage will be something less that 154 but never as low as the 110 we started with -- unless you're trying to blow the fuse by drawing more current than you should be! This is where you'd see a difference in voltage between a half wave and full wave system, because the full wave system recharges that capacitor to the peak voltage twice as often as in the half wave source. So, if the load and the capacitor are the same, then the drop in voltage from our peak of 154V is half as big for a full wave system as for the half wave system. GOT THAT!?? <grin> So here you have it. Rectify your DC and you'll average 55V or 110V for half and full-wave rectifier setups, respectively. If your load is small enough, you can filter either of them with a large enough capacitor that you'll always see the *peak* voltage of 154, or very close to it. Use a heavier load or a smaller capacitor and you'll see your average voltage drop somewhat as it ripples downward between peaks. In this case, the half wave rectifier will get pulled down twice as far as a full-wave bridge would in the same circuit. These relationships apply just the same to a 12 volt wall-wart transformer as to straight line voltage. Just apply your factor of 1.4 to the average voltage you read on your meter to get the peak voltage, and be prepared to use cheaper and smaller capacitors, 'cause they don't have to stand up to as high a voltage. I've left out details like voltage drops accross diodes and resistance in wires and such, but I'm sure you get the idea. <sigh> Thanks for listening, and I hope that helps! Be well, Mike D. (Hoping there's no glaring mistakes in all of this!) [Mike Devour, Citizen, Patriot, Libertarian] [[email protected] ] [Speaking only for myself... ] -- The silver-list is a moderated forum for discussion of colloidal silver. To join or quit silver-list or silver-digest send an e-mail message to: [email protected] -or- [email protected] with the word subscribe or unsubscribe in the SUBJECT line. To post, address your message to: [email protected] Silver-list archive: http://escribe.com/health/thesilverlist/index.html List maintainer: Mike Devour <[email protected]>
