NZ-USA Voltage Converter Wiring Diagram or Schematic

Sorry Barry,

Since I'm leaving for PA/USA this Wednesday, i'm setting up the converted 115/230 for daughter/son-in-law to use for I don't know how long. BTW: the converted unit is great and handles all the loads at once with no sweat.

Thanks all; and I've got to say NZ is great for those living hear and anyone that might visit NZ.

Over and out.

Bob

Glad you enjoyed it - give us a shout next time you're coming down this way! :-)

Cheers.

Ken

Malcolm Moore tossed in:

Malcolm: It may be even cheaper: maybe a 14A DIODE. That'd cut the power in half (sorta) for a hair dryer, but would produce disastrous harmonics for some electronics devices. A TRIAC would be even worse, as it wouldn't trigger 'til near the voltage peak to produce 1/2 power, so you'd have 0V > 240V instantaneous change. Plugging a wall-wart into it would probably fry the wall-wart in short order. Notice they only recommend it for heavily resistive loads? I'd personally run away from any product like that! They can't sell very many of 'em.

...

It could be just as disastrous if you think you're getting only half the power. I went around and around and around on this a year or two ago; I'd seen the diode trick and I have a soldering iron that I put a diode in series with the line cord so I'd get "half power", but it was explained to me with almost infinite patience on the part of the group that if you half-wave rectify a sine wave, you don't get half the RMS power - you get 70.7% of the RMS power of the sine wave. In other words, if you have a 1000W,

120V hair dryer, and you give it half-wave rectified 240V, it will dissipate 1400W RMS.

They tried to explain it with calculus and crap, but I was adamant: "The current's only flowing HALF the TIME - You can only _GET_ half the power!!!!"

Problem is, it doesn't work that way. I wasn't convinced until I did it on a spreadsheet. I made a chart: x = 0 to 359 degrees, y = sin(x), z = y^2, w = avg(z), p = sqrt(w), that's how you get RMS, right? Sure enough, the RMS is 1.O.

Then, I cleared out all of the y terms that would have been negative (i.e., half-wave rectified), and the answer turned out to be .707!!

Turns out, that's right - that's the square root of .5. I learned something that day.

(yes, what's-his-butt is sure to chime in "Yeah, and that's the last time you've learned anything! Haw! Haw!" - never mind)

So, don't half-wave rectify 240VRMS and expect to get .5 of the RMS power - it's .707. Don't trust me - do the math! :-)

Cheers! Rich

RMS Power?? You're trolling again, right?

You were correct.

^^^^^^^^^^^ Reading back through, *THERE IS* your problem. Power is voltage squared, not the square root of the voltage squared! Your 'Y' is voltage (sinX), and 'Y' is the power (SinX^2).

There is no such thing as "RMS Power". Well, maybe there is, but it makes no sense. Average power is what you're "calculating".

I just did exactly what you proposed. Do the following in Excel:

Column A = degrees (make it difficult, if you must) Column B = SIN(A) Column C = C^2 Column D = Integral(C) (running sum of C) Column G = SIN(A) for 0->179 degrees, 0 for 180->359 degrees) Column H = G^2 Column I = Integral (I) (running sum of I)

To make it easy;

1)enter the following equations in row 1: Cell A1

I'm not even to power yet - that's RMS voltage.

And anyway, power is voltage times current. That applies to DC or to sine waves. But the RMS of a half-wave rectified sine wave is _not_ half the RMS of a sine wave.

That's OK - I'm still calculating voltage.

You did the math wrong.

First you square each of the samples.

Then you add up all of the squares, and divide by the total number of samples, to get the mean of the squares.

Then, you take the square root of that.

The sum of the squares of a sine wave is, normalized, 1.0.

The sum of the squares of a half-wave rectified sine wave is, normalized, 0.5.

The square root of 1 is 1. The square root of .5 is .707 approx.

And THAT'S JUST VOLTS!

OK, try it. Get a 1N5404 diode or so, a 120V, 100W light bulb, and put the diode and bulb in series across the 240 mains, and see if you still get 100W of light.

Thanks! Rich

I might toss in the fact that the later generation of 35mm slide projectors put a heavy duty diode in series with the lamp. The lamp is rated at 82 volts.

Operating on 117 volts AC, multiplying by .707 and then subtracting 0.7 volts for diode drop gives... 82 volts!

Well, somebody convinced me that using a diode doesn't halve the applied power. Maybe I was the butt of a big "get the dork" gag.

Hey, Guys? Doesn't anybody remember that? Anybody wanna back me up here?

Thanks, Rich

Nope! You've already been through voltage (V=sinX), and then squared it for power (we are talking a resistive load here). I have no clue why you took the square root of power.'

You'd better try that again. How about P=V^2/R, which is *exactly* what you have here. You have the instantaneous voltage (v=sinX), thus the instantaneous power (P=(sinX)^2). Integrate that.

No, but the power is. There is no "RMS" in power. The square has already been taken (V*I).

No, you're not. You're calculating the square-root of power, for some unknown reason.

Nope.

Yep. I squared the instantaneous voltage, V=sin(X), to get power. Again Power is V^2/R (R is a constant here).

Nope. I summed the power, which is the *square* of the voltage.

Nope. I never take the square root of power. I have no clue why you insist that you must.

Did that. That is the integral of the *POWER*. Divide that by the samples and that is the *average* power.

Nope. The average power of the full wave is .5. The average of tha half-wave is .25, which is half. ...as god intended.

Sure, but you're taking the square root of power which is nonsense.

Nope. *POWER*. You squared the volts already (that means you're dealing with *POWER*). Let me repeat; THERE IS NO SUCH THING AS RMS POWER. Got it?

I'm not going to try, because I'm an engineer and can figure out the simple stuff without such lunacy. Try learning what the difference is between voltage and power is.

Again, do you think it matters if one half a cycle is dropped, or one of every two? How about two of four? Half-hour of an hour?

--
  Keith

"Rich Grise" schreef in bericht news: snipped-for-privacy@example.net...

A purely resistive soldering iron of 120V/100W would consume

400W on 240. With a diode in series, on 240V, it will consume 200W. Don't need a spreadsheet for that.

A 100W bulb would not give 200W, but perhaps 180W. Not for very long of course.

--
Thanks, Frank.
(remove \'q\' and \'.invalid\' when replying by email)

Then where were you when they were convincing me that the power isn't halved?

I guess I've been bagged. Oh, well.

Thanks, Rich

I didn't square volts to get power. I squared volts to get volts squared, which I added up and divided by the number of samples to get the mean of the squares, and then I took the square root of that. Isn't that what "Root-Mean-Square" means? Take the square root of the mean of the squares?

Thanks, Rich

...which *is* power ("instantaneous", but power). P=V^2/R, remember.

Ok, you have RMS *VOLTAGE*. Now square that and you have the power difference.

Again, if a heater is on for one cycle and off for one, is the energy dissipated .707 times that if it was on for two cycles? One second out of two? Why would a half-cycle be different?

--
   Keith

Damfino. I just know that a year or so ago, a gang of folks were throwing calculus at me to prove that it _is_ different, I guess I got dazzled by their bullshit.

I'll just say, I thought it was one answer, then they talked me into changing my answer, now they want me to change it back.

I'll just let it be whatever the majority says. That's how they do "science" these days anyway, right? By consensus? I'm tired of this puzzle anyway.

In fact, a year or two ago, someone asked, "I only have a 240V feed to my pump house, but 240V bulbs are hard to find - can I use just one wire and the dirt?" I said, "Just half-wave rectify the 240, and you can use a 120V bulb," and about a half-dozen people jumped on me and read me the riot act. "NO! He'll blow the bulb! That's 140watts! etc,etc, etc."

I give up.

Thanks, Rich

I read in sci.electronics.design that Clive Tobin wrote (in ) about 'NZ-USA Voltage Converter Wiring Diagram or Schematic', on Wed, 21 Sep

2005:

Why would you multiply the RMS voltage by 1/sqrt(2)?

But this is where the confusion arises; the power is halved, but the voltage isn't.

--
Regards, John Woodgate, OOO - Own Opinions Only.
If everything has been designed, a god designed evolution by natural selection.
http://www.jmwa.demon.co.uk Also see http://www.isce.org.uk

I don't remember that here. But who knows?

Not "they". Me! ;-)

Oh, no! Science by majority rule is no different than religion! In fact it's worse, becasue it is hiding behind science. ...rather like "environmentalism".

I wouldn't recommend such things, but you _were_ right. The voltage across the fixture may have exceeded specs, but the power wouldn't have.

Never give up on the facts. Like opinions (technically, value judgements), facts are never wrong.

--
  Keith

Guess I forgot the tag. :-)

Thanks! Rich

Nope!

-- Keith