Re: NZ-USA Voltage Converter Wiring Diagram or Schematic

In article <pan.2005.09.21.16.08.14.518639@xxxxxxxxxxx>,
rich@xxxxxxxxxxx says...
> On Wed, 21 Sep 2005 10:11:01 -0500, Drifter wrote:
> > Malcolm Moore <abor1953needle@xxxxxxxxxxxxxxxxx> tossed in:
>
> >>"Operates 120VAC heating devices from 240VAC
> >>For irons, lamps, handheld hair dryers
> >>Range 50-1600 watts. Not recommended for products under 50 watts"
> >>
> ...
> > 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.
>
> 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.

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

> 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!!!!"

You were correct.

> Problem is, it doesn't work that way. I wasn't convinced until
> I did it on a spread***. 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
^^^^^^^^^^^
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).

> get RMS, right? Sure enough, the RMS is 1.O.

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 <= 0
Cell B1 <= =SIN(A1*2*PI()/360)
Cell C1 <= =B1*B1
Cell D1 <= 0
Cell G1 <= =SIN(A1*2*PI()/360)
Cell H1 <= =G1*G1
Cell I1 <= 0

2)Now in row 2 enter these equations:
Cell A2 <= =A1+1
Cell B2 <= =SIN(A2*2*PI()/360)
Cell C2 <= =B2*B2
Cell D2 <= D1+C2
Cell G2 <= =SIN(A2*2*PI()/360)
Cell H2 <= =G2*G2
Cell I2 <= I1+H2

3)Copy row 2 and paste to rows 3 thru 360.
4)Zero cell G181
5)Copy cell G181 into cells G182 thru G360
6)Enter the following equations into the following cells

D361<= =D360/360
I361<= =I360/360

You'll find that D361 contains .5 and I361 contains .25, as you first
expected. I don't know what you did before, but if you turn on a
heater for half the time,it will put out half the heat.

> 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!!

Nope. You'll find that D361 contains .5 and I361 contains .25, as you
first expected (*NOT* .707). I don't know what you did before, but if
you turn on a heater for half the time,it will put out half the heat.

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

You may have learned what the square root of .5 was, but that's all.
Power/2 is power divided by two. ;-)

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

Haw! Haw! We're not sure what you learned though. ;-)

> 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! :-)

Would your answer be any different if the hair dryer was on for one
cycle and off the next? Half hour?

--
Keith
.