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NZ-USA Voltage Converter Wiring Diagram or Schematic

B

Barry Lennox

Jan 1, 1970
0
Malcolm, Barry, Terry and Ken,

My issue seems resolved... TBD

My daughter took me to a flea market where I was able to get four NZ
(5vdc - 9vdc) chargers and a rather large 115->230 step-up transformer.
I rewired the step-up for stepdown and it looks like it will work for
my needs. The five items cost me $20 NZD; such a bargin. We all have to
remember to include those flea markets in our problem solving
situations...

I really want to thank all of you again for your input, advice, links,
and offer of equipment loans (Barry, yes I'm in CHC).

Good to hear. Don't suppose you want to get rid of the 115-230vac
transformer when you are done?

Barry
 
B

BibBob5

Jan 1, 1970
0
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
 
K

Ken Taylor

Jan 1, 1970
0
BibBob5 said:
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
 
D

Drifter

Jan 1, 1970
0
Malcolm Moore said:
Hi Bob
Searching the Radio Shack website on the cat # you gave previously
shows that your device is supposed to be capable of 1000W for one
hour, 1600W for 15 mins. Irrespective of the 88V output, the devices
you list below are all low power and wouldn't be expected to cause a
stepdown transformer output voltage to noticably fall.

However, that cat# doesn't seem to still be available. Searching on
travel converters gives a range that includes another 1600W unit cat #
273-1404 that includes the warnings:

"Operates 120VAC heating devices from 240VAC
For irons, lamps, handheld hair dryers
Range 50-1600 watts. Not recommended for products under 50 watts"

Looking at the picture of it shows a device that is quite small
compared to a transformer based unit of that wattage. How big and
heavy is your unit?

My guess is that your Radio Shack unit is a phase control unit (Triac
or SCR) that is not going to work for your use. This would also
explain the 88V reading your meter is giving.

As others have suggested, go and visit DSE or Jaycar.

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.
 
R

Rich Grise

Jan 1, 1970
0
Malcolm Moore <[email protected]> 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.

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
 
K

Keith Williams

Jan 1, 1970
0
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 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
^^^^^^^^^^^
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-butt 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?
 
R

Rich Grise

Jan 1, 1970
0
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!

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.
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".

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


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.

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

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
 
F

Frank Bemelman

Jan 1, 1970
0
Rich Grise said:
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.

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.
 
R

Rich Grise

Jan 1, 1970
0
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.

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
 
C

Clive Tobin

Jan 1, 1970
0
Rich said:
Hey, Guys? Doesn't anybody remember that? Anybody wanna back
me up here?

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!
 
K

keith

Jan 1, 1970
0
I'm not even to power yet - that's RMS voltage.

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.'
And anyway, power is voltage times current.

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.
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.

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

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

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

First you square each of the samples.

Yep. I squared the instantaneous voltage, V=sin(X), to get power. Again
Power is V^2/R (R is a constant here).
Then you add up all of the squares, and divide by the total number of
samples, to get the mean of the squares.

Nope. I summed the power, which is the *square* of the voltage.
Then, you take the square root of that.

Nope. I never take the square root of power. I have no clue why you
insist that you must.
The sum of the squares of a sine wave is, normalized, 1.0.

Did that. That is the integral of the *POWER*. Divide that by the
samples and that is the *average* power.
The sum of the squares of a half-wave rectified sine wave is,
normalized, 0.5.

Nope. The average power of the full wave is .5. The average of tha
half-wave is .25, which is half. ...as god intended.
The square root of 1 is 1.
The square root of .5 is .707 approx.

Sure, but you're taking the square root of power which is nonsense.
And THAT'S JUST VOLTS!

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?
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.

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?
 
J

John Woodgate

Jan 1, 1970
0
I read in sci.electronics.design that Clive Tobin <[email protected]>
wrote (in <[email protected]>) about
'NZ-USA Voltage Converter Wiring Diagram or Schematic', on Wed, 21 Sep
2005:
Operating on 117 volts AC, multiplying by .707 and then subtracting 0.7
volts for diode drop gives... 82 volts!

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.
 
R

Rich Grise

Jan 1, 1970
0
I read in sci.electronics.design that Clive Tobin <[email protected]>
wrote (in <[email protected]>) 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.

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
 
R

Rich Grise

Jan 1, 1970
0
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.'

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
 
K

Keith Williams

Jan 1, 1970
0
I didn't square volts to get power. I squared volts to get volts squared,

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

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?
 
R

Rich Grise

Jan 1, 1970
0
...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?

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
 
K

keith

Jan 1, 1970
0
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 don't remember that here. But who knows?
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.

Not "they". Me! ;-)
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.

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

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

Rich Grise

Jan 1, 1970
0
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".

Guess I forgot the </sarcasm> tag. :)

Thanks!
Rich
 
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