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AC sine wave: What does increasing the frequency do?

T

The Phantom

Jan 1, 1970
0
No, the fat friers used zero-crossing, multi-cycle burst triac
controllers, not phase control like a dimmer.

Of course; I didn't pick up on the zc part.
 
J

John Fields

Jan 1, 1970
0
In the case of the filament, its resistance varies with time (as
well as with current), and this causes the fundamental component of
the current to be shifted (slightly) in phase. If there weren't the
time delay in the heating of the filament, things would be like the
two diode and resistor load above, and there would be no phase shift
of the fundamental component of the current.
 
J

John Popelish

Jan 1, 1970
0
I love this newsgroup because a simple, one sentence question can
generate days of technical exchange that teaches quite a few of us
things we knew but didn't realize. :)
 
T

The Phantom

Jan 1, 1970
0
I love this newsgroup because a simple, one sentence question can
generate days of technical exchange that teaches quite a few of us
things we knew but didn't realize. :)

Yes. And all without any name-calling and foul language. Much
more pleasant.

John Larkin brings up some more interesting points that deserve a
new thread. I'm going to have to think about them and perhaps in a
day or so, start a thread. He said:

"a. For a sinusoidal source, a time-varying resistive load can have a
load current with a non-zero fundamental phase shift, hence a reactive
load component. This load component can be expressed as an equivalent
inductance or capacitance.

b. For a sinusoidal source, a time-varying reactive load can have a
load current with a non-quadrature phase shift, hence a real load
component. This real component can be expressed as a positive or
negative equivalent resistance. This is why a varicap can be used as a
parametric amplifier.


In case a, it takes no power to vary the resistance (as say moving a
pot wiper or switching resistors in or out) because the synthesized
reactance doesn't dissipate power. In case b, power must be involved
in varying the reactance (spinning the shaft of a variable cap, or
pumping a varactor) because we're synthesizing a real resistance."
 
C

Commander Dave

Jan 1, 1970
0
John Popelish said:
I love this newsgroup because a simple, one sentence question can
generate days of technical exchange that teaches quite a few of us
things we knew but didn't realize. :)

It amazes me that my simple question on power spawned this much
discussion! I followed a little of it, but for the most part it made me
realize how much I have to learn... :)

Cheers!
-Commander Dave
 
J

John Popelish

Jan 1, 1970
0
The said:
Yes. And all without any name-calling and foul language. Much
more pleasant.

John Larkin brings up some more interesting points that deserve a
new thread. I'm going to have to think about them and perhaps in a
day or so, start a thread. He said:

"a. For a sinusoidal source, a time-varying resistive load can have a
load current with a non-zero fundamental phase shift, hence a reactive
load component. This load component can be expressed as an equivalent
inductance or capacitance.

b. For a sinusoidal source, a time-varying reactive load can have a
load current with a non-quadrature phase shift, hence a real load
component. This real component can be expressed as a positive or
negative equivalent resistance. This is why a varicap can be used as a
parametric amplifier.

In case a, it takes no power to vary the resistance (as say moving a
pot wiper or switching resistors in or out) because the synthesized
reactance doesn't dissipate power. In case b, power must be involved
in varying the reactance (spinning the shaft of a variable cap, or
pumping a varactor) because we're synthesizing a real resistance."

The wheels are turning.
 
J

John Popelish

Jan 1, 1970
0
It amazes me that my simple question on power spawned this much
discussion! I followed a little of it, but for the most part it made me
realize how much I have to learn... :)

You too?
 
J

john jardine

Jan 1, 1970
0
message >
......
a. For a sinusoidal source, a time-varying resistive load can have a
load current with a non-zero fundamental phase shift, hence a reactive
load component. This load component can be expressed as an equivalent
inductance or capacitance.

b. For a sinusoidal source, a time-varying reactive load can have a
load current with a non-quadrature phase shift, hence a real load
component. This real component can be expressed as a positive or
negative equivalent resistance. This is why a varicap can be used as a
parametric amplifier.


In case a, it takes no power to vary the resistance (as say moving a
pot wiper or switching resistors in or out) because the synthesized
reactance doesn't dissipate power. In case b, power must be involved
in varying the reactance (spinning the shaft of a variable cap, or
pumping a varactor) because we're synthesizing a real resistance.

Also interesting is that, in case a, since we can shift the
fundamental but can't shift the zero crossings, we must also generate
harmonics. There's probably something similar in case b.

I'm not trying so much to win an argument as I am marvelling over a
few things I hadn't given a lot of thought to before. There's some
sort of neat duality going on here. I'm especially impressed by the
requirement to generate harmonics to reconcile the fundamental phase
shift with the zero crossings.

The classes of gyrators here, has made for an interesting and
*understandable* thread.
Sometime this kind of stuff will be written up in depth, (maybe already
has!). I'll bet that during the process 'understanding' will be #1 item to
fall by the wayside :)
regards
john
 
J

John Larkin

Jan 1, 1970
0
message >
.....

The classes of gyrators here, has made for an interesting and
*understandable* thread.
Sometime this kind of stuff will be written up in depth, (maybe already
has!).

Tons of theoretical work has been done on time-varying capacitances,
mostly in the 60's and such when two-terminal devices (varactors,
tunnel diodes, step-recovery diodes) were the rage.

I'll bet that during the process 'understanding' will be #1 item to
fall by the wayside :)
regards
john


I like to try to avoid equations until I can really feel what's going
on. Being able to do the math doesn't mean you understand it, just
that you can push some symbols around. This is risky of course,
because instincts are often wrong about stuff like this. But the guys
who just do the math can make ghastly blunders, too, and they
sometimes don't have the instincts to recognize an absurd result when
they see it.

Things like Fourier transforms can be visualized and sort of done by
inspection, but not many EE courses try to track that along with the
math.

To a creature that was sufficiently intelligent, everything would be
intuitively obvious.

John
 
P

peterken

Jan 1, 1970
0
John Larkin said:
Which brings up the concept that an incandescent lamp appears to have
a capacitive component of impedance, which is itself a function of
frequency.

John

'scuse my ignorance, but doesn't an incandescent lamp behave more like say a
triac or a set of zeners after the startup cycle?
this means, the gas ignites at a specific voltage (think it was about 80V
somewhere) thus lowering the lamps impedance from that point on.
that's also the reason of the coil in the circuit, the impedance reduces
lamps current, it allows the lamps voltage to drop to the burning point.
the gas stays ignited until again a specific voltage where it stops
conducting, and re-ignites again at a specific voltage the other half of the
cycle.
only other thing in the circuit is a coil, so i don't see the "capacitor"
anywhere
(view drawing in notepad using fixed font)

|mains
| . .
| . .
|. .
.-----------.------------.------
| . .
| . .
| . .
|
| ..
| . .
| . .
| . .
|....----- .....-------.......--
| . .
|lamp . .
|current . .
| ..
|
 
R

Roger Johansson

Jan 1, 1970
0
peterken said:
'scuse my ignorance, but doesn't an incandescent lamp behave more like
say a triac or a set of zeners after the startup cycle?
this means, the gas ignites at a specific voltage (think it was about
80V somewhere) thus lowering the lamps impedance from that point on.

What you are talking about is not an incandescent lamp, it is a
fluoroscent lamp.

An incandescent lamp has a thin wire inside which glows, it is the most
common lamp in the world. But fluoroscents with their higher efficiency
are taking increasing parts of the market.

It is difficult to keep all these technical terms apart in english if you
are not born in an english-speaking country.
 
P

peterken

Jan 1, 1970
0
Roger Johansson said:
What you are talking about is not an incandescent lamp, it is a
fluoroscent lamp.

An incandescent lamp has a thin wire inside which glows, it is the most
common lamp in the world. But fluoroscents with their higher efficiency
are taking increasing parts of the market.

It is difficult to keep all these technical terms apart in english if you
are not born in an english-speaking country.

oops, sorry, my mistake :-(
indeed I'm dutch, thereof the mistake
 
K

Khwaj

Jan 1, 1970
0
Jack// ani said:
Oh yes, power has nothing to do with frequency. But I'm in confusion,
during peak power consumption hours; it is observed that there is
little bit decrease in frequency, it's between 48-50Hz. Why it happens
so if power is independent of frequency??

Thanks
This is a mechanical problem. At the most basic level the generators are
unable to maintain their rpm while heavily loaded.
 
J

John Larkin

Jan 1, 1970
0
message >
.....

The classes of gyrators here, has made for an interesting and
*understandable* thread.
Sometime this kind of stuff will be written up in depth, (maybe already
has!). I'll bet that during the process 'understanding' will be #1 item to
fall by the wayside :)
regards
john

So, could a motor-driven variable capacitor, paralleled with an
inductor, become an oscillator? Seems like it.

John
 
D

Don Kelly

Jan 1, 1970
0
Khwaj said:
This is a mechanical problem. At the most basic level the generators are
unable to maintain their rpm while heavily loaded.
 
B

Bill Bowden

Jan 1, 1970
0
Bob Myers said:
Right idea, but the wrong compromise. At the time the
power-line frequency was standardized, flickering fluorescent
tubes weren't a concern (and incandescents don't flicker,
even on the original 24 Hz standard). The choice of 50
or 60 Hz was a compromise between long-distance losses
and the size (and cost) of the magnetics (transformers and
such) required to efficiently deal with the current. (And so
the much higher frequency standard - 400 Hz - for aviation
AC; long-distance losses obviously weren't an issue there,
but you couldn't have bulky transformers at all.)

Bob M.

Wasn't it Tesla that proposed the 60 Hz. standard?
And probably because there are 60 minutes in an hour,
and 60 seconds in a minute, and therefore 60 cycles
in a second, and 60 was high enough to avoid flicker,
and maybe some other reasons.

What was the reasoning for 50 Hz, other than
slightly better transmission efficiency?

-Bill
 
D

Don Kelly

Jan 1, 1970
0
Bill Bowden said:
Wasn't it Tesla that proposed the 60 Hz. standard?
And probably because there are 60 minutes in an hour,
and 60 seconds in a minute, and therefore 60 cycles
in a second, and 60 was high enough to avoid flicker,
and maybe some other reasons.

What was the reasoning for 50 Hz, other than
slightly better transmission efficiency?

-Bill
---------
The difference between 50 and 60 Hz will make some gain in a reduction of
iron in machines at 60 Hz and, on the other side, some increase in
transmission capability (not necessarily efficiency) at 50 Hz. However, it
appears that the areas which originally had longer transmission distances
went to 60 Hz. so where's the logic.
Probably a choice of "the Brits chose 50Hz so we will choose 60Hz" (or the
opposite with Yanks substituted for Brits ).

As for flicker at 25 Hz ( Not 24) with incandescents. It exists and can be
noticed. I have seen it. Subtle but there- somewhat similar to a computer
monitor refresh rate of 55 to 60Hz. It can be annoying if you are not used
to it..
 
R

Rich Grise

Jan 1, 1970
0
Tons of theoretical work has been done on time-varying capacitances,
mostly in the 60's and such when two-terminal devices (varactors, tunnel
diodes, step-recovery diodes) were the rage.


I like to try to avoid equations until I can really feel what's going on.
Being able to do the math doesn't mean you understand it, just that you
can push some symbols around. This is risky of course, because instincts
are often wrong about stuff like this. But the guys who just do the math
can make ghastly blunders, too, and they sometimes don't have the
instincts to recognize an absurd result when they see it.

Things like Fourier transforms can be visualized and sort of done by
inspection, but not many EE courses try to track that along with the math.

To a creature that was sufficiently intelligent, everything would be
intuitively obvious.

About time somebody noticed! ;-)

Actually, I've been following the thread, and about the only contribution
I can make is that I was visualizing the water-hose model of current,
where the lightbulb goes "OOF!!" when it gets hot, stopping up the
current, but when it cools, it goes, "Aaaah!", and lets the current flow
through again.

So you see why I didn't try to contribute to the actual science of the
thing! ;-)

Thanks!
Rich
 
R

Rich Grise

Jan 1, 1970
0
It amazes me that my simple question on power spawned this much
discussion! I followed a little of it, but for the most part it made me
realize how much I have to learn... :)

Oh, this is nothing compared to the thread I instigated a couple of years
ago about running a 120V bulb off 240V mains by just putting a diode in
series!

Cheers!
Rich
 
R

Rich Grise

Jan 1, 1970
0
So, if one did a Fourier analysis of this current waveform, the
fundamental current would be in phase with the voltage?

I don't even know what a Fourier analysis _is_ other than a way of
translating a data set from the time domain to the frequency domain
(transforming?), but my "gut-feeling" is that the zero-crossings would be
identical _of the actual source waves_ - the fundamental would be out of
phase, in the plot of the Fourier-transformed resolved fundamental, but
the zero-crossing would be brought back into sync by the harmonics after
you added them back together.

It's like a sine wave with the half-cycles tilted to one side.

But wouldn't that act more inductive?

Thanks,
Rich
 
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