AC sine wave: What does increasing the frequency do?

[John Fields]

On Sat, 27 Nov 2004 16:53:47 -0800, John Larkin
<jjlarkin@highSNIPlandTHIStechPLEASEnology.com> wrote:

On Sat, 27 Nov 2004 17:52:36 -0600, John Fields
jfields@austininstruments.com> wrote:

but it still looks resistive because current is
staying precisely in phase with voltage, since where resitance is
gonna be or where it was doesn't matter. What does matter is what's
the resistance right now and what's the voltage across it right now.

Phase shift has to be measured over time. No instantaneous measurement
of a circuit can identify a phase shift, even a circuit with real
capacitors. "Gonna be and where it was" is fundamental to a
time-referenced measurement. What matters is how the current waveform
looks compared to the voltage waveform, and a point measurement isn't
a waveform.

---
I don't know why you keep belaboring this point since I'm not
disagreeing with you about the way a phase measurement has to be made.
After all, I did describe my equipment setup and methodology early-on
in this thread and, if you like, I'll post some scope screen shots of
the tests.

What I'm saying, and what you seem loath to agree with is that with
respect to the circuit under discussion it doesn't matter how the
resistance of the load varies, as long as it stays resistive the
voltage and current through the resistance _must_ be in phase. Do you
disagree?
---

There's no Xl or Xc in the circuit, and without a reactance the
impedance will be entirely resistive with no difference in phase
between E and I.

We're not getting anywhere on this, are we.

---
Sure we are!-)
---

Do you propose that a
triac dimmer, driving a resistive load, runing at 50% conduction
angle, has no current-versus-line-voltage phase shift? Even though all
the load current flows in the last half of each cycle? That seems like
a phase shift to me.

---
In the last half of each half-cycle.

That's a totally different proposition from the one that's being
discussed, which is that of the phase relationship between the
voltage across a resistance varying parametrically with the current
through it, _not_ with the phase relationship between voltage and
current in a load caused by an arbitrarily switched voltage waveform
impressed across a load resistance.

However, using your example and assuming that you mean firing angles
of 90° and 270° when you say a conduction angle of 50%, then consider:

With the TRIAC off and a multitude of instantaneous, coincidental
voltage and current measurements made during that quarter cycle, it
will be seen that there is no voltage across the load and no current
through it at any measurement point, so the phase angle between
voltage and current _must_ be 0°. Now, when the TRIAC fires, the
voltage across the load will be at either the positive or negative
peak of the voltage waveform and, neglecting the sign of the voltage,
current will flow in the load according to


E
I = --- (1)
Z

where


Z = sqrt (R˛ + (Xl-Xc))


Assuming an ideal circuit with no stray inductances or capacitances,
the reactance terms drop out and what we're left with is


Z = sqrt R˛


which further reduces to


Z = R


Now, plugging that into (1) gives us the familiar


E
I = ---
R


which means that the current waveform through the resistance will
track the voltage waveform through the quarter cycle, i.e. they will
be in phase.

This can be verified by making a series of instantaneous, coincidental
voltage and current measurements on the load while the TRIAC is on.
It might even be a good idea to fire the TRIAC a little before 90° and
270° just to be able to zero in on the current peaks and verify that
they're coincident with the voltage peaks.

Finally, since we're not talking about the harmonics generated by the
TRIAC turn-on, since the load is resistive, and since the angle
between current and voltage remains at 0° at any point during the
cycle, I can't see where you think a phase shift is coming from.

--
John Fields

 

[John Popelish]

The Phantom wrote:

On 28 Nov 2004 02:12:29 -0600, The Phantom <phantom@aol.com> wrote:

On Sun, 28 Nov 2004 00:51:59 -0500, John Popelish <jpopelish@rica.net
wrote:

Okay, I created data representing a 32 point per cycle representation
of a cosine wave turned on at the peaks.

If I understand the results, the magnitudes and angles of the even
harmonics are:

harmonic magnitude angle
1st .644 29.19 (lagging)
3rd .32 78.76
5th .112 56.25
7th .112 56.25

The last one is pretty ragged because of only 32 samples per cycle.

Using a continuous sine excitation and triac triggered at peak
voltage, I get:

Harmonic Magnitude Angle
1st .5654 -34.06
3rd .31822 84.35
5th .10602 72.8
7th .1058 73.15
9th .0636 61.18

PF, which is real power/apparent power is .6868

real power computed as integral(i(t)*e(t)) over one cycle.

apparent power is (rms voltage)*(rms current)

Well, I got out the big gun (Mathematica) and got exact results (I
used an excitation of a 1 volt peak sine wave, and a 1 ohm resistor
which is connected (with a perfect triac having no voltage drop!) just
when the excitation sine wave's voltage reaches the positive and
negative peaks, and switches off at the next zero crossing:

Harmonic Exact Magnitude Approx Magnitude Angle
1st SQRT(1/4+1/pi^2) .59272 -32.4816
3rd 1/pi .31831 90.0000
5th 1/(3*pi) .1061 -90.0000
7th 1/(3*pi) .1061 90.0000
9th 1/(5*pi) .06367 -90.0000

After I got the results for PF, I realized that a little thought
will give them to you exactly without a high-powered computer algebra
program. If the 1 ohm resistor were connected to the .707 volt (rms)
source all the time, the current (rms) would be .707 amps. Since it's
connected exactly half the time, the current (rms) is .5 amps. The
apparent power is then .707*.5 (SQRT(.5)*.5).
If the resistor were connected all the time, the current would be
SQRT(.5) and so would the voltage giving a real power of 1/2. But
since it's only connected half the time, the real power is 1/4. PF is
(real power)/(apparent power) which in this case is
(1/4)/(SQRT(.5)*.5) = SQRT(.5) = .707

A search of the web turns up a bunch of sites that don't correctly
describe the modern view of Power Factor. The circuit example under
discussion in this thread shows how a non-linear load, without
reactive components, can create a Power Factor less than unity. The
modern way of thinking is to consider Power Factor to be composed of a
Displacement Factor and a Distortion Factor. The Displacement Factor
is the cosine of the phase shift between the *fundamental* component
of the applied voltage and the *fundamental* component of the current.
In our case, that angle is -32.4816 degrees, and the Displacement
Factor is the cosine of that angle, namely .8436. A number of the web
sites I found say that it is the angle between the voltage and
current, without specifying that it is the fundamental frequency
components of voltage and current that should be used. Some PF meters
apparently just look at zero crossings of voltage and current and take
that to be the Displacement Angle. That is wrong.

The other factor involved is the Distortion Factor, which is the
ratio of the fundamental component of current to the total current.
In our case the rms value of the fundamental is .707 * .59272, so the
DF is SQRT(.5) * SQRT(1/4+1/pi^2) / .5 = .83824.

The Power Factor (PF) is given by Displacement Factor * Distortion
Factor; in our case we get PF = .8436 * .83824 = .7071 which is the
same thing we got from (real power)/(apparent power).

If we had a load that generated (mostly) third harmonic distortion
without shifting the fundamental (a metal-oxide varistor or the older
thyrite can do this), the Power Factor would still be less than unity
because of the Distortion Factor.

Thanks for this detailed work. You should produce a tutorial web page
on this subject.

--
John Popelish

 

[John Larkin]

On 28 Nov 2004 06:47:19 -0600, The Phantom <phantom@aol.com> wrote:

A search of the web turns up a bunch of sites that don't correctly
describe the modern view of Power Factor. The circuit example under
discussion in this thread shows how a non-linear load, without
reactive components, can create a Power Factor less than unity. The
modern way of thinking is to consider Power Factor to be composed of a
Displacement Factor and a Distortion Factor. The Displacement Factor
is the cosine of the phase shift between the *fundamental* component
of the applied voltage and the *fundamental* component of the current.
In our case, that angle is -32.4816 degrees, and the Displacement
Factor is the cosine of that angle, namely .8436. A number of the web
sites I found say that it is the angle between the voltage and
current, without specifying that it is the fundamental frequency
components of voltage and current that should be used. Some PF meters
apparently just look at zero crossings of voltage and current and take
that to be the Displacement Angle. That is wrong.

The other factor involved is the Distortion Factor, which is the
ratio of the fundamental component of current to the total current.
In our case the rms value of the fundamental is .707 * .59272, so the
DF is SQRT(.5) * SQRT(1/4+1/pi^2) / .5 = .83824.

The Power Factor (PF) is given by Displacement Factor * Distortion
Factor; in our case we get PF = .8436 * .83824 = .7071 which is the
same thing we got from (real power)/(apparent power).

I've done a number of electronic power meters, mostly for end-use load
surveys (which used to be popular, but aren't much any more.) I just
defined power factor as

PF = true_power / (trueRMSamps * trueRMSvolts)

averaged over a minute maybe, which allowed zero-crossing-burst triac
controls to produce reasonably consistant data. We did one extensive
study of fast-food restaurants which used zc triac controllers for
their deep-fat friers. Such a load has, by my definition, a low power
factor but no phase shift.

Got away with it, anyhow.

John

 

[The Phantom]

On Sun, 28 Nov 2004 10:44:43 -0600, John Fields
<jfields@austininstruments.com> wrote:

On Sat, 27 Nov 2004 16:53:47 -0800, John Larkin
jjlarkin@highSNIPlandTHIStechPLEASEnology.com> wrote:

On Sat, 27 Nov 2004 17:52:36 -0600, John Fields
jfields@austininstruments.com> wrote:

but it still looks resistive because current is
staying precisely in phase with voltage, since where resitance is
gonna be or where it was doesn't matter. What does matter is what's
the resistance right now and what's the voltage across it right now.

Phase shift has to be measured over time. No instantaneous measurement
of a circuit can identify a phase shift, even a circuit with real
capacitors. "Gonna be and where it was" is fundamental to a
time-referenced measurement. What matters is how the current waveform
looks compared to the voltage waveform, and a point measurement isn't
a waveform.

---
I don't know why you keep belaboring this point since I'm not
disagreeing with you about the way a phase measurement has to be made.
After all, I did describe my equipment setup and methodology early-on
in this thread and, if you like, I'll post some scope screen shots of
the tests.

What I'm saying, and what you seem loath to agree with is that with
respect to the circuit under discussion it doesn't matter how the
resistance of the load varies, as long as it stays resistive the
voltage and current through the resistance _must_ be in phase. Do you
disagree?
---

There's no Xl or Xc in the circuit, and without a reactance the
impedance will be entirely resistive with no difference in phase
between E and I.

We're not getting anywhere on this, are we.

---
Sure we are!-)
---

Do you propose that a
triac dimmer, driving a resistive load, runing at 50% conduction
angle, has no current-versus-line-voltage phase shift? Even though all
the load current flows in the last half of each cycle? That seems like
a phase shift to me.

---
In the last half of each half-cycle.

That's a totally different proposition from the one that's being
discussed, which is that of the phase relationship between the
voltage across a resistance varying parametrically with the current
through it, _not_ with the phase relationship between voltage and
current in a load caused by an arbitrarily switched voltage waveform
impressed across a load resistance.

However, using your example and assuming that you mean firing angles
of 90° and 270° when you say a conduction angle of 50%, then consider:

With the TRIAC off and a multitude of instantaneous, coincidental
voltage and current measurements made during that quarter cycle, it
will be seen that there is no voltage across the load and no current
through it at any measurement point, so the phase angle between
voltage and current _must_ be 0°. Now, when the TRIAC fires, the
voltage across the load will be at either the positive or negative
peak of the voltage waveform and, neglecting the sign of the voltage,
current will flow in the load according to


E
I = --- (1)
Z

where


Z = sqrt (R˛ + (Xl-Xc))


Assuming an ideal circuit with no stray inductances or capacitances,
the reactance terms drop out and what we're left with is


Z = sqrt R˛


which further reduces to


Z = R


Now, plugging that into (1) gives us the familiar


E
I = ---
R


which means that the current waveform through the resistance will
track the voltage waveform through the quarter cycle, i.e. they will
be in phase.

This can be verified by making a series of instantaneous, coincidental
voltage and current measurements on the load while the TRIAC is on.
It might even be a good idea to fire the TRIAC a little before 90° and
270° just to be able to zero in on the current peaks and verify that
they're coincident with the voltage peaks.

Finally, since we're not talking about the harmonics generated by the
TRIAC turn-on, since the load is resistive, and since the angle
between current and voltage remains at 0° at any point during the
cycle, I can't see where you think a phase shift is coming from.

Start up your favorite circuit simulator and create a voltage
source of sin(2*pi*60*t) volts. Apply a load consisting of 2 1N4007
diodes in anti-parallel with a 2 ohm resistor in parallel with the two
diodes, for a total of 3 two-terminal devices in parallel. Run the
simulation and look at the current out of the voltage source. That
current has harmonics and a fundamental. The fundamental component of
the current is *in phase* with the voltage source, because the
resistance of the total load is *only* varying with the current
through it, not with time. I'm assuming we can neglect the parasitic
capacitances in the diodes and the rest of the circuit at 60Hz.

Now with the same voltage source, change the load to a single
resistor, but make the resistance equal sin(2*pi*120*t)+2 (assuming
your simulator will allow that). Now run the simulator and look at
the current out of the source. That current waveform has harmonics
and a fundamental, but the fundamental component is *not* in phase
with the source, *because* the resistance is now varying with time.

That is what is happening with the triac load, but the variation of
the load with time has a discontinuity, which somewhat obscures
things. Even though the current has the same waveshape as the voltage
when the triac is on, it (the current) does not have the same
waveshape as the source (a sinusoid) over the *complete* cycle, and
that causes the fundamental component of the current to be shifted in
phase with respect to the source voltage.

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.

 

[John Larkin]

On Sun, 28 Nov 2004 10:44:43 -0600, John Fields
<jfields@austininstruments.com> wrote:

On Sat, 27 Nov 2004 16:53:47 -0800, John Larkin
jjlarkin@highSNIPlandTHIStechPLEASEnology.com> wrote:

On Sat, 27 Nov 2004 17:52:36 -0600, John Fields
jfields@austininstruments.com> wrote:

but it still looks resistive because current is
staying precisely in phase with voltage, since where resitance is
gonna be or where it was doesn't matter. What does matter is what's
the resistance right now and what's the voltage across it right now.

Phase shift has to be measured over time. No instantaneous measurement
of a circuit can identify a phase shift, even a circuit with real
capacitors. "Gonna be and where it was" is fundamental to a
time-referenced measurement. What matters is how the current waveform
looks compared to the voltage waveform, and a point measurement isn't
a waveform.

---
I don't know why you keep belaboring this point since I'm not
disagreeing with you about the way a phase measurement has to be made.
After all, I did describe my equipment setup and methodology early-on
in this thread and, if you like, I'll post some scope screen shots of
the tests.

What I'm saying, and what you seem loath to agree with is that with
respect to the circuit under discussion it doesn't matter how the
resistance of the load varies, as long as it stays resistive the
voltage and current through the resistance _must_ be in phase. Do you
disagree?

I disagree. I contend that

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.

Finally, since we're not talking about the harmonics generated by the
TRIAC turn-on, since the load is resistive, and since the angle
between current and voltage remains at 0° at any point during the
cycle, I can't see where you think a phase shift is coming from.

JP and the Phantom have both done the analysis.

John

 

[The Phantom]

On Sun, 28 Nov 2004 10:27:32 -0800, John Larkin
<jjlarkin@highlandSNIPtechTHISnologyPLEASE.com> wrote:

On 28 Nov 2004 06:47:19 -0600, The Phantom <phantom@aol.com> wrote:


A search of the web turns up a bunch of sites that don't correctly
describe the modern view of Power Factor. The circuit example under
discussion in this thread shows how a non-linear load, without
reactive components, can create a Power Factor less than unity. The
modern way of thinking is to consider Power Factor to be composed of a
Displacement Factor and a Distortion Factor. The Displacement Factor
is the cosine of the phase shift between the *fundamental* component
of the applied voltage and the *fundamental* component of the current.
In our case, that angle is -32.4816 degrees, and the Displacement
Factor is the cosine of that angle, namely .8436. A number of the web
sites I found say that it is the angle between the voltage and
current, without specifying that it is the fundamental frequency
components of voltage and current that should be used. Some PF meters
apparently just look at zero crossings of voltage and current and take
that to be the Displacement Angle. That is wrong.

The other factor involved is the Distortion Factor, which is the
ratio of the fundamental component of current to the total current.
In our case the rms value of the fundamental is .707 * .59272, so the
DF is SQRT(.5) * SQRT(1/4+1/pi^2) / .5 = .83824.

The Power Factor (PF) is given by Displacement Factor * Distortion
Factor; in our case we get PF = .8436 * .83824 = .7071 which is the
same thing we got from (real power)/(apparent power).


I've done a number of electronic power meters, mostly for end-use load
surveys (which used to be popular, but aren't much any more.) I just
defined power factor as

PF = true_power / (trueRMSamps * trueRMSvolts)

This is, of course, the *fundamental* definition of Power Factor
about which there seems to be no dispute.

averaged over a minute maybe, which allowed zero-crossing-burst triac
controls to produce reasonably consistant data. We did one extensive
study of fast-food restaurants which used zc triac controllers for
their deep-fat friers.

John Popelish (we seem to have a lot of John's in this thread)
convinced himself that the current is phase shifted in this triac load
case. I just posted something to one of John Field's replies to you
to explain why the triac current is phase shifted.

Such a load has, by my definition, a low power
factor but no phase shift.

But now I'm confused; I thought your were arguing *for* phase shift
in the triac circuit. Here, you seem to be saying the opposite.

Didn't you say earlier:

"We're not getting anywhere on this, are we. Do you propose that a
triac dimmer, driving a resistive load, runing at 50% conduction
angle, has no current-versus-line-voltage phase shift? Even though all
the load current flows in the last half of each cycle? That seems like
a phase shift to me."

Got away with it, anyhow.

John

 

[John Larkin]

On 28 Nov 2004 13:09:14 -0600, The Phantom <phantom@aol.com> wrote:

averaged over a minute maybe, which allowed zero-crossing-burst triac
controls to produce reasonably consistant data. We did one extensive
study of fast-food restaurants which used zc triac controllers for
their deep-fat friers.

John Popelish (we seem to have a lot of John's in this thread)
convinced himself that the current is phase shifted in this triac load
case. I just posted something to one of John Field's replies to you
to explain why the triac current is phase shifted.

Such a load has, by my definition, a low power
factor but no phase shift.

But now I'm confused; I thought your were arguing *for* phase shift
in the triac circuit. Here, you seem to be saying the opposite.

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

John

 

[The Phantom]

On Sun, 28 Nov 2004 11:30:24 -0800, John Larkin
<jjlarkin@highlandSNIPtechTHISnologyPLEASE.com> wrote:

On 28 Nov 2004 13:09:14 -0600, The Phantom <phantom@aol.com> wrote:


averaged over a minute maybe, which allowed zero-crossing-burst triac
controls to produce reasonably consistant data. We did one extensive
study of fast-food restaurants which used zc triac controllers for
their deep-fat friers.

John Popelish (we seem to have a lot of John's in this thread)
convinced himself that the current is phase shifted in this triac load
case. I just posted something to one of John Field's replies to you
to explain why the triac current is phase shifted.

Such a load has, by my definition, a low power
factor but no phase shift.

But now I'm confused; I thought your were arguing *for* phase shift
in the triac circuit. Here, you seem to be saying the opposite.


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.

John

 

[John Fields]

On 28 Nov 2004 12:41:06 -0600, The Phantom <phantom@aol.com> wrote:

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.

---
OK. I get it.

Thanks,


--
John Fields

 

[John Popelish]

John Fields wrote:

On 28 Nov 2004 12:41:06 -0600, The Phantom <phantom@aol.com> wrote:

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.

---
OK. I get it.

Thanks,

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.

--
John Popelish

 

[Commander Dave]

"John Popelish" <jpopelish@rica.net> wrote:

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.
--
John Popelish

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

 

[John Popelish]

Commander Dave wrote:

"John Popelish" <jpopelish@rica.net> wrote:
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...

You too?

--
John Popelish

 

[john jardine]

"John Larkin" <jjlarkin@highlandSNIPtechTHISnologyPLEASE.com> wrote in
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

 

[John Larkin]

On Mon, 29 Nov 2004 00:54:10 -0000, "john jardine"
<john@jjdesigns.fsnet.co.uk> wrote:

"John Larkin" <jjlarkin@highlandSNIPtechTHISnologyPLEASE.com> wrote in
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!).

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

 

[peterken]

"John Larkin" <jjlarkin@highlandSNIPtechTHISnologyPLEASE.com> wrote in
message news:falfq05oqo9qp57eesnaqh0rcujf4iopt9@4ax.com...

On Fri, 26 Nov 2004 08:56:34 -0800, John Larkin
jjlarkin@highlandSNIPtechTHISnologyPLEASE.com> wrote:

On Fri, 26 Nov 2004 09:51:32 -0600, "Commander Dave"
cmdr-dave@spamcop.net> wrote:

Thanks for the answer... it is exactly what I needed. I was really
looking to see if increasing the frequency increased power. From what I
gather, while it makes it incompatible with things that run on 60 Hz, it
doesn't change the available power... it just cycles faster.


Right. Resistive loads (heaters, light bulbs) won't care; they'll use
the same current and power independent of frequency (except at the far
extremes.) Reactive loads, like motors and transformers, will behave
differently at different frequencies.


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 . .
| ..
|

 

[John Larkin]

On Mon, 29 Nov 2004 00:54:10 -0000, "john jardine"
<john@jjdesigns.fsnet.co.uk> wrote:

"John Larkin" <jjlarkin@highlandSNIPtechTHISnologyPLEASE.com> wrote in
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

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

John

 

[Don Kelly]

"Khwaj" <almostnormal@yahoo.com> wrote in message
news:10qm8a4hs1foi9b@corp.supernews.com...

Jack// ani wrote:

"peterken" <peter273@hotmail.com> wrote in message
news:<eCHpd.1295$pN1.59451@phobos.telenet-ops.be>...
"Commander Dave" <cmdr-dave@spamcop.net> wrote in message
news:i8Hpd.2990$_6.1280@fe40.usenetserver.com...

changing frequency doesn't change available power
however, household appliances (eg with motors, like vacuum cleaners)
wouldn't be able to handle it, since they're built for 60Hz
60Hz is a choice to avoid flickering in lighting (fluorescent tubes)
and
to minimize losses during transport

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.
---------

Of course, this much variation only applies to small independent generators.
In a grid system, a 2Hz frequency deviation is unacceptable. A change in the
order of 0.05 Hz is a fairly extreme transient .
--
Don Kelly
dhky@peeshaw.ca
remove the urine to answer

 

[Bill Bowden]

"Bob Myers" <nospamplease@address.invalid> wrote in message news:<7iRpd.3521$y45.1511@news.cpqcorp.net>...

"peterken" <peter273@hotmail.com> wrote in message
news:eCHpd.1295$pN1.59451@phobos.telenet-ops.be...

changing frequency doesn't change available power
however, household appliances (eg with motors, like vacuum cleaners)
wouldn't be able to handle it, since they're built for 60Hz
60Hz is a choice to avoid flickering in lighting (fluorescent tubes) and
to minimize losses during transport

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

 

[Don Kelly]

"Bill Bowden" <wrongaddress@att.net> wrote in message
news:ad025737.0411292224.43084056@posting.google.com...

"Bob Myers" <nospamplease@address.invalid> wrote in message
news:<7iRpd.3521$y45.1511@news.cpqcorp.net>...

"peterken" <peter273@hotmail.com> wrote in message
news:eCHpd.1295$pN1.59451@phobos.telenet-ops.be...

changing frequency doesn't change available power
however, household appliances (eg with motors, like vacuum cleaners)
wouldn't be able to handle it, since they're built for 60Hz
60Hz is a choice to avoid flickering in lighting (fluorescent tubes)
and
to minimize losses during transport


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
---------

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..
--
Don Kelly
dhky@peeshaw.ca
remove the urine to answer

 

[Rich Grise]

On Sun, 28 Nov 2004 19:06:35 -0800, John Larkin wrote:

On Mon, 29 Nov 2004 00:54:10 -0000, "john jardine"
....
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

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