AC sine wave: What does increasing the frequency do?

Discussion in 'Electronic Basics' started by Commander Dave, Nov 26, 2004.

1. John FieldsGuest

---
OK, but I think that was due to the fact that the filament was used as
a gain-changing element, so when the output frequency got down low
enough for the loop time constant to start looking like an appreciable
part of the output signal's period it wasn't capable of operating so
much like a gain control as a modulator.
---
---
OK, both true, _but_ the fact that the filament resistance lags
voltage doesn't mean that there's a phase difference between the
voltage across and the current through the lamp.

All you've got, in essence, is a pot and a fixed resistor in series
being driven by a sinusoidal source, and no matter where you choose to
look _in that circuit_ voltage and current will be precisely in phase.

2. John FieldsGuest

---
Since there are two heating events per cycle, ISTM like it should be
that resistance is higher in the last half of each _half_ cycle than
in the first half, 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.
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.

3. John LarkinGuest

Right, the resistance varies in half-cycles, at 120 Hz, just like the
temperature and light output do.
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.
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.

John

4. John PopelishGuest

Actually, it represents even harmonics, not phase shift

5. John LarkinGuest

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

John

6. John PopelishGuest

I think that is pretty likely, since the zero crossings between
voltage and current still match. If you collect the data, I will do
the analysis for you.

7. John LarkinGuest

This is getting interesting.

Consider an AC line with a sinusoidal voltage, 1 hz for illustration.
Connect a triac dimmer and a resistive load. Set the dimmer *very*
dim, so only tiny time slices get through to the load, one just before
each zero crossing.

Assume the line voltage is sin(t). So it peaks at 0.5 * pi seconds
(positive) and 1.5 * pi (negative). The load current waveform is a
tiny triangle that peaks at just a hair under pi seconds (positive
glitch) and again just before 2*pi (negative glitch).

Is the Fourier fundamental component of the current waveform in phase
with the line voltage? In other words, if you passed the little load
glitches through an ideal w=1 bandpass filter, would the resulting
waveform peak at 0.5*pi seconds, as does the line voltage?

After all, the voltage and current waveform zero crossings "still
match".

John

8. Jack// aniGuest

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

9. John PopelishGuest

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.

But I have proven myself to be full of it, to my own satisfaction.

No even harmonics, and a net phase shift in the fundamental.

I am not sure how to calculate (means stumped)
the effective power factor of that wave.

10. John PopelishGuest

More specifically, Fourier analysis assumes that the chunk of waveform
you use in the analysis is one chunk of endlessly repeating train of
identical chunks. If you pick an actual representative chunk in a
train of repeating chunks, the analysis works. If you pick something
else (or if the waveform does not actually have a repeating pattern)
then you get an analysis of something other than what is real.

11. The PhantomGuest

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)
But this triac load fails to look like an inductive load is some
important respects.

1. An inductor's current is the integral of the applied voltage; it
would smooth the waveform of the applied voltage, not add
discontinuities. An inductor would conduct a sine wave of current in
this case.

2. An inductor (iron cores not allowed here) will never generate
harmonics that are not present in the applied voltage.

3. An inductor's current would halve for a doubling of the frequency
of the applied voltage.

3. An inductor will store energy, and therefore the phenomenon of
resonance can occur.

12. The PhantomGuest

I'm assuming you are posting in a country where the mains frequency
is a nominal 50Hz. During heavy load times, the frequency decreases a
little, but not anywhere as much as 48-50Hz. Maybe 49.9Hz. Then at
night during light load times, the utilities speed up their generators
to make sure that electric clocks see the right average frequency
during a 24 hour period. You can see this happen if you stay up for
24 hours and watch the second hand of a mains operated electric clock.
Set the clock at the beginning of your vigil with some accurate atomic
reference, such as a GPS receiver, or your national frequency
reference (WWV here in the US; on 5, 10, 15, 20 MHz. You may be able
to receive it overseas). Then turn on late night TV and periodically
compare the second hand indication of the mains operated clock to the
atomic reference. The clock will be slow during the day, and catch up
at night. I saw it get nearly a minute behind 20 years ago. I don't
know how good it is these days.

13. The PhantomGuest

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.

14. John FieldsGuest

---
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?
---
---
Sure we are!-)
---
---
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 (R² + (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 R²

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.

15. John PopelishGuest

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

16. John LarkinGuest

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

17. The PhantomGuest

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.

18. John LarkinGuest

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.

JP and the Phantom have both done the analysis.

John

19. The PhantomGuest

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

20. John LarkinGuest

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

John

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