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AC sine wave: What does increasing the frequency do?

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

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  1. John Fields

    John Fields Guest

    ---
    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 Fields

    John Fields Guest

    ---
    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 Larkin

    John Larkin Guest

    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. Actually, it represents even harmonics, not phase shift
     
  5. John Larkin

    John Larkin Guest

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

    John
     
  6. 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 Larkin

    John Larkin Guest

    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// ani

    Jack// ani Guest

    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. 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. 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 Phantom

    The Phantom Guest

    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 Phantom

    The Phantom Guest

    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 Phantom

    The Phantom Guest

    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 Fields

    John Fields Guest

    ---
    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. Thanks for this detailed work. You should produce a tutorial web page
    on this subject.
     
  16. John Larkin

    John Larkin Guest

    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 Phantom

    The Phantom Guest

    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 Larkin

    John Larkin Guest


    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 Phantom

    The Phantom Guest

    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 Larkin

    John Larkin Guest

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

    John
     
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