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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. I am studying basic electronics and was thinking about AC current. In
    simplistic terms, household electricity in the USA is around 120 VAC
    (rms) at 60 Hz, which looks like a sine wave. I know that if the voltage
    increases, the amplitude of the waveform increases and you have more
    power available. What happens if you increase the frequency but the
    amplitude remains the same? Does power increase or stay the same? What
    effects does this have on AC in theoretical terms?

    I don't think the question has any practical application to my studies,
    but it was something I just can't seem to work out. Anyone care to
    enlighten me? A general answer would be fine.

    -Commander Dave
  2. peterken

    peterken Guest

    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
  3. 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.

  4. John Larkin

    John Larkin Guest

    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.

    But your electric meter will make substantial errors at different

  5. More correctly, the higher the voltage, the less current a given
    amount of power requires. Power is volts times amperes.
    The frequency is not inherently related to power. It is a practical
    matter of losses in different transmission components and it relates
    to things like synchronous or induction motor rotational speed.
  6. John Larkin

    John Larkin Guest

    Which brings up the concept that an incandescent lamp appears to have
    a capacitive component of impedance, which is itself a function of

  7. John Fields

    John Fields Guest

  8. peterken

    peterken Guest

    " ...that an incandescent lamp appears to have a capacitive component of
    impedance... "

    don't think so....
    the resistive part are the filaments, used at startup
    a starter (bimetallic switch) is in series with both filaments
    the starter is open at startup, closing if voltage over the lamp is high
    thus filaments glow (preheat gas)
    together with the (LARGE) coil and the starter, a voltage spike is generated
    to ignite the gas when starter opens again
    from the moment the gas gets ignited, the resistive part of the coil lowers
    the voltage over the lamp, so starter doesn't close again
    (when the lamp is ignited, it's impedance drops)

    as far as i see it, the large coil makes the load of an incandescent lamp
    more inductive
    there is however a capacitor connected to power leads to compensate the
    power factor again from inductive to resistive
  9. John Larkin

    John Larkin Guest

    The filament has a substantial 120 Hz temperature cycle (you can hear
    it with a photocell) and the tungsten has a positive TC. So the
    resistance varies with time. The thermal lag results in the filament
    resistance peaking later than the voltage peak. So the current leads
    the voltage, which looks like a capacitive component.

    There are also harmonics in the current, for the same reasons. GR once
    made a line-voltage regulator that used a motorized variac; the
    voltage sensor was an incandescent bulb, and they sensed the second
    harmonic current (somehow) to servo on.

  10. Bob Myers

    Bob Myers Guest

    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.
  11. John Larkin

    John Larkin Guest

    "incandescent" <> "fluorescent"

  12. John Fields

    John Fields Guest

    Since there's no energy storage in the form of anything other than the
    incidental capacitance and inductance of the filament, I don't see how
    that can happen. That is, whether the resistance is parametric or
    not, it's still just resistance and the current which will be forced
    through the filament will remain in phase with the voltage forcing it

    Seems to me it would be akin to a simple resistive divider where one
    of the resistors is variable, like this:


    Since there's no reactive term in there, then the total impedance of
    the string is simply the resistance, R1+R2, and E2 will always be
    equal to

    E2 = --------

    for any instantaneous value of E1 and RV1 and any value of R2.

    To check, I did this:

    240RMS>----+-----> TO SCOPE VERT A
    +-----> TO SCOPE VERT B
    240RMS>----+-----> TO SCOPE GND

    The lamp was a 120V 25W incandescent, the resistor was 576 ohms worth
    of wirewounds in a Clarostat power decade resistor box, and the scope
    was an HP 54602B. I found a phase shift of about +/- 1.1° max which,
    since it varied randomly about zero seemed to me like it might be
    quantization noise.

    But, there was the inductance of the decade box to consider, so in
    order to rule it out I measured it and it came out to about 6mH, which
    comes out to an Xl of 2.2 ohms at 60Hz, so the angle due to the
    reactance of the box comes out to 0.109° which, being an order of
    magnitude smaller than what the scope measured, puts it way down in
    the noise.
  13. John Larkin

    John Larkin Guest

    But the resistance in question is time-varying at 120 Hz. And phase
    shift is not determined by an instantaneous measurement.

    How did you measure the phase shift? Looking at the zero crossings?
    They will obviously *not* be shifted by a time-varying filament

    The effect is not large; running the lamp at roughly half power will
    further reduce the apparent phase shift, as thermal radiation drops
    severely as voltage falls.

  14. john jardine

    john jardine Guest

    There's even odder things out there.
    First scratched my head over this when designing with Triacs. Only recently
    came across the (messy) analysis ...

    A _/
    .---o o----------o/ o-----------.
    | Triac |
    Switch .-.
    AC Mains | |
    | |
    | '-'
    | B | Rload
    '--o o--------------------------'

    Triac or switch is run at an arbitrary phase angle.
    Looking into points A and B one sees not a resistance but an inductive
    (worst at 90degrees fire angle and not a jot of energy storage anywhere)
  15. Yep. Power factor (energy between the source and load that is not
    consumed by the load) can be produced by storage at the load (e.g..
    capacitive or inductive effect in parallel with the load) or by the
    load generating harmonic currents that convert source energy to
    harmonic energy and send it back toward the source.
  16. John Fields

    John Fields Guest

    Yeah, poor choice of words. You can measure the phase shift by
    measuring the time from the zero crossing of one signal to the zero
    crossing of the other, measuring the direction of crossing, measuring
    which one crossed "first", and all the rest of it...

    Excuse me???

    You stated that there would be a phase shift between the current
    through the lamp and the voltage across it, and my measurement, which
    measured the difference in time between the voltage across the lamp
    and the current through it showed that the voltage and current
    waveforms were congruent, refuting your earlier statement which you
    now seem to be abandoning.

    Perhaps we're talking apples and oranges here, but I'm of the opinion
    that if there's a difference in phase between current and voltage
    their zero crossings will occur at different times.
  17. John Larkin

    John Larkin Guest

    Sure. If the current waveform is "lopsided" in time compared to the
    voltage waveform, the current's Fourier fundamental component is phase
    shifted. For SCR/triac dimmers, the current happens late in the cycle,
    so current lags voltage, as it does for a true inductor.

    Some textbooks flat declare that power factor is undefined for
    non-sinusoidal loads or for unbalanced 3-phase loads.

  18. John Larkin

    John Larkin Guest

    The zero crossings obviously can't move, since there can be no current
    anywhere in this setup when the line voltage is zero. But the time of
    peak current is not simultaneous with the voltage peak, because the
    filament resistance varies with time and doesn't peak at the voltage
    peak. This is not a paradox, because harmonics are present to make
    everything work out. If you were to measure the Fourier fundamental
    component of current, *that* would lag the voltage.

    You can google "incandescent filament harmonics" and such for some

    The old classic HP audio oscillators, the wein briges with
    incandescent lamp amplitude levelers, had increased harmonic
    distortion at low frequencies because of the wobble in the filament
    The light intensity lags the voltage waveform because of the thermal
    lag of the filament. And the filament resistance has a positive tc, so
    it lags too.

  19. john jardine

    john jardine Guest

    You've been there already!, (yet still retain your sanity :).

    Fourier doesn't understand on-off switches. Only handling 'forever' waves
    but a Fourier analysis is used to pin down the lagging current.
  20. John Larkin

    John Larkin Guest


    Oops, lead. Resistance is higher in the last half of each cycle as
    compared to the first half. Current leads: It looks capacitive.

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