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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.
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 impedance. (worst at 90degrees fire angle and not a jot of energy storage anywhere) regards john
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".
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??
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.
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.
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.
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.
An inductor (iron cores not allowed here) will never generate harmonics that are not present in the applied voltage.
An inductor's current would halve for a doubling of the frequency of the applied voltage.
An inductor will store energy, and therefore the phenomenon of resonance can occur.
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.
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:
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.
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.
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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?
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.
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.
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."
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