Re: AC sine wave: What does increasing the frequency do?

From: The Phantom (phantom_at_aol.com)
Date: 11/28/04 

Date: 28 Nov 2004 12:41:06 -0600

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 (Rē + (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 Rē
>
>
>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.

Relevant Pages

  • Re: Will Radio Engineering be QMs worst nightmare?
    ... See what kind of voltage gets produced by 150 watts ... I am thinking of an amplifier that is turned on with no load. ... > don't increase because the load resistance changes. ... What kind of amp. ...
    (sci.physics)
  • Re: Kathode folower question
    ... How do I calculate the cathode follower cathode resistor ... Depends on the tube type and what sort of load you want ... But in simple terms - you know the anode voltage of the ... the tube's plate resistance and this method takes no ...
    (rec.audio.tubes)
  • Re: Question re. shunts to measure amperage
    ... connector between source and load as the voltage or amperage changes due ... direction as the load reduces. ... compression' at high volumes as the voice coil resistance increases. ...
    (sci.electronics.basics)
  • Re: Impedance
    ... For a resistive source and load, model your source as a voltage source ... in series with a resistance and your load as a resistance to ground. ... >cables are too long in unbalanced systems? ...
    (sci.electronics.basics)
  • Re: AC sine wave: What does increasing the frequency do?
    ... >>the resistance right now and what's the voltage across it right now. ... >time-referenced measurement. ... >triac dimmer, driving a resistive load, runing at 50% conduction ...
    (sci.electronics.basics)