AC sine wave: What does increasing the frequency do?
Commander Dave
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.
Thanks!
-Commander Dave
--
Help me get rich... Join PLUS Lotto! (I get a percentage).
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peterken
"Commander Dave" <cmdr...@spamcop.net> wrote in message
news:i8Hpd.2990$_6....@fe40.usenetserver.com...
changing frequency doesn't change available power
however, household appliances (eg with motors, like vacuum cleaners)
wouldn't be able to handle it, since they're built for 60Hz
60Hz is a choice to avoid flickering in lighting (fluorescent tubes) and to
minimize losses during transport
Commander Dave
looking to see if increasing the frequency increased power. From what I
gather, while it makes it incompatible with things that run on 60 Hz, it
doesn't change the available power... it just cycles faster.
Cheers!
-Dave
"peterken" <pete...@hotmail.com> wrote in message
news:eCHpd.1295$pN1....@phobos.telenet-ops.be...
John Larkin
<cmdr...@spamcop.net> wrote:
>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.
>
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
frequencies!
John
John Popelish
>
> 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.
More correctly, the higher the voltage, the less current a given
amount of power requires. Power is volts times amperes.
> 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?
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.
> 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.
--
John Popelish
John Larkin
<jjla...@highlandSNIPtechTHISnologyPLEASE.com> wrote:
>On Fri, 26 Nov 2004 09:51:32 -0600, "Commander Dave"
><cmdr...@spamcop.net> wrote:
>
>>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.
>>
>
>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.
>
Which brings up the concept that an incandescent lamp appears to have
a capacitive component of impedance, which is itself a function of
frequency.
John
John Fields
<jjla...@highlandSNIPtechTHISnologyPLEASE.com> wrote:
---
It may seem that it does if you're referring to the inrush current,
but put a resistor in series with the lamp and the voltage and current
will be in phase across and through them both, I believe, since all
that changes is the resistance of the lamp filament.
--
John Fields
peterken
"John Fields" <jfi...@austininstruments.com> wrote in message
news:iimfq0h2e0f4gv60u...@4ax.com...
" ...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
John Larkin
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.
John
Bob Myers
> 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
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.
>
>
John Larkin
wrote:
"incandescent" <> "fluorescent"
John
John Fields
<jjla...@highlandSNIPtechTHISnologyPLEASE.com> wrote:
>On Fri, 26 Nov 2004 19:42:05 -0600, John Fields
><jfi...@austininstruments.com> wrote:
>
>>On Fri, 26 Nov 2004 17:23:14 -0800, John Larkin
>><jjla...@highlandSNIPtechTHISnologyPLEASE.com> wrote:
>>>Which brings up the concept that an incandescent lamp appears to have
>>>a capacitive component of impedance, which is itself a function of
>>>frequency.
>>
>>---
>>It may seem that it does if you're referring to the inrush current,
>>but put a resistor in series with the lamp and the voltage and current
>>will be in phase across and through them both, I believe, since all
>>that changes is the resistance of the lamp filament.
>
>
>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.
---
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
through.
Seems to me it would be akin to a simple resistive divider where one
of the resistors is variable, like this:
E1
|
[RV1]
|
+---E2
|
[R2]
|
0V
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
E1R2
E2 = --------
RV1+R2
for any instantaneous value of E1 and RV1 and any value of R2.
To check, I did this:
240RMS>----+-----> TO SCOPE VERT A
|
[LAMP]
|
+-----> TO SCOPE VERT B
|
[576R]
|
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.
---
>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.
---
Since I don't have a schematic in front of me... ;)
--
John Fields
John Larkin
<jfi...@austininstruments.com> wrote:
But the resistance in question is time-varying at 120 Hz. And phase
shift is not determined by an instantaneous measurement.
>
>To check, I did this:
>
>
>240RMS>----+-----> TO SCOPE VERT A
> |
> [LAMP]
> |
> +-----> TO SCOPE VERT B
> |
> [576R]
> |
>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.
How did you measure the phase shift? Looking at the zero crossings?
They will obviously *not* be shifted by a time-varying filament
resistance.
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.
John
john jardine
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
John Popelish
> > ---
> 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
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.
--
John Popelish
John Fields
---
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...
---
>>
>>To check, I did this:
>>
>>
>>240RMS>----+-----> TO SCOPE VERT A
>> |
>> [LAMP]
>> |
>> +-----> TO SCOPE VERT B
>> |
>> [576R]
>> |
>>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.
>
>How did you measure the phase shift? Looking at the zero crossings?
>They will obviously *not* be shifted by a time-varying filament
>resistance.
---
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.
---
>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.
---
If what you're talking about is the difference in phase between the
intensity of the radiation emitted by the lamp and the voltage across
the lamp, then I agree with you. If you're not, then we need to find
out where the misunderstanding is.
--
John Fields
John Larkin
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.
John
John Larkin
<jfi...@austininstruments.com> wrote:
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
references.
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
resistance.
>
>>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.
>
>---
>If what you're talking about is the difference in phase between the
>intensity of the radiation emitted by the lamp and the voltage across
>the lamp, then I agree with you. If you're not, then we need to find
>out where the misunderstanding is.
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.
John
john jardine
"John Popelish" <jpop...@rica.net> wrote in message
news:41A8F7E8...@rica.net...
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.
regards
john
John Larkin
<jjla...@highSNIPlandTHIStechPLEASEnology.com> wrote:
>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.
^^^
Oops, lead. Resistance is higher in the last half of each cycle as
compared to the first half. Current leads: It looks capacitive.
John
John Fields
<jjla...@highSNIPlandTHIStechPLEASEnology.com> wrote:
---
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.
---
>>>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.
>>
>>---
>>If what you're talking about is the difference in phase between the
>>intensity of the radiation emitted by the lamp and the voltage across
>>the lamp, then I agree with you. If you're not, then we need to find
>>out where the misunderstanding is.
>
>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.
---
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.
--
John Fields
John Fields
<jjla...@highSNIPlandTHIStechPLEASEnology.com> wrote:
---
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.
---
--
John Fields
John Larkin
<jfi...@austininstruments.com> wrote:
>On Sat, 27 Nov 2004 15:18:08 -0800, John Larkin
><jjla...@highSNIPlandTHIStechPLEASEnology.com> wrote:
>
>>On Sat, 27 Nov 2004 15:01:06 -0800, John Larkin
>><jjla...@highSNIPlandTHIStechPLEASEnology.com> wrote:
>>
>>>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.
>> ^^^
>>
>>Oops, lead. Resistance is higher in the last half of each cycle as
>>compared to the first half. Current leads: It looks capacitive.
>
>---
>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,
Right, the resistance varies in half-cycles, at 120 Hz, just like the
temperature and light output do.
> 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.
>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. 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
John Popelish
> 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.
Actually, it represents even harmonics, not phase shift
--
John Popelish
John Larkin
wrote:
So, if one did a Fourier analysis of this current waveform, the
fundamental current would be in phase with the voltage?
John
John Popelish
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.
--
John Popelish
John Larkin
wrote:
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
Jack// ani
> news:i8Hpd.2990$_6....@fe40.usenetserver.com...
> 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
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
John Popelish
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.
--
John Popelish
John Popelish
>
> "John Popelish" <jpop...@rica.net> wrote in message
> news:41A8F7E8...@rica.net...
> > 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.
> 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.
> regards
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.
--
John Popelish
The Phantom
wrote:
>John Larkin wrote:
>>
>> On Sat, 27 Nov 2004 20:13:48 -0500, John Popelish <jpop...@rica.net>
>> wrote:
>>
>> >John Larkin wrote:
>> >
>> >> 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.
>> >
>> >Actually, it represents even harmonics, not phase shift
>>
>> So, if one did a Fourier analysis of this current waveform, the
>> fundamental current would be in phase with the voltage?
>
>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.
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 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.
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.
The Phantom
wrote:
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.
>
>Thanks
The Phantom
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.
John Popelish
>
> On 28 Nov 2004 02:12:29 -0600, The Phantom <pha...@aol.com> wrote:
>
> >On Sun, 28 Nov 2004 00:51:59 -0500, John Popelish <jpop...@rica.net>
> >wrote:
> >>Okay, I created data representing a 32 point per cycle representation
Thanks for this detailed work. You should produce a tutorial web page
on this subject.
--
John Popelish
John Fields
<jjla...@highSNIPlandTHIStechPLEASEnology.com> wrote:
>On Sat, 27 Nov 2004 17:52:36 -0600, John Fields
><jfi...@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 (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.
--
John Fields
John Larkin
> 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).
>
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
The Phantom
<jfi...@austininstruments.com> wrote:
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.
John Larkin
<jfi...@austininstruments.com> wrote:
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.
>
>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.
JP and the Phantom have both done the analysis.
John
The Phantom
This is, of course, the *fundamental* definition of Power Factor
about which there seems to be no dispute.
>
>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.
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.
>Such a load has, by my definition, a low power
>factor but no phase shift.
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."
>
John Larkin
>>
>>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.
>
> 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.
>
>>Such a load has, by my definition, a low power
>>factor but no phase shift.
>
> 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.
>
No, the fat friers used zero-crossing, multi-cycle burst triac
controllers, not phase control like a dimmer.
John
The Phantom
Of course; I didn't pick up on the zc part.
>
>John
>
>
John Fields
> 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.
---
OK. I get it.
Thanks,
--
John Fields
John Popelish
I love this newsgroup because a simple, one sentence question can
generate days of technical exchange that teaches quite a few of us
things we knew but didn't realize. :-)
--
John Popelish
The Phantom
wrote:
>John Fields wrote:
Yes. And all without any name-calling and foul language. Much
more pleasant.
John Larkin brings up some more interesting points that deserve a
new thread. I'm going to have to think about them and perhaps in a
Commander Dave
> I love this newsgroup because a simple, one sentence question can
> generate days of technical exchange that teaches quite a few of us
> things we knew but didn't realize. :-)
> --
> John Popelish
It amazes me that my simple question on power spawned this much
discussion! I followed a little of it, but for the most part it made me
realize how much I have to learn... :-)
Cheers!
-Commander Dave
John Popelish
>
> "John Popelish" <jpop...@rica.net> wrote:
> > I love this newsgroup because a simple, one sentence question can
> > generate days of technical exchange that teaches quite a few of us
> > things we knew but didn't realize. :-)
> It amazes me that my simple question on power spawned this much
> discussion! I followed a little of it, but for the most part it made me
> realize how much I have to learn... :-)
You too?
--
John Popelish
John Popelish
The wheels are turning.
--
John Popelish
john jardine
"John Larkin" <jjla...@highlandSNIPtechTHISnologyPLEASE.com> wrote in
message >
.....
> 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.
>
The classes of gyrators here, has made for an interesting and
*understandable* thread.
Sometime this kind of stuff will be written up in depth, (maybe already
has!). I'll bet that during the process 'understanding' will be #1 item to
fall by the wayside :-)
regards
john
John Larkin
Tons of theoretical work has been done on time-varying capacitances,
mostly in the 60's and such when two-terminal devices (varactors,
tunnel diodes, step-recovery diodes) were the rage.
> I'll bet that during the process 'understanding' will be #1 item to
>fall by the wayside :-)
>regards
>john
>
I like to try to avoid equations until I can really feel what's going
on. Being able to do the math doesn't mean you understand it, just
that you can push some symbols around. This is risky of course,
because instincts are often wrong about stuff like this. But the guys
who just do the math can make ghastly blunders, too, and they
sometimes don't have the instincts to recognize an absurd result when
they see it.
Things like Fourier transforms can be visualized and sort of done by
inspection, but not many EE courses try to track that along with the
math.
To a creature that was sufficiently intelligent, everything would be
intuitively obvious.
John
Roger Johansson
> 'scuse my ignorance, but doesn't an incandescent lamp behave more like
> say a triac or a set of zeners after the startup cycle?
> this means, the gas ignites at a specific voltage (think it was about
> 80V somewhere) thus lowering the lamps impedance from that point on.
>
What you are talking about is not an incandescent lamp, it is a
fluoroscent lamp.
An incandescent lamp has a thin wire inside which glows, it is the most
common lamp in the world. But fluoroscents with their higher efficiency
are taking increasing parts of the market.
It is difficult to keep all these technical terms apart in english if you
are not born in an english-speaking country.
--
Roger J.
peterken
"John Larkin" <jjla...@highlandSNIPtechTHISnologyPLEASE.com> wrote in
> On Fri, 26 Nov 2004 08:56:34 -0800, John Larkin
> <jjla...@highlandSNIPtechTHISnologyPLEASE.com> wrote:
>
> >On Fri, 26 Nov 2004 09:51:32 -0600, "Commander Dave"
> ><cmdr...@spamcop.net> wrote:
> >
> >>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.
> >>
> >
> >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.
> >
>
> Which brings up the concept that an incandescent lamp appears to have
> a capacitive component of impedance, which is itself a function of
> frequency.
>
> John
>
'scuse my ignorance, but doesn't an incandescent lamp behave more like say a
triac or a set of zeners after the startup cycle?
this means, the gas ignites at a specific voltage (think it was about 80V
somewhere) thus lowering the lamps impedance from that point on.
that's also the reason of the coil in the circuit, the impedance reduces
lamps current, it allows the lamps voltage to drop to the burning point.
the gas stays ignited until again a specific voltage where it stops
conducting, and re-ignites again at a specific voltage the other half of the
cycle.
only other thing in the circuit is a coil, so i don't see the "capacitor"
anywhere
(view drawing in notepad using fixed font)
|mains
| . .
| . .
|. .
.-----------.------------.------
| . .
| . .
| . .
|
| ..
| . .
| . .
| . .
|....----- .....-------.......--
| . .
|lamp . .
|current . .
| ..
|
peterken
"Roger Johansson" <no-e...@home.se> wrote in message
news:Xns95B0727...@130.133.1.4...
oops, sorry, my mistake :-(
indeed I'm dutch, thereof the mistake
Khwaj
This is a mechanical problem. At the most basic level the generators are
unable to maintain their rpm while heavily loaded.
John Larkin
<jo...@jjdesigns.fsnet.co.uk> wrote:
So, could a motor-driven variable capacitor, paralleled with an
inductor, become an oscillator? Seems like it.
John
Don Kelly
"Khwaj" <almost...@yahoo.com> wrote in message
news:10qm8a4...@corp.supernews.com...
---------
Of course, this much variation only applies to small independent generators.
In a grid system, a 2Hz frequency deviation is unacceptable. A change in the
order of 0.05 Hz is a fairly extreme transient .
--
Don Kelly
dh...@peeshaw.ca
remove the urine to answer
Bill Bowden
> "peterken" <pete...@hotmail.com> wrote in message
> news:eCHpd.1295$pN1....@phobos.telenet-ops.be...
> > 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
> 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.
> >
> >
Wasn't it Tesla that proposed the 60 Hz. standard?
And probably because there are 60 minutes in an hour,
and 60 seconds in a minute, and therefore 60 cycles
in a second, and 60 was high enough to avoid flicker,
and maybe some other reasons.
What was the reasoning for 50 Hz, other than
slightly better transmission efficiency?
-Bill
Don Kelly
"Bill Bowden" <wronga...@att.net> wrote in message
news:ad025737.04112...@posting.google.com...
---------
The difference between 50 and 60 Hz will make some gain in a reduction of
iron in machines at 60 Hz and, on the other side, some increase in
transmission capability (not necessarily efficiency) at 50 Hz. However, it
appears that the areas which originally had longer transmission distances
went to 60 Hz. so where's the logic.
Probably a choice of "the Brits chose 50Hz so we will choose 60Hz" (or the
opposite with Yanks substituted for Brits ).
As for flicker at 25 Hz ( Not 24) with incandescents. It exists and can be
noticed. I have seen it. Subtle but there- somewhat similar to a computer
monitor refresh rate of 55 to 60Hz. It can be annoying if you are not used
to it..
Rich Grise
> On Mon, 29 Nov 2004 00:54:10 -0000, "john jardine"
>>The classes of gyrators here, has made for an interesting and
>>*understandable* thread.
>>Sometime this kind of stuff will be written up in depth, (maybe already
>>has!).
>
> Tons of theoretical work has been done on time-varying capacitances,
> mostly in the 60's and such when two-terminal devices (varactors, tunnel
> diodes, step-recovery diodes) were the rage.
>
>> I'll bet that during the process 'understanding' will be #1 item to
>>fall by the wayside :-)
>
> Being able to do the math doesn't mean you understand it, just that you
> can push some symbols around. This is risky of course, because instincts
> are often wrong about stuff like this. But the guys who just do the math
> can make ghastly blunders, too, and they sometimes don't have the
> instincts to recognize an absurd result when they see it.
>
> Things like Fourier transforms can be visualized and sort of done by
> inspection, but not many EE courses try to track that along with the math.
>
> To a creature that was sufficiently intelligent, everything would be
> intuitively obvious.
Actually, I've been following the thread, and about the only contribution
I can make is that I was visualizing the water-hose model of current,
where the lightbulb goes "OOF!!" when it gets hot, stopping up the
current, but when it cools, it goes, "Aaaah!", and lets the current flow
through again.
So you see why I didn't try to contribute to the actual science of the
thing! ;-)
Thanks!
Rich
Rich Grise
Oh, this is nothing compared to the thread I instigated a couple of years
ago about running a 120V bulb off 240V mains by just putting a diode in
series!
Cheers!
Rich
Rich Grise
> On Sat, 27 Nov 2004 20:13:48 -0500, John Popelish <jpop...@rica.net>
> wrote:
>
>>John Larkin wrote:
>>
>>> 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.
>>
>>Actually, it represents even harmonics, not phase shift
>
> So, if one did a Fourier analysis of this current waveform, the
> fundamental current would be in phase with the voltage?
>
I don't even know what a Fourier analysis _is_ other than a way of
translating a data set from the time domain to the frequency domain
(transforming?), but my "gut-feeling" is that the zero-crossings would be
identical _of the actual source waves_ - the fundamental would be out of
phase, in the plot of the Fourier-transformed resolved fundamental, but
the zero-crossing would be brought back into sync by the harmonics after
you added them back together.
It's like a sine wave with the half-cycles tilted to one side.
But wouldn't that act more inductive?
Thanks,
Rich
John Larkin
At the intuitive level, a Fourier series answers the question "how
much does this waveform look like a 60 Hz sine wave? How much like a
120 Hz sine wave?...". The answers are of the form "2 volts, 45
degrees" and such, one answer for DC and one for each harmonic. The
Fourier transform is a math operation that gives these answers. It
produces the same results you could get using a bandpass filter bank
at f, 2f, etc (plus the DC term, the zero frequency Fourier term,
which you'd get using a lowpass filter).
You can do an eyeball Fourier by printing the waveform on a piece of
paper. Suppose some waveform has a basic frequency of 60 Hz. Now plot
a 60 Hz sinewave on another piece of paper and hold it next to the
original waveform. Slide it horizontally until you see the best match,
so that the input waveform "helps" the sinewave template, pushing it
up and pulling it down in the best places. Now make a rough estimate
of how much it helps (amplitude), and how far you shifted the papers
to get the best match (phase.) Repeat for higher harmonics, one at a
time. Fourier!!
The SCR phase control waveform at 50% on has power on the load in the
last half of each half-cycle. That shifts the center-of-gravity of the
waveform later in time from the line voltage wave, so the fundamental
component, the 60 Hz Fourier line, lags. By something like 32 degrees,
some people have calculated in other posts. That does look inductive.
John
John Larkin
Bright. Won't last long.
John
peterken
"John Larkin" <jjla...@highlandSNIPtechTHISnologyPLEASE.com> wrote in
lasts like forever if the diode is selected correctly
(say 1N4007 or so for standard domestic bulbs)
John Fields
wrote:
>
>"John Fields" <jfi...@austininstruments.com> wrote in message
>news:uj4vq0tcpmgolliic...@4ax.com...
>> On Thu, 02 Dec 2004 22:02:33 GMT, "peterken" <pete...@hotmail.com>
>> wrote:
>>
>>
>> >lasts like forever if the diode is selected correctly
>> >(say 1N4007 or so for standard domestic bulbs)
>>
>> ---
>> Really? Why would that be?
>>
>> --
>> John Fields
>
>"standard domestic bulbs" is any type from say 5W upto 250W
>for these types power rating of 1N4007 diode is ok
---
Perhaps you missed this:
"Oh, this is nothing compared to the thread I instigated a couple of
years ago about running a 120V bulb off 240V mains by just putting a
diode in series!"
Notice that it's not a 120V lamp running on 1/2 wave rectified 120V
mains, it's a 120V lamp running on 1/2 wave rectified 240V mains.
That being the case, how much power would a 120V 100W lamp dissipate
if it were being run on 1/2 wave rectified 240V?
Don't forget to consider the tempco and thermal time constant of the
filament... :-)
--
John Fields
John Larkin
peterken
> >(say 1N4007 or so for standard domestic bulbs)
>
John Fields
wrote:
>lasts like forever if the diode is selected correctly
>(say 1N4007 or so for standard domestic bulbs)
---
Rich Grise
> "John Larkin" <jjla...@highlandSNIPtechTHISnologyPLEASE.com> wrote in
>> >Oh, this is nothing compared to the thread I instigated a couple of
>> >years ago about running a 120V bulb off 240V mains by just putting a
>> >diode in series!
>> >
>>
> for standard domestic bulbs)
Uh-oh.