AC sine wave: What does increasing the frequency do?

[John Larkin]

On Fri, 26 Nov 2004 19:42:05 -0600, John Fields
<jfields@austininstruments.com> wrote:

On Fri, 26 Nov 2004 17:23:14 -0800, John Larkin
jjlarkin@highlandSNIPtechTHISnologyPLEASE.com> wrote:

On Fri, 26 Nov 2004 08:56:34 -0800, John Larkin
jjlarkin@highlandSNIPtechTHISnologyPLEASE.com> wrote:

On Fri, 26 Nov 2004 09:51:32 -0600, "Commander Dave"
cmdr-dave@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.

---
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.

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]

"peterken" <peter273@hotmail.com> wrote in message
news:eCHpd.1295$pN1.59451@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.

 

[John Larkin]

On Sat, 27 Nov 2004 01:53:36 GMT, "peterken" <peter273@hotmail.com>
wrote:

"John Fields" <jfields@austininstruments.com> wrote in message
news:iimfq0h2e0f4gv60unkp4rp14rg3u2cpnv@4ax.com...
On Fri, 26 Nov 2004 17:23:14 -0800, John Larkin
jjlarkin@highlandSNIPtechTHISnologyPLEASE.com> wrote:

On Fri, 26 Nov 2004 08:56:34 -0800, John Larkin
jjlarkin@highlandSNIPtechTHISnologyPLEASE.com> wrote:

On Fri, 26 Nov 2004 09:51:32 -0600, "Commander Dave"
cmdr-dave@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.

---
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


" ...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

"incandescent" <> "fluorescent"

John

 

[John Fields]

On Fri, 26 Nov 2004 18:02:46 -0800, John Larkin
<jjlarkin@highlandSNIPtechTHISnologyPLEASE.com> wrote:

On Fri, 26 Nov 2004 19:42:05 -0600, John Fields
jfields@austininstruments.com> wrote:

On Fri, 26 Nov 2004 17:23:14 -0800, John Larkin
jjlarkin@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]

On Sat, 27 Nov 2004 12:48:37 -0600, John Fields
<jfields@austininstruments.com> wrote:

On Fri, 26 Nov 2004 18:02:46 -0800, John Larkin
jjlarkin@highlandSNIPtechTHISnologyPLEASE.com> wrote:

On Fri, 26 Nov 2004 19:42:05 -0600, John Fields
jfields@austininstruments.com> wrote:

On Fri, 26 Nov 2004 17:23:14 -0800, John Larkin
jjlarkin@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.

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]

"John Fields" <jfields@austininstruments.com> wrote in message
news:79bhq010jlqj9e8n55nsqid0gt2r57nnm4@4ax.com...

On Fri, 26 Nov 2004 18:02:46 -0800, John Larkin
jjlarkin@highlandSNIPtechTHISnologyPLEASE.com> wrote:

On Fri, 26 Nov 2004 19:42:05 -0600, John Fields
jfields@austininstruments.com> wrote:

On Fri, 26 Nov 2004 17:23:14 -0800, John Larkin
jjlarkin@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

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]

john jardine wrote:

---

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]

On Sat, 27 Nov 2004 13:11:32 -0800, John Larkin
<jjlarkin@highSNIPlandTHIStechPLEASEnology.com> wrote:

On Sat, 27 Nov 2004 12:48:37 -0600, John Fields
jfields@austininstruments.com> wrote:

On Fri, 26 Nov 2004 18:02:46 -0800, John Larkin
jjlarkin@highlandSNIPtechTHISnologyPLEASE.com> wrote:

On Fri, 26 Nov 2004 19:42:05 -0600, John Fields
jfields@austininstruments.com> wrote:

On Fri, 26 Nov 2004 17:23:14 -0800, John Larkin
jjlarkin@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.



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

---
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]

On Sat, 27 Nov 2004 16:13:06 -0600, John Fields
<jfields@austininstruments.com> wrote:

On Sat, 27 Nov 2004 13:11:32 -0800, John Larkin
jjlarkin@highSNIPlandTHIStechPLEASEnology.com> wrote:

On Sat, 27 Nov 2004 12:48:37 -0600, John Fields
jfields@austininstruments.com> wrote:

On Fri, 26 Nov 2004 18:02:46 -0800, John Larkin
jjlarkin@highlandSNIPtechTHISnologyPLEASE.com> wrote:

On Fri, 26 Nov 2004 19:42:05 -0600, John Fields
jfields@austininstruments.com> wrote:

On Fri, 26 Nov 2004 17:23:14 -0800, John Larkin
jjlarkin@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.



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

---
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 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" <jpopelish@rica.net> wrote in message
news:41A8F7E8.2DB7F60E@rica.net...

john jardine wrote:

---

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

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 Fields]

On Sat, 27 Nov 2004 15:01:06 -0800, John Larkin
<jjlarkin@highSNIPlandTHIStechPLEASEnology.com> wrote:

On Sat, 27 Nov 2004 16:13:06 -0600, John Fields
jfields@austininstruments.com> wrote:

On Sat, 27 Nov 2004 13:11:32 -0800, John Larkin
jjlarkin@highSNIPlandTHIStechPLEASEnology.com> wrote:

On Sat, 27 Nov 2004 12:48:37 -0600, John Fields
jfields@austininstruments.com> wrote:

On Fri, 26 Nov 2004 18:02:46 -0800, John Larkin
jjlarkin@highlandSNIPtechTHISnologyPLEASE.com> wrote:

On Fri, 26 Nov 2004 19:42:05 -0600, John Fields
jfields@austininstruments.com> wrote:

On Fri, 26 Nov 2004 17:23:14 -0800, John Larkin
jjlarkin@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.



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

---
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 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.

---
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]

On Sat, 27 Nov 2004 15:18:08 -0800, John Larkin
<jjlarkin@highSNIPlandTHIStechPLEASEnology.com> wrote:

On Sat, 27 Nov 2004 15:01:06 -0800, John Larkin
jjlarkin@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, 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]

On Sat, 27 Nov 2004 17:52:36 -0600, John Fields
<jfields@austininstruments.com> wrote:

On Sat, 27 Nov 2004 15:18:08 -0800, John Larkin
jjlarkin@highSNIPlandTHIStechPLEASEnology.com> wrote:

On Sat, 27 Nov 2004 15:01:06 -0800, John Larkin
jjlarkin@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]

John Larkin wrote:

On Sat, 27 Nov 2004 20:13:48 -0500, John Popelish <jpopelish@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 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]

On Sun, 28 Nov 2004 00:16:32 -0500, John Popelish <jpopelish@rica.net>
wrote:

John Larkin wrote:

On Sat, 27 Nov 2004 20:13:48 -0500, John Popelish <jpopelish@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 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.

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

 

[John Popelish]

John Larkin wrote:

On Sat, 27 Nov 2004 20:13:48 -0500, John Popelish <jpopelish@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.

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 jardine wrote:

"John Popelish" <jpopelish@rica.net> wrote in message
news:41A8F7E8.2DB7F60E@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]

On Sun, 28 Nov 2004 00:51:59 -0500, John Popelish <jpopelish@rica.net>
wrote:

John Larkin wrote:

On Sat, 27 Nov 2004 20:13:48 -0500, John Popelish <jpopelish@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]

On 27 Nov 2004 21:37:36 -0800, nospam4u_jack@yahoo.com (Jack// ani)
wrote:

"peterken" <peter273@hotmail.com> wrote in message news:<eCHpd.1295$pN1.59451@phobos.telenet-ops.be>...
"Commander Dave" <cmdr-dave@spamcop.net> wrote in message
news:i8Hpd.2990$_6.1280@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??

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]

On 28 Nov 2004 02:12:29 -0600, The Phantom <phantom@aol.com> wrote:

On Sun, 28 Nov 2004 00:51:59 -0500, John Popelish <jpopelish@rica.net
wrote:

John Larkin wrote:

On Sat, 27 Nov 2004 20:13:48 -0500, John Popelish <jpopelish@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)

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.

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.