AC sine wave: What does increasing the frequency do?

[Commander Dave]

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

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

Thanks!
-Commander Dave

--
Help me get rich... Join PLUS Lotto! (I get a percentage).
http://tinyurl.com/4mpf4 It is for charity & they DO pay out!

 

[Commander Dave]

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.

Cheers!
-Dave

"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

 

[John Larkin]

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.

But your electric meter will make substantial errors at different
frequencies!

John

 

[John Larkin]

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.

John

 

[John Larkin]

On Sat, 27 Nov 2004 21:30:32 -0000, "john jardine"
<john@jjdesigns.fsnet.co.uk> wrote:

"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

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]

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.

John

 

[John Larkin]

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?

John

 

[Jack// ani]

"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??

Thanks

 

[The Phantom]

On Sun, 28 Nov 2004 16:08:50 -0500, John Popelish <jpopelish@rica.net>
wrote:

John Fields wrote:

On 28 Nov 2004 12:41:06 -0600, The Phantom <phantom@aol.com> wrote:

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,

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.

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
day or so, start a thread. He said:

"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."

 

[John Popelish]

The Phantom wrote:

On Sun, 28 Nov 2004 16:08:50 -0500, John Popelish <jpopelish@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.

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
day or so, start a thread. He said:

"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."

The wheels are turning.

--
John Popelish

 

[Roger Johansson]

"peterken" <peter273@hotmail.com> wrote:

'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.

 

[Khwaj]

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??

Thanks
This is a mechanical problem. At the most basic level the generators are

unable to maintain their rpm while heavily loaded.

 

[John Popelish]

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

--
John Popelish

 

[peterken]

"Roger Johansson" <no-email@home.se> wrote in message
news:Xns95B072740791286336@130.133.1.4...

"peterken" <peter273@hotmail.com> wrote:

'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.

oops, sorry, my mistake :-(
indeed I'm dutch, thereof the mistake

 

[Rich Grise]

On Thu, 02 Dec 2004 22:02:33 +0000, peterken wrote:

"John Larkin" <jjlarkin@highlandSNIPtechTHISnologyPLEASE.com> wrote in
On Thu, 02 Dec 2004 08:54:22 GMT, Rich Grise <rich@example.net> wrote:
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!

Bright. Won't last long.

lasts like forever if the diode is selected correctly (say 1N4007 or so
for standard domestic bulbs)

Uh-oh.

 

[John Larkin]

On Thu, 02 Dec 2004 22:02:33 GMT, "peterken" <peter273@hotmail.com>
wrote:

"John Larkin" <jjlarkin@highlandSNIPtechTHISnologyPLEASE.com> wrote in
message news:sdhuq0hafh8bojb8tb8m4a9gailopqid86@4ax.com...
On Thu, 02 Dec 2004 08:54:22 GMT, Rich Grise <rich@example.net> wrote:

On Sun, 28 Nov 2004 17:54:39 -0600, Commander Dave wrote:

"John Popelish" <jpopelish@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. --
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...


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


Bright. Won't last long.

John


lasts like forever if the diode is selected correctly
(say 1N4007 or so for standard domestic bulbs)

Not bright.

John

 

[peterken]

"Commander Dave" <cmdr-dave@spamcop.net> wrote in message
news:i8Hpd.2990$_6.1280@fe40.usenetserver.com...

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

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

Thanks!
-Commander Dave

--
Help me get rich... Join PLUS Lotto! (I get a percentage).
http://tinyurl.com/4mpf4 It is for charity & they DO pay out!

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

 

[John Popelish]

Commander Dave wrote:

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 Fields]

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

 

[peterken]

"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