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

Discussion in 'Electronic Basics' started by Commander Dave, Nov 26, 2004.

1. ### Commander DaveGuest

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

2. ### peterkenGuest

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

3. ### Commander DaveGuest

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

4. ### John LarkinGuest

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

5. ### John PopelishGuest

More correctly, the higher the voltage, the less current a given
amount of power requires. Power is volts times amperes.
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.

6. ### John LarkinGuest

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

8. ### peterkenGuest

" ...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

9. ### John LarkinGuest

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

10. ### Bob MyersGuest

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.

11. ### John LarkinGuest

"incandescent" <> "fluorescent"

John

12. ### John FieldsGuest

---
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.
---

13. ### John LarkinGuest

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

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

14. ### john jardineGuest

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 | |
| |
| '-'
'--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

15. ### John PopelishGuest

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.

16. ### John FieldsGuest

---
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...
---

---
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.
---

17. ### John LarkinGuest

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

John

18. ### John LarkinGuest

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 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

19. ### john jardineGuest

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

20. ### John LarkinGuest

^^^

Oops, lead. Resistance is higher in the last half of each cycle as
compared to the first half. Current leads: It looks capacitive.

John