AC sine wave: What does increasing the frequency do?

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

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

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

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

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

"incandescent" "fluorescent"

John

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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.

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.

--
John Popelish

it

" ...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

to

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.

--- 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.

---

--- Since I don't have a schematic in front of me... ;)

-- John Fields

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

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

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

^^^

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

John

--
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...

Right, the resistance varies in half-cycles, at 120 Hz, just like the temperature and light output do.

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.

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

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

--
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.