Re: Interesting dimmable CFL

In art. <rojbv1d8ju6olehfqi28kgah5m2ddrpugk@xxxxxxx>, Victor Roberts wrote:

However, since the watt-hour meters on individual houses, at least at
that time, could not measure the DC component of the power line current,
it was still possible for one homeowner to select and use lamps with all
the diodes oriented in the same direction and thus get part of his power
for free. The power companies didn't like the diode at all, because of
the lost revenue

Are you sure about this?

Although an incandescent lamp with a diode has a DC voltage component
that the voltage coil in an electric meter does not see and so the
meter cannot multiply that DC voltage by the DC current, I did not think
it was so easy to beat an electric meter.

I thought that an electric meter did a good job of integrating through
time the product of instantaneous voltage times instantaneous current,
which would mean good integration of instantaneous power consumption and
this provides for measurement of actual power consumption by reactive
loads, nonlinear loads, and I thought also loads that drew a DC component.
I thought the only way to beat an electric meter would be to get it to
inaccurately integrate over time the product of instantaneous voltage and
instantaneous current.

Now, I am aware that an electric meter metering power consumption by a
resistor with a diode in series with it will (at least ideally) have its
voltage coil seeing no DC at all, while a DC voltage component is across
the resistor.
And there is a DC component for current.

And for the fundamental AC frequency and for every harmonic thereof, the
AC current's component for each such frequency should be the same through
the load as through the current coil of the electric meter.

Now, for why I suspect the resistor having a diode in series with it can
have less AC power than is "read" by the electric meter to an extent that
the resistor has DC power that the electric meter cannot read:

What the meter reads has voltage coil having only sinewave at
fundamental frequency. Let's say for the sake of argument that this is
120 volts RMS value. No DC and no harmonics exist, at least ideally, in
the voltage waveform presented to the voltage coil in the wattmeter.
And not that I want to bog down into details thereof, but power
consumption can be broken down into volts at DC and specific frequencies
times for each amps only of the same. For example, DC volts times
amps-excluding-DC and vice-versa, as well as AC voltage of a specific
spectrum component (frequency) times AC amps of a different spectrum
component, equals no power consumption. 180-Hz spectral component of the
current waveform only accounts for power consumption if the voltage
waveform has a 180-Hz spectral component, and only to the extent that
these voltage and current 180-Hz spectral components are in-phase.

Now, back to the resistor-diode combo:

The resistor has DC voltage and DC current so has power attributable to
DC, and the electric meter sees no DC voltage so has no billing effect on
the DC.

But countering this: The 60-Hz component of the voltage (assuming 60 Hz
120V RMS sinewave line voltage) is 120V RMS across the electric meter's
voltage coil and assuming an ideal diode 84.85V across the resistor.
60-Hz component of current is the same through the meter and the resistor.

Now for math assuming a 100 ohm resistor and an ideal diode and 60 Hz
120V RMS sinewave line voltage:

Total power consumption: Half of 120^2/100, which is 72 watts.

Load voltage: 84.85V RMS
Load amps: .8485 amp RMS

DC component:

Current is steady (due to considering "pure DC component") by the
average value of .54 amp, and corresponding voltage is 54 volts,
accountinmg for about 29.2 watts.
Electric meter counts zero here out of the total of 72 watts.

60-Hz component:

I have not bothered to do Fourier analysis of a halfwave rectified
sinewave for amount of content at each harmonic and at fundamental
frequency and at DC, so how about I assume the 60-Hz component of the
current waveform is the power divided by the pure-sinewave voltage (RMS),
which is 72/120 or .6 amp. Multiply this by the resistor's resistance and
if I am not going wrong so far the 60-Hz component of the resistor voltage
(of the 84.85V RMS total) is 60 volts. Makes for 36 watts of
60-Hz-spectral-component at/into the resistor.

If I have not gone wrong so far, then the resistor is getting 29.2 watts
of DC and 36 watts of 60 Hz (65.2 watts for these two components), while
the electric meter is billing for 72 watts and entirely accountable
through the 60-Hz spectral component (sole spectral component in the line
voltage seen by the voltage coil of the electric meter).
6.8 watts of resistor power dissipation remain to be accounted for. I
expect that sums of products of RMS values of specific-harmonic-components
of voltage waveform (across the resistor) and current waveform (negative
for any harmonics where these are 180-degrees-out-of-phase) will add up to
this remaining 6.8 watts or 9.44% of power consumption.

Assuming I have not made a bad assumption, the 60-Hz component of power
consumption is half as much at the dioded resistor as is "read" by the
electric meter, while the dioded resistor has DC and harmonic voltage
components not seen by the meter but multipliable by these spectral
components of voltage see by the resistor and not the meter, and
"assumably" making up the difference.

I expect the main complaints of power companies about diodes would be DC
screwing up transformers and DC/harmonic amps (to same extent as any
out-of-phase line-frequency amps drawn by reactive loads) requiring
current carrying capacity while being unbillable by a wattmeter, rather
than achieving actual energy consumption downstream of an electric meter
unbilled by the meter.

Any comments?

- Don Klipstein (don@xxxxxxxxx)
.

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