Although it is true that the DC component of voltage is 0, the DC
component of current is not. Since the current is non-sinusoidal, it is
not possible to analyze it using only the fundamental frequency. This is
the whole issue of power supply design.
Consider an old-school power supply using a half wave rectifier (diode)
a smoothing capacitor, and a load. This supply produces output power,
even though the DC component of voltage at the input is zero. Where does
this power come from, if not from the AC outlet?
The "fundamental component" of a non-sinusoidal waveform represents only
a fraction of that waveform. The rest of it is represented by the
harmonics. If you are interested in power supply design, the following
text book is excellent:
http://www.amazon.com/Switching-Power-Supply-Design-ebook/dp/B001AO0GDG/ref=dp_kinw_strp_1
Duncan
On 5/27/2013 12:17 PM, David Roberson wrote:
That is a good try. I agree with all that you say except for one key
item. 1). No negative current flows due to the diode. 2). The
instantaneous power being delivered to the resistor is I^2*R as you
suggest. 3). The DC rated clamp on meter should measure the total RMS
current provided it can handle distorted AC waveforms. Now, here is
where you have a problem with the measurement. You say to multiply
the true RMS current by the voltage and that is where the problem
arises. I am confident that you realized that as soon as you said it!
All of the power that is applied to the resistor comes from the wall
socket. The voltage at this location is a sinewave at the frequency
supplied by the electrical service. There is no DC voltage component,
so the DC power being supplied is 0. The AC component of the input
frequency is the only one that can have power supplied and that can
only be given to current at its fundamental frequency. So, to
determine how much power the resistor absorbs you must take the
fundamental current component and multiply it by the fundamental
voltage supplied by the wall socket. This needs to be corrected for
phase shift if any exists with the product by the cosine of the
difference in phase of the two components.
Therefore, the DC flowing through the rectifier does not contribute to
the measurement.
Dave
-----Original Message-----
From: Duncan Cumming <[email protected]>
To: vortex-l <[email protected]>
Sent: Mon, May 27, 2013 2:56 pm
Subject: [Vo]:Re: [Vo]:Torbjörn Hartman describes power measurments
I will give it my best shot.
Consider a diode in series with a resistor, and connected to an AC
outlet. For the first half of the cycle the diode conducts, and a
positive current flows. For the second half, the diode does not
conduct and NO NEGATIVE CURRENT FLOWS, even though a negative voltage
is present. This is the function of a diode. So what you have is an
intermittent flow of positive current, which delivers power to the
load resistors. The magnitude of this power is given by I^2*R.
If you were to measure the current using a clamp on ammeter which is
DC rated, then the current would be determined accurately and the RMS
value determined by the digital voltmeter (which must be a true RMS
type, of course). Multiplying this by the voltage gives the power
dissipated in the resistor.
If you were to measure the current using a clamp on ammeter which is
NOT DC rated, then only the fluctuations in current would be measured,
which fluctuations would be a lot less than the true value of current.
So a misleadingly low value of current would be measured, leading to a
substantial under estimate of the power dissipated in the resistor.
This is electrical engineering 101, but it shows some of the problems
involved in measuring AC power with a non-sinusoidal waveform. There
are those who assume that a commercially available "power meter"
measures just that, but in fact this is a difficult task that should
be undertaken by a qualified engineer with knowledge of the waveform
that he is measuring. The specs of such an instrument clearly indicate
the limitations to which it is subject, and one must be careful not to
exceed these limitations. An absence of DC sensing capability is one
such limitation.
In the diode example above, the diode itself does not provide any
power whatever. It merely confuses some types of power meters.
On 5/26/2013 9:51 PM, David Roberson wrote:
It does not make any difference whether or not the instrument
measures DC current through the input power cables. That issue is
dead unless someone wants to insist that Rossi or one of his partners
hid a DC supply inside the wall, or in some other place which allows
the DC to appear at the power input terminals. This would have been
obvious to anyone looking at the voltage.
Andrew or Duncan please explain how the DC current through the input
power cable is able to deliver a large power to the load resistors?
It can not be done with any type of diode hidden within the blue
box. Are you ready to concede the point?
Dave
-----Original Message-----
From: Alan Goldwater <[email protected]>
To: vortex-l <[email protected]>
Sent: Mon, May 27, 2013 12:19 am
Subject: Re: [Vo]:RE: [Vo]:Re: [Vo]:Torbjörn Hartman describes power
measurments
That is a different instrument. The one used in the tests (PCE-830
<http://www.industrial-needs.com/technical-data/power-anlayser-PCE-830.htm>)
does not measure DC.
On 5/26/2013 7:53 PM, Jones Beene wrote:
Did you actually check the PCE site?
It looks to me like all the current clamps on the PCE power analyzer
site measure both AC and DC
http://www.industrial-needs.com/technical-data/current-detector-PCE-DC-3.htm
