As a follow-up to this, and as part of my appropriate public recanting, here are the equations for power during a sample time. This shows that if the current is made a constant, then only the average voltage need be acquired. Between samples, a capacitance on the voltage measurement node will cause the voltage to be averaged and the resulting voltage sample will be an average voltage (RMS is explicitly NOT needed). Here are the equations. I hope I got them right this time and I hope the image gets through to Vortex (it is small).
On Mon, Oct 27, 2014 at 8:58 PM, Bob Higgins <rj.bob.higg...@gmail.com> wrote: > Well Dave, you have made a good and convincing argument. My hat is off to > you and I need to eat it with a big public helping of crow. > > It seems like if we go back to basics, the average power is integral((I > dot V)dt)/integral(dt). If I is a constant, then you can pull I outside > the integral and you get the average power as I x > integral(Vdt)/integral(dt), which means the average power is I times the > average voltage. > > Thank you for taking the challenge, making me rethink, and putting me > straight! > > Regards, Bob > > On Mon, Oct 27, 2014 at 5:53 PM, David Roberson <dlrober...@aol.com> > wrote: > >> Bob, I take that as a challenge. I am not offended my friend, but find >> this a great opportunity to prove what I am saying is correct. I predict >> that you will agree with me once you have an opportunity to dig deeper into >> the subject. >> >> It is not clear to me what you are showing in your example, perhaps due >> to a problem with my display. Let me choose an example for you to >> consider. Again, we can assume that the current being delivered into the >> load is exactly 1 amp. If we further assume that the load resistance is 1 >> ohm, then under DC conditions we will measure precisely 1 volt across the >> load resistor. >> >> I and I assume you would calculate the power as being 1 watt delivered to >> the load resistor under this static condition. Now, suppose that the >> resistance changes to .5 ohms. In that case the voltage becomes exactly .5 >> volts. This results in a power being delivered to the resistor of .5 >> watts. For the other half of the AC square waveform the resistor becomes >> 1.5 ohms. In that case the power delivered becomes 1.5 watts since 1 amp x >> 1.5 volts = 1.5 watts. Since we are assuming a symmetrical AC waveform, >> this is a pretty good example of that with numerous harmonics that also get >> into the act. The assumed waveform is therefore a 1 volt peak to peak >> square wave that is riding upon a 1 volt DC average. >> >> So the total power average becomes (.5 watts + 1.5 watts) / 2 = 1 watt. >> Each half of the waveform makes its contribution and they balance each >> other out about the normal DC average of 1.0 watt. This is true for all AC >> waveforms, regardless of the harmonic content provided that the current >> retains a constant DC value. >> >> I have stated this on numerous occasions and it is a general concept. >> Power can only be extracted from a source current that flows at the same >> frequency as the source voltage. In this case the current is at a DC >> frequency, so no power can be extracted from the source except into a DC(0 >> Hertz) voltage related load. >> >> Dr. McKubre essentially made the same statement with respect to his >> experimental setup. Another feature of a constant current environment is >> that the power delivered into the load varies directly with the load >> voltage and not proportional to the square of the voltage as is normally >> encountered. That is what allows the average to be used in this case >> instead of having to deal with the messy RMS waveform additions. >> >> If you have reservations about what I have stated I strongly suggest that >> you put together a Spice model. That will prove that what I am saying is >> right on target. >> >> Dave >> >
