ISOTEK EVO3 Aquarius Power Conditioner Review

[DualTriode]

I am peddling as fast as I can to power the generator but the light bulb refuses to go on.

Perhaps you should slow down and catch up on the people skills.

There may be a bit of illumination.

 

[audio2design]

Perhaps you should slow down and catch up on the people skills.

There may be a bit of illumination.

When are you starting your lessons maybe I can tag along?

 

[DualTriode]

When are you starting your lessons maybe I can tag along?

People are always more important than technology.

Take a deep breath and listen to your client.

You asked for you first lesson.

 

[Ingenieur]

If you apply the math wrong, you will get the wrong answer. The power factor, as seen by the source (V1) in my example, is 0.529. That is not debatable. That is what it is. You are totally missing the function of the diodes.

If you do not understand the math you will never understand the system:

Real world example
120, 1, 5% THD, triplets
Load 8 Ohm
Diodes have a bias V and R
xfmr 2 kVA with suitable L and R
I ran it with 0% THD, no measurable difference. xfmr dealt with it.

 

[Ingenieur]

If you apply the math wrong, you will get the wrong answer. The power factor, as seen by the source (V1) in my example, is 0.529. That is not debatable. That is what it is. You are totally missing the function of the diodes.

It is because you claim the xfmr had no inductance. And the load is primarily C.
Did you add L
Does your source have a phase angle?

 

[Ingenieur]

I am peddling as fast as I can to power the generator but the light bulb refuses to go on.

You are the bulb. If you wired, no surprise.
That is all you have are insults and a dim bulb.

 

[Ingenieur]

Maybe you two should get a room

The only room he needs is a classroom.

A 0% THD input's output is no different than one with 5% THD.
The harmonics do not increase/decrease output.
Drop across the filter may.
No different than a high Z feeder

Btw, your 200 W will not do 200 J/sec of work. e, heater, motor, etc.

 

[audio2design]

QUOTE="Ingenieur, post: 1069140, member: 30993"]
The only room he needs is a classroom.

A 0% THD input's output is no different than one with 5% THD.
The harmonics do not increase/decrease output.
Drop across the filter may.
No different than a high Z feeder

Btw, your 200 W will not do 200 J/sec of work. e, heater, motor, etc.
[/QUOTE]

I provided the bloody simulation. Almost 200 real watts. It's a resistor. How much more can that work be? If I added the leakage / magnetizing inductance of the transformer it will provide a filter function and only prove my point more that the rectified voltage on the output caps will change with load. You show a schematic. Now use it.

200w in a resistor. That is joules/sec. It's not open to interpretation. Your analysis is totally off the rails at this point.

NO! The load is not primarily C! The diode acts as an envelope demodulator. It changes everything. And with that I am done. The mute function is there for a reason.

 

[audio2design]

Drop across the filter may.

And what do you think that drop comes from?

 

[Ingenieur]

And what do you think that drop comes from?

It could come from pure R
Long feeder cable, hi PU Z xfmr, etc.
It is not harmonic related

If you have a low pass series R/Shunt C drop will be minimal.
You will filter harmonics and V will only drop based on R which cab be very small, mOhm, pf correction at the load.

But since utility power is only on the order of 5% V THD, moot

 

[amirm]

Doesn't PS Audio's "Power Plant" do that? It's a big power amplifier fed with a clean 60Hz sine wav signal.

That's its design goal. In reality like any power amplifier, it has distortion. I tested my PS Audio 300 before: https://www.audiosciencereview.com/...ac-distortion-noise-on-audio-equipment.25501/

So much cleaner than mains by itself but nowhere near a noise and distortion-free power.

 

[amirm]

But since utility power is only on the order of 5% V THD, moot

It amazes me that he still doesn't get this simple point. There is so little power in the harmonics of mains. No way does it contribute to correct operation of anything.

 

[Ingenieur]

QUOTE="Ingenieur, post: 1069140, member: 30993"]
The only room he needs is a classroom.

A 0% THD input's output is no different than one with 5% THD.
The harmonics do not increase/decrease output.
Drop across the filter may.
No different than a high Z feeder

Btw, your 200 W will not do 200 J/sec of work. e, heater, motor, etc.

There is a reason power is rated 'average' based on RMS values.

RMS Power vs. Average Power | Analog Devices

It depends on how you define rms power.You do not want to calculate the rms value of the ac power waveform. This produces a result that is not physically meaningful.You do use the rms values of voltage and/or current to calculate average power, which does produce meaningful...

www.analog.com

How much power is dissipated when a 1 V rms sinusoidal voltage is placed across a 1 Ω resistor?

This is well understood1 and there is no controversy here.

Now, let’s see how this compares with the value from an rms power calculation.

Figure 1 shows a graph of a 1 V rms sinusoid. The peak-to-peak value is 1 V rms × 2 √2 = 2.828 V, swinging from +1.414 V to –1.414 V.2
The power dissipated by a sinusoidal 1 V rms across a 1 Ω resistor is 1 W, not 1.225 W. Thus, it is the average power that produces the correct value, and thus it is average power that has physical significance. The rms power (as defined here) has no obvious useful meaning (no obvious physical/electrical significance), other than being a quantity that can be calculated as an exercise.

It is a trivial exercise to perform the same analysis using a 1 A rms sinusoidal current through a 1 Ω resistor. The result is the same.

Power supplies for integrated circuits (ICs) are generally dc, so rms power is not an issue for IC power. For dc, average and rms are the same value as dc. The importance of using average power, as opposed to rms power as defined in this document, applies to power associated with time-varying voltage and current—that is, noise, RF signals, and oscillators.

Use rms voltage and/or rms current to calculate average power, resulting in meaningful power values.

 

[Ingenieur]

Average power

rms values
P = V I, average power

Peak values
P = V/sqrt2 I/sqrt2 = V I / 2 ... AVERAGE

Peak power (no real meaning)
Using RMS values
V sqrt 2 x I sqrt2 = 2 V I, 2 x average

That is why was at 100 and you 200

 

[audio2design]

It amazes me that he still doesn't get this simple point. There is so little power in the harmonics of mains. No way does it contribute to correct operation of anything.

Your lack of knowledge is showing. Big difference between measuring and doing. The voltage may be at 60Hz. The current and power is not. Those are defined by the load, not the source. What if the source was DC? If you put a filter inline with the load what would happen? That is entirely dependent on the load. That should be obvious but apparently not.

I have shown many actual simulations of real world implementations simplified that illustrate the effect of the load on what the filter will do. It's time to go back to some basic circuit theory. In addition to audio I have done a lot of power supply work, everything from a few watts to KW. I know how this stuff works, I know how it effects the line, I know what effect filters have, I know how it effects what it is connected to.

If you filter down to 120hz on the AC input of linear supply you break it. To expect that is wrong and using it as a criteria of good or bad is poor engineering and misleads others less knowledgeable. The voltage will be all over the place as the load changes. Period. Not debatable. That is EE201. I showed it clearly in simulations using realistic values.

 

[Ingenieur]

Your lack of knowledge is showing. Big difference between measuring and doing. The voltage may be at 60Hz. The current and power is not. Those are defined by the load, not the source. What if the source was DC? If you put a filter inline with the load what would happen? That is entirely dependent on the load. That should be obvious but apparently not.

I have shown many actual simulations of real world implementations simplified that illustrate the effect of the load on what the filter will do. It's time to go back to some basic circuit theory. In addition to audio I have done a lot of power supply work, everything from a few watts to KW. I know how this stuff works, I know how it effects the line, I know what effect filters have, I know how it effects what it is connected to.

If you filter down to 120hz on the AC input of linear supply you break it. To expect that is wrong and using it as a criteria of good or bad is poor engineering and misleads others less knowledgeable. The voltage will be all over the place as the load changes. Period. Not debatable. That is EE201. I showed it clearly in simulations using realistic values.

V 60 Hz
Line I 60 Hz basis (120 if you consider the rectified flow when the C is discharged.
But not really since it does not cross zero.
Your examples are bogus since the line supply does not have a pf, ie, lagging and absorb reactive power.

Power in 120 Hz

A perfect 60 Hz 0% THD line source and one with 5% will show no difference.

If it did PS would not work on high quality UPS units with <1% THD.

You are way out of line with the demeaning nd insults. I'm done with your BS
MSEE my @$$

 

[audio2design]

Just an FYI, the diodes in the circuit means it is no longer a linear circuit hence you can't use simple linear circuit analysis any more.

 

[Ingenieur]

It amazes me that he still doesn't get this simple point. There is so little power in the harmonics of mains. No way does it contribute to correct operation of anything.

He lives in a theoretical bubble.
He lacks the fundamentals.

If I cared I'd model a PS with a filter.

 

[Ingenieur]

Just an FYI, the diodes in the circuit means it is no longer a linear circuit hence you can't use simple linear circuit analysis any more.

Really?
You better explain that to the author of numerous text books.
It is linear, as in LINEAR power supply.
It has a discontinuity.
It is a nonlinear device but can be drawn as an equivalent and analyzed with linear methods and graphically.

Your 200 W is NOT 200 W
It is 100 W
Grasp that

 

[Ingenieur]

It's amazing that they figured out how to handle these things.