ISOTEK EVO3 Aquarius Power Conditioner Review

[amirm]

One of the lithium ones?

Yes. it is this one: https://www.amazon.com/gp/product/B00DW2J7SY/ref=ppx_yo_dt_b_search_asin_title?ie=UTF8&psc=1

 

[jam]

Another great eye-opener of a review on power contitioners Amir. I've been so close to pulling the trigger on these kind of expensive PCs many times over the years. I just was hoping to see some empirical hard data on them at some point. ASR has made this wish come through, thank you!

What I find really funny in retrospect is all the great accolades we've read throughout the years from the hi-fi press, like the ones you've listed in your review from some highly respected professional reviewers.

 

[DonR]

Isn't there a way to disable that beeping alarm? Some settings I didn't bother to read about? I have 2 of those CyberPower units from Costco I'm using with a couple PC's and most of my main AV system.

I surgically remove the transducers with a pair of sidecutters. Worked on my CyberPower and my APC UPS's.

Now for my review of my tweak: Without the transducers, stereo imaging is greatly enhanced, it's like a veil has been lifted, much more air and space and the bass is notably tighter. Vocalists are much more present without being shrill and something about sibilance... did I get that right?

 

[elvisizer]

My home is in the state of Jefferson

JFC NO IT ISN'T you live in california ya looney

 

[DualTriode]

JFC NO IT ISN'T you live in california ya looney

If you do not have a cognitive disorder you might already know that the proposed State of Jefferson is made up of the rural counties of NorCal and southern Oregon.

If there is a 4-wheel drive pickup truck in most every driveway you might be in the State of Jefferson.

 

[ousi]

AC filter not so much. Likely the only one that's even worth considering are the Pure-Sine-Wave online UPS which regenerates 50/60Hz 120VAC from its battery. I don't know if it is true that those heavy and expansive PS Audio power "regenerator" do the same too. Would be interesting to see that getting tested

 

[Billy Budapest]

Anybody know what the upright square thing in the front right corner is? Some sort of circuit breaker, maybe?

 

[DonR]

View attachment 182598Anybody know what the upright square thing in the front right corner is? Some sort of circuit breaker, maybe?

Going by the instruction manual it appears to be a 16A breaker (step 4). I am surprised that no reviewer has mentioned how the breaker sonically affects equipment hooked up to the unit. Does it "restrict" the bass? Perhaps it veils the highs? Maybe it affects the soundstage moving instruments around like a poltergeist does to chairs. Who knows.

 

[Ingenieur]

No, low impedance or your DC voltage droops. The diode bridge and capacitor effectively forms an envelope demodulation function (AM). If all you pass is 60Hz, then the output of the full wave bridge into a capacitor under load will just be the average value of the rectified voltage not the peak. I.e. a 30V RMS AC transformer will yield (1.414*30-0.7)*0.637 volts under heavy load as opposed to close to 1.414*30-0.7

1 phase full wave Vavg will always be:
2 sqrt(2) / Pi x Vrms
Current harmonics have no role
0.90 x Vrms
Your 0.637 = 2/Pi
You will still have 2 'pulses' of Pi radian each when integrated

 

[Ingenieur]

I'd like to see the line/primary V vs the secondary V out of a good audio PS xfmr..
At rated load, eg, 500 VA xfmr, 30 Ohm.

 

[audio2design]

1 phase full wave Vavg will always be:
2 sqrt(2) / Pi x Vrms
Current harmonics have no role
0.90 x Vrms
Your 0.637 = 2/Pi
You will still have 2 'pulses' of Pi radian each when integrated

Are you still going on about this? You seem to have some fundamental disconnect in how a linear power supply works. Hint, the harmonics are a factor of the capacitors, load, transformer, filtering inductance, etc.

Rectified vout will vary between (sqrt(2) * Vrms - 2*Vdiode) and something lower depending on those factors. If you had a perfect filter that passed only 60Hz and under (on AV side) the output voltage rectified on the caps will be (2/pi* Vrms - 2*Vdiode) assuming no other losses.

Here, maybe this will help you:

ISOTEK EVO3 Aquarius Power Conditioner Review

So an UPS is functionally similar to an audio amplifier? I mean, it makes sense, since an UPS uses batteries for backup power, so the AC-DC-AC conversion happens in both devices. But I'd still like to see an actual amp power a HiFi system, just for the fun :) I think that’s what PSAudio...

audiosciencereview.com

 

[Ingenieur]

Are you still going on about this? You seem to have some fundamental disconnect in how a linear power supply works. Hint, the harmonics are a factor of the capacitors, load, transformer, filtering inductance, etc.

Rectified vout will vary between (sqrt(2) * Vrms - 2*Vdiode) and something lower depending on those factors. If you had a perfect filter that passed only 60Hz and under (on AV side) the output voltage rectified on the caps will be (2/pi* Vrms - 2*Vdiode) assuming no other losses.

Here, maybe this will help you:

ISOTEK EVO3 Aquarius Power Conditioner Review

So an UPS is functionally similar to an audio amplifier? I mean, it makes sense, since an UPS uses batteries for backup power, so the AC-DC-AC conversion happens in both devices. But I'd still like to see an actual amp power a HiFi system, just for the fun :) I think that’s what PSAudio...

audiosciencereview.com

Ignore the diode forward bias.
It will ALWAYS be 2 x 2^1/2) / Pi x Vrms or
2 x Vpk / Pi
Power may compromised if fx > fo are filtered

Please do not speak down nor condescending to me.

I don't need links to mfgs. websites, I had a quality education.

You can see from the Fourier series the harmonics are accounted for..
Without them still Vim x 2 / Pi
Vrms = Vim/sqrt(2) rearrange & substitute
Vim = sqrt(2) x Vrms
Vl = sqrt(2) x Vrms x 2 / Pi

 

[audio2design]

Please do not speak down nor condescending to me.

I don't need links to mfgs. websites, I had a quality education.

You can see from the Fourier series the

View attachment 182747

I am going to if you keep posting to me and not use that education. Now put a large capacitor in parallel with that resistor. You keep writing stuff that is barely related to the actual situation.

 

[Ingenieur]

I am going to if you keep posting to me and not use that education. Now put a large capacitor in parallel with that resistor. You keep writing stuff that is barely related to the actual situation.

You are making yourself look bad.
In character and knowledge.

What do you think produces the harmonics? And why are they ignored in the EQUIVALENT circuit?

 

[audio2design]

What do you think produces the harmonics? And why are they ignored in the EQUIVALENT circuit?

Now do the equivalent full power circuit with diodes and storage capacitors and a load. There are significant current harmonics. This is not "controversial".

 

[Ingenieur]

Now do the equivalent full power circuit with diodes and storage capacitors and a load. There are significant current harmonics. This is not "controversial".

You never answer a question, only deflect.
That is the PS filter. As you can see the series is the same with the exception of full wave. They show 1/2 first in the text to increase complexity as comprehension increases.

It is ignored in the EQUIVILENT circuit.
What produces the harmonics?
Why is it ignored?

The answers are in the pics.

 

[Ingenieur]

You do not want the harmonics/ripple, they are filtered to a low level. They will be produced regardless if present or not on the line supply side.

Even if all produced harmonics are removed from the rectified output Fourier series you STILL have
2 x sqrt(2) / Pi x Vrms

THAT is not 'controversial'
You can lead a horse to water, but, .....

 

[audio2design]

You do not want the harmonics/ripple, they are filtered to a low level. They will be produced regardless if present or not on the line supply side.

Even if all produced harmonics are removed from the rectified output Fourier series you STILL have
2 x sqrt(2) / Pi x Vrms

THAT is not 'controversial'
You can lead a horse to water, but, .....

View attachment 182763

I have lost so much interest in this i am not sure even what you are going on about any more.

All your pictures and formulas don't take into account the load which absolutely is part of the circuit nor the rectification which together impacts what current harmonics show up on the AC line.

The output of rectified AC will be sqrt(2)*Vrms with no loading and no filtering, but if you add filtering and load, then the rectified voltage on those capacitors drop. If that is controversial to you then you don't understand the basic premise. May I suggest prototyping or simulation as I showed you above.

 

[Ingenieur]

I have lost so much interest in this i am not sure even what you are going on about any more.

All your pictures and formulas don't take into account the load which absolutely is part of the circuit nor the rectification which together impacts what current harmonics show up on the AC line.

The output of rectified AC will be sqrt(2)*Vrms with no loading and no filtering, but if you add filtering and load, then the rectified voltage on those capacitors drop. If that is controversial to you then you don't understand the basic premise. May I suggest prototyping or simulation as I showed you above.

Again, you digress, obfuscate. Those 'pictures and formulas' are what engineers use to depict systems like this.
VL is the load V
Load determines the I, not V, if properly designed, ie, if the engineer did their job.

The DC component will be 2 sqrt(2) / Pi

2 points
1 the analysis ASSUMES perfect 60 Hz (or whatever) and 0 harmonics. So if the line is is filtered, it does not matter contrary to your opinion that line filtering impacts DC output.

2 harmonics do not matter
Input, as noted above in 1

Output, they are undesirable, you filter/attenuate them.
In the text they assume RC to be 100/wo
n > 1 ~ harmonic
And ripple amplitude to be VL/(100 x n)
For 2nd harmonic ~ 0.5%
3rd 0.33
4th 0.25
....

I'm done, the reader can decide.

 

[Ingenieur]

I have lost so much interest in this i am not sure even what you are going on about any more.

All your pictures and formulas don't take into account the load which absolutely is part of the circuit nor the rectification which together impacts what current harmonics show up on the AC line.

The output of rectified AC will be sqrt(2)*Vrms with no loading and no filtering, but if you add filtering and load, then the rectified voltage on those capacitors drop. If that is controversial to you then you don't understand the basic premise. May I suggest prototyping or simulation as I showed you above.

With no filtering it will be the integral of
1/2 wave (x 2 full wave)
1/(2 Pi) Int[(V df]evaluated +/- Pi/2
= Vl/Pi = sqrt(2) Vrms / Pi 1/2 wave
x 2 for full wave
Or the average area under 1/2 cycle