ISOTEK EVO3 Aquarius Power Conditioner Review

[audio2design]

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

I will let you think on why these simulations mean the DC rectified voltage will vary w.r.t. filtering.

 

[Ingenieur]

I will let you think on why these simulations mean the DC rectified voltage will vary w.r.t. filtering.

View attachment 182791

View attachment 182792

View attachment 182793
View attachment 182794
View attachment 182796

That means nothing.
Include xfmr L
Show output V rms across R2
Show I at R2
Show scales
Give xfmr specs
The load is too large for that tc 0.75 sec
~13.5 A

You are displaying I thru R5, not the load
Magnitude at each harmonic
Summed as a % of distortion

The first 2 an ideal supply
I assume the last one shows a supply with harmonics.
It has higher I distortion, this is contradictory to your assertion.

 

[audio2design]

That means nothing.
Include xfmr L
Show output V rms across R2
Show I at R2
Show scales
Give xfmr specs
The load is too large for that tc 0.75 sec
15 A
You are depleting the caps


You are displaying I thru R5, not the load
Magnitude at each harmonic
Summed as a % of distortion

The first 2 an ideal supply
I assume the last one shows a supply with harmonics.
It has higher I distortion, this is contradictory to your assertion.

- The load is 8 ohms approx on a 4:1 off 120VAC, so call it 200 watts. Very appropriate loading for 10,000 or 100,000uF.
- The resistor is the resistor that was put in the path to show the current after the bridge that will go into the C\load (which of course you can reflect back to the input)
- If I include transformer L, that would be a FILTER, but I was trying to illustrate the real harmonic content of the load, because if you filter that content, the voltage must drop, as that power is lost
- These are all pure 60Hz, 0 distortion AC source. I am not distorting or contradicting anything. I am showing real world linear power supply (sans the perfect AC line and perfect transformer). I could add leakage inductance to the transformer, and some eddy losses, and model a source impedance and resistance on the AC line if you would like, but that will just prove my point even more about DC voltage at the caps dropping
- These are all the same by the way, all I changed was the capacitance, so how are the first 2 any different? The only difference in the last simulation is the simulation is not long enough so the large capacitor charging time impacts the displayed spectrum as there is too much not steady state.

The simulations are right there. They give you everything you need to know about what MUST happen to the DC voltage IF YOU FILTER.

 

[Ingenieur]

Assume xfmr 1:1
Vc ~ 108 V
I load ~ 13.5 A
P ~ 1460 W or 1480 J/sec

C energy = 1/2 C V^2 ~ 580 J

Make the load 80 Ohm 145 W (J/Sec)
Or the caps 2 F

Get back to me

 

[Ingenieur]

- The load is 8 ohms approx on a 4:1 off 120VAC, so call it 200 watts. Very appropriate loading for 10,000 or 100,000uF.
- The resistor is the resistor that was put in the path to show the current after the bridge that will go into the C\load (which of course you can reflect back to the input)
- If I include transformer L, that would be a FILTER, but I was trying to illustrate the real harmonic content of the load, because if you filter that content, the voltage must drop, as that power is lost
- These are all pure 60Hz, 0 distortion AC source. I am not distorting or contradicting anything. I am showing real world linear power supply (sans the perfect AC line and perfect transformer). I could add leakage inductance to the transformer, and some eddy losses, and model a source impedance and resistance on the AC line if you would like, but that will just prove my point even more about DC voltage at the caps dropping
- These are all the same by the way, all I changed was the capacitance, so how are the first 2 any different? The only difference in the last simulation is the simulation is not long enough so the large capacitor charging time impacts the displayed spectrum as there is too much not steady state.

The simulations are right there. They give you everything you need to know about what MUST happen to the DC voltage IF YOU FILTER.

Nothing happens if you filter on the primary.
That is EXACTLY what the xfmr does.

Vdc ~ 27 V, I ~ 3.4 A, P ~ 90 W ???

Harmonics have nothing to do with it.
Only voltage drop across the line filters.
The can be sized appropriately to not attenuate the 60 Hz fundamental.

V THD is usually 5% on the grid
Moot

 

[Ingenieur]

Do you see the issue with you model?
The pf ~ 0.0035, purely reactive.

If the load is ~ 90 W (J/sec)
C energy 36 J (or W Sec)
tc ~ 0.75 sec
In 1 sec, 90 W (or J)
2.5 x C
In 1 sec 75% recharge

 

[audio2design]

Nothing happens if you filter on the primary.
That is EXACTLY what the xfmr does.

Vdc ~ 27 V, I ~ 3.4 A, P ~ 90 W ???

Harmonics have nothing to do with it.
Only voltage drop across the line filters.
The can be sized appropriately to not attenuate the 60 Hz fundamental.

V THD is usually 5% on the grid
Moot

No really you cannot. Simulate it and prove it.

10,000-100,000 uF on a 4:1 transformer (120VAC) with 200W load are very practical realistic values for real amplifiers properly designed. If you don't accept those values as realistic I can't help you.

Apply formulas without understanding the problem is not going to get you any farther ahead. I have shown you voltage droop with filtering, I have shown you the harmonics in the current waveform. Not much else I can do if you don't want to understand the actual problem not the one you think you are solving.

 

[Ingenieur]

No really you cannot. Simulate it and prove it.

10,000-100,000 uF on a 4:1 transformer (120VAC) with 200W load are very practical realistic values for real amplifiers properly designed. If you don't accept those values as realistic I can't help you.

Apply formulas without understanding the problem is not going to get you any farther ahead. I have shown you voltage droop with filtering, I have shown you the harmonics in the current waveform. Not much else I can do if you don't want to understand the actual problem not the one you think you are solving.

The load is not 200 W
27 V / 7.4 Ohm = 3.65 A, P = 100 W total
C is only 35 J
Change the load to 24 Ohm
C to 0.2 F

Reflected to the primary ~ 4^2 x -0.027j ~ -0.43 j Ohm
You do know what that means?
I = 120/-0.43 j = 280j A
S = -5.5j MVA
Continuous inrush current bursts
As I said, pf 0.00xx, purely reactive.

What engineering school doesn't teach the math representation of a system?
You derive the system from the math to understand it. Then you confirm in labs.

Please, gain, do not demean me.
One of us 'understands'.
Respect wold be more coming face to face.

 

[audio2design]

Do you see the issue with you model?
The pf ~ 0.0035, purely reactive.

If the load is ~ 90 W (J/sec)
C energy 36 J (or W Sec)
tc ~ 0.75 sec
In 1 sec, 90 W (or J)
2.5 x C
In 1 sec 75% recharge

Actually, the power factor of this circuit, from the AC line view of course, is 0.53

 

[Ingenieur]

Actually, the power factor of this circuit, from the AC line view of course, is 0.53

View attachment 182853

The load pf is 0.0035
You've made an oscillator

 

[audio2design]

The load is not 200 W
27 V / 7.4 Ohm = 3.65 A, P = 100 W total
C is only 35 J
Change the load to 24 Ohm
C to 0.2 F

Like I said, you don't seem to really know how this works, and as opposed to trying to understanding it, you are dead set of being right. It is ~200W at 120VAC (nominal). at 108V, it is ~150W

 

[audio2design]

The load pf is 0.0035
You've made an oscillator

No, I made a linear power supply. The load PF, which includes the transformer, is 0.529 with an ideal transformer and the load as stated.

 

[Ingenieur]

Like I said, you don't seem to really know how this works, and as opposed to trying to understanding it, you are dead set of being right. It is ~200W at 120VAC (nominal). at 108V, it is ~150W

View attachment 182857

View attachment 182858

You do not understand power
W vs VA
some of the power goes to C, it is not power in W, does no work.

nor RMS = sqrt(peak/2)^2
If we use peak V 38 V I = 5.14, P = 196 W

Lol

 

[Ingenieur]

No, I made a linear power supply. The load PF, which includes the transformer, is 0.529 with an ideal transformer and the load as stated.

No
The pf of the bridge's load is 0.0035
Follow the math

Make the load 24
C 0.2 F

The load is > C

 

[audio2design]

You do not understand power
W vs VA
some of the power goes to C, it is not power in W, does no work.

nor RMS = sqrt(peak/2)^2
If we use peak V 38 V I = 5.14, P = 196 W

Lol

I understand power. I also understand how this circuit works, something you are apparently lacking.

I am showing you the circuit and the ACTUAL voltage on the load, not the one in your head. With a beefy transformer, it will be quite close to this too.

 

[Ingenieur]

I understand power. I also understand how this circuit works, something you are apparently lacking.

I am showing you the circuit and the ACTUAL voltage on the load, not the one in your head. With a beefy transformer, it will be quite close to this too.

It is not I who lack understanding


Again
24 Ohm load
0.2 F

 

[audio2design]

No
The pf of the bridge's load is 0.0035
Follow the math

Make the load 24
C 0.2 F

The load is > C

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.

 

[voodooless]

Maybe you two should get a room

 

[egellings]

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

 

[audio2design]

Maybe you two should get a room

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