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Yes. it is this one: https://www.amazon.com/gp/product/B00DW2J7SY/ref=ppx_yo_dt_b_search_asin_title?ie=UTF8&psc=1One 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=1One of the lithium ones?
I surgically remove the transducers with a pair of sidecutters. Worked on my CyberPower and my APC UPS's.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.
JFC NO IT ISN'T you live in california ya looneyMy home is in the state of Jefferson
JFC NO IT ISN'T you live in california ya looney
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.View attachment 182598Anybody know what the upright square thing in the front right corner is? Some sort of circuit breaker, maybe?
1 phase full wave Vavg will always be: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
Ignore the diode forward bias.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
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
You are making yourself look bad.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.
What do you think produces the harmonics? And why are they ignored in the EQUIVALENT circuit?
You never answer a question, only deflect.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 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
Again, you digress, obfuscate. Those 'pictures and formulas' are what engineers use to depict systems like this.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.
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.