Why do I need power management if I have a great power cord?

The problem with this kind of pissing contest is that everybody gets wet and nobody's happy.

@raysmtb1,

All good!😊

@Bruce19

Basic physics.  Everytime you double the size of a conductor, you reduce the resistance of the conductor by half.  Output impedance is the product of LCR.  Reducing any aspect of the impedance allows the available current to increase (the current well).  Instantaneous power demands in audio are not often spoken about.  Take the Telarc recording of the 1812 Overture as an example.  Someone smarter than me did the calculations on the reproduction of the cannon shots in that recording.  To accurately reproduce that event considering the average listening level of 2 watts would require 10,000 watts of instantaneous power.  Point of the exercise was to illustrate the power demands on a musical system.  I don't know of any presently available amplifier to produce that type of current dump.  If there was an amp capable of  that considering that most amplifiers (not talking class D here) are somewhere between  50%, - 78% efficiency.  Do the math.  The power distribution cable from the transformer to the entry service is #2 wire.  It just makes sense, logically and mathematically, to supply a dynamic system with huge instantaneous current demands as much current as possible.  A Class A PASS XS300 monoblock will dump 48A on demand.  You need two of these.  See where this is going?  

Solid state systems are current hogs.  Ever feed a hog? 

@jumia,

My two cents. If you have clock radio as you source of music any power cord will work and everything will sound the same. If you have something more resolving and you have good hearing, then everything makes a difference. My ears are definitely not what they were, but with my equipment, I can hear the difference between lamp cord and 12 gauge romax and a $300 power cord that was designed hifi amps. And then there comes the point of diminishing returns, but that’s a whole other story.

All the best.

Basic physics. Everytime you double the size of a conductor, you reduce the resistance of the conductor by half. Output impedance is the product of LCR. Reducing any aspect of the impedance allows the available current to increase (the current well). Instantaneous power demands in audio are not often spoken about. Take the Telarc recording of the 1812 Overture as an example. Someone smarter than me did the calculations on the reproduction of the cannon shots in that recording. To accurately reproduce that event considering the average listening level of 2 watts would require 10,000 watts of instantaneous power.

The capacitors only get filled up when they are at or below the voltage coming out of the transformer-rectifier-regulator. That physically is happening in a rectified sine wave that operating at 120 Hz. So there is a great deal of time (maybe a 1/4 of the time), where the AC is NOT feeding the capacitors any DC.

The power supply in an amplifier is holding DC and buffering the AC to provide that energy reserve..

This is pretty much analogous to how a toilet works. The thing flushes very quickly, but takes many seconds to fill the bowl. The pressure and size of the pipes coming into the house do not go directly into the flushing… they merely keep the bowl filled for subsequent flushes, thereby buffering the demand for water.

The power distribution cable from the transformer to the entry service is #2 wire.

For the entire house… running an electric oven, AC, etc. It is not for a single (0r pair) of MB amps that draw 900W (~9A) each.

It just makes sense, logically and mathematically, to supply a dynamic system with huge instantaneous current demands as much current as possible.

Do the lights in the house flicker whenever the bass drum hits?
If not, then the amp is not pulling that output power directly from the input AC.

It is buffering the energy in the capacitors.

A Class A PASS XS300 monoblock will dump 48A on demand.

How do we arrive at 48A? Are you using 8 ohm speakers? Or what is their impedance?
300W at 8 ohms implies “rails” at ~50V, and at 8 ohms that is about 6 amperes.
(Driving a 1 ohm load can get us to 48A - so I guess you have 1 ohm speakers?)

In any case that amp is a device providing voltage amplification and impedance matching. It is not forcing 48A into the speaker, it is only providing up to 50V. The speakers get whatever ohms law translates into in terms of current.

You need two of these. See where this is going?

Not exactly… no.