Headroom Loss for 1600w on 14ga/120v


Headroom loss for 14ga vs 10ga is less than ¼db @ ≈14A / ≈1600W
See http://ielogical.com/Audio/CableSnakeOil.php#ACWiring
128x128ieales
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it was obvious from the OP alone that this is all about a guy with zero listening skills and even less understanding of audio
@millercarbon: I've been listening for more than ½ century. No one who has heard my HiFi in all that time has ever had a negative comment. One fellow recording engineer once said "It sounds too good. It exposes just how awful are pop records."

I was nominated for a Grammy, invented the Vocal Splicer, designed and built 6 & 8 channel film monitors, wired some of the largest LA studios, have a patent for digital patch recording, was head of analog development for Neve, had my equipment installed in no name places like Sony, Sound One, Hit Factory, Abbey Road, NHK, CBS, Paramount, etc. So if you have any recorded analog material from the early '80's through the '00's, there is a chance that somewhere in the line, it passed through something I designed.

There is so much B.S. in HiFi postulated by people that haven't got a clue:
I mean, headroom, when what he probably means is loss.
Headroom is the delta between rated and peak output. Power line resistance reduces the voltage available to the transformer which charges the power supply capacitors. The amplifier rated power does not change. Hence, headroom decreases.

For a 100w/8Ω amp, output voltage is ≈28V. [V=(PR)^½]. A ±30v transformer will give about 35V on the filter caps. Assuming the transformer can supply the current, head room is 20*log(35/28) or +1.94db. If the line voltage drops 3v from 120 to 117 due to wire resistance, the cap voltage is going to drop 117/120*35=34.125. The headroom is now +1.72db. 20log(117/120)=-0.22db. 1.94-1.72=-.22db. 

It's simple math.

Increasing to 10ga reduces the voltage drop to ≈1.2v. 20log(118.8/120)=-.09db. A 0.13db improvement!

Our ±30v transformer has  ≈4:1 turns ratio. At 90% efficiency, that gives a 3.6:1 current increase from the supply side. Load current 100WPC x 2 into 8Ω is about 2A from the wall. [28/8*2/3.6=1.94] A 2Ω load draws ≈7.75A. Note that this is continuous. Most program isn't. Class AB amps, which is what power most HiFi, are ≈50% efficient at full power, so there is still plenty of current available. 
 
What many fail to comprehend is the powerline is only supplying current a fraction of the time. See http://sound.whsites.net/power-supplies.htm#s51 and scroll down to Figure 3. Note that the current is not a sharp transient. In the image, the current wave shape is about 150Hz [50Hz/≈0.33]. For typical music program, the duty cycle is lower. Even the most compressed pop still has about a 10db dynamic range, which is a 10:1 power ratio. Hence, most of the time, the line draw is well below maximum.

Imagine an amplifier putting out line frequency out of phase with the line. This is an acid test for a power supply as the capacitor ripple current will be maximum. The transformer will only charge the capacitors on the opposite phase. The transformer supplies ZERO current to the load.

Transients are supplied by the capacitors ALONE except when the transient occurs when the line voltage is in the same phase and sufficiently high to forward bias the power supply diodes.

If the system requires more than 15A, i.e. > ≈1800W, both heavier wire and breakers should be installed. Most systems do not fall in that range.
OOPs - just realized I had my 240v hat on doing the calcs above. ±30V is 2:1 turns ratio for 120v. Wall current draw is double.
Mea Culpa.