@simon7000, I'm pretty sure the mains voltage spec would swamp any drop on a severely under rated line cord.
The numbers didn't help illustrate what your point is, if any.
The numbers didn't help illustrate what your point is, if any.
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At a 50% voltage drop you get 1/4 power if the amplifier were a LINEAR LOAD.
Do you have 100% ripple on your power supply capacitors?
OK, let's look at this from a different angle. How can a standard, good quality 14 guage power cord that is drawing from a standard 15 amp circuit that is wired with the standard 14 guage romex cause a 50% power loss all by itself? It would not create any more loss than the house wiring. And if the house wiring is causing that much loss, how can changing the power cord fix it? The numbers just don't support your claim.
Mike
Mike
Ed is apparently postulating someone hooking up their 500 watt amp through a 24 gauge linecord which loops over several rooms.😀
The mains resistance is under .3 ohms. Read the electrical code or figure it out for yourself. At a residential 200 amp service what must the resistance be for less than a 5% drop? (5% 0f 240 V = 12V 12V/200A= .06 Ohms!)
Or just measure your AC voltage and then plug in a 750 watt electric Iron and measure the drop. Now plug the iron into an extension cord and see how much more drop there is at the end of the cord. (750 W at 120 V = 6.25 A. 5% of 120 V = 6 V 6 V/20=.3 ohms so .36 ohms at 6.25 A = 2.25 Volts drop. Extension cord at 6' 16 Ga with 2 connectors at .03 ohms ea. = .108 ohms or .675 more volts drop.)
Now measure the ripple on the power supply capacitors in your power amplifier and let me know what amount of the AC sine wave will charge them. Multiply that times your extension cord drop, and then square it for power loss! (500 Watts into 8 ohms = 63 volts, rail =1.414*63 +5 (Ripple and transistor losses)=94 volts arcsin 89/94 =71 degrees 90-71=19 degrees of charging time. 180/19 =9.2 so peak current should be 9.2 times the average current for a 500 watt amplifier!)
I had a fellow work for me who had just gotten his EE degree, after it became obvious he had managed to avoid learning the basics, I mentioned this to one of the faculty. They reviewed his course of study and then saw to it the the department changed the requirements so that no one would ever be able to get such a degree again without having at least being exposed to the basics!
This ain't rocket science, it is very basic AC analysis.
The electrical code suggests a maximum of 5% drop from the pole to the panel and again 5% from the panel to the outlet. So even with a uniform current draw you would get 10*Log((.95*.95)squared) = .89 db power loss.
Now tell me the amplifier current draw is uniform across the entire AC cycle!
Most houses have less than the maximum AC line drop and well designed linear power supplies have a peak current of around 3 times what the average should be. Run the math and you should get the loss of full power is 3db. Now for short peaks the stored charge in the capacitor bank will help, but as you get a peak past 8.33 mS then the power line loss comes into play.
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You need to learn how these power supplies work. Add resistance to the supply line and watch the charging pulses flatten out but still charge the caps to almost the same voltage. The pulse voltages arent as high but there duration increase so they deliver almost the same power. Do I need to show you some sims?
No show me some measurements! I have done them. The inductance of the transformer and other losses do extend the charging time that is why the calculation shows 9 x when the measured is 3 x for decent supplies and maybe 4.5 x for "Audiophile" designs.
But try simulating 100,000+ uf filter banks with 1 volt of ripple and let me know what peak currents you get.
well yeah, but the grid voltage varies more than what the amplifier cable would cause (unless it is under rated). I just bump started my whole amp rack (1750w) and the line drop was .2 volts, although that could have been caused by the neighbors air conditioner kicking on.
5% drop with resistive load. Now assume 1:5 pulse ratio, and get more like 25% drop. Ohm's law still works, but you have to correctly apply it. Its the same process as in a power supply: an ohm in series with the secondary/rectifier drops much more voltage than an ohm after the reservoir cap, even though the average current is the same.
One possible solution is to use a choke input PSU, as that spreads the current more evenly through the mains cycle.
One possible solution is to use a choke input PSU, as that spreads the current more evenly through the mains cycle.
Now load your amplifiers to 1750 watts and measure the voltage. Of course you should blow the breaker.
They were fully loaded, (i was wearing ear protection, scared the heck out of the dog though) The run for the amps is 12 AWG, outlet is hospital grade 20amp on a dedicated 20amp circuit. The run from drop is 14ft. 🙂 average voltage is 120.4v
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With a sine wave? Or music with 10 db or more of headroom? (Don't try it with a sine wave you will blow up your loudspeakers.) (Also loudspeakers when run at full power often double their impedance.)
Did you measure the drop across the power cord? That is the issue here. So if you are crazy enough to try it see what the instantaneous peak voltage drop across the power cord is. You will need a storage scope for this.
music (Enya if it matters), the system was set at -12db. I measured the drop at my Furman PL-PRO DMC because it would be hard to measure 6 amplifier power cords simultaneously.
So allowing for 12 db gain room and only 20 db dynamics your .2 volts would increase to 7.96 volts. Assuming the .2 was the peak and not an RMS reading. So for a very tightly wired system your peak power loss would be .6 db. Can you hear this? Don't know.
(Of course if you were measuring the drop as RMS then there is a good chance the actual peak power loss is close to 3db!)
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Measurements mean nothing if not done correctly (with a scope) or applied correctly.
You still dont understand the basic cap charging cycle. The pulse duration increase is not because of the inductance etc. its because the current keeps flowing untill the cap voltage is a diode drop away from the line voltage, the less current flow the longer it takes for this to occur. Do some homework instead of falling back on misinformation (over and over).
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