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17th January 2013, 04:45 PM  #8151 
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I'm using 6volt rms when playing loud , ok its nominal 1ohm (30022k) , only man amplifiers need apply ...
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17th January 2013, 04:56 PM  #8152 
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let's see, Apogee?
(look, it rhymes!)
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17th January 2013, 05:21 PM  #8153 
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Close , but no apogee , This is what I'm currently using ...
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17th January 2013, 05:22 PM  #8154 
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DIY? really have no idea.
I'm considering the La Folia, just out of masochism. obviously, my girly classD amp won't do
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17th January 2013, 05:29 PM  #8155 
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Yes DIY, 3way direct drive ribbon (no transformers) hybrid.
Bel Canto classd mono's did not work , as well as others . Krell,Threshold ,PS audio, adcom(565/555) are the few that has worked , i did not like the sonics of the adcoms .. The PS audio works amazingly well considering 3pr output ....
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17th January 2013, 06:00 PM  #8156 
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Ncore never got back with me on their NC400 , so i will pass ...
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17th January 2013, 11:03 PM  #8157  
diyAudio Member

Quote:
Frank 

18th January 2013, 02:21 AM  #8158 
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Then it was not a monster amp
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18th January 2013, 03:53 AM  #8159  
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Quote:
The calculation example that you quoted was not calculating the required reservoir capacitance for a power supply. It was merely to help illustrate what types of calculations could be performed, with an equation that had been derived. In that particular example, the result meant that IF you needed a capacitance to supply all of the current for the first zerotopeak rise of a 5AmpPeak 30 Hz sine wave, then if that capacitance was not at least 53050 uF, the voltage across the capacitance would drop by at least one volt. The result was subsequently verified with an LTSpice simulation. That equation was actually just an approximate result; a sort of "worstcase estimation" equation: (13b): C ≥ a / (πfΔv_MAX) where a is the 0topeak amplitude of the capacitor's sinusoidal current waveform, f is frequency in Hertz, and C is capacitance in Farads. The morecomplete picture, for the scenario being considered in that derivation, is this one: (17): C ≥ Δi / ( 2πf∙(Δv  (ESR∙Δi))) Equation (17) gives the capacitance value, C, that would be required in order to supply the current for the first quartercycle of a sine signal of frequency f (in Hz), with 0topeak amplitude Δi Amperes, while causing the voltage across the capacitor to dip by no more than your choice of Δv Volts. To use equation (17), it will be easier to first set ESR to zero, calculate a C value, find an estimate for ESR for that C value at the frequency being used, and then recalculate the C value with the ESR value. If we still want to just account for the whole sine wave, we should be able to simply double the C value given by equation (17), since we're considering only the positive or negative halfcycle, but not both, and the other half of the halfcycle of a sine wave is symmetrical and thus encloses the same area (its integral), i.e. the same ampseconds value, as the first half. [Edit: Note, too, that positive capacitor current was defined as current flowing into the positivevoltagedesignated lead of the capacitor. So, typically, for this scenario, both Δi and Δv will be negative.] For more of this exciting story, including some of the considerations involved when Δv  (ESR∙Δi) gets close to zero, go to Power Supply Resevoir Size Tom Last edited by gootee; 18th January 2013 at 04:19 AM. 

18th January 2013, 04:20 AM  #8160 
diyAudio Member

What's in a name ...?
I remember a hulking GAS being quite a disappointment ... Krell was OK, but tends to sound fairly rough. ME, an Australian brand with an intelligent power supply for its time, a vast array of paralleled small value smoothing caps, did pretty well ... Frank 
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