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 A40 Clipping Point
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 23rd May 2014, 01:08 AM #1 diyAudio Member   Join Date: Jun 2007 A40 Clipping Point Can someone help with understanding Clipping in Pass A40? Feeding a 1khz signal into a 5.5 ohm resistor. The amp puts out 19.8 vrms when it justs starts clipping. The voltage rails are at +/- 31vdc. Don't think anything is wrong just want to understand the math.
 23rd May 2014, 01:27 AM #2 diyAudio Member   Join Date: Nov 2009 E squared over load or 19.8 x 19.8 = 392 divided by 5.5 = 71.28 watts
 23rd May 2014, 02:56 AM #3 diyAudio Member   Join Date: Jun 2007 Yes I understand that. I'm running dual mono with 650VA xfmr's per side. So surely my supply voltage isn't sagging at the 19.8 volts correct? I realize I'm dropping some across the emitter resisters but only about 6.5vdc per channel. Again just want to understand the specifics. BTW power supply is CLC. Might need to look across inductors.
 23rd May 2014, 03:11 AM #4 diyAudio Member     Join Date: Nov 2007 Location: Near Dallas Texas USA 19.8 VRMS is 28V Peak. So it seems your amp is operating normally.
 25th May 2014, 08:58 AM #5 diyAudio Member   Join Date: Jul 2004 Location: Scottish Borders find info on Vpk (peak voltage of an AC signal) Vpp (peak to peak voltage of an AC signal, measured by reading the oscilloscope waveform) Vac (the measured voltage using an AC voltmeter) Vrms (the heating effect of the AC waveform) Vdc (the steady voltage from direct current supply or load) Vripple (the variation of voltage on a nominal DC supply). Then read about sinewave signal and how to convert from/to Vac, Vrms, Vpk, Vpp __________________ regards Andrew T. Last edited by AndrewT; 25th May 2014 at 09:00 AM.
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Quote:
 Originally Posted by Freecrowder Can someone help with understanding Clipping in Pass A40? Feeding a 1khz signal into a 5.5 ohm resistor. The amp puts out 19.8 vrms when it justs starts clipping. The voltage rails are at +/- 31vdc. Don't think anything is wrong just want to understand the math.
lazy to calc exact losses across mosfet-Rs combo in peak , but assume that you're having 60V clean PSU , including losses resulting in some 55Vpp sine

then divide with 2 to have Vp

then divide with 1.41 (root of 2) to have Vrms

P=(Vrms^2)/R

so , P = (((55/(2 x 1.41))^2)/8=47W

incidentally , that's exactly around 19.5Vrms

disclaimer - I really don't care how much Wpeak is that

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Join Date: Jan 2003
Location: ancient Batsch , behind Iron Curtain
Quote:
 Originally Posted by Zen Mod ......exact losses across mosfet-Rs......
meaning , of course , about bjt-Re , but I'm sure that everyone is already accustomed on Mighty ZM's brainfarts

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