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.
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.
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
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
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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
meaning , of course , about bjt-Re , but I'm sure that everyone is already accustomed on Mighty ZM's brainfarts
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