How Much RMS can be obtained from a class D amplifier with 12V @ 60A? approx CHEERS
I icecoolwas Member Joined 2008 2009-01-31 3:47 pm #1 2009-01-31 3:47 pm #1 How Much RMS can be obtained from a class D amplifier with 12V @ 60A? approx CHEERS ••• More options Share Saturnus Member Joined 2005 2009-01-31 4:17 pm #2 2009-01-31 4:17 pm #2 Depends on impedance and effeciency. Maximum Power Output = voltage ^ 2 / impedance * effeciency This gives the theoretical maximum output at 100% distortion. To get a realistic value at roughly 1% distortion multiply that number with 0.8. To get the RMS output double impedance. ••• More options Share A Anthony C Smith Member Joined 2005 current-thinking.com 2009-01-31 4:27 pm #3 2009-01-31 4:27 pm #3 power (rms) = (V^2/2R)*efficiency for a sine wave since Vrms = Vpeak/root2. regards ••• More options Share I icecoolwas Member Joined 2008 2009-01-31 4:57 pm #4 2009-01-31 4:57 pm #4 Sorry I dont understand this.. But the Impendance is 2ohm. ••• More options Share Saturnus Member Joined 2005 2009-01-31 5:54 pm #5 2009-01-31 5:54 pm #5 Ok then, let's also assume that effeciency is 80% which is typical for a class D playing into 2 ohms impedance Then ... Watt (peak) = 12 * 12 / 2 * 0.8 * 0.8 = 46.08 and Watt (rms) = 12 * 12 / 4 * 0.8 = 28.8 ••• More options Share Show hidden low quality content Status This old topic is closed. If you want to reopen this topic, contact a moderator using the "Report Post" button. Share: Facebook Twitter Reddit Pinterest Tumblr WhatsApp Email Share Link
Saturnus Member Joined 2005 2009-01-31 4:17 pm #2 2009-01-31 4:17 pm #2 Depends on impedance and effeciency. Maximum Power Output = voltage ^ 2 / impedance * effeciency This gives the theoretical maximum output at 100% distortion. To get a realistic value at roughly 1% distortion multiply that number with 0.8. To get the RMS output double impedance. ••• More options Share A Anthony C Smith Member Joined 2005 current-thinking.com 2009-01-31 4:27 pm #3 2009-01-31 4:27 pm #3 power (rms) = (V^2/2R)*efficiency for a sine wave since Vrms = Vpeak/root2. regards ••• More options Share I icecoolwas Member Joined 2008 2009-01-31 4:57 pm #4 2009-01-31 4:57 pm #4 Sorry I dont understand this.. But the Impendance is 2ohm. ••• More options Share Saturnus Member Joined 2005 2009-01-31 5:54 pm #5 2009-01-31 5:54 pm #5 Ok then, let's also assume that effeciency is 80% which is typical for a class D playing into 2 ohms impedance Then ... Watt (peak) = 12 * 12 / 2 * 0.8 * 0.8 = 46.08 and Watt (rms) = 12 * 12 / 4 * 0.8 = 28.8 ••• More options Share Show hidden low quality content Status This old topic is closed. If you want to reopen this topic, contact a moderator using the "Report Post" button. Share: Facebook Twitter Reddit Pinterest Tumblr WhatsApp Email Share Link
Depends on impedance and effeciency. Maximum Power Output = voltage ^ 2 / impedance * effeciency This gives the theoretical maximum output at 100% distortion. To get a realistic value at roughly 1% distortion multiply that number with 0.8. To get the RMS output double impedance.
A Anthony C Smith Member Joined 2005 current-thinking.com 2009-01-31 4:27 pm #3 2009-01-31 4:27 pm #3 power (rms) = (V^2/2R)*efficiency for a sine wave since Vrms = Vpeak/root2. regards ••• More options Share I icecoolwas Member Joined 2008 2009-01-31 4:57 pm #4 2009-01-31 4:57 pm #4 Sorry I dont understand this.. But the Impendance is 2ohm. ••• More options Share Saturnus Member Joined 2005 2009-01-31 5:54 pm #5 2009-01-31 5:54 pm #5 Ok then, let's also assume that effeciency is 80% which is typical for a class D playing into 2 ohms impedance Then ... Watt (peak) = 12 * 12 / 2 * 0.8 * 0.8 = 46.08 and Watt (rms) = 12 * 12 / 4 * 0.8 = 28.8 ••• More options Share Show hidden low quality content Status This old topic is closed. If you want to reopen this topic, contact a moderator using the "Report Post" button. Share: Facebook Twitter Reddit Pinterest Tumblr WhatsApp Email Share Link
I icecoolwas Member Joined 2008 2009-01-31 4:57 pm #4 2009-01-31 4:57 pm #4 Sorry I dont understand this.. But the Impendance is 2ohm. ••• More options Share Saturnus Member Joined 2005 2009-01-31 5:54 pm #5 2009-01-31 5:54 pm #5 Ok then, let's also assume that effeciency is 80% which is typical for a class D playing into 2 ohms impedance Then ... Watt (peak) = 12 * 12 / 2 * 0.8 * 0.8 = 46.08 and Watt (rms) = 12 * 12 / 4 * 0.8 = 28.8 ••• More options Share Show hidden low quality content Status This old topic is closed. If you want to reopen this topic, contact a moderator using the "Report Post" button.
Saturnus Member Joined 2005 2009-01-31 5:54 pm #5 2009-01-31 5:54 pm #5 Ok then, let's also assume that effeciency is 80% which is typical for a class D playing into 2 ohms impedance Then ... Watt (peak) = 12 * 12 / 2 * 0.8 * 0.8 = 46.08 and Watt (rms) = 12 * 12 / 4 * 0.8 = 28.8 ••• More options Share
Ok then, let's also assume that effeciency is 80% which is typical for a class D playing into 2 ohms impedance Then ... Watt (peak) = 12 * 12 / 2 * 0.8 * 0.8 = 46.08 and Watt (rms) = 12 * 12 / 4 * 0.8 = 28.8
