Audio Amplifier, Current Limiting Voltage Supply problem
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 Class D Switching Power Amplifiers and Power D/A conversion

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 13th January 2013, 03:27 PM #11 DUG   diyAudio Member   Join Date: Jan 2003 Location: mississauga ontario canada ChocoHolic, like you said: C=Q/V I=Q/t.......Q=I*t C=(I*t)/V....or...C=(I*dt)/dV If we assign values we can get a cap value to start and see if it is a practical size. The Ipk we need is 1.8A...lets make that a constant load. The deltat (dt) we will use is one half cycle of 20Hz...0.025S The deltaV (dV) we will assign is 1V. That gets us C=(1.8*.025)/1 C=45,000uF So that is a good size cap and we don't need to go larger than that. It is a good starting point and we can work from there. Sacrifice the 20Hz to 40Hz and you can get away with 23,000uF Allow a 2V drop and you can go to 12,000uF. Not assigning a constant 1.8A and the value can be reduced further. So somewhere between 10,000uF and 45,000uF. If the amplifier needs to produce only 6W then even a 5,000uF would suffice. efevi, I hope that helps Good luck __________________ Doug We are all learning...we can all help "You can't stop the signal, Mal. Everything goes somewhere..."
ChocoHolic
diyAudio Member

Join Date: Dec 2003
Location: Munich
Quote:
 Originally Posted by DUG The Ipk we need is 1.8A...lets make that a constant load. The deltat (dt) we will use is one half cycle of 20Hz...0.025S The deltaV (dV) we will assign is 1V.

...pretty tough worst case estimation.

When calculating more detailed, you will see that 2x4700uF is enough for normal use and will allow by far more than the 6W.
Anyway, you highlighted the key information that the requirements can be easily fulfilled with standard components.

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