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Determining truth of 555 watt amplifier board

  1. Jan 4, 2014 #1
  2. jcsd
  3. Jan 4, 2014 #2
    The power is always directly proportional to the voltage.
    So for +/-36V at 8 ohm we can get theoretical
    P_max = 36^2/(2*8) = 81W RMS power. Of course in real life the power will be much lower in range of a 50W ...60W

    And if you shot the output, the power dissipated in the amp will be equal to:

    Ptot = 36V * 7*6.5A = 1638W (234 per TDA).
  4. Jan 4, 2014 #3
    What do you get with all of them in parralel?

    I found the internal resistense of the ic at max power, found what it is all at parralel and added the 8ohms and used v=ir and iv=w and got bout 140
  5. Jan 4, 2014 #4
    Current, you increase output current capability.
    But this is wrong. This 140 has nothing to do with RMS power (continuous power).
    To get 140W of a RMS power at 8 ohm you need √(8 * 140) = 33.47V but this is RMS voltage.
    The peak voltage is 1.41 times larger. So your power supply voltage must be larger than +/-47.3V
  6. Jan 4, 2014 #5
    Ok thanks, I think thatsall I need for now, thanks.
  7. Jan 4, 2014 #6


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    Why are you fixated on 8 ohms impedance?

    555W from 7 chips is about 80W per chip. That is consistent with P0, in table 2.3 of the datasheet.
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