Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    AlephZero

    User Avatar
    Science Advisor
    Homework Helper

    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Determining truth of 555 watt amplifier board
  1. Watt / Watt-Hour (Replies: 5)

  2. 555 as Clock (Replies: 19)

  3. 555 timer (Replies: 5)

Loading...