# Determining truth of 555 watt amplifier board

1. Jan 4, 2014

2. Jan 4, 2014

### Jony130

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).

3. Jan 4, 2014

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

4. Jan 4, 2014

### Jony130

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

5. Jan 4, 2014