# Problems with power calculation in amp.

Status
This old topic is closed. If you want to reopen this topic, contact a moderator using the "Report Post" button.

#### Duo

I've got a problem here...

I wanted to know the power O/T of a push pull amp, by ohm's law, I calculated the full swing of the amplier in to 4 ohms, like this.

I=E/R so I = 55volts/4ohms thus I=13.75amps

P=E*I so P = 55volts * 13.75amps thus P=756.25 watts

but first of all, that seems like a lot of power for 55 volt rails

next, a person told me another formula for this which is

vcc^/4/2 where you have supply(55) squared divided by the speaker imp divided by two, which gave me 378 watts for the same amp! Can anyone tell me whats going on here and confirm the proper way to calculate power output??

#### Duo

Help, anybody, please????? #### AudioFreak

Peak power output is V^2/R less losses

RMS power output is (V^2/R)/2 less losses

Where V is the rail voltage.

#### hugobross

yep, that's right.

Duo, an example for the calculation:

imagine you'd like to have 200W RMS, so 400W peak power.
the load is 4 ohm, so:
P = I*I / R

I(max)^2= 400W/4 => I(max) = 10A

Now, you'll have to calculate the supply voltage:

U = I * R => U = 40V

this is the DC-voltage of the supply, now you want to calcute the needed AC-voltage of the transformer:

u = U / "square of 2" = 28V => + 2 times diode voltage (rectifier) = 30V

Best regards,

HB.

#### Shaun

Convert output swing to RMS volts

Hi

First convert your 55V swing (which would be peak voltage) to RMS volts. 55/sqrt, then V^2/4, which gives 378W.

Shaun

#### e96mlo

The calculations you made, Duo, are correct if you are only talkning DC voltages and currents and only resistive loads. Since all signals in an amp are AC, you will have to use RMS values as the other guys here just described. Remember that there are voltage losses in the O/P transistors, which means that even if you convert your rail voltages to RMS you will still get a too optimistic number. But in the right direction.

/Marcus

Yes.

#### Shaun

BUT....

In my latest experience, this is the power region where you want to pay close attention to the SOA of the output transistors, and the current sourcing capability of the driver stage. I'm busy with a Load Invariant amp which has 60V rails at present. I'm not so sure that this is ideal. Further tests will tell...

Shaun

Status
This old topic is closed. If you want to reopen this topic, contact a moderator using the "Report Post" button.