RMS Power

[tab30]

I tried to calculate the RMS-Power of an amp at given DC-Driver voltage.

But I always get higher values than in the description of the (DIY) amp is given.
Well reasonable I must reduce the output Umax by the "Saturation" voltage of the amp but I don't think that an 400W amp working with 70V only give out 56 V.

I added my calculations.

This Probelm hounded me the last day's I hope someone can help.


Andi

[Thoru]

Hi,

#1 is correct, #2 is not. RMS means 'root mean square' and you ommit the root.
I don't know what you mean with #3 nor the rest, but perhaps that adding that root solves your problem.
If I remember correctly than the Vrms of a sinus is 1/2*sqrt(2)*sin(x), and you can than square that voltage, divided by the number of Ohms should give you the output power.

Remco Poelstra

[tab30]

In #1
I mean not U=Umax I mean U=Urms

now you get #3 with Urms=Umax*1/sqr(2)

sin(x)*Umax is the voltage at the "angel" 0-360 deg respectively 0-2pi.

After the first equation I have to square the voltage -> (sin(x)*Umax)² and divide it with the impendance r.

now I add all pieces Power together and divide it with 2pi.

right ?

The Education should be right ? !

[sreten]

Simple fact is you are disregarding losses in the amplifier,
power supply average voltage droop and the the extra
ripple below the average voltage droop.

You cannot estimate 4 ohm output power from nominal rail voltage.

(nor 8 ohms for that matter but its nearer)

What you can do is estimate losses for 8 ohms max power,
which again will be lower than predicted by rail voltages
and double them for 4 ohms.

sreten.

[Thoru]

It is correct for the most part, but #2 is wrong. It should be something like the image below. But that boils down to P=Vrms over r if I'm correct.
That for the math part.

Remco Poelstra

[tab30]

there is no way to estimate the Power of an amp, before you have measured it ?

Isn't it possible to calculate/ name circa values for all the loss that occur ?

and take that in the estimation.

Well,
if I use a verry good/ powerfull supply, with big capacitances.
What loss will occur except the Saturation and some mV in the connection lines ?

[Thoru]

Class B amp's only have a efficiency of +-60% no matter what powersupply or wires you use, it's just the way they are build. Class A gets closer to 30%. Thus for every Watt you put in you'll only get 0.6 Watt back. 612*0.6=367 which is close to enough to 400 to rate the amp at 400 I think, taking into consideration that the calculations are very rough.

Remco Poelstra

[tab30]

@ Thoru

but with this education you get much lower values as given in the DIY-Amps.

And I thik my solution should be right because you get the same as you get if you use #1 and Urms=Umax*1/sqr(2)

[sreten]

Originally posted by tab30
there is no way to estimate the Power of an amp, before you have measured it ?

Isn't it possible to calculate/ name circa values for all the loss that occur ?

and take that in the estimation.

Well,
if I use a verry good/ powerfull supply, with big capacitances.
What loss will occur except the Saturation and some mV in the connection lines ?

A perfect supply is hardly the point.

You want output with a sensible cost effective supply.

And yes intelligent estimations can be made.

But they are not based on nominal no load rail voltage.

sreten.

[tab30]

@ Thoru second message


But the efficiency of 60% means only you need an PowSup witch delivers 1,6 * (Pwanted out)

isn't it so ?