Average vs rms current in EF output stage

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when you multiply rms volts x rms current, you get power in watts average...simple as this...no need to get confused...

and watts average is the same heating when you used equivalent dc voltages and currents...this is the convention...

ac sine waves have peak values, rms x 1.4, and peak to peak values, 2 x peak or rms x 2.8..
in an amplifier, maximum peak to peak is theoretically +- Vcc,
your amps' rail voltages in case of bipolar rails,
in the case of single rails, then it is still 1/2 Vcc...
dc is converted to ac at the output of your power amp and fed to your speakers...
 
Ketje said:
Well no, it's RMS, so √(5x5+3.5x3.5)=6.1V !
Yes, you are right. I must have had a 'senior' moment!

AJT said:
when you multiply rms volts x rms current, you get power in watts average...simple as this...no need to get confused...
Only true if the voltage and current waveforms are identical in shape and time i.e. the load is a pure resistance. If the load is a capacitor then RMS voltage and RMS current are still well-defined but the average power is zero.
 
Thanks Sreten & AJT - what's puzzling me is why the calculation uses average current and not rms current. I'm wondering is it because you presumably can't have rms of a half sine wave?

I think it is because the average value over time determines the, well, average current that the power supply must deliver, and that was the purpose of Bob's calculation.

You'd use RMS if you would want to calculate how much power a signal delivers into a load: the RMS value squared / load.
By definition, the RMS value is the value of a DC voltage that would deliver the same power as the signal of which you took the RMS value.

jan
 
Ah yes, the famous cos φ 🙂
Mona

watts average causes heating in a resistor, this is true power..
in cases where load contains reactive components like caps and inductors,
then, there is reactive power, energy is stored in these elements...
true power is only dissipated in resistors where currents and voltages are in phase.....

and then someone will say, but inductors and capacitors have resistive components....
the discussions then become even more complicated...

if you add the sum of the true power and reactive power vectorially, then you get apparent power...
 
@DF96 -"Elementary introductions to electricity teach the difference between AC and DC. Like much early stuff, this is a simplification but not always flagged as such. Later on, people get confused (and argue in threads about - see the search function) about whether 'AC' has to cross the zero line and whether 'DC' can vary or not. The best option is to regard AC and DC as stories we tell newbies, and as useful descriptions in many contexts, but not definitive terms - Fourier theory is better."

Thanks DF96. I find that the most useful comment on this thread. The point that it's a simplification of something much more complex helps me make sense of things, and even helps me towards the conclusion that I'm not just totally mad - just lacking in the differential calculus! 😀

Thanks to all who have commented. I'm forever learning! 🙂
 
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