Amplifier Analysis: Average vs. RMS

[Gregory]

Hello guys! I was studing amplifiers and classes of operarion. This videos about it are incredible https://youtu.be/2GijhTUQTbw

In his equations he used the average current to calculate the input power. Why not the RMS current?

P = Vrms * Irms right?

Thanks

[Neganur]

probably since the load is resistive.
Average power is \$ V_{rms} \cdot  I_{rms} \cdot cos\phi\$ and since current and voltage are in phase with the resistive load, the power factor \$cos\phi\$ is equal to 1.

[Andy Watson]

He is talking about the input power to the amplifier (not the load) and it is supplied at a constant DC value (28V ?) The current consumption is sinusoidal (actually, half-sines), so using the mean current in this case does make sense.

[Neganur]

Ah gotcha, the input power is DC ofc. That's why it's more accurate to use the mean power.

[Andy Watson]

That's why it's more accurate to use the mean power.

It is the correct method.

[TimFox]

Note that when measuring power into a resistive load, the value given by Vrms X Irms = (Vrms^2)/R is the mean power, not the rms power as has been mis-stated by audio companies since day 2.

[Gregory]

So, Vdc*Imean is the active input power.
If we did Vdc*Irms we would get the aparent power, as the current is not constant and Vdc=Vrms for DC input.

Right?

[danadak]

From https://en.wikipedia.org/wiki/Root_mean_square

[Gregory]

"However, if the current is a time-varying function, I(t), this formula must be extended to reflect the fact that the current (and thus the instantaneous power) is varying over time. If the function is periodic (such as household AC power), it is still meaningful to discuss the average power dissipated over time, which is calculated by taking the average power dissipation:"

As the colector current is time variant (has a AC component) the aparent power is calculated using the RMS current. The Class C amplifiers has a greater aparent power than active power, because the colector current is not in phase and only the first harmonic that transfer active power to the output.

As he was calculating the efficiency witch is about the transferency of active power from the input to the outputs, he used the Mean current. This is my conclusion

[Neganur]

Exept that in the video there is no talk of class C. Only A, B and AB.

[Don Hills]

More on the topic of why the term "Watts RMS" is misleading and, arguably, nonsense:

http://www.hifi-writer.com/he/misc/rmspower.htm

[Gregory]

It's a series of videos

[danadak]

Here is a pretty good explanation -


http://opencourses.emu.edu.tr/mod/resource/view.php?id=142&redirect=1


Regards, Dana

[3db]

I find the guy a bit confusing but get what he's trying to teach.
I don't want to get into the rms/average debate.
If you want a power amp to deliver a sine wave and produce 10 watts into 10 ohms then.
Power *R= V² ie square root 100 = 10 Volts rms
10 * square root 2 = 14.142 *2 = 28.28 Volt pk-pk.
In his circuit the emitter is going sit at 14.14 volts
All the rest of the calculations are just biasing the transistor and showing a class A amplifier.
The circuit is NOT really practical unless you want  a sine wave across that 10 ohm resistor.
Drive it +ve and the volts across the 10 ohm R will go up drive it -ve (ie below 14.14 volts and the volts across R will go down.

Sorry for the rubbish presentation I'm not used to posting comments yet.

PS how do I use the emotes ?

3DB

[Neganur]

It's a series of videos

sorry! not going to watch all of those and guess what parts of it you're referring to.
(the one linked in the first post is already 50 minutes and you didn't give a minute mark that referred to your question)

I also think that this is hardly a debate about Watt RMS since this was about the power dissipated by the amp, not the 'audio power output'.

The powerpoint Dana linked is excellent.