Amplifier Analysis: Average vs. RMS - Page 1
Forum section: electronics / beginners

author: Gregory
author_rank: Regular Contributor
timestamp: April 09, 2016, 08:10:15 pm
content: Hello guys! I was studing amplifiers and classes of operarion. This videos about it are incrediblehttps://youtu.be/2GijhTUQTbwIn his equations he used the average current to calculate the input power. Why not the RMS current?P = Vrms * Irms right?Thanks

author: Neganur
author_rank: Unknown
timestamp: April 09, 2016, 08:58:11 pm
content: 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.

author: Andy Watson
author_rank: Super Contributor
timestamp: April 09, 2016, 09:13:57 pm
content: 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.

author: Neganur
author_rank: Unknown
timestamp: April 10, 2016, 04:39:11 am
content: Ah gotcha, the input power is DC ofc. That's why it's more accurate to use the mean power.

author: Andy Watson
author_rank: Super Contributor
timestamp: April 10, 2016, 11:38:20 am
content: Quote from: Neganur on April 10, 2016, 04:39:11 amIt is thecorrectmethod.

author: TimFox
author_rank: Super Contributor
timestamp: April 10, 2016, 12:29:20 pm
content: Note that when measuring power into a resistive load, the value given by Vrms X Irms = (Vrms^2)/R is themeanpower, not thermspower as has been mis-stated by audio companies since day 2.

author: Gregory
author_rank: Regular Contributor
timestamp: April 12, 2016, 09:41:17 pm
content: 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?

author: danadak
author_rank: Super Contributor
timestamp: April 12, 2016, 10:32:31 pm
content: Fromhttps://en.wikipedia.org/wiki/Root_mean_square

author: Gregory
author_rank: Regular Contributor
timestamp: April 12, 2016, 11:33:52 pm
content: "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

author: Neganur
author_rank: Unknown
timestamp: April 12, 2016, 11:57:42 pm
content: Exept that in the video there is no talk of class C. Only A, B and AB.

author: Don Hills
author_rank: Regular Contributor
timestamp: April 13, 2016, 12:17:11 am
content: More on the topic of why the term "Watts RMS" is misleading and, arguably, nonsense:http://www.hifi-writer.com/he/misc/rmspower.htm

author: Gregory
author_rank: Regular Contributor
timestamp: April 13, 2016, 12:38:42 am
content: It's a series of videos

author: danadak
author_rank: Super Contributor
timestamp: April 13, 2016, 12:51:25 am
content: Here is a pretty good explanation -http://opencourses.emu.edu.tr/mod/resource/view.php?id=142&redirect=1Regards, Dana

author: 3db
author_rank: Frequent Contributor
timestamp: April 13, 2016, 01:37:32 am
content: 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= V2ie square root 100 = 10 Volts rms10 * square root 2 = 14.142 *2 = 28.28 Volt pk-pk.In his circuit the emitter is going sit at 14.14 voltsAll 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

author: Neganur
author_rank: Unknown
timestamp: April 13, 2016, 04:57:24 am
content: Quote from: Gregory on April 13, 2016, 12:38:42 amsorry! 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.