EveryCircuit - Inefficient amp 2

9 years ago

Yeah but it works well

9 years ago

The efficiency is 4% which is indeed veeery inefficient

9 years ago

Wrong buggzy, its even less then 4%. You are misstaken by input power calculation, again and again. Check topic RMS to understand.

thebugger

9 years ago

It's exactly 4.097006501% at this power level. Output power is 1.7W input power is constant (around 43W)

hurz

9 years ago

43W is the average power you compare with output RMS power. This is as wrong as Khz is not kHz

thebugger

9 years ago

It's wrong according to you. That doesn't mean it's really wrong. So yeah, don't take offense if I don't take notice of your opinion this time. I've been measuring it the same way for years. And BTW according to Rod Elliot, this is exactly the way to calculate it. RMS vs Avg http://sound.whsites.net/efficiency.htm and him I trust more than you, because sometimes you correct people out of pure spite,

hurz

9 years ago

First, nothing what Rob is writting does fit to this amplifier here. Rob demonstrates correctly how to do it for a class B amp, and after that for a class A amp with constant current source as working resistor. I see you dont even understand this. BTW, I gave up some time ago to help you to understand. You are realy to stumbborn. I just do that to help other who you continue to irritate with your "statements" how our univers has to function in your head only. Dont know if I ever told you, but stop pulling this shit out of your ass and try to be a little more precise. So tell us all, which formular you have used to calculate 43W.

thebugger

9 years ago

I used approximation. Here I'll just show you http://everycircuit.com/circuit/5151560569192448

thebugger

9 years ago

And yes I noticed the amp in rod's example is in class B, but the power he uses is still averaged.

hurz

9 years ago

What you do is an average only without any formfactor. So you completly ignore the AC part. BTW, there is no need for CantFeedSolution hacks to measure the average current. (2.74A-0.0955A)/2+0.0955A=1.41775A! However, this is only half the story and wrong! You have to apply the correct formfactor (as Rob is doing correctly for its amp topology) for this kind of amp topology to the avg and you get 1.417*1.225=1.737A which then results in 52Watt input Power.

thebugger

9 years ago

The current is right there, basically rectified and it shows 1.4A. I don't see why you need to average it.

hurz

9 years ago

Were do you see in this circuit here or in you example above a rectifier? Rod is having a rectified waveform and applies correctly the right formfactor 0.636 (2/pi) for its class B example. But you applied no formfactor and just averaged the current. So No, "the" current is not there. 1.42A is ONLY the average which is the base to apply the correct formfactor for this special waveform only! Here its sqrt(3/2)=1.2247 ...(for a DC shifted sine wave form)... avg * 1.22 = Irms = 1.74A and power is then P=Irms * U= 52W. I know, this is difficult to understand.

thebugger

9 years ago

But the current draw when smoothened by the capacitor is 1.41A at 40V. It stays the same?

hurz

9 years ago

Now its getting interessting, right? I will turn it around to make it not to easy for you. Unwire the input and explain the quitesense power 42.3W. And this is without 1.7Watt output. Were is it?

thebugger

9 years ago

http://everycircuit.com/circuit/5151560569192448 26.2W dissipated as heat from the transistors, and 16.8W dissipated from the 8ohm active load.

thebugger

9 years ago

And yes, I noticed Rod's warning that typically the rms output value is somewhat wrong to use...but he also said, that it has become a widely accepted practice as to present a given amp's efficiency parameter this way

hurz

9 years ago

Rod said, don't use the Term "RMS power", because power a product out of rms values and its RMS by default. If not, its peak power or average power or whatsoever.

hurz

9 years ago

Back to math. You think drawn quitesense power and under maximum loaded drawn power condition are equal ~43W?

thebugger

9 years ago

It's right there. I just see the absolute values. The power draw is 43W, even when the current is rectified (with the capacitor example). The output is 1.7Wrms, and yes while he said it's technically wrong to use the rms value, it's become a common practice to do so.

hurz

9 years ago

Silence 43W anplifier 0W load. In case of maxload amp 43W-1.7W=41.3W and load 1.7W?????? Does this make sense?

hurz

9 years ago

And NO, he does not said its wrong to use RMS. Actually he is using only RMS for in and out and never an averaged power!

thebugger

9 years ago

No for drawn power he uses the average current vs the output rms value.

hurz

9 years ago

This is a missinterpretation you made. Quote please were he said that?

hurz

9 years ago

You think he gets the same stupid results like you have with your wrong methode, cool.

thebugger

9 years ago

The bias voltage of the amp in Figure 2 must be high enough to ensure that the transistor's quiescent current is about equal to (or slightly greater than) the peak speaker current. The efficiency of this amp can be easily calculated by the same method as described above. A Class-A amp conducts at all times and the DC is at least equal to the peak output current. A 20V peak signal will still create a 2.5A peak current in the load, so the DC quiescent current must be equal to (or greater than) this figure. Note that the supply is not bipolar - a single +40V supply is the same as a ±20V supply for the purposes of calculation. The output to the load will still be 25W, but the DC input power is now 100W (2.5A * 40V). Efficiency now (from equation 4) is ...

thebugger

9 years ago

This is for class A, he doesn't use any power corrections

hurz

9 years ago

Ok, to stop this before you get nasty and rude like in the past to many times I will lift the magical for you. What you did is the so called arithmetic average, which is not quadratic as RMS and does not help you to see the AC swing correctly (or better said, you remove AC completly from your power drawn analysis!) It would need a little more affort to get the so called "quadratic average" I know, your math skills are far from perfect, but what you understand for sure is: double the voltage at a constant load R cause 4 times higher power, cuz the current also doubles and Power = Irms * Urms. Arithmetic average doesn't take care about this at all! Hope this helps.

thebugger

9 years ago

Yeah I just went through Rod's explanation on class A efficiency, and he uses exactly the same formula. The rms output vs the average drawn power. The average drawn power is equal to the supply voltage multiplied by the peak load current. So for a 40V supply and a 20Vmax voltage in an 8ohm load the efficiency is Pout=(20*0.707)^2/8=25W. The power draw is Pin=Imax*Vin. Pin=2.5A*40=100W. The efficiency is then (25W/100W)*100=25%

hurz

9 years ago

"Yeah" what? Any conclusion?

thebugger

9 years ago

Yeah, I've went through the method he uses to measure class A amp efficiency, and he doesn't take any consideration of power factors and such.

thebugger

9 years ago

He just multiplies peak load current by the supply voltage for the drawn power, and the rms powee for the output and uses the standard efficiency formula.

hurz

9 years ago

And you see that your methode is completly wrong?

thebugger

9 years ago

No, it's identical... Da fuq

hurz

9 years ago

So if you went through Rod's methode and you understand it, why dont you go yourself through this darlington amp here and explain it like Rod does? But be careful, I already told you, this amp here is different then both Rod is having! When can we expect a break down?

alekkorange

9 years ago

Honestly, I just got bored and tossed a Darlington emitter-follower onto a basic voltage gain stage.