Hugh,
Can you confirm how we get to the nominal 39w for the AN?
I recently posted what I thought was the power output of my AN4Rs ... and I wonder if I had made a mistake with the calculation?
I thought it was:
Which is near enough to the nominated power output!
But I saw someone had posted another way of calculating power for a Class A amp - which is: bias current^2 * R.
In which case, my power output would be: 2.5^2 * 4 ... which equals only 25w???
Can you confirm how we get to the nominal 39w for the AN?
I recently posted what I thought was the power output of my AN4Rs ... and I wonder if I had made a mistake with the calculation?
I thought it was:
- ptp voltage at clipping = 38v
- therefore peak = 19v
- therefore rms = 19 x 0.707 = 13.4v
- power (rms) = 13.4^2 / 4
- = 45w?
Which is near enough to the nominated power output!
But I saw someone had posted another way of calculating power for a Class A amp - which is: bias current^2 * R.
In which case, my power output would be: 2.5^2 * 4 ... which equals only 25w???
Andy,
You will need 3.2A quiescent current to attain 39W pure class a into 4ohm load. With yours just 2.5A quiescent current you will have just 25W of class A. Above that value you are in the AB region. Selection of quiescent current isn't just choice of some arbitrary value. It is ridiculous to see how people beleive that they have class A amplifier just because it is, let's say "Krell", and running "cool". There must be considerable ammount of amperes to attain full class A operation.
You will need 3.2A quiescent current to attain 39W pure class a into 4ohm load. With yours just 2.5A quiescent current you will have just 25W of class A. Above that value you are in the AB region. Selection of quiescent current isn't just choice of some arbitrary value. It is ridiculous to see how people beleive that they have class A amplifier just because it is, let's say "Krell", and running "cool". There must be considerable ammount of amperes to attain full class A operation.
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Nice one, and in sixteen ohms it would be 100Watt
. If you drove 600 ohm headphones their cones will meet inside your head somewhere?
Even worse: such power could fry his brains.
Fortunately for him that was just wrong interpretation of Ohms laws.
Wrong conclusion, as you surely know very well ;-).Nice one, and in sixteen ohms it would be 100Watt
U/R_load = I
(13Vrms)^2 / 16Ohm = 10,6W
(13Vrms)^2 /600Ohm = 282mW
13V/4Ohm = 3,25A
!
Could you please provide the equation to back up your claims, B?
These were:
* 3.2a quiescent ==> 39w into 4 ohms
and
* 2.5a quiescent ==> 25w.
The AN amp is purely Class A - it doesn't have a 'Class B' component.
Imax = 13V/4Ohm = 3,25A_peak
Imax / 2 = 1,625A
times 1/sqrt(2) = 1,15A
21,16W Class A into 4Ohm
42,32W Class A into 8Ohm (18,4Vrms ---> 26Vpeak)
!
Rails +/-26Vdc
u_peak_max <23V
equals 33Wrms_8
!
Let us view this in purely intuitive way, through a logic experiment:
Reduce the quiescent current to zero. Is it still class A?
This amplifier isn't class A per se. It provides means to attaining parameters needed to drive the output devices into the class A operation - by applying adequate current to render the output devices always opened.
wrong ;-)
See Nelson Pass Aleph's series.
The negativ leg is an active, modulated current source /sink. So 1,7Adc Idlecurrent lead a little bit to the SE-World. Ultimately, however, the output stage is in push-pull mode.
Bye,
HBt
But of course you're right. Would you bother to explain us. I'm eager to learn from you. 😀
There was lots of discussion about se vs ppse, current vs double current...posts by daanve.
https://www.diyaudio.com/community/...-8ohm-class-a-amp.344540/page-20#post-6004893
https://www.diyaudio.com/community/...-8ohm-class-a-amp.344540/page-20#post-6004893
Andy,
Any amp either clips for lack of current or by hitting the rail for lack of headroom.
Let us design our 4R amp for 39W.
The math tells that you square the peak to peak output and divide by 32.
If we plug in 35Vpp, the watts are 38.3W into 4R. Into 8R load the rating is half, 19.1W.
To be sure you have sufficient current in the AN negative half cycle, divide the negative half cycle peak voltage (17.5Vp) by the load impedance to get peak current - 17.5/4 which gives you 4.37A. Since the peak neg current supported by the AN39 is double the quiescent current (and assuming your impedance is exactly 4R) then your quiescent should be set at 2.19A.
Hugh
Any amp either clips for lack of current or by hitting the rail for lack of headroom.
Let us design our 4R amp for 39W.
The math tells that you square the peak to peak output and divide by 32.
If we plug in 35Vpp, the watts are 38.3W into 4R. Into 8R load the rating is half, 19.1W.
To be sure you have sufficient current in the AN negative half cycle, divide the negative half cycle peak voltage (17.5Vp) by the load impedance to get peak current - 17.5/4 which gives you 4.37A. Since the peak neg current supported by the AN39 is double the quiescent current (and assuming your impedance is exactly 4R) then your quiescent should be set at 2.19A.
Hugh