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- Thread starter xrk971
- Start date

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.

Last edited:

Nice one, and in sixteen ohms it would be 100Watt . If you drove 600 ohm headphones their cones will meet inside your head somewhere?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???

Even worse: such power could fry his brains.If you drove 600 ohm headphones their cones will meet inside your head somewhere?

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 . If you drove 600 ohm headphones their cones will meet inside your head somewhere?

U/R_load = I

(13Vrms)^2 / 16Ohm = 10,6W

(13Vrms)^2 /600Ohm = 282mW

13V/4Ohm = 3,25A

!

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.

Could you please provide the equation to back up your claims, B?

These were:

and

The AN amp is purely Class A - it

Imax = 13V/4Ohm = 3,25A_peakMaty,

19.1W into 4R, clipping at negative 13Vpk. This is not designed for 4R; it's specifically for 8R with dips to 6R. For big power, say >50W into 4R, I would change outputs and rejig some of the output stage values to maintain stability, THD and power into lower loads.

Thanks for your interest!

Hugh

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:The AN amp is purely Class A - ita 'Class B' component.doesn't have

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 ;-)This amplifier isn't class A per se.

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. 😀wrong ;-)

https://www.diyaudio.com/community/...-8ohm-class-a-amp.344540/page-20#post-6004893

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

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