# Alpha Nirvana 39w 8ohm Class A Amp

#### xrk971

Paid Member
Very nice work Rdb64. Glad the build was fun and interesting. Nice measurements too!

1 user

#### AKSA

Excellent output, Bobby!
And very nicely built - congratulations and thank you for creating that AN39!
Hugh

1 user

#### andyr

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:
• 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???

1 user

#### Berlusconi

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.

Last edited:

#### Nico Ras

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

1 users

#### Berlusconi

Even worse: such power could fry his brains.
Fortunately for him that was just wrong interpretation of Ohms laws.

#### hbtaudio

Nice one, and in sixteen ohms it would be 100Watt . If you drove 600 ohm headphones their cones will meet inside your head somewhere?
Wrong conclusion, as you surely know very well ;-).

(13Vrms)^2 / 16Ohm = 10,6W
(13Vrms)^2 /600Ohm = 282mW

13V/4Ohm = 3,25A
!

#### hbtaudio

See post one!

Idc =1.7A

(3,25A / 2 = 1,625A) So in push-pull: 42Wrms with Load equal to 4Ohm.

#### andyr

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:
* 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.

1 user

#### hbtaudio

Maty,

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.

Hugh
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

!

#### Berlusconi

The AN amp is purely Class A - it doesn't have a 'Class B' component.
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.

#### hbtaudio

This amplifier isn't class A per se.
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

#### Berlusconi

wrong ;-)
But of course you're right. Would you bother to explain us. I'm eager to learn from you. 😀

#### Berlusconi

Meanwhile, I've made some calculations and have decided not to (re)-open that A/AB can of worms.
Lets leave the subject of Iq and move on.

Paid Member
It's been discussed ad nauseam. Just go back in the thread. No need to repeat that nonsense.
Important thing is it has glorious sound.

3 users

#### hbtaudio

What is the "nonsense"?

Paid Member
It isn't class A per se.

#### poseidonsvoice

Paid Member
It's been discussed ad nauseam. Just go back in the thread. No need to repeat that nonsense.
Important thing is it has glorious sound.

@hbtaudio

I discussed it here.

Best,
Anand.

1 user

#### AKSA

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

3 users