Maximum sinusoidal output power of the Aleph 2

Hi everyone,

I wonder why the rated output power of the Aleph 2 @ 4 ohms is a 160 watts.

The output stage basically consists of a high-side 3A current source and a common-source circuit.

In case we want to have a negative current flowing through the load the transistor is an "electrical short" and the current is just limited by the rail negative rail voltage.

But if the output current is supposed to be positive, all of the bias current will flow through the load and none through the transistor.

Since the bias current is 3A, that would mean that the peak current through a resistive load would be exactly those 3A without clipping. Assuming a sinusoidal signal, that's a 2.12A RMS and at 4 Ohms the effective power of that current is just 18W and a 36W at 8 Ohms.

I've actually been thinking for the last few days where the heck could be the error in my thinking!?

Thanks in advance,
Thomas
 
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Thanks a lot. I didn't really look at the capacitors in the aleph schematic. This modification is actually quite simple, but seems to be very efficient..

Since there's not much explanation in the "Zen Variations 9" I'll try to explain how it works (which is actually pretty hard for me, because I don't know many technical terms in englisch)

No signal present: Q3 and Q5 form a Ube-constant current source. If the voltage across R4 is lower than the forward voltage of the base-emitter-diode of Q5, the collector current will become zero and the gate of Q3 will be loaded by the current flowing through R10, R11 and R12. When the current through R4 is so large, that it reaches the base-emitter-diode of Q5's forward voltage, the transistor will become conducting and it's Uce will lower as long as the voltage across R4 is too large. So it's a self regulating system, that will keep the base-emitter voltage of Q5 at around 0.5V.
So the 0.5V across R4 will show up across the capacitor C9. The voltage across R18 is zero, because there's no output current.

But what happens if a positive current is drawn by the load?

The voltage across R18 will rise. Let's assume it rises instantly to 0.25V, that wold be 3A output current. Since C9 is quite large, that means, that those 0.25V would be dropping across R13 and R19. Any voltage drop across R13 would cause, that the base-emitter voltage of Q5 gets lower if voltage across R4 remained constant. But since it's base-emitter voltage ist regulated (see above), the voltage across R4, and therefore the current through R4 will need to become larger.

It may have been pretty stupid that I mixed the values of the Aleph 2 with the reference designatiors of the Zen Variations 9 :) Despite this, I hope that I didn't do any mistakes there..

Thomas
 
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To be more precise: (Describing the case of the Aleph 2 with R19=680R and R13=1k)

In this case descriped above, at 3A output current: The voltage across R4+R13 that is regulated to 0.5V. Adding the voltage across R18 the voltage between the base of Q5 and the ooutput will be 0.75V in total. That same voltage will have to drop across R19+C9. V(C9) is a constant 0.5V. So the whole 0.25V will drop across R19. That causes a current of 368µA flowing through R19, and since the base current of Q9 is negligable, the same current will be flowing through R13, causing a 0.368V across R13. So the voltage across R4 will have to rise the exact same voltage. And 0.868V across R4 means 5.2A current delivered by the source. So the common-source transistor is still biased at 2.2A.

The same thing will happen when the output current is negative: But in that case the voltage across R4 is negative and will cause a decreasing current.

Yeaah, now that I know how it works, I like my Aleph 2 clone even a lot more :)
 
@Admin: In my last post I mistyped a reference designator. Could you please alter my previous post and replace the first "large" paragraph by this:

In this case descriped above, at 3A output current: The voltage across R4+R13 that is regulated to 0.5V. Adding the voltage across R18, the voltage between the base of Q5 and the ooutput will be 0.75V in total. That same voltage will have to drop across R19+C9. V(C9) is a constant 0.5V. So the whole 0.25V will drop across R19. That causes a current of 368µA flowing through R19, and since the base current of Q5 is negligable, the same current will be flowing through R13, causing a 0.368V across R13. So the voltage across R4 will have to rise the exact same voltage. And 0.868V across R4 means 5.2A current delivered by the source. So the common-source transistor is still biased at 2.2A.
 
Because I don't really like patents and don't want to be sued for patent infringement, I just came up with a simple circuit based on a differential amplifier that should do about the same thing as current source in the Aleph. Of course I've got no idea how well it performs in practice.
But it's free and can be used by anybody :)

current_controlled_cu6iuo6.png


Here the circuit is part of (the part of) a headphone amplifier, with Q1 being the actual current source and Q5 being the transistor to be biased.

Thomas