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29th April 2013, 11:35 PM  #1 
diyAudio Member
Join Date: Sep 2010
Location: Erlangen/Selb, Germany

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 highside 3A current source and a commonsource 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 
30th April 2013, 12:01 AM  #2 
diyAudio Member
Join Date: Jan 2003
Location: ancient Batsch , behind Iron Curtain

what if peak current is double Iq ?
edit : I forgot that nastiness called Aleph CCS , having some gain , too .......
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30th April 2013, 12:08 AM  #3 
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........
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30th April 2013, 12:17 AM  #4 
diyAudio Member
Join Date: Sep 2010
Location: Erlangen/Selb, Germany

If the current is doubled, it would be four times the output power, of course.. But where is that current supposed to come from, when the current source just supplys 3A?

30th April 2013, 12:31 AM  #5 
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Join Date: Jan 2003
Location: ancient Batsch , behind Iron Curtain

it obviously supplies more than 3A
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30th April 2013, 01:47 AM  #6 
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Join Date: Nov 2007
Location: Near Dallas Texas USA

Go here: https://www.passdiy.com/project/amplifiers
Download and read Zen Variations 9. It explains the Aleph current source and how it works. 
30th April 2013, 07:45 AM  #7 
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Join Date: Sep 2010
Location: Erlangen/Selb, Germany

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 Ubeconstant current source. If the voltage across R4 is lower than the forward voltage of the baseemitterdiode 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 baseemitterdiode 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 baseemitter 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 baseemitter voltage of Q5 gets lower if voltage across R4 remained constant. But since it's baseemitter 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 Last edited by mrchecker; 30th April 2013 at 08:12 AM. 
30th April 2013, 08:49 AM  #8 
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Join Date: Sep 2010
Location: Erlangen/Selb, Germany

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 commonsource 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 
30th April 2013, 09:29 AM  #9 
diyAudio Member
Join Date: Sep 2010
Location: Erlangen/Selb, Germany

@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 commonsource transistor is still biased at 2.2A. 
30th April 2013, 05:29 PM  #10 
diyAudio Member
Join Date: Sep 2010
Location: Erlangen/Selb, Germany

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 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 
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