batee said:
Actually, that's exactly what we are protecting against - a
short between + and - outputs is limited by the positive
current source.
😎
r13, r13, r13.......
Ok, now that I have learned that my understanding of the function of the networks composed of R10, R13, Q4, Q9, R35, R39 is completely wrong, (which is ok, r13 never fit into the theory, and working that out was on the "to do" list) -- I am left with one big question---
How
does
my
aleph mini manage to have 0V plus or minus 50mV absolute DC offset? The aleph current source delivers app. 2 amps DC plus some AC with a .33 ohm resistor occupying the position equivalent to R5 in the original Aleph x schematic posted on page 1, but how does the other side know to deliver 2 amps from the negative rail???????
Ok, now that I have learned that my understanding of the function of the networks composed of R10, R13, Q4, Q9, R35, R39 is completely wrong, (which is ok, r13 never fit into the theory, and working that out was on the "to do" list) -- I am left with one big question---
How
does
my
aleph mini manage to have 0V plus or minus 50mV absolute DC offset? The aleph current source delivers app. 2 amps DC plus some AC with a .33 ohm resistor occupying the position equivalent to R5 in the original Aleph x schematic posted on page 1, but how does the other side know to deliver 2 amps from the negative rail???????
Re: r13, r13, r13.......
The 390-ohm load to the IRF9610 drain pin calculates the necessary bias Vgs of the lower IRF240, based on the dc current given by the CCS that is located above the IRF9610 source pin. I hope that I understood your question correctly and answered right near to your heart.
jupiterjune said:. . . but how does the other side know to deliver 2 amps from the negative rail???????
The 390-ohm load to the IRF9610 drain pin calculates the necessary bias Vgs of the lower IRF240, based on the dc current given by the CCS that is located above the IRF9610 source pin. I hope that I understood your question correctly and answered right near to your heart.
I'm no expert on this - I'm still working through this design. I figured I should get it done before the next design comes out...🙂
Something that might help is to draw a diagram where you replace the Aleph current source (R14, V1, R11, R15, R7, R8, Q1, Q3, R5) with a Current source where I=2). Ignore R2, R3, R12, and C1 for now. These modify the constant current source so that it becomes a variable current source called the Aleph current source. This current source is what's causing constant current of ~2A to flow through Q2. If you add back the additional components R2/3/12 and C1, it changes from a Constant Current Source to the variable current source (called Aleph Current Source) that changes its output based on current flow to the load. The CCS version I understand, but I have no clue how the Aleph current source version works. I expect that the transfer function for this will look like something out of my Quantum Physics text.
No significant current is flowing through the 'X' wires that connect R25 to the junction of Q4/R9. This is a signal path connection between the dif pair and the source follower? comprised by Q2 and R6, and probably also has a DC component that causes the gate of Q2 to be above the turn-on voltage, so that Q2 is always conducting and doesn't go into cutoff.
Good luck,
Something that might help is to draw a diagram where you replace the Aleph current source (R14, V1, R11, R15, R7, R8, Q1, Q3, R5) with a Current source where I=2). Ignore R2, R3, R12, and C1 for now. These modify the constant current source so that it becomes a variable current source called the Aleph current source. This current source is what's causing constant current of ~2A to flow through Q2. If you add back the additional components R2/3/12 and C1, it changes from a Constant Current Source to the variable current source (called Aleph Current Source) that changes its output based on current flow to the load. The CCS version I understand, but I have no clue how the Aleph current source version works. I expect that the transfer function for this will look like something out of my Quantum Physics text.
No significant current is flowing through the 'X' wires that connect R25 to the junction of Q4/R9. This is a signal path connection between the dif pair and the source follower? comprised by Q2 and R6, and probably also has a DC component that causes the gate of Q2 to be above the turn-on voltage, so that Q2 is always conducting and doesn't go into cutoff.
Good luck,
Wow, that magic half million is coming up real fast, in fact at about 22 reads a minutes for the last hour or so. Mightly popular thread this. 
Yes, you understand my question. But is it truly as simple as this?
I am puzzled because when I removed the .33 ohm resisors and replaced them with 1 ohm resistors, the amplifiers DC offset remained at app. 0V plus or minus a few mV. The voltage across the 1 ohm resistors remained at .66 volts, so then the bias current became .66 amps through each resistor......
It would seem that the basic instructions to keep "X" amps of bias flowing generated by q3 are fed to both the gain and CCS output devices (Q1 and Q2)??
JJ
still
Here are my predictions for what we'll see in Aleph X Mk II
Top 10 Upgrades to the Aleph X
10) More Folded Cascodes. I think I read something like that in the patent.
9) Less heatsink. Grey has hinted that his next project will be water cooled.
8) More Rail Voltage. 170V rails = no transformer needed.
7) JFET frontend (by popular demand).
6) A really bitchin' chassis. A thicker front panel, some of those spiky feet, more Blue LEDs and heatsinks that go all the way down.
5) JFET backend. Grey has to get rid of all those Lovotechs somehow.
4) More efficient. Nelson is headquartered in Calif, after all.
3) Less parts. Lets start with potentiometers and other things that have to be adjusted.
2) A headphone jack on the front, and an iPod jack on the back. Business in the front, party in the back - just like a mullet.
1) We find out that Grey and Nelson are the same person, and he gets into a big fight with himself.
0) More cowbell.
Most of this is a joke. Please don't email me telling me that you don't see the benefit of an iPod jack in an audiophile amp, because it's going in my version anyway...🙂
Bryan A. Thompson
bryan@batee.com
Top 10 Upgrades to the Aleph X
10) More Folded Cascodes. I think I read something like that in the patent.
9) Less heatsink. Grey has hinted that his next project will be water cooled.
8) More Rail Voltage. 170V rails = no transformer needed.
7) JFET frontend (by popular demand).
6) A really bitchin' chassis. A thicker front panel, some of those spiky feet, more Blue LEDs and heatsinks that go all the way down.
5) JFET backend. Grey has to get rid of all those Lovotechs somehow.
4) More efficient. Nelson is headquartered in Calif, after all.
3) Less parts. Lets start with potentiometers and other things that have to be adjusted.
2) A headphone jack on the front, and an iPod jack on the back. Business in the front, party in the back - just like a mullet.
1) We find out that Grey and Nelson are the same person, and he gets into a big fight with himself.
0) More cowbell.
Most of this is a joke. Please don't email me telling me that you don't see the benefit of an iPod jack in an audiophile amp, because it's going in my version anyway...🙂
Bryan A. Thompson
bryan@batee.com
jupiterjune said:
Yes, you understand my question. But is it truly as simple as this?
I am puzzled because when I removed the .33 ohm resisors and replaced them with 1 ohm resistors, the amplifiers DC offset remained at app. 0V plus or minus a few mV. The voltage across the 1 ohm resistors remained at .66 volts, so then the bias current became .66 amps through each resistor......
It would seem that the basic instructions to keep "X" amps of bias flowing generated by q3 are fed to both the gain and CCS output devices (Q1 and Q2)??
JJ
still
You're right - the voltage remains at .66 Volts and the current went down. Q3 isn't generating significant current, it's regulating gate voltage on Q1 which causes Q1 to generate a constant current.
First lets look at the case where there is no input. If we apply KCL at the node where Q1/Q3/R5/C1/C3 all connect, we see that if 2A is flowing out of the bottom of R5, it's flowing into the drain of Q2, out of the drain of Q2 and into R6 and back to the negative rail. As long as you replace all of the source resistors (R5/6/40/41) at the same time, as long as they're the same value, the differential voltage should stay close to 0.
Next lets look at the case where input signal is applied. Load current will only flow if there is a potential difference between the + and - speaker terminals. This happens when a difference between the + and - input terminals (or + and gnd in the case of unbalanced inputs). The differential circuitry formed by Q5/6, R23/25, and the constant current source (all the stuff above Q5/6) decyphers the difference between the inputs, adds a DC component to place the bottom power transistor Q2 into its linear region where it's acting like a variable resistor whose resistance is controlled by the input signal voltage (Vgs), and sends the signal, DC and all, to the gate of Q2.
The top transistor current source source is responsible for maintaining a mostly constant current. It slightly modifies its behavior based on how much of that current is flowing through the load, but mostly it just tries to achieve a constant current through Q1 which is then divided between the load and the bottom transistor Q2 and resistor R6. The voltage at the gate of Q2 is ultimately what's setting the voltage at the drain of Q2.
The difference in the + and - input signal voltage is ultimately what's causing the potential difference between the + and - speaker terminals, and therefore causing the load current to flow.
Looking at the original Aleph circuit might help you to visualize what's happening without adding the confusion of the Aleph X full differential amp.
Bryan A. Thompson
bryan@batee.com
Tweaking..tweaking...must beat that Krell..trim, trim, other diff-pair..trim, trim, up that bias...tweak...CL37...boom...damn...must beat.
/Hugo
/Hugo
jupiterjune said:
I am looking at Aleph-M. The current-limiting set of BC550C, 100R and 221R works only when there is excessive current through the source R of 0.33--i.e. about 6A. So, they are useless in normal condition. Remove them to make the circuit simpler.
Then, we earn the gate voltage to the lower IRF240 of about 4-5V from the 390R. As this gate voltage is a fixed value, the mosfet is self-biased with the 0.33R. If we increase the source R to 1 ohm, the self-bias current should be getting lower.
I think I hear a creaking sound coming from clear over on the other side of the country. Somebody better Super Glue Baja to the rest of the continent.
There seems to be some confusion about the the output stage. I thought this had been covered before, but maybe I'm wrong.
There are several ways to look at it, but let's start with the original Zen amp. It's simple enough to grasp fairly easily, but at the same time, it's a good stepping stone to the Aleph current source.
So...What is the sound of one transistor with the clap?
Er...
Take two: What is the sound of two transistors in love?
Better, but still not quite right...
Third time pays for all: What is the sound of one common Source transistor working into a current source?
Hmmm...lacks elegance, but it'll do until more people learn what a koan is.
The gain device in a Zen is identical to the gain device in an Aleph. It's an ordinary MOSFET in the most ordinary configuration; to wit: common Source. Common Source means that the Source is the "unused" pin. Why in the world they chose this particular nomenclature, I will leave to the philosophers. Seems goofy to me. I mean, really, name the hookup after the odd man out? Whatever. That's what it's called, and rather than say "Gate/Drain Hookup" or something along those lines, we'll leave the name alone. There are more important fish to fry.
Common Source means that the signal enters through the Gate and leaves via the Drain. There are advantages...and disadvantages. On the plus side of the scales, the circuit has both voltage and current gain. Pretty handy if you're setting out to design a single stage amplifier. On the minus side, common Source inevitably has a high Zout. Bummer. But there's no such thing as a free lunch.
The circuit needs a load, elseways the signal going out the Drain disappears into the power supply, never to be seen again. You could use a resistor as the load (see the Son of Zen), but for the Zen, Nelson chose to use a current source. It makes things more efficient and keeps the N-ch MOSFET on the bottom from getting lonely at night.
Now to the nut of the matter. The bias is set by the current source. More particularly by the small value/higher wattage resistor above the current source's Source (R1, for those keeping score). The Vbe of the small bipolar (Q3) plays against R1, and since bipolar transistors are pretty consistent about such things, you can count on something along the lines of .65V showing up across R1. With a properly dazzling display of mathematical prowess, I will hereby invoke Ohm's Law, crank the handle, and...presto!...out pops the answer.
V/R=I
.65V/.33 Ohms= 1.9A
So what about the MOSFET on the bottom? You use P1 to set the DC voltage at the Gate of Q1 so that it is in balance with the current source. You can use an oscilloscope to go for symmetrical clipping or you can do a pretty fair job by using a voltmeter to set the DC between Q1 and Q2 to half of the rail voltage.
Fast forward to the Aleph current source. The current source has been flipped upside-down. This really unnerves some people because they think that a current source must always present its high-Z node to the intended load, but in reality the current is the same whether you take it from the Source or the Drain. What you gain by using the Source is a lower impedance, not to mention the ability to use a device that matches your gain device. In the usual Aleph configuration, you see an N-ch MOSFET working into an N-ch MOSFET, which helps because they have the same characteristics. This is good because it lowers distortion (not to mention saving money--something that never goes out of fashion).
Like the Zen current source, the Aleph current source's current is set by the resistor under the MOSFET's Source.
But what to do with the MOSFET on the bottom? How does it know when it's in balance with the current source?
The coarse adjustment is comprised of the front end bias current, half of which goes through the load resistor for one side of the front end differential. Again, Ohm's Law:
IR=E
(20mA/2) * 392 Ohms= 3.9V
Where 20mA is the nominal front end bias (divided in half--half goes up each side of the differential), 392 Ohms is the front end load, and the 3.9V is a pretty close approximation of the voltage needed to light up an IRF MOSFET.
The fine adjustment occurs within the differential itself. It looks at the DC offset at the output, relative to ground, and adjusts itself to fine tune the lower MOSFET's bias such that it's always in balance with the current source.
Yes, there are other things that you can throw in, like the extra parts in the Aleph current source, the fact that the front end's current source is adjustable, and so forth, but that's enough to be getting along with.
Besides, people get fussy when I start making long-winded posts.
And my fingers are tired.
And I'm thirsty.
Grey
There seems to be some confusion about the the output stage. I thought this had been covered before, but maybe I'm wrong.
There are several ways to look at it, but let's start with the original Zen amp. It's simple enough to grasp fairly easily, but at the same time, it's a good stepping stone to the Aleph current source.
So...What is the sound of one transistor with the clap?
Er...
Take two: What is the sound of two transistors in love?
Better, but still not quite right...
Third time pays for all: What is the sound of one common Source transistor working into a current source?
Hmmm...lacks elegance, but it'll do until more people learn what a koan is.
The gain device in a Zen is identical to the gain device in an Aleph. It's an ordinary MOSFET in the most ordinary configuration; to wit: common Source. Common Source means that the Source is the "unused" pin. Why in the world they chose this particular nomenclature, I will leave to the philosophers. Seems goofy to me. I mean, really, name the hookup after the odd man out? Whatever. That's what it's called, and rather than say "Gate/Drain Hookup" or something along those lines, we'll leave the name alone. There are more important fish to fry.
Common Source means that the signal enters through the Gate and leaves via the Drain. There are advantages...and disadvantages. On the plus side of the scales, the circuit has both voltage and current gain. Pretty handy if you're setting out to design a single stage amplifier. On the minus side, common Source inevitably has a high Zout. Bummer. But there's no such thing as a free lunch.
The circuit needs a load, elseways the signal going out the Drain disappears into the power supply, never to be seen again. You could use a resistor as the load (see the Son of Zen), but for the Zen, Nelson chose to use a current source. It makes things more efficient and keeps the N-ch MOSFET on the bottom from getting lonely at night.
Now to the nut of the matter. The bias is set by the current source. More particularly by the small value/higher wattage resistor above the current source's Source (R1, for those keeping score). The Vbe of the small bipolar (Q3) plays against R1, and since bipolar transistors are pretty consistent about such things, you can count on something along the lines of .65V showing up across R1. With a properly dazzling display of mathematical prowess, I will hereby invoke Ohm's Law, crank the handle, and...presto!...out pops the answer.
V/R=I
.65V/.33 Ohms= 1.9A
So what about the MOSFET on the bottom? You use P1 to set the DC voltage at the Gate of Q1 so that it is in balance with the current source. You can use an oscilloscope to go for symmetrical clipping or you can do a pretty fair job by using a voltmeter to set the DC between Q1 and Q2 to half of the rail voltage.
Fast forward to the Aleph current source. The current source has been flipped upside-down. This really unnerves some people because they think that a current source must always present its high-Z node to the intended load, but in reality the current is the same whether you take it from the Source or the Drain. What you gain by using the Source is a lower impedance, not to mention the ability to use a device that matches your gain device. In the usual Aleph configuration, you see an N-ch MOSFET working into an N-ch MOSFET, which helps because they have the same characteristics. This is good because it lowers distortion (not to mention saving money--something that never goes out of fashion).
Like the Zen current source, the Aleph current source's current is set by the resistor under the MOSFET's Source.
But what to do with the MOSFET on the bottom? How does it know when it's in balance with the current source?
The coarse adjustment is comprised of the front end bias current, half of which goes through the load resistor for one side of the front end differential. Again, Ohm's Law:
IR=E
(20mA/2) * 392 Ohms= 3.9V
Where 20mA is the nominal front end bias (divided in half--half goes up each side of the differential), 392 Ohms is the front end load, and the 3.9V is a pretty close approximation of the voltage needed to light up an IRF MOSFET.
The fine adjustment occurs within the differential itself. It looks at the DC offset at the output, relative to ground, and adjusts itself to fine tune the lower MOSFET's bias such that it's always in balance with the current source.
Yes, there are other things that you can throw in, like the extra parts in the Aleph current source, the fact that the front end's current source is adjustable, and so forth, but that's enough to be getting along with.
Besides, people get fussy when I start making long-winded posts.
And my fingers are tired.
And I'm thirsty.
Grey
thanks!