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Old 22nd June 2004, 10:22 PM   #1
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Default GC SuperSymmetry pt II

Here's another one for you guys who would rather put
the discretes on the output instead of the input.

www.passlabs.com/np/GC-SS-2a.pdf
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Old 23rd June 2004, 04:45 AM   #2
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Confused:

The amp output is disconnected from the output of the chip. How come??? I hope this is a drawing mistake. Otherwise, I have to admit that I should have studied myself more.

JH
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Old 23rd June 2004, 05:12 AM   #3
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In this case the op-amps are at the inputs, feeding the bipolar transistors which are the output devices.
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Old 23rd June 2004, 05:15 AM   #4
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The drawing is correct. The output of each chip connects to
the output of the other chip. You can view each chip amp
as the first transistor in a complementary "Sziklai" stage,
also known as conjugate complementary.
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Old 23rd June 2004, 05:47 AM   #5
azira is offline azira  United States
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Quote:
Originally posted by jh6you
Confused:

The amp output is disconnected from the output of the chip. How come??? I hope this is a drawing mistake. Otherwise, I have to admit that I should have studied myself more.

JH
I agreed with you when I first looked at it... But I decided to bite my tongue and defer to the wisdom of NP and look at it in a different way.

Try working backwards. Firstly, we'll assume this is some kind of amplifying device right? If you note that the GCs are NOT connected to the outputs, so what is? It looks a lot like a complementary CE amplifier in bridge mode (well besides the X feedback). So working backwards from there, if you were to take the upper left transistor and assume it's a CE amplifier, you'll get gain by modulating the base voltage. How would the base voltage get modulated? Well, the only way is by varying the current through the associated base resistor... or in otherwords through current draw in the GC...
So, the amplification action happens when the GCs push/pull current at eachother and thusly take it from the supply rails.

One thing I notice is that you need the chipamps to pull quiescent current to set a bias but the more current the GCs pull, the lower their rails go which limits the amount of current they can output... it's pretty likely that ideally you would want very little current to shift between the GCs which also helps stabilizes their rail voltages and most of your gain is going to come from a complementary VAS action. It's already a high-gain configuration.

If I am right then actually any decent audio OPamp would likely do the job because 40mA swing would probably be enough. Although the higher bias current of a GC would be easier to work with.

Just my thoughts...
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Danny
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Old 23rd June 2004, 06:04 AM   #6
azira is offline azira  United States
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Well... either I'm working in the right direction or gonna go down burning so lets add more fuel..

3 more thoughts.

Firstly, if I was right about the opamps, then a suitable replacement might use something that was unity gain stable instead and not necessarily need the input attenuator. A higher feedback factor so less stage distortion. The downside is that the PSRR of the GC is so good.

Secondly, since the pushpull action of the 2 opamps puts a virtual ground in the middle of the resistor, why not split it to gnd on both ends... because it would require 2 matching elements instead of one unmatched device... ok answered my own question there..

Thirdly, say we use a 8-ohm load, same virtual ground applies to the complementary VAS so in reality, each gain pair sees a 4-ohm load. That means that their gain is actually pretty low. In order to compensate for this, we'd have to use more input signal. So it seems that the real gain mechanism is current pull over the VCC/VSS resistors of the opamps... in which case we would really want 4 nicely matched devices there.
And then following this observation, since the gain of the BJTs isn't critical, perhaps choosing a fairly high Re (so more local feedback) might help reduce distortion in that stage...

The question that remains then is... if the BJTs are more of a current gain device, why not use an EF instead of a CE?
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Danny
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Old 23rd June 2004, 06:11 AM   #7
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I see that differences in bias current and DC offset which is essentially non-symmetric bias current would require that the VCC/VSS resistors be adjustable, but once you do that you would then be disrupting atleast some of the gain, much more signifcantly than any minor stage distortion... very well matched devices with very low DC offset is a requirement...
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Old 23rd June 2004, 10:12 AM   #8
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Happy:

Thanks, Danny. I wrote it down.
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Old 23rd June 2004, 04:42 PM   #9
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Danny,

Well said and makes sense to me. By the way, my brain hurts, does yours?
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Old 23rd June 2004, 05:16 PM   #10
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It would hurt except that I've had lots of episodes with friends where we would come up with some totally absurd idea that would almost seem like it would work... talk about it for a while and then dismiss it.
one example would be using a couple of diving tanks to make a water cannon on the back of my Jeep to siphon water from the river and shoot it back at passing rafters...
something like "lets take the current draw from the supply rails to modulate the gain of an amplifier" seems a lot like that, I'm conditioned to the shock...
Just one problem, NP actually wrote it down...
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Danny
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