Current feedback - Voltage feedback, how do I see the difference? - Page 43 - diyAudio
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Old 3rd March 2006, 06:09 PM   #421
mikeks is offline mikeks  United Kingdom
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Quote:
Originally posted by jcx
there appears to be more technically succinct language(s) for this discussion:

http://people.ee.ethz.ch/~hps/publications/2000cas.pdf

ran across this while adding nordholt to search:

http://www.s2.chalmers.se/graduate/c...c/lecture2.pdf


I donít think the level of exchange we manage here is really getting at the subject matter
I am not sure about this JCX....it would appear this ground has already been covered here:

http://www.diyaudio.com/forums/showt...719#post422719

http://www.diyaudio.com/forums/showt...077#post425077
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Old 28th May 2011, 10:06 AM   #422
kees52 is offline kees52  Netherlands
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Default current feedback (alexander type)

Current feedback is independent of bandwith so it is interesting, using a opamp like in the schematic is a clever way to overcome tthe disadances of current feedback be low impedance, using a opamp as current suplier through its voltage pins.

Special the TIM is very low, this distortion gives music a very raw sounding character in most amps, it is the most important distortion, that have to be as low as possible, voltage feedback gives high TIM if not properly used and is why the most amps not giving open sound. as always feeback is the bad thing in tube world but is not always true, I have now a hybride without feedback, it sounds very good and open, but current feedback is also very nice sounding en fast...

A Paul Kemble web page - SSM 2131 (power amp design).

somehwere here i have posted a schematic with alexander thinking with whole bipolairs.

Last edited by kees52; 28th May 2011 at 10:21 AM.
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Old 15th March 2013, 11:29 PM   #423
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Quote:
Originally Posted by Charles Hansen View Post
It is a current feedback amplifier when the feedback is applied to the emitter (cathode, source), because then the current of the feedback signal is added to (or subtracted from) the current of the input stage.
No: the shunt derived series applied feedback voltage is applied to the emmiter of the input stage and therefore subtracted from the input voltage apllied to its base.

That the current through the feedback network is non-negligible is irrelevant.

Last edited by michaelkiwanuka; 15th March 2013 at 11:33 PM.
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Old 15th March 2013, 11:44 PM   #424
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This is probably the least necessary topic to revive. Go back to Middlebrook, you are not simply "sampling" the voltage at the input and there is significant current into the inverting input so "breaking the loop" is fundamentally different in a major way.
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Old 22nd March 2013, 08:41 AM   #425
RNMarsh is offline RNMarsh  United States
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Phew! Glad that is settled. Finally. "It is fundamentally different in a major way".

A short study, sponsored by IEEE, on the subject can be obtained as a UK Oxford booklet produced in 1994 by Chris Toumazou, John Lidgey and Alison Payne.
"Emerging Techniques For High Frequency BJT Amplifier Design: A Current-Mode perspective".

-RNM

Last edited by RNMarsh; 22nd March 2013 at 08:59 AM.
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Old 22nd March 2013, 08:57 AM   #426
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Quote:
Originally Posted by michaelkiwanuka View Post
No: the shunt derived series applied feedback voltage is applied to the emmiter of the input stage and therefore subtracted from the input voltage apllied to its base.

That the current through the feedback network is non-negligible is irrelevant.
I was always under the impression that there would be non-negligable current into a low-impedance (emitter) feedback node in a 'CFA' opamp. In that sense, I could understand where the term came from, even if it was completely opposite the terminology since the middle of last century.

But if you think about it, the node impedance may be low, but the current into it is also determined by the voltage difference between that node and the input. Feedback doing its job makes that voltage difference very, very small (zero in the limiting case of infinite feedback). I simmed a simple gain-of-10 amp with an AD844 and found that the current into the low-impedance (emitter) node was less than 500nA! Another perspective...

jan
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Old 22nd March 2013, 10:32 AM   #427
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Quote:
Originally Posted by scott wurcer View Post
This is probably the least necessary topic to revive. Go back to Middlebrook, you are not simply "sampling" the voltage at the input and there is significant current into the inverting input so "breaking the loop" is fundamentally different in a major way.
We are not sampling the voltage at the input. Rather, we have a voltage divider in parallel with the load sampling the voltage at the output, and, self evidently, not the output current.

The fraction of the output voltage dropped across the resistor in the voltage divider connected to ground is fed into the emitter of the input stage of the input device. The current through the divider is completely irrelevant.

Moreover, Middlebrook's method of section 5 in the paper below can be used in LTSpice at any point in the loop, so breaking the loop is not different in any way from that executed in a VFA which uses a differential pair for its input:

http://www.ele.tut.fi/teaching/ele-3...dleBrook75.pdf
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Old 22nd March 2013, 10:41 AM   #428
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Quote:
Originally Posted by jan.didden View Post
I was always under the impression that there would be non-negligable current into a low-impedance (emitter) feedback node in a 'CFA' opamp.
But all transistors are voltage controlled devices, so its the difference between the input voltage and the voltage at the emitter (or source as the case may be) that controls the device's collecter current (or drain current) in a so-called "CFA".

As noted above, the current through the voltage sampling divider is completely irrelevant as it does not drive the input transistor.
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Old 22nd March 2013, 11:35 AM   #429
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Also dont forget that a typical CFA input stage is symmetrical and thus the error current distortion or output voltage distortion is the difference between the distortion of the n and p side currents (and that these base emitter voltages divide equally, less effects from the error current).

Thanks
-Antonio
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Old 22nd March 2013, 02:27 PM   #430
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Quote:
Originally Posted by michaelkiwanuka View Post

The current through the divider is completely irrelevant.
Mike this is just wrong. EE102 1969 series-shunt feedback, the classic example is the three npn video triple. Breaking the loop involves loading the inverting input with the feedback network, for a VFA this is almost irrelevant but for a CFA this varies the input transconductance with closed loop gain. The important properties then fall out. Your statement elsewhere that the open-loop GBW of a CFA is constant is also wrong. As you can see my esteemed colleagues from Texas agree (picture below). This view point only arises when the loop is simply broken at the inverting input and it is considered a voltage input point, this is simply not valid.

Folks find an EE text that introduces the four basic feedback configurations as two ports with Y parameters and see for yourself. I mentioned Middlebrook because his technique avoids these mistakes.
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