Kees, as I noted, there really isn't any current to speak of going into that emitter. The current going in is determined by the feedback factor more than the low emitter resistance.
How much curent you think is going into it?
jan
Yet, Richard, that's not what I see. The voltage at the inverting input (the low impedance node) swings just as wide as the voltage on the non-inverting input. And the current into the inverting (low impedance) node is much lower than you'd expect.
I checked with a G=10 amp with an AD844 and the inverting input current was just a few 100 nA.
I do agree with you that the term seems to be cast in concrete now and no amount of arguing is going to change that. Not even Mike will pull that off
jan
Well that is one of my points -- where the name came from and that is what it is now called. How it works is easy to explain and measure. I developed the circuit and measured it and learned things from it and published an app for it..... then went off to do other new things. Seems like such a waste of time when you can just measure it to decide what's what.
However, in some of the newest CMFB opamps the reasoning has evolved...... but still depends on actual design circuitry as to what you get with a particular version being called Current-Mode. Now they are hybrids of VM and CM and neither exactly explains them... whole new stuff with same old names.
Thx-RNMarsh
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Sounds intriguing Richard, do you have any part numbers or something?
Edit: I know hybrid opamps as in having discrete dies on the chip but I guess that's not what you meant.
jan
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Yet, Richard, that's not what I see. The voltage at the inverting input (the low impedance node) swings just as wide as the voltage on the non-inverting input. And the current into the inverting (low impedance) node is much lower than you'd expect.
I checked with a G=10 amp with an AD844 and the inverting input current was just a few 100 nA.
I do agree with you that the term seems to be cast in concrete now and no amount of arguing is going to change that. Not even Mike will pull that off
jan
Jan let me summarize one way of looking at it. A typical analysis starts with breaking the feedback loop into a forward path (A) and feedback path (f) to state the problem with the classic feedback equation A/(1+Af). With a VFA the break is easy and the forward path is essentiallv independent of the feedback. I learned using Y parameter two ports. The answer is easy by equating voltages. You cannot solve via input currents you simply get 0=0.
With a CFA you use H parameters and the forward path contains the feedback network. Point number one, the voltage across input to the amplifier is no longer the driving function of the forward path. You now can easily solve by setting the net input CURRENT to 0. That is by superposition the current in the input loop (out the -input) from the input source equals the current into the -input from the feedback network.
Surprise, surprise the answer is the same. Vin=Vo*Rg/(Rf+Rg). So the CFA case can be solved by only considering the -input CURRENT to be forced to 0.
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Thanks Scott, I get it.
I had come to the same conclusion empirically, in the limiting case the 'feedback current' approaches zero.
In hindsignt, since the inverting input current is determined by the delta V between Vin and Vfb across that low impedance, that necessarily leads to very low current since that delta V is basically Vout/OLgain.
Edit: Scott, I do understand and accept the difference in behaviour of a 'VFA' and a 'CFA', to be clear.
I believe I've said that already several times.
One of those differences are the reason I like those AD844's so much ;-)
jan
I had come to the same conclusion empirically, in the limiting case the 'feedback current' approaches zero.
In hindsignt, since the inverting input current is determined by the delta V between Vin and Vfb across that low impedance, that necessarily leads to very low current since that delta V is basically Vout/OLgain.
Edit: Scott, I do understand and accept the difference in behaviour of a 'VFA' and a 'CFA', to be clear.
I believe I've said that already several times.
One of those differences are the reason I like those AD844's so much ;-)
jan
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It can sometimes help to understand some circuits by knowing the meaning of 'virtual' as opposed to actual, physical. Check your understanding of virtual and a lot of things fall into place.
For example, an active node may be called a virtual ground node but doesnt actually measure that way with a volt/current/ohm instruments. Also, a port labelled low Z may not actually measure that way but behaves that way... as if.
Thx-RNMarsh
For example, an active node may be called a virtual ground node but doesnt actually measure that way with a volt/current/ohm instruments. Also, a port labelled low Z may not actually measure that way but behaves that way... as if.
Thx-RNMarsh
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I just used your post to put out that viewpoint. I think the constant BW for low gains and the much overlooked clean pulse response are most important. You might not have seen Nelson's comments about how he saw CFA's at military houses used for high speed signal processing and that got him going.
My main problem with Cherry is that he takes a narrow somewhat audio-centric view. He criticizes CFA's at high closed-loop gains (virtually no one uses them that way). He also shows basicly a 741 with variable compensation as an alternative. This is a poor choice, an undegenerated diff-pair is suboptimal for input linerarity and slew rate.
OTOH when all is said and done some of Mike's criticisms of CFA's for audio are perfectly good points, casting them as "bad" VFA's in disguise is not necessary.
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I guess I owe an explanation for my statement "a CFA is a VFA where the loading at the input (series) node can't be ignored". Here's the math:
Consider a standard topology as in http://www.analog.com/static/imported-files/seminars_webcasts/36692482463115527495427664003sect1.pdf fig. 1.11. For the moment, consider R2->oo. The amplifier gain is (after the usual approximations and a little algebra) Ao(jw)=(Rt/R1)/(1+jw*Rt*Cp). Consider now R2 finite, but much larger than R1. By breaking the feedback loop and calculating the open loop gain, we note that R1||R2 ~ R1, while R2 loading at the output is also negligible. h21=B=R1/(R1+R2) and under these circumstances, the closed loop gain is exactly the voltage feedback gain, 1+R2/R1, while the 3dB point given by RtCp is swept away with the same factor of 1+B*Ao.
Now, what's happening if we decrease R2 so that when breaking the feedback loop we can't ignore the effect of R2 in R1||R2? In this case, Ao(jw)=(Rt/(R1||R2)/(1+jw*Rt*Cp). Interesting enough, now B*Ao(jw) becomes exactly the expression of Ao in the "VFB" case above and after a few algebra manipulations the closed loop gain results the same 1+R2/R1, however the 3dB point is now controlled by R2, as in (1+jw*R2*Cp). We just got the fundamental property of CFAs, the gain independent bandwidth, as we got the other fundamental result, the feedback resistor is "compensating" the amp, as it controls the -3dB frequency.
That's about all I can put here without having an equation editor. Please note that I did not make use of any "low impedance node" or "short circuit" current. I used the canonical voltage feedback formalism that Michael loves to prove (for a certain topology) the current feedback amplifier properties.
Conclusion: this topology behaves strictly as a classic VFA, as long as the feedback resistor R2 is much larger than the emitter resistor R1. As soon as R2 and R1 become comparable, the closed loop behaviour transitions to a CFA, with gain independent bandwidth. This happens mathematically because R2 (the feedback resistor) affects the open loop gain, both the DC gain and the 3dB frequency.
Bottom line, "A CFA is a VFA where the loading at the input (series) node can't be ignored" Q.E.D.
Exactly! The low impedance node 'looks' like a low impedance but doesn't really act like one because there's a kind of 'bootstrapping' because the signal 'at the other side' (the non-inverting input) closely follows whatever is on the low impedance node.
jan
Where to use CMFB circuits.
As I said earlier, learn the pro-con (strength and weakness) of each and choose the one or optimize it/them to your priorities.
High speed pulse/data and wide band, high freqs is its best place in the line-up. OTOH -- they can work very well in audio as well. But I wouldnt use them in the lowest noise app if that were a priority (vinyl/LP). LIne stages (which I chose to apply it while using VFB for phono) and I/V converters, current-conveyor, signal processing and other areas in audio find good app for CMFB circuits. Or headphone amps where full power, at wide BW is needed.... that would include power amps where wideband and high current are needed at low distortion.
SW-- your answeres are always great... succinct, acccurate and conclusive.
Thx-RNMarsh
As I said earlier, learn the pro-con (strength and weakness) of each and choose the one or optimize it/them to your priorities.
High speed pulse/data and wide band, high freqs is its best place in the line-up. OTOH -- they can work very well in audio as well. But I wouldnt use them in the lowest noise app if that were a priority (vinyl/LP). LIne stages (which I chose to apply it while using VFB for phono) and I/V converters, current-conveyor, signal processing and other areas in audio find good app for CMFB circuits. Or headphone amps where full power, at wide BW is needed.... that would include power amps where wideband and high current are needed at low distortion.
SW-- your answeres are always great... succinct, acccurate and conclusive.
Thx-RNMarsh
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No one said "there was no such thing as a CFA". On the contrary, there is such a thing as a CFA, and it must necesarily be a current amplifier with a small input impedance and a large ouput impedance.
It must also possess series (current) derived-shunt (current) applied negative feedback, which further reduces the input impedance and increases the output impedance of the amplifier.
Clearly, your so-called "CFA" does not fall into the above category, but is, instead, a voltage amplifier because of its shunt (voltage) derived-series (voltage) applied negative feedback.
True. With a VFA the feedback network is buffered from the emitter of the input transistor.
True.
I am afraid I have to disagree; on the contrary, the resistor in the feedback network connected to ground has a finite voltage across it which wouldn't exist if the feedback were driving a near zero input impedance at the inverting input of the so-called "CFA".
In otherwords, if the inverting input of a so-called "CFA" were a virtual ground, one would be completely unable to vary the closed loop gain of the amplifier by altering the value of the feedback resistor connected to ground. In fact, the feedback resistor to ground carries a fraction of the output voltage (since the feedback network is a simple voltage divider) and applies it to the inverting input.
This implies the input device in a so-called "CFA" amplifies the difference in voltage between its base and emitter. This difference is expressed in the collector current of the input device (scaled up by the transconductance of the input transistor) and appears as a voltage at the output node.
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So we disagree, there are more important battles to fight. Mike you are a smart guy this issue is trivial in the stream of things. You would serve yourself by focusing on some bigger problems.
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This is 100% true as the feedback components appear in the forward path gain of the so-called "CFA".
This is not the case with VFAs with a differential pair for their input because the feedback network does not load the input device. Which is why the bandwidth of a VFA does not vary with the value of the feedback components.
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