It would be great if this lengthy thread with great resurses could end up in propositions of a few new topologi names and abbreviations to distinktion between the different uses. These could be presented to AES and other organisations and maybe end up in easier communication for future generations.
Yes, because they do not understand the different of internal operation between VFA and CFA. They need back to university
Even, someone need to learn how to read all comments and not only read what he like.
Baxandall, like Douglas Self, also thought complementary balanced amplifiers were ‘flawed’, and that JFET input power amplifiers were suboptimal.
Both clever, practical men, but not always right.
The original reason the military was building discrete versions. A variable gain amplifier with invariant pulse fidelity (i.e. constant BW and rise time).
No correlation with Cfas whose input stages can be made with a single device greatly facilitating the basic analysis.
That implies you are.
It's clear from the attached that Baxandall's objections were not based on circuit analysis or an understanding of how a CFA works. Rather, he bemoaned the appropriation of a term that was (improperly, in my view) assigned to describe a different circuit (see attached).
To see why I say "improperly", we have to agree on a definition of feedback, about which there actually seems to be some controversy (could this be related to our disagreement?) Here's my definition:
Feedback is the application of a signal derived from a first point in a circuit and applied to a second in a manner that influences behavior at the first.
The circuit Baxandall describes fails the test. As long as feedback is applied to a triode grid, a FET gate, or the base of a high-beta BJT, the only place that "current feedback" goes back to is ground (which it thankfully fails to influence.) This circuit is a classic VFA in which an element of the feedback network is the circuit's load. What is truly unfortunate is not that an historical term was used to describe the operation of a different circuit, but that the term was a misnomer in the first place.
Nothing in the citation of Baxandall indicates that his understanding of CFA operation differs from that of those of us who hold that a CFA employs current feedback. Baxandall's argument is etymological rather than technical.
To see why I say "improperly", we have to agree on a definition of feedback, about which there actually seems to be some controversy (could this be related to our disagreement?) Here's my definition:
Feedback is the application of a signal derived from a first point in a circuit and applied to a second in a manner that influences behavior at the first.
The circuit Baxandall describes fails the test. As long as feedback is applied to a triode grid, a FET gate, or the base of a high-beta BJT, the only place that "current feedback" goes back to is ground (which it thankfully fails to influence.
) This circuit is a classic VFA in which an element of the feedback network is the circuit's load. What is truly unfortunate is not that an historical term was used to describe the operation of a different circuit, but that the term was a misnomer in the first place.Nothing in the citation of Baxandall indicates that his understanding of CFA operation differs from that of those of us who hold that a CFA employs current feedback. Baxandall's argument is etymological rather than technical.
Attachments
I'd like to ask a question that is meant not to challenge, but rather to elucidate. (Oooh, big word! Yeah, but if fits my intent perfectly.)
I'm addressing those who hold that a CFA does not employ current feedback: Can you give us any examples of circuits that employ current feedback? If not, is the very concept (as feedback is defined in the prior post) physically unrealizable?
Who knows, the answers might improve the understanding of the rest of us and move the debate forward!
Naaaah.
I'm addressing those who hold that a CFA does not employ current feedback: Can you give us any examples of circuits that employ current feedback? If not, is the very concept (as feedback is defined in the prior post) physically unrealizable?
Who knows, the answers might improve the understanding of the rest of us and move the debate forward!
Naaaah.
That's easy you can use a complementary current mirror to convert an op-amp to current out. I've used this in servos where you need a true integrator, an integrator made out of an op-amp eventually has a zero and can add noise.
"Both clever, practical men, but not always right. "
That implies you are.
What, clever? Practical? Not always right? Absolutely!
But, you still have not answered me on the thought experiment I posted earlier.
You believe a CFA is a VFA because the inverting input and non-inverting input voltages (as in a PNP-NPN DB pair) must be equal when its operating in the linear mode.
The inverting input voltage can only equal the non-inverting input when sufficient current flows through the feedback network resistors - and this is very frequency dependent at HF, and the network impedance is confined to a narrow range of values unlike a VFA, where the current in the feedback network is essentially independent of frequency and can take on a very wide range.
That implies you are.
What, clever? Practical? Not always right? Absolutely!
But, you still have not answered me on the thought experiment I posted earlier.
You believe a CFA is a VFA because the inverting input and non-inverting input voltages (as in a PNP-NPN DB pair) must be equal when its operating in the linear mode.
The inverting input voltage can only equal the non-inverting input when sufficient current flows through the feedback network resistors - and this is very frequency dependent at HF, and the network impedance is confined to a narrow range of values unlike a VFA, where the current in the feedback network is essentially independent of frequency and can take on a very wide range.
Thanks, Scott. Let's see what those who say CFAs do not employ c.f. have to say.
You obtain the same results by varying the emitter resistor in the input stage of a VFA.
The customer only wants to change the feedback resistors. We've been through this several times, in practice it's an applications nightmare. General purpose kitted out parts are not very popular.
Here are excerpts of a discussion between people of the Sofia university.
There are very pertinent observations, notably a reference to a circuit showing how a very low impedance device can measure a voltage impedance without involving current. There are also a few general comments which can apply very well to this discussion.
Sorry, I did not reduce the huge size of the pictures, so you'll have to use the links... if you are interested. Or better, follow the link.
What is the truth about the exotic current feedback amplifier? Is it something new or just well known old? Is it really a current feedback device?
Lutz von Wangenheim And in the present case, we have the situation that (a) there is a current out of (or into) the low-resistive node
Therefore, in this case, I think it is appropriate to follow the terminology
commonly used and speak about "current feedback" (although it is the voltage
at the emitter node which is the physical cause of the effect).
Cyril Mechkov Seen from the side of Vout, the inverting input has extremely
high resistance (open circuit)... it is not really but virtually increased
input resitance...
Well Lutz, but how do we explain the fact that, if we are always in the
steady-state mode, the negative feedback is both voltage (determined by
Rg/(Rg+Rf and Vout) and current type (determined only by Rf and Vout)? Does a current enter/exit the common emitters? If so, the feedback is current; if no,
it is a voltage type...
At low enough frequency, the "input" voltage VIN- of the inverting input
manages to follow the output voltage of the voltage divider RG-Rf. So, the
"input" impedance is high (due to the mystic bootstrapping effect), and the
voltage divider is practically unloaded... it has the illusion that it is not
connected to anything This explains why the gain is determined by the
voltage divider... and this is the well-known voltage-feedback amplifier. At
high enough frequency, the "input" voltage VIN- of the inverting input does
not manage to follow the output voltage of the voltage divider RG-Rf, it is
"stiff". So, the "input" impedance is low (there is no the fancy bootstrapping
effect), and the voltage divider is practically shorted. This explains why in
this case the gain is determined only by the resistor Rf... and this is the so
called current-feedback amplifier.
I think that one thing is to change the emitter voltage of an emitter
follower when the base and accordingly, the emitter voltage, is fixed (as in
the ordinary common-base configuration) and other - when the base and emitter
voltages change in the same direction... The latter is a sort of a common-base
stage with varying base voltage and we can see here the effect of
bootstrapping...
For example, imagine the input voltage of the non-inverting input
increases... the emitter voltage also increases... the output voltage of the
buffer increases. . and the output voltage of the voltage divider also
increases... The result should be a decreased input current and accordingly,
an increased input impedance, of the inverting input... All this is valid at
low (below 3 dB) frequency.
You try to express and explain circuits by means of expressions and laws. . .
while my main goal is to grasp the basic idea behind the specific circuit
implementation and to express it by words.
For this purpose, I try to see well-known configurations, devices, analogies,
principles in the unfamiliar circuit... to reduce it to a new combination of
old building blocks... Formulas do not help me almost nothing to achieve my
goal .... I do not need them... they even hinder me obscuring and watering
down what is achieved by other means... Maybe it is because formulas show the
quantative side of the circuits... but, at this stage, I need qualitative
means. . . I need something else that is not written in sources. . .
Simply put, I realized that we speak different languages, and nobody listens
to what the other said. I wrote huge masses of explanations in all these nine
pages, but no one wants to think that way. You also wrote detailed
explanations, but I do not want to use them...
But puzzling to me is that I noted (from the first time, in 90's) my circuit
insights irritate and even enrage (even not conventionally thinking)
professionals. What is most interesting, they do not refute my specific
allegations but deny them by conclusions of a general nature. . .
To illustrate the written above about my way of thinking, I will say that I
have obtained a perfect notion about the operation of the CFA non-inverting
amplifier by the help of the humble 18th century electromechanical analogy -
measuring a voltage by a compensating voltage. On the page 5 of this dicussion
I have written something like this: "In this 'manual servo system', we change
the attenuated voltage VRG to make it equal to the input voltage and the
indicator shows zero. The problem is that the indicator (it was the so-called
galvanometer) and the input voltage source shows extremely low resistance; so
during the transition the whole current is diverted from Rg and it flows
entirely through the input source..."
And then, "This 'manual negative feedback' makes the system move towards its
equilibrium. Finally, during the steady-state, the input voltage is almost
equal to the unaffected output voltage of the voltage divider VRg = (Rg+Rf)
/Rg.V... and the whole current flows through Rg. Seen from the side of V (via
the voltage divider), the input voltage source has extremely high impedance
(open circuit)."
So, I had managed to realize the circuit idea by thinking of it as of a kind
of a servo system.
https://www.researchgate.net/profil...08125794/download/CFA+idea+-+steady-state.JPG
IMO if "current" in "current-feedback amplifier" means "current flowing
between the negative feedback network and the inverting input", this name is
misleading because no current flows (excluding the small erroneous current).
In steady state, what is the normal condition of the circuit, the feedback
network (the voltage divider) produces following voltage that is equal to the
emitter voltage; so, (almost) no current enters/exits the inverting input.
This is a unique property of any bridge circuit including the "manual servo"
above where no current flows through the bridge element (regardless of how it
is low-resistive), and is named "bootstrapping". The manual servo (I have
attached it again for convenience) is an excellent illustration of this
phenomenon - if you vary the input voltage and I keep the compensating voltage
equal to it, no current will flow through the galvanometer as much as
low-resistive it is... as though its resistance has become infinite. . .
So, in the steady state, we have a voltage-type negative feedback,
implemented by a classical voltage-to-voltage converter (voltage divider) that
continuously adjusts its output voltage R1/(R1+R2)*Vo to make it equal to the
input (and emitter) voltage VIN. As a result, the output voltage Vo is (R1+R2)
/R1*Vo... and this is exactly the same as in any "voltage-feedback" op-amp
configuration. . . Thus we have every right to call this device
"voltage-feedback amplifier".
So, in the steady state, we have a voltage-type negative feedback,
implemented by a classical voltage-to-voltage converter (voltage divider) that
continuously adjusts its output voltage R1/(R1+R2)*Vo to make it equal to the
input (and emitter) voltage VIN. As a result, the output voltage Vo is (R1+R2)
/R1*Vo... and this is exactly the same as in any "voltage-feedback" op-amp
configuration. . . Thus we have every right to call this device
"voltage-feedback amplifier".
https://www.researchgate.net/profil...61272860/download/CFA+idea+-+steady-state.JPG
If voltage (I suppose this will happen above the cutoff frequency), the
voltage-type negative feedback will become a current-type. . . the
voltage-to-voltage converter (the voltage divider) will transmute into a
voltage-to-current converter (the resistor R2). In our manual servo, this
situation means that I cannot follow you and a significant current begins
flowing through the galvanometer... the role of the lower resistor RG is less
(it is shunted) by the low emitter resitance... and the voltage divider on the
right acts as a simple resistor RS... Now we have every right to call this
device "current-feedback amplifier" since a significant current flows between
the feedback network and the emitters.
https://www.researchgate.net/profil...5162191271/download/CFA+idea+-+transition.JPG
There are very pertinent observations, notably a reference to a circuit showing how a very low impedance device can measure a voltage impedance without involving current. There are also a few general comments which can apply very well to this discussion.
Sorry, I did not reduce the huge size of the pictures, so you'll have to use the links... if you are interested. Or better, follow the link.
What is the truth about the exotic current feedback amplifier? Is it something new or just well known old? Is it really a current feedback device?
Lutz von Wangenheim And in the present case, we have the situation that (a) there is a current out of (or into) the low-resistive node
Therefore, in this case, I think it is appropriate to follow the terminology
commonly used and speak about "current feedback" (although it is the voltage
at the emitter node which is the physical cause of the effect).
Cyril Mechkov Seen from the side of Vout, the inverting input has extremely
high resistance (open circuit)... it is not really but virtually increased
input resitance...
Well Lutz, but how do we explain the fact that, if we are always in the
steady-state mode, the negative feedback is both voltage (determined by
Rg/(Rg+Rf and Vout) and current type (determined only by Rf and Vout)? Does a current enter/exit the common emitters? If so, the feedback is current; if no,
it is a voltage type...
At low enough frequency, the "input" voltage VIN- of the inverting input
manages to follow the output voltage of the voltage divider RG-Rf. So, the
"input" impedance is high (due to the mystic bootstrapping effect), and the
voltage divider is practically unloaded... it has the illusion that it is not
connected to anything
This explains why the gain is determined by thevoltage divider... and this is the well-known voltage-feedback amplifier. At
high enough frequency, the "input" voltage VIN- of the inverting input does
not manage to follow the output voltage of the voltage divider RG-Rf, it is
"stiff". So, the "input" impedance is low (there is no the fancy bootstrapping
effect), and the voltage divider is practically shorted. This explains why in
this case the gain is determined only by the resistor Rf... and this is the so
called current-feedback amplifier.
I think that one thing is to change the emitter voltage of an emitter
follower when the base and accordingly, the emitter voltage, is fixed (as in
the ordinary common-base configuration) and other - when the base and emitter
voltages change in the same direction... The latter is a sort of a common-base
stage with varying base voltage and we can see here the effect of
bootstrapping...
For example, imagine the input voltage of the non-inverting input
increases... the emitter voltage also increases... the output voltage of the
buffer increases. . and the output voltage of the voltage divider also
increases... The result should be a decreased input current and accordingly,
an increased input impedance, of the inverting input... All this is valid at
low (below 3 dB) frequency.
You try to express and explain circuits by means of expressions and laws. . .
while my main goal is to grasp the basic idea behind the specific circuit
implementation and to express it by words.
For this purpose, I try to see well-known configurations, devices, analogies,
principles in the unfamiliar circuit... to reduce it to a new combination of
old building blocks... Formulas do not help me almost nothing to achieve my
goal .... I do not need them... they even hinder me obscuring and watering
down what is achieved by other means... Maybe it is because formulas show the
quantative side of the circuits... but, at this stage, I need qualitative
means. . . I need something else that is not written in sources. . .
Simply put, I realized that we speak different languages, and nobody listens
to what the other said. I wrote huge masses of explanations in all these nine
pages, but no one wants to think that way. You also wrote detailed
explanations, but I do not want to use them...
But puzzling to me is that I noted (from the first time, in 90's) my circuit
insights irritate and even enrage (even not conventionally thinking)
professionals. What is most interesting, they do not refute my specific
allegations but deny them by conclusions of a general nature. . .
To illustrate the written above about my way of thinking, I will say that I
have obtained a perfect notion about the operation of the CFA non-inverting
amplifier by the help of the humble 18th century electromechanical analogy -
measuring a voltage by a compensating voltage. On the page 5 of this dicussion
I have written something like this: "In this 'manual servo system', we change
the attenuated voltage VRG to make it equal to the input voltage and the
indicator shows zero. The problem is that the indicator (it was the so-called
galvanometer) and the input voltage source shows extremely low resistance; so
during the transition the whole current is diverted from Rg and it flows
entirely through the input source..."
And then, "This 'manual negative feedback' makes the system move towards its
equilibrium. Finally, during the steady-state, the input voltage is almost
equal to the unaffected output voltage of the voltage divider VRg = (Rg+Rf)
/Rg.V... and the whole current flows through Rg. Seen from the side of V (via
the voltage divider), the input voltage source has extremely high impedance
(open circuit)."
So, I had managed to realize the circuit idea by thinking of it as of a kind
of a servo system.
https://www.researchgate.net/profil...08125794/download/CFA+idea+-+steady-state.JPG
IMO if "current" in "current-feedback amplifier" means "current flowing
between the negative feedback network and the inverting input", this name is
misleading because no current flows (excluding the small erroneous current).
In steady state, what is the normal condition of the circuit, the feedback
network (the voltage divider) produces following voltage that is equal to the
emitter voltage; so, (almost) no current enters/exits the inverting input.
This is a unique property of any bridge circuit including the "manual servo"
above where no current flows through the bridge element (regardless of how it
is low-resistive), and is named "bootstrapping". The manual servo (I have
attached it again for convenience) is an excellent illustration of this
phenomenon - if you vary the input voltage and I keep the compensating voltage
equal to it, no current will flow through the galvanometer as much as
low-resistive it is... as though its resistance has become infinite. . .
So, in the steady state, we have a voltage-type negative feedback,
implemented by a classical voltage-to-voltage converter (voltage divider) that
continuously adjusts its output voltage R1/(R1+R2)*Vo to make it equal to the
input (and emitter) voltage VIN. As a result, the output voltage Vo is (R1+R2)
/R1*Vo... and this is exactly the same as in any "voltage-feedback" op-amp
configuration. . . Thus we have every right to call this device
"voltage-feedback amplifier".
So, in the steady state, we have a voltage-type negative feedback,
implemented by a classical voltage-to-voltage converter (voltage divider) that
continuously adjusts its output voltage R1/(R1+R2)*Vo to make it equal to the
input (and emitter) voltage VIN. As a result, the output voltage Vo is (R1+R2)
/R1*Vo... and this is exactly the same as in any "voltage-feedback" op-amp
configuration. . . Thus we have every right to call this device
"voltage-feedback amplifier".
https://www.researchgate.net/profil...61272860/download/CFA+idea+-+steady-state.JPG
If voltage (I suppose this will happen above the cutoff frequency), the
voltage-type negative feedback will become a current-type. . . the
voltage-to-voltage converter (the voltage divider) will transmute into a
voltage-to-current converter (the resistor R2). In our manual servo, this
situation means that I cannot follow you and a significant current begins
flowing through the galvanometer... the role of the lower resistor RG is less
(it is shunted) by the low emitter resitance... and the voltage divider on the
right acts as a simple resistor RS... Now we have every right to call this
device "current-feedback amplifier" since a significant current flows between
the feedback network and the emitters.
https://www.researchgate.net/profil...5162191271/download/CFA+idea+-+transition.JPG