I have just realized that your claims lead to the conclusion that negative current feedback is a physical impossibility.
Consider the load-free CFA we have been discussing. You agreed that the AC portion of ie1 is the AC portion of ic2. Here, one current flows into the inter-stage junction as the other flows out of it. A little thought reveals that this is regenerative behavior. An increase in output stage current leads to an increase in input stage current which leads to an increase in output stage current... (Fortunately, the loop gains in our circuits must be less than unity, or we’d have oscillation.) If you wish to deny that this is an example of positive current feedback, then please supply one. Alternatively, declare that positive current feedback is a physical impossibility. If impossible, either give an example of negative current feedback, or declare that it too is physically impossible.
But if positive current feedback does occur when input and output stage currents flow in opposite directions with respect to some inter-stage point, then negative current feedback, if it exists, must occur when the currents flow in the same direction with respect to that point (such as with a practical, adequately loaded CFA). However, you have declared that because CFA currents do this, a CFA cannot employ current feedback!
There are only two sensible conclusions I can draw from this situation: (1) you are asserting that negative (and maybe even positive) current feedback is physically impossible, or (2) your reasoning is flawed.
Gads. Some of you are correct, some of you are wrong and some both and quit a few are confused.
Try this..... why is the resistor from output to low z port (emitter etc) of a specific value or limited to a narrow range of values? See CMA IC apps.
-RNM
Try this..... why is the resistor from output to low z port (emitter etc) of a specific value or limited to a narrow range of values? See CMA IC apps.
-RNM
It has to do with transistor h sub o.
Actually, I now think your question is related to a different thread here. As Emily Littela once said, "Never mind."
Actually, I now think your question is related to a different thread here. As Emily Littela once said, "Never mind."
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I repeated the problem today and I suddenly remembered that I already met it last year. It comes from a bug of the jumpers in my TINA simulator.
Work of the day http://iodau.pagesperso-orange.fr/forr_CFA%20Sziklai%20currents%20&%20phases.png
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I am not sure of a physical impossibility and that what I said can imply this.
But I think and try to show that the CFA topology as we know does not rely on a difference of currents, while the VFAs undoubtly rely on a difference of voltages.
So you deny that CFA input stage current equals the difference between load plus network ground current and output stage current?
First amplifiers named CFAs had a diamond input stage caracterised by a low impedance inverting input. Like all amplifiers without a differential input. So now, with the large acceptance of the term, the logic implies that a push-pull inverting input is not mandatory and that single input device - tube or transistor - feedback amplifiers can be considered as CFAs.
In my simulation of the day with the Sziklai pair, we have
ie1 + ic2 = iload
So
ie1 = iload - ic2
So then in your simulation, the CFA input stage current is the difference between these currents. Do you deny that this is true for the more general CFA, in which the feedback network ground current is added to the load current?
If it is true, how is this not current feedback?
😀
Yeah. It is not what you are talking about and trying to understand. But, it should be.
-RNM
Cryptic. Whether or not I'm justified in my certitude, I'm not aware of "trying to understand" anything.
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Try this... anyone .... might help you more.... what happens and why when you place a Cf across the Rf -- in a CMA? Compared to VMA?
(CFB vs VFB if you prefer).
THx-RNMarsh
(CFB vs VFB if you prefer).
THx-RNMarsh
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On the off chance that anyone is actually interested to understand these issues, I can recommend Toumazou and Lidgey:
Analogue Ic Design: The Current-Mode Approach (I E E CIRCUITS, DEVICES AND SYSTEMS SERIES): C. Toumazou: 9780863412158: Amazon.com: Books
Old, but the basics. Targeted at IC design but the issues, principles and insights are applicable to general circuit design.
Jan
Analogue Ic Design: The Current-Mode Approach (I E E CIRCUITS, DEVICES AND SYSTEMS SERIES): C. Toumazou: 9780863412158: Amazon.com: Books
Old, but the basics. Targeted at IC design but the issues, principles and insights are applicable to general circuit design.
Jan
Demonstrably untrue. Please see attached.
Please see the equation in the attached. To deny its validity would be to deny Kirchoff's current laws.
The output and input currents must have the relative phases shown for there to be negative current feedback. If their relative phases were reversed, the currents would add and there would be positive feedback. To say that there is no feedback because of the depicted relative phase is to say that negative current feedback is impossible.
CFA input current is the difference between the ground and output stage currents. Because output and input stage currents are related linearly, it is correct to say that output stage current is fed back to the input stage. Depending on the resistances, output stage current could augment (positive current feedback) or diminish (negative current feedback) the input stage current.
Attachments
Demonstrably false as shown and properly formulated in post 837.
Please note that the possibility of a valid, more more complex, voltage plus transistor transconductance plus voltage divider explanation of circuit operation that you might prefer does not invalidate the claims of post 837.
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Cpaul, the arrows in your diagram seem to go in opposite direction
which may give a false impression. Put vertically, they will go in the
same direction.
Here is my demonstration with a diagram, DC and AC currents and
voltages, phases, and a short text which I copied below :
http://iodau.pagesperso-orange.fr/forr_Cfa%20NPN-PNP%20bootstrap%20in-.png
The voltage following property in the inverting input of CFAs
Ïn this circuit, T1 acts an emitter follower. When the feedback
in not connected, T1 emitter is only loaded by Rg, the current
it delivers being fundamentally tied to T1 base emitter voltage.
T1 collector current is equal to T1 emitter current less T1 base
current. The collector laod is R2 in parallel with T2 emitor-base junction,
the voltage of which controls the current across this transistor.
When the feedback is connected, T1 emitter keeps its status of
voltage follower. However T2 now provides most of current
into Rg through Rf, so the amplitudes of the AC current and voltage
Vbe of T1 are much reduced,
The load of T1 emitter becomes much highly resistive, as currents
in both transistors are in phase and share the same load Rg.
All that is under the ultimate control of T1 base-emitter voltage
What happens when a non-linearity introduced by T2 does not
provide enough current and causes a drop of the AC voltage
on its collector ?
-> Vload is too low
-> the value of the resistive load seen by T1 emitter lowers,
-> T1 emitter must deliver more current
-> Vbe1 and Ae1 inscrease
-> VR2 (Vbe of T2) increases
-> Ac2 increases
-> Vout increases
and this until, the equilibrium is obtained.This is almost
instantanenous. If there is too much current in T2 collector,
a similar but inverse process happens. The real command
of the whole is Vbe1 which is a difference of potential
(voltage for short) between the two inputs, often named
in+ et in- of the input stage.
There is an approach of current feedback in CFAs which considers
the in- of the input stage as a current input. It is difficult to be in
agreement since, by definition, a current input - common base
transistors in bipolar technology, for instance - does not influence
its emitter input current. It not the case in amplifying circuits
having a low impedance inverting input because it not a
current input.
which may give a false impression. Put vertically, they will go in the
same direction.
Here is my demonstration with a diagram, DC and AC currents and
voltages, phases, and a short text which I copied below :
http://iodau.pagesperso-orange.fr/forr_Cfa%20NPN-PNP%20bootstrap%20in-.png
The voltage following property in the inverting input of CFAs
Ïn this circuit, T1 acts an emitter follower. When the feedback
in not connected, T1 emitter is only loaded by Rg, the current
it delivers being fundamentally tied to T1 base emitter voltage.
T1 collector current is equal to T1 emitter current less T1 base
current. The collector laod is R2 in parallel with T2 emitor-base junction,
the voltage of which controls the current across this transistor.
When the feedback is connected, T1 emitter keeps its status of
voltage follower. However T2 now provides most of current
into Rg through Rf, so the amplitudes of the AC current and voltage
Vbe of T1 are much reduced,
The load of T1 emitter becomes much highly resistive, as currents
in both transistors are in phase and share the same load Rg.
All that is under the ultimate control of T1 base-emitter voltage
What happens when a non-linearity introduced by T2 does not
provide enough current and causes a drop of the AC voltage
on its collector ?
-> Vload is too low
-> the value of the resistive load seen by T1 emitter lowers,
-> T1 emitter must deliver more current
-> Vbe1 and Ae1 inscrease
-> VR2 (Vbe of T2) increases
-> Ac2 increases
-> Vout increases
and this until, the equilibrium is obtained.This is almost
instantanenous. If there is too much current in T2 collector,
a similar but inverse process happens. The real command
of the whole is Vbe1 which is a difference of potential
(voltage for short) between the two inputs, often named
in+ et in- of the input stage.
There is an approach of current feedback in CFAs which considers
the in- of the input stage as a current input. It is difficult to be in
agreement since, by definition, a current input - common base
transistors in bipolar technology, for instance - does not influence
its emitter input current. It not the case in amplifying circuits
having a low impedance inverting input because it not a
current input.
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