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Old 11th June 2013, 02:53 PM   #3461
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Quote:
Originally Posted by manso View Post
Hi Bob

Indeed your results may be somewhat different because of topology and compensation used.
My results were obtained through simulation of the regular blameless topology. If no-one posts the Self reference before Ill post it tommorrow. If I remember correctly it is in his fifth edition where he took actual measurements of this technique.
Thanks, manso.

I should add that I like the replica technique because it does not hang any components off of the LTP tail. Anything hanging there may degrade common mode rejection at high frequencies and add to EMI susceptibility.

Cheers,
Bob
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Old 11th June 2013, 03:15 PM   #3462
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Originally Posted by Samuel Groner View Post
One thing to consider is that bootstrapping with a significantly bandwidth-limited signal leads to negativ input capacitance of the bootstrapped transistor. Unless adressed, this may cause instability.

Samuel
Hi Samuel,

Good point. However, I always put 100 ohms in the emitter of each cascode device, even when not doing any driving of the cascode. Nevertheless, this whole stability thing is worth a closer look.

Cheers,
Bob
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Old 11th June 2013, 05:39 PM   #3463
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Originally Posted by HarryDymond View Post
Not at all. Inductors are used as effective current sources in switch-mode circuits all the time, especially the ubiquitous clamped inductive load test circuit. What Bob said: Is 100% true.

Once current is set up in a large inductor it behaves a lot like a current source. You know, just how a very large capacitor that has been previously energised (aka "charged") behaves a lot like a voltage source.

Hi Harry,

That this view is incorrect may be established by running a simulation of the loop gain of the amplified negative feedback current source with the very large inductor removed from the collector of the source transistor so that said source transistor is loaded exclusively by the output impedance of the transimpedance stage's transistor. The latter is then effectively an active current sink for the ANF current source. See figure below.

Now, if the large inductor (used in the circuit on the left to isolate the ANF current source from the rest of the circuit) behaved like a current source, as you and Bob contend, then the loop gain in both instances (with and without the inductor) would be virtually the same.

In fact the loop gain local to the ANF current source without the inductor to isolate it from the TIS current sink is over 40dB greater within the audio band than that obtained with the inductor in situ. (Blue trace: without inductor; green trace: with inductor).

This suggests that far from behaving like a current source, as you aver, the inductor in this case merely acts to isolate one circuit node from another at the frequencies of interest as Rosenstark indicates in his paper “Loop Gain Measurement in Feedback Amplifiers" Int. Journal of Electronics, Vol. 57, No.3., pp. 415-421, 1984 and in his book "Feedback Amplifier Principals".

From the figure below it is apparent that the true loop gain local to the ANF current source is obtained with its collector isolated from the rest of the circuit at the frequencies of interest by the large inductor.
Attached Images
File Type: png 01.png (31.9 KB, 106 views)
File Type: png 02.png (52.9 KB, 102 views)
Attached Files
File Type: asc ANF_loop_gain2.asc (9.6 KB, 3 views)
File Type: txt BJT_models.txt (7.4 KB, 4 views)

Last edited by michaelkiwanuka; 11th June 2013 at 05:49 PM.
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Old 11th June 2013, 08:04 PM   #3464
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Quote:
Originally Posted by michaelkiwanuka View Post
Hi Harry,

That this view is incorrect may be established by running a simulation of the loop gain of the amplified negative feedback current source with the very large inductor removed from the collector of the source transistor so that said source transistor is loaded exclusively by the output impedance of the transimpedance stage's transistor. The latter is then effectively an active current sink for the ANF current source. See figure below.

Now, if the large inductor (used in the circuit on the left to isolate the ANF current source from the rest of the circuit) behaved like a current source, as you and Bob contend, then the loop gain in both instances (with and without the inductor) would be virtually the same.

In fact the loop gain local to the ANF current source without the inductor to isolate it from the TIS current sink is over 40dB greater within the audio band than that obtained with the inductor in situ. (Blue trace: without inductor; green trace: with inductor).

This suggests that far from behaving like a current source, as you aver, the inductor in this case merely acts to isolate one circuit node from another at the frequencies of interest as Rosenstark indicates in his paper “Loop Gain Measurement in Feedback Amplifiers" Int. Journal of Electronics, Vol. 57, No.3., pp. 415-421, 1984 and in his book "Feedback Amplifier Principals".

From the figure below it is apparent that the true loop gain local to the ANF current source is obtained with its collector isolated from the rest of the circuit at the frequencies of interest by the large inductor.
Hi Mike,

You are finally beginning to get something that resembles the right answer, but you are drawing the wrong conclusion based on a wrong understanding of the VAS circuit in the closed loop amplifier and with the Miller compensation.

Your statement that the current source loop gains should be the same whether the inductor is in or out is wrong. The VAS node DOES NOT look like a current source for AC. It looks like a fairly low impedance because the feedbacks in the amplifier keep it that way. This allows the feedback current source to function normally. This is why your loop gain analysis came up with the larger loop gain number that all of the rest of us have been asserting all along.

Cheers,
Bob
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Old 11th June 2013, 08:07 PM   #3465
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Quote:
Originally Posted by michaelkiwanuka View Post
Hi Harry,

That this view is incorrect
It is not a "view", it is fact.

Do you deny that a charged very large capacitor behaves like a voltage source? Very large inductors are the dual of very large capacitors. Very large inductors require a very large volt-second product to be applied in order to change the current flowing; very large capacitors require very large current-second product to be applied in order to change the voltage across them. Which of these facts do you deny?

When you run a frequency analysis in LTspice, the first step is the determination of a DC operating point; this allows all non-linear circuit elements to be replaced with linearised models at their respective operating points and for the initial voltage or current in energy storage elements (respectively capacitors and inductors) to be determined.

The DC operating point analysis of the circuit with the very large inductor will determine that a certain DC current is flowing through it. Do you deny this?

V=L * di/dt Do you deny this?

di = V * dt/L Do you deny this?

According to the equation above, if L is very large, this requires the V*dt product to be similarly large in order for the current to change a non-negligible amount. Do you deny this?

In the frequency analysis, an effectively infinitely small ac signal is used to analyse the circuit's frequency response. Do you deny this?

From where is the very large voltage required to change the inductor's current going to come?

Answer - it is not going to come from anywhere and the inductor is behaving like a current source.

A large inductor is not always the correct component to "isolate" one part of a circuit from another.

Last edited by HarryDymond; 11th June 2013 at 08:10 PM.
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Old 11th June 2013, 08:22 PM   #3466
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Originally Posted by Bob Cordell View Post
Your statement that the current source loop gains should be the same whether the inductor is in or out is wrong. The VAS node DOES NOT look like a current source for AC. It looks like a fairly low impedance because the feedbacks in the amplifier keep it that way. This allows the feedback current source to function normally. This is why your loop gain analysis came up with the larger loop gain number that all of the rest of us have been asserting all along.

Cheers,
Bob

This was what I thought initially; viz. that because of minor loop feedback the TIS couldn't funtion as a current sink. I was wrong and so are you.

I established this fact by removing the miller compensation capacitor on the amplifier on the left and removing the inductor in both cases. If the minor loop feedback were to make a difference, then the loop gain local to the ANF current source would be radically different in both cases. It isn't. See below.
Attached Images
File Type: png 01.png (31.6 KB, 94 views)
File Type: png 02.png (52.7 KB, 92 views)
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Old 11th June 2013, 08:26 PM   #3467
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Mike,

I'm not entirely sure why you even want to measure the current sources' loop gain when it's isolated from the rest of the circuit. Surely what matters is the stability of the loop when the circuit is operating normally, i.e. current source and TIS are not isolated from one another?
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Old 11th June 2013, 08:26 PM   #3468
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Originally Posted by HarryDymond View Post
A large inductor is not always the correct component to "isolate" one part of a circuit from another.
Suffice it to say that it is used both in practice and in simulation to isolate circuit nodes from each other at the frequencies of interest by a great many luminaries in the field who would disagree with your averment.
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Old 11th June 2013, 08:31 PM   #3469
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Originally Posted by HarryDymond View Post
Mike,

I'm not entirely sure why you even want to measure the current sources' loop gain when it's isolated from the rest of the circuit. Surely what matters is the stability of the loop when the circuit is operating normally, i.e. current source and TIS are not isolated from one another?
As you've noticed by now the loop gain of the ANF current source is highly dependent on its collector load.

Thus, the purpose of isolating it from extraneous circuitry is to establish its intrinsic loop gain.
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Old 11th June 2013, 08:33 PM   #3470
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Originally Posted by michaelkiwanuka View Post
Suffice it to say that it is used both in practice and in simulation to isolate circuit nodes from each other at the frequencies of interest by a great many luminaries in the field who would disagree with your averment.
No, it will not suffice to say that, because they use large inductors to isolate a voltage source (the output of a feedback amplifier) from a medium-impedance load (the input to the feedback network). They don't use the large inductor to isolate a current source from a low impedance load.

An inductor is not always the right component to use to isolate parts of a circuit from one another.

You have not provided any rebuttal of post #3465. Perhaps you are working on this. I hope in the process you realise that we are correct about the fact that very large inductors behave like current sources.
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