can any one help me to find gain bandwidth product of the bode plot i attached

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I prefer the usual view that treats the OP as black box with specified transfer characteristic and restricts the analysis to feedback in the external circuit.

Referring to the previous discussion it should be underlined, that a complex pole pair can be generated without multiple feedback pathes, as you can see from a second order Sallen Key filter with ideal amplifier.
 
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zorro said:
Assuming that the dominant pole in the OA is obtained with a feedback capacitor, then this is a sort of multiple feedback circuit (there are two nested loops), but not like the ones swown in posts #6 and #18.
Click to expand...
LvW said:
I think, following this philosophy, even the simplest opamp amplifier (inverting, non-inverting) would be classified as multiple feedback circuits.
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Yes. I just mean that the circut we are discussing about has multiple-feedback in that sense, not with the topology of posts #6 and #18.

LvW said:
What about feedback caused by the collector-base capacitance of a BJT?
Click to expand...
As a matter of fact, this is the usual way the dominant pole is created in OA's since the 741, including the TL082.

FvM said:
I prefer the usual view that treats the OP as black box with specified transfer characteristic and restricts the analysis to feedback in the external circuit.
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Me too, of course.

FvM said:
Referring to the previous discussion it should be underlined, that a complex pole pair can be generated without multiple feedback pathes, as you can see from a second order Sallen Key filter with ideal amplifier.
Click to expand...

Right.
Regards
Z
 

zorro said:
As a matter of fact, this is the usual way the dominant pole is created in OA's since the 741, including the TL082.
Click to expand...
I was, of course, referring to the internal C-B capacitance of a BJT (not to the compensation capacitor).
 

LvW said:
I was, of course, referring to the internal C-B capacitance of a BJT (not to the compensation capacitor).
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Yes, of course I undertstood even if I didn't clarify the point.

If we want to go farther, let's say that an inverting lossy integrator (ideal OA with a parallel RC in feedback path) could be considered like an ideal integrator (with only C) with another "external" feedback loop formed by the R. An academic exercise, I think.
Regards

Z
 


Hi Zorro,
Did you check it by calculation? What is a "feedback loop formed by the R"? A loop consisting of one single resistor? I don`t think it works - even if considered as an academic exercise only. Do the calculation - and you will be convinced.
Regards
LvW
 

If you assume a gain block with an ideal first order low-pass transfer function (the said lossy integrator). Then external positive feedback can shift the pole towards zero (turn it into an ideal integrator). Or in other words, compensate the loss R with a negative resistor.

But in practice, the integrator won't be first order, and you'll have difficulties to compensate it excatly.
 


But - the above is not an answer to Zorro's last contribution, right?

- - - Updated - - -

More than that, I think this principle (1st order lowpass with positive feedback) is exploited in the non-inverting NIC integrator (Deboo integrator).
 

But - the above is not an answer to Zorro's last contribution, right?
Click to expand...
I'm describing the condition under which you can compensate a lossy integrator. As far he's suggesting negative feedback (a resistor across an inverting OP circuit), I'm saying that it doesn't work.
 


this is the ckt i am analysing
 

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Yes, yadhu - that was my first assumption: A (non-inverting) differentiator circuit.
Coming back to your original question - as mentioned in post#2 - you only can ask for the GBW of the opamp used, that means: not for the whole circuit.
The definition of the gain-bandwidth product needs a lowpass response.
 

then gbw for opamp is how much ? how to calculate it from the plot
 

yadhu said:
then gbw for opamp is how much ? how to calculate it from the plot
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As I have mentioned in post#2 already: The GBW is identical to the 0dB crossing of the open-loop gain function (approx. 2 MHz).
Note: This applies for all opamps exhibiting unity gain compensation (that means: first-order behaviour for open-loop gains > 0 dB).
In your case, this can be assumed.
 

Different from a simple N.I. differentiator, the present circuit has a voltage divider. At the 0 dB point of the transfer function, the OP has still a gain of 3.2 dB. Ends up in a GBW of 3 to 3.5 MHz, as expectable for TL082.
 
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Ohh yes, you are right of course! I forgot the factor introduced by the voltage divider. Thank you for correcting me.
 

Hi,

I'm back in order to answer this point.

This is the circuit. I consider the output voltave vs. input current because in this way we have an input node and an output node (see 2nd figure below).
The total transfer function is the transimpedance V2/I1.



We can consider this circuit as a basic transfer function (in this case it is an ideal current-to-voltage integrator) with a feedback resitor, like this:



The topology of the feedback is node-node or voltage-current, i.e., a sample of the output voltage is taken at the output node and a current proportional to it is injected into the input node.
We will take the direct transfer funcion A and the feedback β, and then we will find the closed loop transfer funcion AF=A/(1+Aβ). The circuit for the transimpedance A, taking into account the load of the feedback part at both the input and the output is as follows:



In this particular case (ideal OA) R has no effect on either of the sides. The direct transfer function is A=V2/I1=-1/sC .
The feedback transfer function is the transconductance β=IF/V2|V1=0 in this two-port:



i.e. β=-1/R.

As said, the closed-loop transfer function is AF=A/(1+Aβ).
Replacing the obtained A and β, we get AF=-R/(1+sRC) .

Please let me know if it is clear or not.
Regards

Z
 

zorro said:
Please let me know if it is clear or not.
Click to expand...

Hi Zorro - Yes, it’s clear, of course. Good exercise.


However, recalling the sentence as cited above (from your post #24) my remark in post#25 was related to an opamp-based circuit that is able to integrate voltages (classic MILLER integrator).
Your calculations show that the classic opamp-based circuit for current-to-voltage conversion (assumption: ideal opamp) with capacitive feedback has a transfer characteristic -1/sC and, thus, performs current integration (with sign inversion) - in principle, like each single grounded capacitor. Therefore, it is clear that another resistor in parallel will turn the circuit into a 1st-order current-in and voltage-out lowpass(damped integration). No doubt about it.

Regards
LvW
 

Yes, the inverting voltage integrator is what I've shown preceded by a two-port that performs voltage-to-current conversion (i.e. a resistor that "see" a very low load impedance, that in this case is the virtual ground).
An accurate analysis for a nonideal OA would take into account the input impedance of the second two-port, that is the impedance of the open-loop two-port divided by (1+Aβ).
Regards

Z
 
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