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

[yadhu]

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

 

[LvW]

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

The gain-bandwidth product is given for an amplifier with a lowpass response only.
In contrast, your BODE plot has a bandpass shape.
However, one can assume that the plot represents an opamp circuit that has differentiating properties up to 10 kHz and that the falling part of the
curve shows the open-loop characteristics of the opamp used.
Therefore: The used opamp has a GBW of approx. 2 MHz (0 dB crossing frequency) .
Does this answer your question?

 

[yadhu]

i rigged the circuit by using TL082 opamp. in its datasheet they given GBW as 3Mhz

 

[Audioguru]

It is probably a Multiple-Feedback-Bandpass-Filter circuit that has perfectly straight sides attenuating at 6dB per octave.
But notice that the right side begins bending down at about 2MHz, probably caused by the 3MHz gain-bandwidth product of the opamp.

 

[LvW]

It is probably a Multiple-Feedback-Bandpass-Filter circuit that has perfectly straight sides attenuating at 6dB per octave.
But notice that the right side begins bending down at about 2MHz, probably caused by the 3MHz gain-bandwidth product of the opamp.

No audioguru, I don't think that the response belongs to a second-order bandpass (why do you think of a MFB filter ?). Again: I think it's a circuit that differentiate (very poor phase margin as a typical property).
However, I think it's funny that we are guessing here - and the questioner (yadhu) does not tell us what kind of circuit he is investigating.

 

[Audioguru]

No audioguru, I don't think that the response belongs to a second-order bandpass (why do you think of a MFB filter ?). Again: I think it's a circuit that differentiate (very poor phase margin as a typical property).
However, I think it's funny that we are guessing here - and the questioner (yadhu) does not tell us what kind of circuit he is investigating.

Of course it is a Multiple Feedback Bandpass filter with a high Q giving the sharp narrow peak and gradual single-order lowpass and gradual single-order highpass slopes for most of its response.

 

[FvM]

Of course it is a Multiple Feedback Bandpass filter with a high Q

You mean, the transfer function looks similar.

But what does your example tell about the bode plot in the original post and the circuit producing it?

As a trivial point, a TL082 isn't able to produce 80 dB gain at 15 kHz without a transformer or a LC resonator. We can further assume, that the term GBW has been used erroneously in the question.

 

[LvW]

Of course it is a Multiple Feedback Bandpass filter with a high Q giving the sharp narrow peak and gradual single-order lowpass and gradual single-order highpass slopes for most of its response.

Audioguru, perhaps you have overlooked that the "midband" gain is 80 dB.
More than that, are you able to detect a certain type of filter topology based on a BODE diagram only?

 

[Audioguru]

The Multiple Feedback Bandpass Filter is the only filter I have seen with a sharp and narrow high-Q peak then gradual single-order slopes.
The high-Q of the circuit produces a peak that has an amplitude much higher than the opamp can do by itself.

EDIT: I am just guessing because the filter is so bad that I have never made one.

 

[LvW]

The Multiple Feedback Bandpass Filter is the only filter I have seen with a sharp and narrow high-Q peak then gradual single-order slopes.

Each of the various filter topologies is able - in principle (!) - to produce high Q values.
For ideal conditions (amplifier, parts tolerances) ALL topologies exhibit the same properties. However, the behaviour of real circuits depends on the selected pole frequency and the type of opamp used (GBW).

The high-Q of the circuit produces a peak that has an amplitude much higher than the opamp can do by itself.

Your opamps can produce an output voltage amplitude "much higher than the opamp can do by itself" ? That means: Higher than the supply voltage ?

 

[FvM]

The high-Q of the circuit produces a peak that has an amplitude much higher than the opamp can do by itself.

I assume you are talking about filter gain. That's true e.g for an oscillator which achieves infinite gain. Similarly, the gain peaking of the present circuit can be explained by a near to zero phase margin.

My previous comment about the impossibility of a filter with 80 dB gain is only valid for an ideal filter. But a it's well possible utilizing the peaking of a real OP circuit.

In fact, the shown transfer characteristic can be achieved by simple inverting differentiator, as already assumed by LvW in post #2.

 

[Audioguru]

In fact, the shown transfer characteristic can be achieved by simple inverting differentiator, as already assumed by LvW in post #2.

A simple single-order differentiator is not a active bandpass filter. If it is followed by a simple single-order (second-order?) integrator then the "peak" is not sharp and narrow.

 

[FvM]

Please take a look at this simulation result. It's using an "ideal" OP with first order transfer function, open loop gain of 1e5 and 2 MHz GBW.



The peaking is created by nearly zero phase margin of the differentiator circuit.

 

[LvW]

A simple single-order differentiator is not a active bandpass filter. If it is followed by a simple single-order (second-order?) integrator then the "peak" is not sharp and narrow.

Who has claimed that? Who spoke about an integrator device?
Don't you remember that the problem of realizing an opamp based (inverting) differentiator always consists of a phase margin approaching zero - causing a peak as shown at the beginning of the thread (and repeated by FvM, showing the corresponding circuit) ?

 

[Audioguru]

The peaking is created by nearly zero phase margin of the differentiator circuit.

Thank you.
I have never made a differentiator like that. I see that as the output phase of the opamp increases at a high frequency then its output peaks.
so it is a Multiple Feedback Bandpass filter circuit using only one capacitor and one resistor.

 

[FvM]

I have never made a differentiator like that.

Me neither, you see why. Although the circuit is apparently acting as a high Q second order filter, I won't rely on the exact Q value, or at least circuit stability. You only need to add some capacitive load and make the circuit oscillating.

I think, the original post gives a nice interview question.

 

[LvW]

so it is a Multiple Feedback Bandpass filter circuit using only one capacitor and one resistor.

A "multiple feedback bandpass filter" with just one feedback path?
Audioguru, I suppose you know that there are some other bandpass topologies available than MFB?

- - - Updated - - -

Me neither, you see why. Although the circuit is apparently acting as a high Q second order filter, I won't rely on the exact Q value, or at least circuit stability. You only need to add some capacitive load and make the circuit oscillating.

Yes, a filter circuit relying on the open-loop opamp response is certainly not an option.
In this context, I remember the so called "R-filter" exploiting the frequency characteristics of (at least) two opamps only (without any external capacitors).
There are lot of papers on this subject - however, I never have heard about practical use of this concept. It is more or less an academic exercise.

 

[Audioguru]

There is an almost useless 3-band audio spectrum analyser project that used hopeless MFB filters.
I would never make something as bad as that.

The bass filter peaks at about 170Hz and cuts off any real bass, the high frequency filter peaks at about 3.3kHz and cuts off the top two highest octaves of sound and the mid filter peaks at 700Hz and does not show much else.

 

[zorro]

The Bode plot shows clearly a transfer function with a zero at s=0 and a high-Q pair of poles. An additional pole at high frequency produces an increased slope at the right of the plot.
As far as I can see, the exact topology of the circuit can not be completely inferred.

FvM's circuit in post #13 produces practically the same response.
The "open loop" configuration of this circuit has two poles: one of them is the real dominant compensation pole of the OA at 20 Hz, and the other is at -1/(R1*C2). Closing the loop with R1 (think in the root locus) they move forming a high-Q pair. The high frequency pole of the OA must be responsible of the higher frequency pole .
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.
Regards

Z

 

[LvW]

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.
Regards
Z

I think, following this philosophy, even the simplest opamp amplifier (inverting, non-inverting) would be classified as multiple feedback circuits.
I doubt, if this is a definition that makes sense. What about feedback caused by the collector-base capacitance of a BJT?