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# Integrator question

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#### electroRF

##### Member
Op Amp question

Hi,

I'm trying to understand how the following circuit works.
View attachment 65175

I simulated it with the following results:
View attachment 65176

I don't manage to understand:
1. Why is it that when Vctrl = 467mV, then Vcap0=1.035V? (for Vset=0V).

2. What does the "Gain=100dB" (next to the OpAmp) mean?
I thought that differential OpAmp charges the output to either Vcc or Vee, no matter what the Gain is.

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I'm trying to understand how the following integrator works.

Why do you call this circuit "integrator"? Have you convinced yourself that the circuit is able to integrate?

Hi,
I might be wrong of course.
The parallel RC network in the feedback confuses me; If the feedback was purely capacitive, it was a simple integrator.
That's why I don't manage to understand the circuit operation.

Hi,
I might be wrong of course.
The parallel RC network in the feedback confuses me; If the feedback was purely capacitive, it was a simple integrator.
That's why I don't manage to understand the circuit operation.

1.) Where does this circuit come from?
2.) The operation called "integration" involves an input-output relationship. Therefore my question: What is your signal input (Vset or Vctl ?)

1.) Where does this circuit come from?
2.) The operation called "integration" involves an input-output relationship. Therefore my question: What is your signal input (Vset or Vctl ?)

Hi,

1.)
The circuit is part of a larger circuit.
The larger circuit is an Automatic Gain Control unit.
I'm trying to understand that part of the unit - i.e. the circuit I attached in the thread.

2.)
Vctl is the input.
Vset is a reference value.
Vcap is the output.

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Hi,

That circuit looks more like a low pass filter which cleans up the signal a bit before passing it along.
The reference voltage provides for some output offset adjustment.

Above the RC feedback frequency (1/2πRC), the circuit functions as an integrator.

Below the RC feedback frequency (1/2πRC), the circuit functions as an amplifier.

You don't necessarily want an integrator's theoretical infinite gain given a very small input voltage difference.

At the same time, you don't necessarily want an amplifier amplifying what is essentially noise on the Vctrl input.

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ElectroRF,
with Vctl as input and parts values as given in the diagram the circuit represents a lowpass with a corner frequency at approx. 100 kHz.
The phase shift approaches 90 deg for frequencies above 10 MHz. The maximum gain is 0 dB.
That means: You can use the circuit as an integrator for f>10 Mhz assuming an opamp that has a transit frequency (GBW) of at least 100 Mhz.

Question: Why don't you use a parallel resistor larger than 1 kohm?

Remark: For split supply the reference is Vset=-1.5 volts.

2nd remark: If the circuit - intended for integration puposes - is part of a stabilizing closed loop you can completely cancel the resistor in parallel to the feedback capacitor.
The feedback loop will stabilize the operating point of the integrator. However, this applies to the closed-loop case only.
This does not work for a separate test of the unit.

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MrAl, cachehiker, Winterstone
Thank you very much!
I very appreciate your kind help.

As seen in the simulation, the input (Vctl) is nearly DC, Therefore I understand that the circuit is simply LPF for such an input.

Question: Why don't you use a parallel resistor larger than 1 kohm?
I just got into a system which I need to learn.
This circuit is just small part of the system.
Currently I'm trying to understand things as they are.

Last edited:
Hi,

I'm trying to understand how the following circuit works.
View attachment 65175

I simulated it with the following results:
View attachment 65176

I don't manage to understand:
1. Why is it that when Vctrl = 467mV, then Vcap0=1.035V? (for Vset=0V).

2. What does the "Gain=100dB" (next to the OpAmp) mean?
I thought that differential OpAmp charges the output to either Vcc or Vee, no matter what the Gain is.

Hello again,

1.
You should see 1.033v on the output (1.035 is close enough) because for DC the circuit functions as a subtractor where it subtracts part of the two applied voltages vctl and vset. There are two gains to consider, the gain for vset and the gain for vctl. When vset is 0v that puts one half of the 1.5v battery voltage at the non inverting terminal. Since the gain for the non inverting terminal is R1/R2+1, the gain is 2, so with vset equal to 0v that puts 0.75 volts at the non inverting terminal and with a gain of 2 that puts 1.5v at the output for that input only. But then the input at vctl is 0.467v and the gain for the inverting part is only 1, so that puts -0.467 at the output for the inverting part only. Adding the non inverting part and the inverting part gives us 1.5-0.467 which comes out to exactly 1.033 volts. With a little input offset it will be slightly different.
So in finding the gain for the two inputs first calculate the voltage at the non inverting terminal, then multiply that by 2 and that is Vout1. Then find the output due to the vctl input, and that will be -1 times whatever vctl is, and call that Vout2. Then add Vout1 and Vout2 and that is the actual output.
A differential amplifier does NOT make the output equal to either supply rail unless one or both inputs are over driven. You might be thinking of a comparator circuit, which does exactly what you thought this does.

2.
The text, "Gain=100dB" means that the internal gain of the op amp is 100dB. That's the open loop again of the op amp.

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Hello again,

1.
You should see 1.033v on the output (1.035 is close enough) because for DC the circuit functions as a subtractor where it subtracts part of the two applied voltages vctl and vset. There are two gains to consider, the gain for vset and the gain for vctl. When vset is 0v that puts one half of the 1.5v battery voltage at the non inverting terminal. Since the gain for the non inverting terminal is R1/R2+1, the gain is 2, so with vset equal to 0v that puts 0.75 volts at the non inverting terminal and with a gain of 2 that puts 1.5v at the output for that input only. But then the input at vctl is 0.467v and the gain for the inverting part is only 1, so that puts -0.467 at the output for the inverting part only. Adding the non inverting part and the inverting part gives us 1.5-0.467 which comes out to exactly 1.033 volts. With a little input offset it will be slightly different.
So in finding the gain for the two inputs first calculate the voltage at the non inverting terminal, then multiply that by 2 and that is Vout1. Then find the output due to the vctl input, and that will be -1 times whatever vctl is, and call that Vout2. Then add Vout1 and Vout2 and that is the actual output.
A differential amplifier does NOT make the output equal to either supply rail unless one or both inputs are over driven. You might be thinking of a comparator circuit, which does exactly what you thought this does.

2.
The text, "Gain=100dB" means that the internal gain of the op amp is 100dB. That's the open loop again of the op amp.

Got it!
Thank you very much MrAl!

I entered what you said into an equation:
View attachment 65185

Last edited:
Got it!
Thank you very much MrAl!

I entered what you said into an equation:
View attachment 65185

Hi again,

Oh ok, very good! If you like you can also use this:
V+ = vset/2+V_DC/2

which for V_DC=1.5 it comes out to:
V+ = vset/2+0.75

(V+ is the voltage at the non inverting terminal due to the vset voltage and the battery voltage)

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Quote ElectroRF: As seen in the simulation, the input (Vctl) is nearly DC, Therefore I understand that the circuit is simply LPF for such an input.

I rather would say: For such an input it is not an LPF but simply an amplifier (unity gain for equal feedback resistors) - as long as the signal frequencies are well below the low pass corner.
And - are you satisfied with this result? Is this what you need? Don't you need an integrating device?

Quote ElectroRF: As seen in the simulation, the input (Vctl) is nearly DC, Therefore I understand that the circuit is simply LPF for such an input.

I rather would say: For such an input it is not an LPF but simply an amplifier (unity gain for equal feedback resistors) - as long as the signal frequencies are well below the low pass corner.
And - are you satisfied with this result? Is this what you need? Don't you need an integrating device?

Thanks for the clarification.

As for whether I'm satisfied with it, I'll discuss it tomorrow with my mentor.

Hi again,

As to the nomenclature, it matters more that the circuit works as the application requires than what we call it exactly. With no parallel feedback resistor it is more of an integrator, and with some resistance it is still a pseudo integrator, but with low resistance it is not as much an integrator as low pass filter, and the low pass filter has its merits too being required in some circuits for various reasons.
We call capacitors capacitors even though there is some series and parallel resistance in each capacitor, but we dont go around worrying what we should call the capacitor, a capacitor or a resistor.

Hi again,

As to the nomenclature, it matters more that the circuit works as the application requires than what we call it exactly. With no parallel feedback resistor it is more of an integrator, and with some resistance it is still a pseudo integrator, but with low resistance it is not as much an integrator as low pass filter, and the low pass filter has its merits too being required in some circuits for various reasons.
We call capacitors capacitors even though there is some series and parallel resistance in each capacitor, but we dont go around worrying what we should call the capacitor, a capacitor or a resistor.

Yes, that's true - and everybody (at least engineers) should always be aware of this.
Here are some further examples:
* We speak of "linear" systems - knowing that all systems are non-linear.
* The same applies to "sinusoidal" waveforms - THD of zero?
* Has a"squarewave" puls really a risetime of zero seconds?

In summary - in the field of electronics nothing is correct by 100%. But that's not a real problem - as long as we know about these deviations from the ideal.

Hi,

And the flip side of the coin is the theory. In the theoretical model we want it to be just the opposite: perfect in as much as we can make it in the math.

ElectroRF,
with Vctl as input and parts values as given in the diagram the circuit represents a lowpass with a corner frequency at approx. 100 kHz.
The phase shift approaches 90 deg for frequencies above 10 MHz. The maximum gain is 0 dB.
That means: You can use the circuit as an integrator for f>10 Mhz assuming an opamp that has a transit frequency (GBW) of at least 100 Mhz.

Hi again Winterstone.

Can you explain please how did you conclude that the LPF has a corner frequency of approx. 100KHz?
Can you specify the formula you used please?

Why does it mean that the op amp integrator acts as an integrator for frequencies above 10MHz?

I read on it in Wiki, but didn't manage to figure out the answers to my above questions.
I didn't quite understand their graph:
View attachment 65316

View attachment 65317

Thank you again!

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Hi electroRF,

here is my answer, based on the circuit as given in your post#10:

(1) Neglecting the influence of C1 (because it is in prallel) to the source the transfer function is

[2]H(s)=(R/R2)/(1+sR2C2)[/B]

with: R2C2=1E3*1E-9=1E-6

the 3-dB corner is at wo=1E6 equivalent to fo=1E6/2*Pi=160 kHz.

(2) To act as an integrator in a certain frequency range, the phase shift for these frequencies must be as exact as possible in the 90 deg region.
As a general rule this requirement can be approx. met for frequencies larger than the 3dB frequency by a factor of (at least) 100 - provided that the opamp phase shift due to its second pole not yet have a remarkable influence. Therefore the mentioned GBW requirement.

However, a further remark to you:

Are you really interested in an integrator function?. I suppose you are primarily interested in a 2nd order filter, aren't you?

Additional remark: With respect to your last figure (magnitude vs. frequency): The 90-deg requirement is identical to a gain slope of -20 dB/dec. (due to Bode's fundamental rule for the gain-phase relationship)

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