How Do You Solve an Op-Amp Integrator Circuit with a DC Offset?

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In summary, The problem is to apply a square wave signal to an op-amp integrator with a resistor across the capacitor to correct for small DC signal. The capacitor has an initial charge of 1V. Three methods were attempted to solve the problem, with the first and second methods resulting in the same expression for the output in the case of 500ms < t < infinity. However, it is uncertain if this expression is correct for the other two intervals as it does not account for the capacitor charging. The third method uses a formula for a simple STC circuit, but the answers do not match up with the second method. It may be simpler to write an equation for the voltage across the capacitor using the virtual ground property of the
  • #1
rhysticlight
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Homework Statement


The problem is shown in the attached PDF, it is basically applying a square wave signal to an op-amp integrator which has a resistor across the capacitor to correct for small DC signal.

In this case it is given that the capacitor has an initial charge of 1 V on it.


Homework Equations


Relevant equations are the transfer function of such an op-amp:
H(S) = [tex]\frac{-R2}{R1}[/tex] [tex]\frac{1}{1+R2_{2}SC}[/tex]

As well as the time domain equation for the output voltage:
Vout = (-1/R1C)*[ integral from 0 to t (Vindt] - Vc

where Vc is the initial voltage across the capacitor



The Attempt at a Solution


So basically I have found three separate ways to solve this problem, but am not sure which one is correct:

First Way
Recognize the fact that the corner frequency of the low pass filter is 0.5 rad/s while the frequency of our input waveform (from 0 to 500 ms) is 4 pi, which is much greater than the corner frequency and we can therefore neglect the shunt resistor and treat this as a simple integrator.
This would leave the same 1 V charge on the capacitor at the end of the waveform, and for 500 ms < t < inifinity, we simply have an exponential decay of this 1 V charge with a time constant of 2 seconds.

Second Way
Draw the circuit in the Laplace domain, which results in us having to add a voltage source in series with the capacitor to account for the initial charge on the capacitor. The expression for the output in the s-domain then becomes:
(see attachment 2)

This reduces to the same expression for my case of 500 ms < t < inifinity, but I don't think it is correct for the other two intervals, but am not quite sure why (I feel like this expression isn't properly accounting for the fact that the capacitor is charging)

Third Way
Use a formula that I got out of the textbook that basically has the form of the final equation in the attached thumbnail. This basically treats the circuit as a simple STC circuit, I feel like the answers I get with this equation should match up with my second method above, but so far they are not.

Clearly I am pretty confused as to how to best approach this problem and would appreciate any advice you can give as everything I have found online or in books doesn't really examine this type of op-amp in this manner (generally just stops after getting the transfer function and noting its characteristics). Sorry for the long and possibly confusing post!

Thanks!
 

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  • #2
rhysticlight said:

Homework Statement


The problem is shown in the attached PDF, it is basically applying a square wave signal to an op-amp integrator which has a resistor across the capacitor to correct for small DC signal.

In this case it is given that the capacitor has an initial charge of 1 V on it.


Homework Equations


Relevant equations are the transfer function of such an op-amp:
H(S) = [tex]\frac{-R2}{R1}[/tex] [tex]\frac{1}{1+R2_{2}SC}[/tex]

As well as the time domain equation for the output voltage:
Vout = (-1/R1C)*[ integral from 0 to t (Vindt] - Vc

where Vc is the initial voltage across the capacitor



The Attempt at a Solution


So basically I have found three separate ways to solve this problem, but am not sure which one is correct:

First Way
Recognize the fact that the corner frequency of the low pass filter is 0.5 rad/s while the frequency of our input waveform (from 0 to 500 ms) is 4 pi, which is much greater than the corner frequency and we can therefore neglect the shunt resistor and treat this as a simple integrator.
This would leave the same 1 V charge on the capacitor at the end of the waveform, and for 500 ms < t < inifinity, we simply have an exponential decay of this 1 V charge with a time constant of 2 seconds.

Second Way
Draw the circuit in the Laplace domain, which results in us having to add a voltage source in series with the capacitor to account for the initial charge on the capacitor. The expression for the output in the s-domain then becomes:
(see attachment 2)

This reduces to the same expression for my case of 500 ms < t < inifinity, but I don't think it is correct for the other two intervals, but am not quite sure why (I feel like this expression isn't properly accounting for the fact that the capacitor is charging)

Third Way
Use a formula that I got out of the textbook that basically has the form of the final equation in the attached thumbnail. This basically treats the circuit as a simple STC circuit, I feel like the answers I get with this equation should match up with my second method above, but so far they are not.

Clearly I am pretty confused as to how to best approach this problem and would appreciate any advice you can give as everything I have found online or in books doesn't really examine this type of op-amp in this manner (generally just stops after getting the transfer function and noting its characteristics). Sorry for the long and possibly confusing post!

Thanks!

First of all, the resistor across the integration cap forms a highpass filter, not a lowpass filter, IMO.

Second, it would seem to be simpler for this problem to just write an equation for the voltage across the cap based on the virtual ground property of the opamp. You are given the input voltage as a function of time, and you know that the left side of the cap is "grounded", and the current through the source resistor has to all flow through the cap and resistor. You should be able to write a KCL or similar equation, for the voltage across the R//C as a function of time... Maybe try that approach to see if it works for you.

Remember, the circuit is just an integrator, with some leakage off of the cap afforded by the parallel resistor. So the output waveform will look like parts of triangle waves (ramps), with some leakage toward zero volts opposing the ramps away from zero volts (and helping the ramps toward zero volts...).
 

Related to How Do You Solve an Op-Amp Integrator Circuit with a DC Offset?

1. What is an Op-Amp Integrator and how does it work?

An Op-Amp Integrator is a type of operational amplifier circuit that integrates the input signal over time. It is commonly used in analog computers and other applications that require the integration of a signal. The circuit consists of an operational amplifier, a feedback resistor, and a capacitor. The input signal is integrated by the capacitor, and the operational amplifier amplifies the integrated output signal. The output voltage of an Op-Amp Integrator is proportional to the integral of the input voltage.

2. How do I choose the right values for the feedback resistor and capacitor in an Op-Amp Integrator?

The values of the feedback resistor and capacitor in an Op-Amp Integrator depend on the desired integration time constant and the input frequency range. The time constant can be calculated by multiplying the resistance and capacitance values. A longer time constant will result in slower integration, while a shorter time constant will result in faster integration. It is important to choose values that are appropriate for the specific application.

3. What are the advantages of using an Op-Amp Integrator?

One of the main advantages of an Op-Amp Integrator is its ability to accurately integrate an input signal over time. It also has a simple circuit design and can be easily implemented with few components. Additionally, it has a high input impedance, meaning it does not draw much current from the input signal, which helps to prevent loading of the input source.

4. Can an Op-Amp Integrator be used for non-linear signals?

No, an Op-Amp Integrator is designed to integrate linear signals. It cannot accurately integrate non-linear signals, as the output will not be proportional to the integral of the input. In order to integrate non-linear signals, a non-linear circuit or a digital integrator is needed.

5. What are some common applications of an Op-Amp Integrator?

Op-Amp Integrators are commonly used in analog computers, electronic filters, and in signal processing applications. They are also used in instrumentation and control systems, where they can be used to measure and integrate signals from sensors. Additionally, they can be used in audio applications to create special effects such as reverberation and echo.

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