Op-Amp Integrator: Understand 10MOhm Feedback Resistor

  • Thread starter fatalenergy
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In summary: The 10MOhm feedback resistor is there to make sure that the output signal is the integral of the input signal. Without the feedback resistor, the theoretical gain at DC would be infinite.
  • #1
fatalenergy
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Hi,
I was assigned a lab that requires me to build the attached circuit and show how the output signal is the integral of the input signal. Basically I can't understand what the function of the 10MOhm feedback resistor is. Its giving me a headache when I try to derive my proof for the integrator, and part of the lab is to explain its significance and what would happen if it were removed.

I don't know if I should be posting this in homework help but this isn't really a math question so I figured I could try here 1st.

Thanks
 

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  • #2
Try tackling the problem systematically...

Convert the circuit to the Laplacian equvilent (i.e. C becomes 1/(Cs) and derive the transfer function. The only special things you have to remember are:

a. For an ideal op-amp, the currents entering the inverting and non-inverting terminals is zero.
b. For an ideal op-amp, the voltage between the inverting and non-inverting terminals is zero.

Clue: In the s domain, a 1/s denotes an intergral.

Try this and see if it makes more sense to you.
 
  • #3
Or that the voltage across a cap:

[tex]Vc = \frac{1}{C} \int i_c dt [/tex]
 
  • #4
The [tex]R(f)=10M\Omega[/tex] resistor is there to make sure you do not have a small constant voltage on the inverted input (this would be integrated and you don’t want that). You can show that there is a frequency where the impedance of the capacitor equals the impedance of the R(f) resistor. Over this frequency the circuit acts as an integrator, so you can try to neglect that resistor in your calculations
 
  • #5
The feedback resistor limis the gain at DC to 100. If it was not there the theoretical gain at DC would be infinit. Any small input offset would cause an integration that would eventually saturate the op amp. Then all the usual assumptions about op amps go to H@$^*.
Joe
 

Related to Op-Amp Integrator: Understand 10MOhm Feedback Resistor

1. What is an Op-Amp Integrator?

An Op-Amp Integrator is a circuit that integrates an input signal over time, producing an output voltage that is proportional to the integral of the input signal. It is commonly used in applications such as analog-to-digital converters and audio filters.

2. How does a 10MOhm feedback resistor affect the Op-Amp Integrator?

The 10MOhm feedback resistor is a key component of the Op-Amp Integrator, as it determines the gain of the circuit. With a higher resistance value, the gain will be lower, resulting in a slower integration of the input signal. This can be beneficial in reducing noise and improving stability.

3. How does the feedback loop work in an Op-Amp Integrator?

The feedback loop in an Op-Amp Integrator helps to stabilize the output voltage by adjusting the input voltage as needed. The feedback resistor is connected between the output and the inverting input of the Op-Amp, creating a negative feedback loop. This ensures that the output voltage is always in balance with the input voltage.

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

Op-Amp Integrators are commonly used in analog circuits for signal processing and filtering. They are also used in audio circuits, such as in audio mixers and amplifiers. Additionally, they are used in instrumentation and control systems for measuring and converting analog signals into digital signals.

5. How can I optimize the performance of an Op-Amp Integrator?

To optimize the performance of an Op-Amp Integrator, it is important to choose the right values for the feedback resistor and the input and output capacitors. Additionally, proper grounding and shielding techniques can help reduce noise and improve stability. It is also important to select an Op-Amp with the appropriate specifications for the desired application.

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