Electronic – the purpose of the opamp in an integrator circuit
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What is the exact role of opamp in an integrator circuit….?
Best Answer
Here comes another explanation: For integrating purposes we need a passive RC lowpass with a very low cut-off frequency. This would require excessive large capacitor values.
Therefore - why not using the capacitance multiplication offered by the MILLER effect? Do you remember the MILLER effect in a simple transistor stage? Hence, we connect the integrating resistor to the inverting input of an amplifier which has a capacitor between inverting input and output.
Now - the capacitor value is amplified by the open-loop gain of the opamp (as seen from the input).
Example: 1nF*1E5=100µF. Together with an integrating resistor of R=10kOhm we would have a time constant of 1sec (cut-off at f=1/6.28 Hz). As another advantage - the output is available at the low-resistive opamp output.
By the way: That is the reason the shown circuit is called "MILLER integrator".
The answer is simple: For an ideal opamp your circuit provides a lowpass function with max gain of 40 dB and a cut-off (-3dB) at app. 100kHz. However, the real opamp (GBW=1MHz) allows a gain of 40 dB only up to app. 10kHz. Hence, it is the open loop gain of the opam that dominates and determines the frequency response of the whole circuit (3dB cut-off at app. 10kHz)
The resistor serves no purpose if the opamp were ideal. However, ideal opamps are hard to find. All the ones I've seen have some non-zero input bias current.
When the input bias current is significant relative to the external circuit, then having the same impedance drive both opamp inputs cancels out the offset voltage caused by the input bias current that is common to both inputs.
Bipolar input opamps have significant enough input bias current that it is usually specified broken down into the common and independent parts per input.
Modern CMOS input opamps have so little input current, and its polarity isn't predictable since it's due to leakage, that usually only its absolute value is specified. In that case, matching the impedances is less useful, and the worst case error from unmatched impedance is much less to begin with anyway.
Best Answer
Here comes another explanation: For integrating purposes we need a passive RC lowpass with a very low cut-off frequency. This would require excessive large capacitor values.
Therefore - why not using the capacitance multiplication offered by the MILLER effect? Do you remember the MILLER effect in a simple transistor stage? Hence, we connect the integrating resistor to the inverting input of an amplifier which has a capacitor between inverting input and output. Now - the capacitor value is amplified by the open-loop gain of the opamp (as seen from the input).
Example: 1nF*1E5=100µF. Together with an integrating resistor of R=10kOhm we would have a time constant of 1sec (cut-off at f=1/6.28 Hz). As another advantage - the output is available at the low-resistive opamp output.
By the way: That is the reason the shown circuit is called "MILLER integrator".