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I understand how an integrator works with just a capacitor in the feedback path:

When the input pulse width is much smaller than \$RC(1+A_{VOL})\$, the constant current charges the capacitor and the output is a linear ramp.

However, I don't get what the resistor is for. Why does its value have to be \$\ge 10R\$ ? An infinite resistor (remove resistor) satisfies this condition. What is the feedback resistor doing?

enter image description here

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2 Answers 2

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All real-world circuits have some DC offsets, and a pure integrator has infinite DC gain. The result is that a pure integrator will inevitably drift over time.

That drift can be managed in various ways.

  1. The integrator may be included in a larger feedback loop, which corrects for the drift.
  2. The integrator may be designed with a "reset" function which is triggered periodically.
  3. The integrator may be made deliberately impure.

The resistor makes the integrator impure. At low frequencies it behaves like a regular amplifier with a gain set by the resistors, at high frequencies it behaves like an integrator.

The value of the resistor is a compromise, a smaller resistor will result in less DC offset in the output, but at the price of more distortion of the slope.

I don't know where the author of your book got their "10R" figure from. If I have to guess I would say that they probably assumed that the circuit would be operated in a configuration where the input and output voltages were roughly the same and then made some assumption about what level of error was acceptable.

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  • \$\begingroup\$ oh so the feedback resistor is there to reduce DC gain! the smaller the resistance, the better it is then. But it also reduces the effective miller RC time constant. so it is a compromise. I get it now thank you so much for clearly explaining! you're awesome! your last paragraph: yes, in one example problem 8V input pulse gives a ramp peak of 4V.. \$\endgroup\$
    – across
    Feb 28 at 9:26
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    \$\begingroup\$ @across It's more complicated than that. It doesn't reduce the integration time constant. But an ideal integrator has a memory for the whole history of the signal. The feedback resistor adds another RC time constant that causes that memory to decay. \$\endgroup\$
    – John Doty
    Feb 28 at 17:15
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If you want the function "integrator" over some decades (6), then one can use a schematic like this (with some good op-amp) with a high feedback resistor.
The feedback resistor fixes the "DC" gain (with the other resistor, 80 db) which would be "lower" than the open-loop gain (~ 120 db).

enter image description here

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