I need help understanding what the highlighted section means. I mainly want to understand how it determines the current drain when using difference resistance to get the same gain, and how it affects the op-amp.

I am reading this from Understanding Basic Analog – Ideal Op Amps from ti on page 3 at this link: https://www.ti.com/lit/an/slaa068b/slaa068b.pdf?ts=1607026009319

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Conceptually, the output of an op-amp is connected to a variable-voltage source through a resistance equal to its output impedance. Given that the output is at some voltage and the negative input (the feedback) pin is at some other voltage, naturally a current will flow between those two points. How much current flows between those two points, naturally, depends on the resistor values according to Ohm's Law (V = I * R). The lower the resistance values the more current will flow, and the more power will dissipate as heat in those resistors accordingly. So you don't want to use resistors that are "too small."

If you go the other extreme, your resistors start to look and act like open circuits. Notably stray capacitance causes some small amount of current to flow in a circuit node because of AC coupling and what some people refer to pick-up. You might also hear some people describe this effect as a circuit 'acting like an antenna.' If the current flowing through the feedback path is comparable in magnitude to those small coupling currents, then the feedback is more influenced than it would otherwise be.

There is some application dependent level that strikes a compromise of noise-immunity and power dissipation to achieve a required level of performance. As a rule of thumb, I personally try and stay in the kOhm to 10s of kOhm range as a starting point.


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