Opamp resistor magnitude

Gone Postal · 2014-04-28T04:44:23+00:00

I have a question about the magnitude of the two feedback resistors on opamps. I am familiar with how to determine the ratio of the resistors to get the desired gain, whether inverting or not. What I'm trying to figure out is what determines the magnitude of the resistors involved? I've seen different circuits with resistors of, say thousands, tens or hundered of thousands and even one that used Megohm-range resistors. Why the variation in magnitude of the resistors, assuming a fixed ratio? Does it have anything to do with the overall impedance of the opamp circuit block? I thank you for whatever illumination that can be provided. GP

Apr 28, 2014 41 Replies

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Phil Hobbs 12 years ago

Nope, not so far. Recreational reading is mostly re-reading at the moment, because I'm pretty well fried from work.

Cheers

Phil Hobbs

Dr Philip C D Hobbs Principal Consultant ElectroOptical Innovations LLC Optics, Electro-optics, Photonics, Analog Electronics 160 North State Road #203 Briarcliff Manor NY 10510 hobbs at electrooptical dot net http://electrooptical.net

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Bill Bowden 12 years ago

Interesting. It looks like that input capacitance helps form a bandpass filter. I set up a inverting op-amp in LTspice with 10K input and 100K feedback for a gain of 10 and input impedance of 10K and it looks good with a gain of 10. Then I added a .02uF cap from the inverting pin to ground to exaggerate the input capacitance and the gain peaks at around 8Khz and 40dB and then rolls off to about 12dB at 15KHz.. I guess it's the phase shift of the RC values that determine the gain. Never did figure out active filters other than looking up formulas.

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