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Old 12th December 2012, 12:20 AM   #11
Jmac222 is offline Jmac222  United States
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I appreciate but don't get it. I've read that the opamp has a flat response but I did not take that to mean that adding external components would not effect it's bandwidth. My original question was does the attenuating feedback resistor effect the lower frequency response because There is a virtual path between the V+ and V - inputs.
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Old 12th December 2012, 12:37 AM   #12
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The bandwidth of the circuit and the gain bandwidth product are two different things. Increasing gain decreases bandwidth. The term gain bandwidth product is self explanatory; the product of the circuit gain and the circuit bandwidth is theoretically constant for a given device. So a device with a GBP of 5.5 mHz (like the 1875) can provide a gain of 10 with a 550 kHz bandwidth, a gain of 20 with a 275 kHz bandwith, etc, theoretically. Of course for an audio circuit we might limit that bandwidth by using a snubber on the output, which decreases loop gain with increasing frequency and helps to prevent oscillation.
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Old 12th December 2012, 01:00 AM   #13
Jmac222 is offline Jmac222  United States
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The rail splitting network should go to ground , correct?
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Old 12th December 2012, 02:23 AM   #14
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Quote:
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The rail splitting network should go to ground , correct?
For your AC Thevenin equivalent circuit? Yes.
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Old 12th December 2012, 02:28 AM   #15
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Quote:
Originally Posted by Jmac222 View Post
My original question was does the attenuating feedback resistor effect the lower frequency response because There is a virtual path between the V+ and V - inputs.
I'm not sure what you mean by 'virtual path' here. There's quite a high differential input impedance in an opamp, above 100k typically. Its true that they stay at the same voltage by virtue of feedback, but I can't see how that's a 'virtual path'. Are you getting 'virtual path' from the notion of 'virtual earth' in the inverting gain case? In which case there's not a path between the + and - inputs in that case, just because the -input looks like a virtual ground. You could put a resistor in the +ve input and the -ve input would still look like ground, the resistor in the +ve input wouldn't have any effect on the characteristics of the -input.
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