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Old 1st March 2005, 07:20 PM   #11
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Location: Taipei
During testing, I choose that circuit (picture)as feedback

Figure (1) is shown the waveform at (C) when only a diff amp was connected to A and B.(EA ,comarator were disconnected, signals provided function gen.)

After, EA and Comparator were added.
Fig(2) is shown the waveform at negative input of EA.
Fig(3) is shown the negative input at comparatorLM361.
Both fig(2)&(3) were failed to get the proper input to LM361 which need +/- 5V input signal applied.

I replaced R15 with 1.5K but the negative input at LM361 nearly the same as Fig(3). The DC offset was still around -8V. The sin wave always seem disappeared.

I am quite confused.
What is the principle of feedback using diff amp (why need do that way? I build this feedback just by reading information searching from webs and old posts and some hand calculations ¡Kbut I still cannot really understand ¡K )

Also what is the role of EA in this case? What should be possible correct signal at negative input of EA?

More important¡K
Subwo1 mentioned that ¡§The best way I know to eliminate the chance of it is feedback referenced to a fixed point between power supply rails.¡¨

In both fig(2)(3), there always a dc offset. I am afraid ..How the feedback cirucit can eliminate the dc offset?
How can I change the circuit(feedback, ..,even power stage)?

Thanks...all experts
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Old 1st March 2005, 07:20 PM   #12
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waveform (fig1,2,3)
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Old 2nd March 2005, 12:02 AM   #13
bjorge is offline bjorge  Norway
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You should measure the offset between A and B. As already mentioned, with a single rail you will of course have a DC offset. Ideally it should be 6V with 12V rails, but you haven't added a "virtual" ground that would serve as a reference point.

It could be achieved by biasing the +input of U9B to +6V and feeding the input signal via a capacitor, maybe.
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