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Help Understanding Electronics Theory (Op Amps & Open/Closed Loop Gain)

Discussion in 'Electronic Basics' started by Adrian Jackson, May 26, 2005.

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  1. I'm in an electronics based job & did my training nearly 4 years ago.
    Although to complete my Modern Apprenticeship I need to complete 4 BTEC
    Assignments, one of which is an Electronic Principles one. As my training
    what a while ago I'm struggling to remember some of the theory & was
    wondering if anyone could help.

    Part of the assignment is on Operational Amplifiers.

    The circuit is an op amp with a voltage on the positive leg & a feedback
    loop fed from the output into the negative leg of input. It then gives you
    an input voltage & open loop gain then asks what the output voltage is.

    Whats I've got is:-

    Open loop gain (Ao) = Vout/Vin

    So I transposed that to Vout = Ao*Vin

    Is that right or would they want the output voltage including the feedback?

    Also it gives me a few bits of data that I'm unsure of:-

    Common Mode Rejection Ratio (CMMR) = 85db
    Slew Rate = 0.5v/us

    What are these?

    As you can probably tell I'm very rusty on this.
  2. Rich Grise

    Rich Grise Guest

    Well, what's the circuit presented along with the question? What was the
    exact wording of the question? I'm just guessing, but if they _gave_ you
    the open-loop gain, they're probably asking, what's the gain of the entire
    CMRR is, in laymens' terms, how much a common-mode voltage affects the
    output. Common-mode voltage is, c'mon. I know this is "basics", but ...
    Common-mode voltage is the voltage that's applied to both input pins
    "in common". Like, as if they were strapped directly in parallel.

    Slew rate is just, how fast can the output slew, or change. Very much
    like rise-time.
    Yes, but as I've noted, this is the "basics" group - please forgive
    me if I've been a little harsh with you here. (there _is_ kind of
    an ambience amongst us old farts that we don't like answering
    homework questions very much, as you've probably been able to tell.)

    Here's a hint that really, really helped me when I was just learning
    opamps. The opamp (assuming infinite open-loop gain) does whatever it
    needs to do to its output to make the voltage at the inverting input
    (by way of the feedback network) equal to the voltage presented at its
    non-inverting input.
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