Math interpretation

Discussion in 'Math' started by Distort10n, Mar 26, 2007.

  1. Distort10n

    Thread Starter Active Member

    Dec 25, 2006
    429
    1
    I need a quick interpretation on a particular phrase. I am putting together a lenghty document on common-mode rejection for op-amps, and I am stuck on the wording on Jiri Dostal's amplifier book.

    Common-mode gain should be 0 in an ideal amplifier. With the inputs shorted to Vcm there should be no change in output voltage with any given change for Vcm.

    The part I am getting stuck on is the wording. "Since the gain A and the rejection ratio X are usually of hte same order with well-designed operational amplifiers, the common-mode gain Acm is usually on the order of 1."

    What does "order of 1" imply in Math Speak?

    See attached for scans of this section.
     
  2. DrNick

    Active Member

    Dec 13, 2006
    110
    2
    "On the order of 1" means that there is no gain or attenuation (what you put in is what you get out).

    Hope this helps.
     
  3. Papabravo

    Expert

    Feb 24, 2006
    10,135
    1,786
    "order of 1" includes a range of numbers whose logarithms are close to zero. If we say "order of 100" we mean numbers whose logarithms are close to 2. This phrase is related to the term "order of magnitude"
    http://en.wikipedia.org/wiki/Order_of_magnitude

    0 dB is another way of expressing "order of 1"
     
  4. Distort10n

    Thread Starter Active Member

    Dec 25, 2006
    429
    1
    Well, 0dB would imply a gain of 1 or unity. Looking at figure 2-11c in the PDF the output voltage would be 0, so Acm would be 0 (ideally) when we ignore offset etc. To me, this is not 0dB or "an order of 1." Any number raised to the 0 power is 1 or unity. A 0 value is not unity.
     
  5. Distort10n

    Thread Starter Active Member

    Dec 25, 2006
    429
    1
    Nevermind. It clicked. After a beer or two. :cool:
     
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