Common Voltage and Differential Voltage

Discussion in 'Homework Help' started by Schmitt_trigger, Feb 11, 2013.

  1. Schmitt_trigger

    Thread Starter New Member

    Feb 5, 2013
    20
    0
    Hi,

    I have the following question

    Calculate the common voltage [Vc] and diferential voltage [Vd] in the input of the opamp considering that the error is 10% when Vin=1.5 V


    I know that  Vd=Vp-Vn<br />
                 Vc=(Vp+Vn)/2<br />
                 Vo=Ad(Vp-Vn)+Ac(Vp+Vn)/(2))


    The information that the error is 10% when Vin=1.5 V what does that tell me about the opamp?
    What is considered to be the error of an ampop?Is the common voltage that is amplified and it shouldn't?

    Does that tell me that  0.15V (10% of 1,5V)=(Vp+Vn)/2<br />
                              0.15V=(Vp+1,5)/2


    Thank you
     
  2. WBahn

    Moderator

    Mar 31, 2012
    17,788
    4,808
    As I was reading your question I was wondering exactly the same thing -- what the hell do they mean by the error being 10% at a particular Vin? And what is Vin referring to? The common mode or the differential component?

    Whenever one talks about error two things need to be adequately specified: the error in what? and the error compared to what?

    So your question is a very valid one, but one that I don't see enough information to even take a good guess. You might look at the problem again and see if there is any further information provided that might answer those two questions.
     
  3. Schmitt_trigger

    Thread Starter New Member

    Feb 5, 2013
    20
    0
    I made a mistake stating the problem:

    They say

    "Calculate the common voltage [Vc] and diferential voltage [Vd] in the input of the opamp considering that the error is 10% when Vout=1.5 V"

    I guess that means that the output voltage can be in a range of values from 1.5V+10% to 1.5V-10%

    I think i just have to use the expressions Vc=vout/ac and Vd=vout/ad and use the maximum and mininum values of Vout to determine the range of common and differential voltages that are being put in the opamp
     
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