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Op Amp Circuit problem

  1. Feb 13, 2013 #1
    2. Relevant equations
    I think I figured out the 1st question. The result I got is Vout=2*Vin+2V.
    Which means twice as big wave as the Vin and shifted with 2V.
    Is this enough or I have to write equations like 2nd -> Vout = 2*Vin+2V?
    or
    2) 2kV/s is the same as 2V/ms, or 20V in 10ms

    3) Vout “clips” at ±8VIn fact it would be rather worse than this, as the output would not quite make it to the ±8V supplies.


    4) Short-circuit across R1 means that negative feedback has gone and the Op-Amp is simply comparing its input to 0V, multiplying the difference by the raw Op-Amp gain (very large) and thus saturating Vout at the (positive) power supply of +15V

    5) Short-circuit across R2 makes the circuit into a unity-gain buffer with R1 as its load resistor, so Vout = Vin
     

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    Last edited: Feb 13, 2013
  2. jcsd
  3. Feb 13, 2013 #2

    rude man

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    see above in red
     
  4. Feb 13, 2013 #3
    shows a non-inverting Op-Amp circuit. *Its input waveform, Vin is a square wave with minimum and maximum values of 0V and +5V respectively. *R1 and R2 are both 1kΩ resistors. *For all of this assignment, apart from (3) below, assume that the Op-Amp’s power supplies are ± 15V.
    The resistors values are 1kOhm
    and the power supply is +-15V
     
  5. Feb 14, 2013 #4

    rude man

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    I think I figured out the 1st question. The result I got is Vout=2*Vin+2V.
    Which means twice as big wave as the Vin and shifted with 2V.
    Is this enough or I have to write equations like 2nd -> Vout = 2*Vin+2V?
    or
    That depends on whether the bias current flows into the + or - input. If into the + there is no effect. If into the - input your output expression is correct.

    2) 2kV/s is the same as 2V/ms, or 20V in 10ms
    So draw the input and output waveforms.

    3) Vout “clips” at ±8VIn fact it would be rather worse than this, as the output would not quite make it to the ±8V supplies.
    Correct.

    4) Short-circuit across R1 means that negative feedback has gone and the Op-Amp is simply comparing its input to 0V, multiplying the difference by the raw Op-Amp gain (very large) and thus saturating Vout at the (positive) power supply of +15V

    With an ideal op amp and the input going from exactly 0 to +5V, the output will obviously be hard-over +15V when the input is at +5V. But what about when the input is at 0V?

    5) Short-circuit across R2 makes the circuit into a unity-gain buffer with R1 as its load resistor, so Vout = Vin
    That is correct.
     
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