Inverting Op-Amp

Discussion in 'Homework Help' started by need_help, Jan 27, 2010.

  1. need_help

    Thread Starter New Member

    Sep 7, 2009
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    It has been a while since I learned Op-Amps. Can someone please direct me how to solve this problem.

    Vin= +1V
    Find Vout, I1, I2, I3,I0
     
  2. beenthere

    Retired Moderator

    Apr 20, 2004
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  3. StayatHomeElectronics

    Well-Known Member

    Sep 25, 2008
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    For an ideal op-amp, the voltage at the V+ and V- terminals are equal and the current into the amplifier at terminals V+ and V- is zero.

    Knowing the voltage at V+ gives you the voltage at V-. Now you can find the current I1. And so on...
     
  4. need_help

    Thread Starter New Member

    Sep 7, 2009
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    Since V- and V+ are O volts and no current.

    I1=I2=(Vin-0V) / 1K = 1mA

    Is this also equal to I0?

    And how about I3?
     
  5. StayatHomeElectronics

    Well-Known Member

    Sep 25, 2008
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    Calculate the voltage across the resistor with I2 running through it. Find Vo.

    Knowing Vo also lets you calculate I3. Now, what is I0?
     
  6. need_help

    Thread Starter New Member

    Sep 7, 2009
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    So Vout= 10K * 1mA = 10V ???

    I0 is the current coming out of the op-amp. does it equal to 0V ???

    If Vout is 10V, then I3= (10V-0) / 10K = 1 mA ???
     
  7. StayatHomeElectronics

    Well-Known Member

    Sep 25, 2008
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    Getting closer...

    Be careful with the direction of the current and the voltage that are derived from it.

    HINT: don't forget this is an inverting amplifier...

    What is Vo?

    Be careful with the sign on the current for I3 as well.
     
  8. t06afre

    AAC Fanatic!

    May 11, 2009
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    IO has to be the sum of I2+I3. And I2 and I1 must have opposite sign
     
  9. StayatHomeElectronics

    Well-Known Member

    Sep 25, 2008
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    Following the convention in the diagram, I1 = I2 and I3 = I2 + I0.

    Now, the sign on the calculated currents must be maintained to the same convention for the answer to be correct.
     
  10. need_help

    Thread Starter New Member

    Sep 7, 2009
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    V(R2) = Vi * (R2/R1) = 1V*( 10K/1K) = 10V

    Vout= -V(R2) = -10V Is this OK?
     
  11. Paulo540

    Member

    Nov 23, 2009
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    Yes, you have a gain of '-10'. due to the 10:1 ratio of resistors, and - because its an inverting input.
     
  12. StayatHomeElectronics

    Well-Known Member

    Sep 25, 2008
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    Vout does equal -10V.

    I1 = (Vin - 0) / 1k = 1mA

    I2 = I1 -- since current into a node must equal current out of the node and the opamp draws no current at V- terminal.

    V(R2) = 10k * 1mA = 10V

    Vout = -10V -- based on defined current from positive to negative voltage, and left side of resistor is at 0V.

    Now I3 and I0 are the only things left. Be careful of which way the current is defined in the diagram.
     
  13. need_help

    Thread Starter New Member

    Sep 7, 2009
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    I3= (0-(-10V)) /10K = 1mA

    I0 = I2+ I0 = 2mA

    Is this correct?
     
  14. StayatHomeElectronics

    Well-Known Member

    Sep 25, 2008
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    Since Vout = -10V

    I3 = ((-10V) - 0)/10K = -1 mA -- based on the direction of the current arrow designated in the problem.

    I3 = I2 + I0
    which is equivalent to :
    I0 = I3 - I2

    I0 = - 1mA - 1mA = -2 mA

    The minus sign on the current indicates that it is in the opposite direction of the arrow, or in this case, being drawn into the amplifier. The amplifier is sinking the current and not sourcing it.
     
  15. StayatHomeElectronics

    Well-Known Member

    Sep 25, 2008
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    Once you draw a current arrow from let's say A pointing to B, make sure you do the current calculation the same way.

    I = (V(a) - V(b)) / R

    A - sign means the current is in the opposite direction. As long as you keep the same convention throughout the problem, the + and - signs will work themselves out and you don't have to remember which way the current is defined for each part of the problem.
     
  16. need_help

    Thread Starter New Member

    Sep 7, 2009
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    Thank you for the explanation.
     
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