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    op amp output voltage derivation

    I need some help in deriving the output voltage of the following circuit
    My calculations does not seem to match the simulations

    Click image for larger version. 

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    I have computed the output voltage as
    (K1/K3)*V2 - (K2/K3)*V1
    K1=R4/(R3+R4)
    K2=( R2//R5 ) / ( R1 + R2//R5 )
    K3=( R1//R2 ) / ( R5 + R1//R2 )
    Numerically it leads me to Vo=4.58*V2-4.21*V1 ...But it does not match the simulation

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    Re: op amp output voltage derivation

    Hi,

    For an about ideal OPAMP the input current is considered to be zero, therefore there is no voltage drop across R2.
    --> you don´t need to care about R2.

    R3 and R4 are a voltage divider.
    --> Calculate the voltage at IN+

    --> IN- has the same voltage as IN+

    Current in R1: I_R1 = (V1- V_IN-) / R1
    The same current is through R5.

    Voltage across R5: V_R5 = - I_R1 x R5

    Output voltage: V_out = V_IN- + V_R5

    Klaus


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  3. #3
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    Re: op amp output voltage derivation

    R2 does not affect any voltage gain.


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    Re: op amp output voltage derivation

    Your opamp does not have a power supply so it will not work.
    Your output voltage is shown with no polarity.



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    Re: op amp output voltage derivation

    @KlausST : Thank you, My mistake was to consider R2 as being tied to GND

    @Audioguru: That it is not the schematic of a real circuit, just a simulation. If the power supply it is not shown it does not mean that it is inexistent. And by convention 5V means +5V.
    I don't know anywone who will asume that 5V it is in fact -5V. In the simulation the op amp supply is +12VDC



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    Re: op amp output voltage derivation

    Quote Originally Posted by fionut View Post
    I need some help in deriving the output voltage of the following circuit
    My calculations does not seem to match the simulations

    Click image for larger version. 

Name:	Untitled.png 
Views:	23 
Size:	10.6 KB 
ID:	135269

    I have computed the output voltage as
    (K1/K3)*V2 - (K2/K3)*V1
    K1=R4/(R3+R4)
    K2=( R2//R5 ) / ( R1 + R2//R5 )
    K3=( R1//R2 ) / ( R5 + R1//R2 )
    Numerically it leads me to Vo=4.58*V2-4.21*V1 ...But it does not match the simulation
    Here are the node calculations. vx is the node between r1 and r2. It is also the voltage of the op-amp inputs. vo is, of course, the op-amp output voltage. An infinitesimal amount of current is assumed to exist in R2 to sustain electrical conductivity so that vx and the op-amp inputs are at the same voltage. The result for vo is a complicated expression, but it matches quite closely with Falstad. If I knew what current Falstad assumed through r2, I could give an exact result.

    Click image for larger version. 

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    Ratch
    Hopelessly Pedantic


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    Re: op amp output voltage derivation

    Start with + input, the easy one.
    VIN+ = 2.5V*(100k/(25.9k+100k)) = 2.0V.

    Now ignore R2 since nobody's telling you input bias
    current. Though depending on op amp this could
    matter slightly. VIN- must equal VIN+. So now
    I(V1) must equal (2.0-1.3)/23.7K or 29.5uA.

    This must also be the current through R5, to 2.0V
    (on VIN-). So VO=2.0+29.5E-5*1E5 or 4.95V.

    There is a ~2% discrepancy. You might look at
    whether there is actually input bias current on the
    op amp subcircuit (probe R2).

    There may also be the "inputs" of limited gain (gain
    error), common mode rejection and so on. On a good
    real op amp these should be on the order of 0.01%
    each. But we know nothing about the particular one
    here.



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    Re: op amp output voltage derivation

    Quote Originally Posted by dick_freebird View Post
    Start with + input, the easy one.
    VIN+ = 2.5V*(100k/(25.9k+100k)) = 2.0V.

    Now ignore R2 since nobody's telling you input bias
    current. Though depending on op amp this could
    matter slightly. VIN- must equal VIN+. So now
    I(V1) must equal (2.0-1.3)/23.7K or 29.5uA.

    This must also be the current through R5, to 2.0V
    (on VIN-). So VO=2.0+29.5E-5*1E5 or 4.95V.

    There is a ~2% discrepancy. You might look at
    whether there is actually input bias current on the
    op amp subcircuit (probe R2).

    There may also be the "inputs" of limited gain (gain
    error), common mode rejection and so on. On a good
    real op amp these should be on the order of 0.01%
    each. But we know nothing about the particular one
    here.
    If you use the correct value for r3, you will get the same answer I did.

    Ratch
    Hopelessly Pedantic


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  9. #9
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    Re: op amp output voltage derivation

    Quote Originally Posted by fionut View Post
    Numerically it leads me to Vo=4.58*V2-4.21*V1 ...But it does not match the simulation
    The correct expression (neglecting R2) is: Vout=4.179*V2-4.2194*V1


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