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Transfer function of an ideal OP amp

  1. Nov 18, 2011 #1
    1. The problem statement, all variables and given/known data

    I have to find the transfer function of a linearly and ideally operating OP Amp.


    2. Relevant equations

    3. The attempt at a solution

    I know that the current through the capacitor, R1, and R2 is the same, so i=C(dv/dt) = Vin/R1 = (VIN-V0)/R2, where do I go from here? Is the transfer function just V0/VIN in terms of R1, R2, C and dv/dt? When I equate the second and third equations I get V0/VIN = 1-(R2/R1) but i'm not sure if this is right.

    For the second part, I assume I just find a value of R1 that gives a value of A = 20-& in the equation VIN=A(gain) V0
  2. jcsd
  3. Nov 18, 2011 #2


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    Staff: Mentor

    Welcome to the PF.

    You are correct that the same AC current flows through C, R1 and R2. But the current is not Vin/R1. Vin is across both R1 and the capacitor, so that must show up in your equations.
  4. Nov 18, 2011 #3

    Sorry I didn't realise that, thanks.

    So via KCL I can assume that Cdv/dt + Vin/R1 = (VIN-V0)/R2.

    This results in a fairly awkward equation if I want to solve for V0(w)/VIN(w):

    VIN(1/R2-1/R1) - Cdv/dt = V0/R2

    I'm not sure where to go from here, I have no initial or final values for v(t) so i can't really do anything with the dv/dt.

    thanks in advance.
  5. Nov 18, 2011 #4


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    Staff: Mentor

    No no no no no no no. :smile:

    Vin is NOT across R1. It is across both the cap and resistor. So the LHS of your first equation above needs to change...
  6. Nov 18, 2011 #5

    Wow, i'm not on the ball at all today.

    Right, so i'll let the voltage across the capacitor be Vc.

    via KCL I have:

    I(cap) + I(R1) = I(R2)

    Cdv/dt + (v0-vc)/r1 = (v0-vIN)/r2

    And the RHS is the same

    This leads to:

    Cdv/dt = V0(1/R2 - 1/R1) + Vin(1/R1 - 1/R2)

    thanks in advance.
    Last edited: Nov 18, 2011
  7. Nov 18, 2011 #6


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    Staff: Mentor

    No. The current through the cap equals the current through each resistor. They are all in series (assuming the ideal opamp needs no input bias current).
  8. Nov 18, 2011 #7
    Sorry I meant equals, I had it in the first post I was just getting confused.

    According to this Cdv/dt = (V0 - VIN)/R1 = (V0 - VR1)/ R2

    where VR1 = (R1)Cdv/dt
  9. Nov 18, 2011 #8


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    Staff: Mentor

    Two things. First, use the "virtual ground" property at the - input node of the opamp. Since the + terminal is connected to ground, what is the voltage always going to be at the - input (due to the negative feedback and infinite gain of the ideal opamp)? That helps simplify the equation for the current through R2.

    Second, you really do need to combine the cap and R1 together as the LHS of an equation. When you try to say things individually about them, you make errors (because the voltage at the point between them would need to be an additional variable if you want to treat them separately. Use the vitural ground property of the - input node to write an equation for the voltage Vin across both C and R1 and the current that flows through both of them.
  10. Nov 18, 2011 #9
    If I remember correctly it's V input = V output = 0 in an ideal OP Amp so the potential at the node above the - op amp terminal is 0. This means that the current through I2 is

    V0 - 0/R2 = V0/R2.

    The Capacitor and resistor in series are a basic RC circuit, and the voltage across them is the same

    Vc = Vin[1 -exp[-t/RC]]where R is R1 and C is the capacitance.

    I know that this is the voltage across the capacitor but I'm not sure how exactly to combine the resistor and capacitors voltages and even if I do would dividing by the resistance R1 suffice to find the total current flowing into the node?

    Regardless, I now have

    Cdv/dt = (Vin-0)/R1 = (V0-0)/R2

    Cdv/dt = Vin/R1 = V0/R2

    So I can conclude that V0/VIN = R1/R2.
    Last edited: Nov 18, 2011
  11. Nov 18, 2011 #10


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    Staff: Mentor

    You have the output equation correct.

    The input equation is not right still. It's probably best to devine a new voltage at the node between the cap and R1. That way you can write the Cdv/dt using the voltage just across the capacitor. The voltage across the resistor is then related to Vin and Vc.
  12. Nov 18, 2011 #11


    I'll call the node potential between the capacitor and the resistor E.

    Cdv/dt = (E - 0)/R1 = V0/R2

    Edit: Sorry it's very long winded.
    Cdv/dt = E/R1 = V0/R2

    You then said to relate the voltage across the resistor (E) to Vin and Vc. I'm not sure at how to do this but i'd guess that

    Vin - E = Vc
    Vc - Vr1 = E

    E = Vin - Vc = Vc - VR1

    2Vc = Vin+VR1

    but VR1 = E-0 = E

    2Vc = Vin + E

    but from Vin-E=Vc we get E = Vin - Vc


    2Vc = Vin + Vin - Vc
    3Vc = 2Vin

    Vc = 2/3 Vin

    E = Vin - Vc = Vin-2/3Vin = 1/3 Vin


    Cdv/dt = E/R1 = V0/R2

    (1/3Vin)/R1 = V0/R2


    V0/Vin = R2/3R1

    Edit: Sorry it's very long winded
    Last edited: Nov 18, 2011
  13. Nov 18, 2011 #12


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    Staff: Mentor

    I'm sorry, I didn't track your last post at all.

    If you call the voltage between the cap and R1 "Ve", then the v(t) across the capacitor is Vin(t)-Ve(t). And the voltage across R1 is Ve(t)-0V. Does that change your equations?
  14. Nov 19, 2011 #13

    Vin(t) - Ve(t) = Vc(t) (eq1)

    Ve(t) - 0 = Vr1(t) (eq2)

    Another equation that should work is

    Vc(t) - Vr1 = Ve(t) (eq3)

    using eqs 2 and 3

    Ve(t) = Vr1(t)
    Ve(t) = Vc(t) - Vr(1)


    Vr1(t) = Vc(t) - Vr1(t)
    Vc(t) = 2Vr1(t) (eq4)

    Does up to here make sense?

    from eqs 1 and 3

    Ve(t) = Vin(t) - Vc(t) = Vc(t) - Vr1(t)

    from eq 4, Vc(t) - Vr1(t) = 2vr1(t) - vr1(t) = vr1(t)


    Vin(t) - Vc(t) = Vr1(t)

    since vc(t) = 2vr1(t)

    vin(t) = 3vr1(t)

    Vin(t)/3 = vr1(t)


    Vc(t) = 2/3 Vin(t)

    ALSO V(E) = Vin(t) - Vc(t)
    Vin(t) - 2/3 Vin(t)
    Ve(t)= 1/3 Vin(t)

    So going back to the original equations:

    Ve(t)/R1 = V0/R2

    Giving V0(t)/Vin(t) = R2/3R1

    Yes it yielded the same equations.

    is up to here correct?
    Last edited: Nov 19, 2011
  15. Nov 19, 2011 #14
  16. Nov 21, 2011 #15


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    Staff: Mentor

    Equation 3 looks wrong to me. The voltages across the capacitor and R1 add up to Vin...
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