OP Amps in electric circuits

Discussion in 'Homework Help' started by uraza, May 24, 2014.

  1. uraza

    Thread Starter New Member

    May 24, 2014

    The circuit in question 4 and they ask to express v_p and v_q as a function of v_g.

    For v_q I am thinking of using node analysis and since we know that there is no current that flows through the bottom OP amp we get from Ohm's law that the current that goes out to the 10R resistor is the same that comes in from v_g/R = I_g.

    So v_q is:

    10R*I_g = v_g*10/R.

    I don't have any clue how to solve the function for v_p though.

    Is my line of reasoning right for v_q and could anyone help me how I should go about to find the function for v_p?
  2. Jony130

    AAC Fanatic!

    Feb 17, 2009
  3. uraza

    Thread Starter New Member

    May 24, 2014
  4. crutschow


    Mar 14, 2008
    You can derive the voltage for Vp quite easily with simple math.

    You know that, due to negative feedback, the positive and negative voltages at the op amp input are ideally the same.

    Therefore, the negative input voltage must equal the positive input voltage (Vg). The current through R is then Ig = Vg/R.

    Since the same current must also flow through 9R, due to the output voltage Vp, then voltage across 9R is V9r= Ig*R9.

    Substituting from the previous paragraph gives V9r = Vg/R * 9R = 9Vg.

    The output voltage is then the sum of the voltages across R and 9R or Vp = Vg +9Vg =10Vg.
  5. Fibonacci

    New Member

    May 23, 2014
    It is an easy problem if you can identify the op-amp circuit. First, V0=Vp-Vq. The op-amp at top is a noninverting amplifier, the closed-loop voltage gain is (1+9R/R), then, the output voltage is given by Vp=10Vg. In the bottom side of the circuit, you have an inverting amplifier with closed-loop voltage gain equal to -10R/R=-10. Thus, Vq=-10 Vg. Finally, the load voltage is given by V0=10Vg-(-10Vg)=20Vg.