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Solving op amp circuits.

  1. Apr 11, 2005 #1
    I have to solve an op amp circuit for a lab I'm doing but the circuit they have given me looks confusing (it's on page 46, http://filebox.vt.edu/users/oshekari/Manual_Student.pdf). Why does the 4.7k ohm resistor have an arrow pointing to the middle of the 100 ohm resistor? I know the 100ohm is a varying resistor but if I were to actual solve the circuit for output and leave the varying resistor at 100ohms would the 4.7k be connected to node 3 or to node 1? And if it is indeed connected to node 3 then how would I go about doing nodal analysis at that node since I don't know the current to the left of the node, I just have a voltage source.

    Correct if I'm wrong but at node 3 the nodal equation is: 5/x+(v3-v1)/4.7k+v3/100. Where x is the resistance, v3 is the voltage at node 3, and v1 is the voltage at node 6.
     
    Last edited: Apr 11, 2005
  2. jcsd
  3. Apr 11, 2005 #2
    Now are we assuming this op-amp is ideal? For laboratory purposes I would suspect that it is not--- however I did not read your lab to find out the Rin, A and Rout values.
     
  4. Apr 11, 2005 #3
    Yes, it's an ideal op-amp.
     
  5. Apr 13, 2005 #4
    Hi,

    I forgot my nodal analysis learnt during 1st year, but I can give you some light on the 100 ohms varying resistor.

    Whether the varying resistor is 100 ohms or 1kohms is of no concern in the amplifier circuit because it effectively acts as a potentiometer OR potential divider. Whatever the voltage at node 2 depends on the position of the dial: If it is at node 3, the voltage at node 2 is the full 5V : If it is at node 1, the voltage at node 2 is the full 0V : If it is halfway in between, it is 1/2 * 5V = 2.5 V.
    This is because as you turn the dial downwards, the resistance with respect to node 1 decreases, and the resistance with respect to node 3 increases.
    The voltage at node 2 can be found using the potential divider formula:
    R1/(R1+R2) * Vcc.
     
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