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Inverting operational amplifier circuit won't work

  1. Jan 8, 2012 #1
    I'm making a simple circuit on the breadboard with a 741 op amp. The feedback resistor is 150 kohms, and the input resistor is 100 kilo ohms, means the gain is 1.5. The input voltage is about 5.33. The power supply is ±18 V. For some reason the output voltage is 0.98 V instead of the amplified voltage. Also when I try the LF356N it doesn't show any output voltage at all.

    Below is a picture of my circuit

    Attached Files:

  2. jcsd
  3. Jan 8, 2012 #2


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    I can only see three batteries. How are you getting +/- 18 volts?

    How are you getting 5.33 volts input? I can't see any divider resistors. I would make this 1 volt for test purposes anyway.

    The red wire on the top battery doesn't seem to go anywhere.

    Are you using this circuit?

  4. Jan 8, 2012 #3
    yah I'm using that circuit.
    But the battery I used for the input signal is 5.33 because I burnt it out, cuz I don't have 1.5 volt batteries around
  5. Jan 9, 2012 #4


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    Do you have a meter?
  6. Jan 9, 2012 #5
    I don't see any connection between the midpoint of the two supply batteries, pin 3 and your input's "zero" side. Those three things need to be connected together.
    As has been pointed out, your supply only seems to be +/-9V. That should be just about enough though.
  7. Jan 10, 2012 #6
    No, you're not using that circuit (inverting amp). You're using this circuit. Its a non-inverting amplifier. Gain = 1 + R2/R1 ( = 2.5 in your case)


    Also give the circuit a common ground by connecting the middle of +-18V battery to the negative terminal of your input. (i.e. at the other end of the resistor you took out from pin 2)
  8. Jan 10, 2012 #7
    I look at you breadboard. The top battery is supposed to be the input voltage. The black wire is connected to the 100K which connect to pin 2 as input. Problem like Vk6kro said, the red wire of the battery is not connected to the breadboard!!!! If I see you picture correctly, you are not going to get anything meaningful out of this. I can't see where the clip on the black wire goes to as your picture don't show. You need to have connection to the ground reference to establish the potential. The ground reference is the junction of the red wire of the lower battery on the left side and the black wire of the lower battery on the right side. You have no connection of the ground reference to any other points.
    Last edited: Jan 10, 2012
  9. Jan 10, 2012 #8
    He has only +/-9V. He used an empty clip to hold the red of the left battery to the black of the right battery.
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