Differential amplifier not functioning properly

Discussion in 'The Projects Forum' started by Tyler Bradbury, Apr 30, 2015.

  1. Tyler Bradbury

    Thread Starter New Member

    Mar 18, 2015
    13
    0
    I UNDERSTAND THAT THIS IS JUST AN INVERTING AMP. I AM JUST TRYING TO GET THE DIFFERENTIAL PART WORKING BEFORE I PUT THE OTHER SOURCE ON THE POSITIVE TERMINAL OF THE AMP.

    I am attempting to create a differential amplifier using an LM324, and four 100k resistors. The schematic is below.

    upload_2015-4-30_17-19-24.png

    I am taking the output at Node3, and getting the correct SIMULATION results.

    upload_2015-4-30_17-19-50.png

    The problem is coming when I put it all together and physically simulate it. I have the hardware set up correctly, but I am not getting what I want out of the circuit.

    Node 1 is performing correctly, but I am getting nothing out of the output. I am using an LM324n. The simulated circuit is an ideal opamp. Could this be the problem? I have the hardware hooked up correctly (as inspected by two other engineers, as well). I cannot figure out why it will not give me the output I desire. Is there something obvious that I am doing incorrectly?

    The one part of the circuit I do not understand is that when, on my Agilent 33250A Function Generator, I switch from "High Z" output impedance to 50ohm impedance, I am getting effectively double. Why is this?
     
  2. tindel

    Active Member

    Sep 16, 2012
    568
    193
    Have you put decoupling caps on your breadboard?
    Have you replaced the IC to make sure it's not damaged?
    Have you connected your function generator ground to your power supply ground?
    Be sure to properly terminate unused op-amps on the chip.

    These are the most common errors.

    As far as your output of your function generator driving twice the voltage - function generators have 50 ohm output impedance ... if you drive a high impedance load (>>50ohms) then you will get twice the voltage because it is not being loaded with 50 ohms.
     
  3. Tyler Bradbury

    Thread Starter New Member

    Mar 18, 2015
    13
    0
    Thank you for your response.

    - Where would the decoupling caps go? Near Vcc pin?
    - I did not try replacing the IC, but I will try when I get back to the lab.
    - I will try to connect the function generator ground to my power supply ground. An additional question on this: where do I ground the oscilloscope ground probe? I had it going to the gnd of my +5v power supply. Is this correct?
    - I will terminate the additional op amps. Would you mind telling me why this might be necessary?

    Thank you for the explanation. You have really cleared things up for me.
     
    Last edited: Apr 30, 2015
  4. Tyler Bradbury

    Thread Starter New Member

    Mar 18, 2015
    13
    0
    Also, in looking for an answer, I came across this configuration:

    upload_2015-4-30_19-26-29.png

    I tried this, and got a different result, but not the one that I desired. I thought this might work because of the explanation about the impedance. Any explanation would be greatly appreciated.
     
  5. crutschow

    Expert

    Mar 14, 2008
    12,977
    3,221
    Explanation of what? It's hard to know what's wrong without a description. :confused:
    What is a "different result"?
     
  6. tindel

    Active Member

    Sep 16, 2012
    568
    193

    - yes as close to the Vcc and Gnd pins as possible. 0.1uF ceramic is pretty common for this sort of thing.
    - The ground of your scope should be connected to circuit ground, typically.
    - Unused amplifiers can mess up the operation of the other op-amps on the chip - especially if there is shared circuitry on the op-amp. It can also damage the part.

    A couple other common problems:
    -Not operating within the common-mode input range of the op-amp.
    -Too much gain saturating the output of the op-amp.
     
  7. Reloadron

    Active Member

    Jan 15, 2015
    963
    232
    Because when you go to 50 Ohms you have a 50 Ohm output impedance designed to work into a 50 Ohm load. Think a 50 Ohm, 50 Ohm voltage divider. So when the load is high like 1 Meg or 10 Meg Ohm what becomes of your divider? :) Your source goes to 50 Ohms and the load is still very high. That is a quick answer, obviously there is more to it. You could read up a little on Thévenin's theorem.

    Ron
     
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