Finding input resistance of uA741 in PSpice

Discussion in 'Homework Help' started by geft, Nov 12, 2013.

  1. geft

    Thread Starter New Member

    Dec 8, 2011
    19
    0
    I'm having difficulty finding input resistance because for some reason there is no virtual ground across the inputs, so the current flowing through each is not the same. What am I doing wrong?
     
  2. LvW

    Active Member

    Jun 13, 2013
    674
    100
    I am afraid you have forgotten to give the power supplies a value.
     
  3. geft

    Thread Starter New Member

    Dec 8, 2011
    19
    0
    Wouldn't it be more difficult to isolate Vin if there is output voltage? I mean, it's still drawing current as there is a 1V source... Regardless, there is still no virtual ground because input currents are still different.
     
  4. WBahn

    Moderator

    Mar 31, 2012
    17,743
    4,795
    geft likes this.
  5. LvW

    Active Member

    Jun 13, 2013
    674
    100
    Hi geft,

    it would be wise to reduce the gain or the input voltage - or to increase the power supply voltages.
    Why?
    Because with 1 Volt input signal and a gain of (1+14)=15 you operate the device outside of its linear amplification range.
     
  6. geft

    Thread Starter New Member

    Dec 8, 2011
    19
    0
    VCC and VEE were 15V and -15V originally, as can be seen by my previous post above. But I don't really see how the output is any way related to input resistance.

    I'm trying to work out offset current to see if it's what's causing the reading to be off...
     
  7. Jony130

    AAC Fanatic!

    Feb 17, 2009
    3,957
    1,097
    Can you tell me how you define this input resistance ?
     
  8. geft

    Thread Starter New Member

    Dec 8, 2011
    19
    0
    I'd assume it to be the resistance across the input nodes of the op-amp. It should be simple to measure, but I must be doing something wrong since I'm not getting the correct number (about 2MΩ for the uA741).
     
  9. t06afre

    AAC Fanatic!

    May 11, 2009
    5,939
    1,222
    Your approach is very wrong as other have said. In any real world situation this would not have been the way to go. Your circuit is an inverting standard opamp stage. As you have wired it up. The opamp will drive the negative input to a virtual ground. If you then connect a DC source to the other end of R1. What will the current in R1 be then ;) In opamp circuits. The input impedance for the system, do not have to be the same as the opamp input impedance.
     
    geft likes this.
  10. Jony130

    AAC Fanatic!

    Feb 17, 2009
    3,957
    1,097
    Try to use AC sweep analysis. And Rin = Vin/Iin = 1Meg.
     
    • 1.PNG
      1.PNG
      File size:
      8.8 KB
      Views:
      56
    • 2.PNG
      2.PNG
      File size:
      7.3 KB
      Views:
      38
    geft likes this.
  11. geft

    Thread Starter New Member

    Dec 8, 2011
    19
    0
    Does that mean I should just short the feedback loop and remove R1? Or short both inputs together?

    What do you mean? Which one should I be looking for?

    Thanks, but why does it work and mine doesn't? Shouldn't an op-amp work with DC as well?
     
  12. LvW

    Active Member

    Jun 13, 2013
    674
    100
    Geft, in post#8 you have no feedback - in contrast to your circuit in post#1.
    Of course, you can do both - find the input resistance with or without feedback.
    But you should know what you really want!
    In case of feedback it is important - as I have mentioned before - that the output is not in saturation.
     
    geft likes this.
  13. geft

    Thread Starter New Member

    Dec 8, 2011
    19
    0
    Sorry, I thought virtual ground only exists in op-amp circuits with feedback, so I assumed it's a given. The picture is something I took from Google to illustrate input resistance...

    Ah, now I see what you meant. I didn't know this before. Does that mean 16V voltage rails will be sufficient?
     
Loading...