1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Circuits: cascading op-amps, general question

  1. Mar 3, 2015 #1
    1. The problem statement, all variables and given/known data
    Solve for final Vout.
    2. Relevant equations

    3. The attempt at a solution
    So 1 and 2 are easy. I'm just going to skip to where I get confused.
    There is a 50k resistor between where I solved for Vout after the 2nd Op Amp.

    I found the voltage to be 12V. Now the solutions manual keeps moving on, ignoring the 50k resistor. It calculates the gain as -2 and the final voltage as -24V.

    It seems to me that there should be a voltage drop across the 50k resistor and that that Vin for the 3rd Op Amp should be less than the 12V that was Vout for the 2nd Op Amp.
  2. jcsd
  3. Mar 3, 2015 #2


    User Avatar

    Staff: Mentor

    Yes indeed. In fact, all 12 volts should be dropped across it. Recall that the op-amp is going to maintain the potential difference between its inputs at as close to zero volts as as it can thanks to the feedback arrangement and high gain. Since the + input is tied to ground, the - input will be likewise at (virtually) ground potential. That means the output has to swing to a value that can make this happen.
  4. Mar 4, 2015 #3


    User Avatar

    I am not quite sure what you mean with "voltage drop across the 50k resistor"; of course, there is a voltage across this resistor.
    But what does this mean? There is a current through both resistors (50k and 100k) - leading to a corresponding output voltage.
    In short: The last stage has a gain of "-100/50=-2".
    This stage is nothing else than the well-known inverting opamp circuit.
    As a consequence, all three gain stages provide a gain of 2*3*(-2)=-12 and the output voltage is Vout=-24 Volt.
  5. Mar 4, 2015 #4


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

  6. Mar 4, 2015 #5


    User Avatar

    I would NOT recommend the "explanations" as given in these "tutorials". They have many errors.
    Example from the linked text:
    This then produces and effect known commonly as Negative Feedback, and thus produces a very stable Operational Amplifier based system.

    My comment: Exactly the opposite is true. Each negative feedback degrades the stability properties of the opamp. This can lead to oscillations.
  7. Mar 4, 2015 #6


    User Avatar

    Staff: Mentor

    What do you base that on? Positive feedback causes oscillations, not negative feedback. That's why phase margin is important in feedback applications.
  8. Mar 4, 2015 #7
    I got it. Thanks guys!
  9. Mar 5, 2015 #8


    User Avatar

    Unfortunately, it is a common misconception that negative feedback would improve (dynamic) stability. It`s simply false.

    The explanation is simple: Each negative feedback (for lower frequencies, including DC) will turn into positive feedback for higher frequencies because of undesired but unavoidable phase shift of the active devive (opamp). And the system will remain stable only if the LOOP GAIN is already below 0 dB for the so-called cross-over frequency (phase shift of -180deg, not taking the phase inversion at the summing node into account). That`s common knowledge to be found in each relevant textbook.

    By the way: That is the only reason for phase-compensating opamps - with the consequence of very small open-loop cut-off frequencies below 100Hz !
    It is very easy to verify this effect: Most opamps with unity-gain compensation will show a magnitude peaking for strong feedback (unity gain voltage follower). In the time domain such a reduced stability margin resiults in a step response with a certain overshoot (or even ringing). These observations are a clear indication for degraded stability.
    Another consequence: When using an uncompensated (or partly compensated) opamp you must not apply too much negative feedback because of stability reasons. That is the reason the opamp manufactureres specify a lower limit for the closed-loop gain (e.g." stable for gains above 5 only").
  10. Mar 5, 2015 #9
    That's not necessarily true, i.e. you can improve the stability margins of a system using negative feedback. It is, however, true that you introduce the possibility of instability when you use negative feedback, but that's a design issue.

    Also note that, in the context of the tutorial @CWatters linked to, I don't think they mean to use 'stability' in the sense that you would in control theory. Rather, they're just describing how negative feedback can accurately control the gain of an opamp that has high finite gain, but is otherwise thought to be ideal.

    I've found that this idea of "negative feedback at low frequencies, but positive feedback at high frequencies"-explanation of stability to be suggestive, for some, of something that isn't true, e.g. if you apply a tone to the system in the "negative-feedback range", then you'd get a bounded signal out of it, but that's nonsense.

    This is only true for systems that are stable in open loop, but opamps usually are :wink:.
  11. Mar 5, 2015 #10


    User Avatar

    Sorry, but this sounds to me contradictory (improve stability margins.....possibility of instability)

    Yes - perhaps you are right. On the other hand, I think the term "stability" is well defined for systems with feedback and should not be used if "fixed gain value" is meant.
    Otherwise, misunderstandings cannot be avoided (as I have experienced very often).

    Is that an "idea"? BODE and NYQUIST have defined the threshold which show when negative feedback turns into positive.

    I think, open-loop systems are always stable, don`t they?
  12. Mar 5, 2015 #11
    If you define 'stability margins' as gain and phase margin, then I interpret your comment as you saying that applying negative feedback around some system P will degrade those margins for the equivalent system of P combined with its feedback network. I'm saying that's not necessarily true, but there will be trade-offs, of course.

    You could, however, turn a stable system into an unstable one, but that depends on how you implement the feedback network.

    I take the view that this:
    is just a special case of evaluating the Nyquist criterion for open-loop stable systems.

    I dislike the whole positive-feedback explanation, since, in addition to what I already mentioned, not all positive-feedback systems are unstable. They're just usually very impractical, since they have a strong tendency to be sensitive to process variations.

    Take a classic example like the inverted pendulum: Its linearization around the upright position is unstable in open loop, but you can stabilize it, for instance, by feeding back its angular position through a carefully designed network.

    You can't determine its stability from a Bode plot alone, though. You'll have to know how many poles it has in the RHP (as per the Nyquist criterion).
  13. Mar 5, 2015 #12


    User Avatar

    Yes - agreed. Because everything (my experience) in electronic design is a trade-off between some conflicting requiremets.

    Yes - of course. It is nothing else than a transfer of Nyquist`s theorem to the BODE diagram.

    Yes - I also agree. I hope I didn`t say that positive feedback would always lead to instability. But it reduces the stability margin. Best example: High-Q active filters. Increasing the Q factor up to infinite, we have an oscillator (some oscillator circuits work that way).

    Yes - of course, there are more complex systems than "simple" opamps. In control theory, for example, there are systems which are unstable without overall feedback (due to some internal, hidden feedback loops). Here, we apply negative feedback to stabilize the system. But these are special applications.
    To summarize: It was only my intention to point to the fact that negative feedback for opamps is connected with many advantages (that`s the reason we make use of it) - but we have to pay for the advantages: Tendency to gain peaking, overshoot or (worst case) even oscillations. Insofar, this is a good example for my "golden rule of electronic design: Everything is a trade-off". And students or other beginners should know about this situation, I think.

Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted