Two simple questions, but I'm troubled by them.

Discussion in 'General Electronics Chat' started by lajka, Jan 14, 2012.

  1. lajka

    Thread Starter New Member

    Jan 14, 2012
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    Hi,

    I was hoping someone could help me resolve a little difficulty I have.

    ------
    Question number one: Say you have an op amp with a negative feedback. I've been taught that the principle of 'virtual ground' must hold. But what happens for a circuit for like this?
    [​IMG]
    As you can see, I intentionally made the other resistor zero in this configuration (invertible amplifier), so I could force voltage directly into the op amps input. It also makes the gain of this circuit infinite. I understand this is not something that can happen in reality, but from a theoretical point of view, how would you analyze this problem? Is the concept of the virtual ground still valid? If it is, I don't see how.
    I still consider this to be the case of a negative feedback, as the portion of the output is brought back into the negative input.

    ------
    Question number two: I have a problem with the way people analyze positive feedback. For example, consider this slide from the presentation. I see absolutely no reason why would anything change if I would just switch the plus and the minus sign on the op amp, and thus go from negative to positive feedback. Following that logic, I would get my virtual ground even with a positive feedback.

    Conceptually, I understand how both negative and positive feedback work. However, I would like to prove myself mathematically that negative feedback forces the inputs to be equal, and positive feedback forces the op amp to go into saturation. I don't know how to do that. If I use the general model, something like:
    [​IMG]
    I could write that:
    A = A/(Aβ + 1) for negative feedback;
    A = A/(Aβ - 1) for positive feedback.
    I don't what to make of that, though. I see no insight in these equations, just new values for gains, respectively.

    So, any help about these two questions would be immensely appreciated. Thanks!
     
  2. Wendy

    Moderator

    Mar 24, 2008
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    Question #1 is a current to voltage converter. You are dealing with a virtual ground, the negative input is matched to the positive input by the op amp. The result is the negative input looks exactly like ground, to the best of the op amps ability. If you were to measure it (and the op amp could handle it) you could not distinguish the difference.

    [​IMG]

    The resistor is a multiplier for the voltage out vs. the current in. You treat the negative input as ground, and any current going through the negative input is measured. It has extremely low ohmage, in theory 0 ohms, reality is good but not perfect.
     
  3. jimkeith

    Active Member

    Oct 26, 2011
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    To be a true virtual ground, the op amp must be operating within its limits (linear mode)--in this case, it cannot because the output has no influence upon the input
     
  4. Wendy

    Moderator

    Mar 24, 2008
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    Dealing with question #1:

    Look again. If the positive input were an input it would be a unity amplifier. There is negative feedback, in either mode, through the resistor R.

    With the positive input grounded and the negative input used the currents through R and input are the same. Iin = IR

    The math is straightforward, and the concept is simpler than most. It goes without saying we are in the linear region, other wise it would be a simple comparator.

    The formula is extremely simple:

    Vout = Iin R
     
    Last edited: Jan 14, 2012
  5. Adjuster

    Well-Known Member

    Dec 26, 2010
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    If there is no input series resistance, and the signal comes from a pure voltage source, the first circuit in your posting has no effective negative feedback. With the full gain of the amplifier effective, only a tiny input is needed to drive the amplifier output to its limits.

    The action then becomes similar to that of a level comparator, the sort of thing that could be used to detect AC zero crossings.

    In a practical case the signal source will itself have some internal resistance (or impedance). If the internal resistance is fairly large the circuit will still behave as a feedback amplifier, where the signal source delivers a current signal into a virtual short-circuit: circuits of this kind are sometimes used when it is important to minimise the impedance that the signal feeds into. This configuration may be described as a "transimpedance" amplifier. http://en.wikipedia.org/wiki/Current-to-voltage_converter

    -----------------------------------------------------------

    In the second part of your posting, you are mistaken in thinking that with positive feedback the gain equation only changes slightly. In the negative feedback case, we have a system which is expected to settle to a given value, and we can write and solve equations to find what this will be - which is how the negative feedback gain equation is derived.

    For positive feedback of any loop gain beyond unity, the system does not converge in this way, and it is not possible to make the same sort of gain analysis made with negative feedback. You cannot therefore represent the positive feedback gain just by making the sign change in the equation that you have assumed.
     
  6. thatoneguy

    AAC Fanatic!

    Feb 19, 2009
    6,357
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    This might help:

    An Op Amp will produce an output required to keep the "+" and "-" inputs at the exact same potential, up to the limit of the rails in reality, and up to infinity in theory. This is often called "servo action" of an op amp.
     
  7. Adjuster

    Well-Known Member

    Dec 26, 2010
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    Note that with positive feedback, the action is utterly different: only a minuscule input is required to send the output flying off to one of its limit values. Once the output reaches that level, it stays there unless the input changes to a value a tiny bit beyond it, at which point the output promptly shifts to its other limit value... and so on.

    Negative and positive feedback are like chalk and cheese.
     
    thatoneguy likes this.
  8. Wendy

    Moderator

    Mar 24, 2008
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    OK, I see where some confusion lay, I'm talking current (I did say it was a current to voltage converter, right?) and other people are thinking voltage.

    Don't forget though, the input current could be from a simple resistor, or a complex circuit. It doesn't matter in either case, it is the current that matters.

    As for question #2, think in terms of a simple inverting amplifier and a Schmitt Trigger, those being the classic examples of what you are talking about.
     
  9. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    I think that the main problem is that you don't understand how op-amp work.
    The real, physical op-amp (without any feedback loop) is nothing more them just a simply differential amplifier.
    The output voltage is the difference between the + and - inputs multiplied by the open-loop gain:

    Vout = (V"+" − V"−") * G_open_loop
    .

    [​IMG]

    So when the voltage at the input "+" (non-inverting) rise,output voltage is also increases.
    Increase voltage at "-" input (inverting input) causes, decrease the output voltage.
    Decrease voltage at input "-" increases the output voltage.

    [​IMG]

    And now let's back to your original first question circuit.
    But instead of ideal op-amp which has infinitely large open loop gain.
    We will use op-amp with a finite value of open loop gain.
    So now we force input voltage 1V directly into the op amps input and let's see what will going to happen.

    [​IMG]


    Now Question 2

    This circuit has negative feedback apply by R2

    [​IMG]

    We apply 2V at inverting input so the op amp see the potential difference between his inputs. So the output voltage start to decrease. And it will be decrease until V"+" = V"-". As you can see nothing magical is happening here.
    When we apply positive voltage at inverting input, the op amp output voltage start to decrease.
    But thanks to R2 some part of the output voltage is feedback to inverting input. And when V"+" = V"-" = VD = 0.
    The op amp see 0V difference between his inputs terminals. And output voltage stop changing.
    From the observer perspective op amp "magically" brought his input terminal differential voltages to 0V (virtual short).

    The more accurate explanation:

    Now let us understand how the negative feedback returned through R2 affects the amplifier operation.
    To begin our discussion, let us momentarily freeze the input signal as it passes through 0 volts.
    At this instant, the op amp has no input voltage (i.e., VD = 0 = voltage between the (+) and (-) input terminals (VD)).
    It is this differential input voltage that is amplified by the gain of the op amp to become the output voltage.
    In this case, the output voltage will be 0. Now suppose the output voltage tried to drift in a positive direction. Can you see that this positive change would be felt through R2 and would cause the inverting pin (-) of the op amp to become slightly positive Since essentially no current flows in or out of the op amp input and the (+) input of the op amp is at ground potential. This causes VD to be greater than 0 with the (-) terminal being the most positive. When VD is amplified by the op amp it appears in the output as a negative voltage (inverting amplifier action). This forces the output, which had initially tried to drift in a positive direction, to return to its 0 state. A similar, but opposite, action would occur if the output tried to drift in the negative direction. Thus, as long as the input is held at 0 volts, the output is forced to stay at 0 volts.


    Now suppose we allow the input signal to rise to at +2 volt instantaneous level and freeze it for purposes of the following discussion.
    With +2 volts applied to R1 and 0V at the output of the op amp, the voltage divider made up of R2 and R1 will have two volts across it. Since the (-) terminal of the op amp does not draw any significant current, the voltage divider is essentially unloaded. We can see, even without calculating values, that the (-) input will now be positive. Its value will be somewhat less than 2 volts because of the voltage divider action, but it will definitely be positive.
    The op amp will now amplify this voltage (VD) to produce a negative-going output.
    As the output starts increasing in the negative direction, the voltage divider now has a positive voltage (+2 volts) on one end and a negative voltage (increasing output) on the other end.
    Therefore the (-) input may still be positive, but it will be decreasing as the output gets more negative.
    If the output goes sufficiently negative, then the (-) pin (VD) will become negative. If, however, this pin ever becomes negative then the voltage would be amplified and appear at the output as a positive going signal.
    So, you see, for a given instantaneous voltage at the input, the output will quickly ramp up or down until the output voltage is large enough to cause VD to return to its near-0 state.
    All of this action happens nearly instantaneously so that the output appears to be immediately affected by changes at the input.


    Now let as change the type of the feedback from negative feedback to positive feedback and see what will happen.

    [​IMG]

    We apply positive input voltage and see what happen.
    As the input voltage increases, the voltage on the (+) input also increases.
    Once the (+) input goes above the voltage on the (-) input, even momentarily, the output of the op amp will go toward positive direction. This rising potential, through R2, further increases the potential on the (+) input pin, in our case, the output will drive all the way to its positive saturation limit (positive supply voltage).
    I hope that now you see difference between positive and negative feedback.
     
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