Explanations that suck

Discussion in 'Electronics Resources' started by Distort10n, Mar 12, 2007.

  1. Distort10n

    Thread Starter Active Member

    Dec 25, 2006
    429
    1
    Through all my reading of op-amp literature, I think the explanation for common-mode voltages and common-mode rejection *suck*. The definitions, in my opinion, overlap and fail to relate those definitions of common-mode voltages to the op-amps 'common-mode rejection'.

    A typical inverting or non-inverting amplifier does not 'reject' the common mode voltage. For example, there is no common-mode voltage on an inverting amplifier. The non-inverting input is grounded, and the summing junction is held at virtual ground. There is nothing to 'reject.'
    The common-mode voltage of a non-inverting amplifier is the voltage on the non-inverting input. This will be dynamic if one is amplifying an AC signal. The common-mode voltage follows the input sinewave. That same sine-wave will pass on through to the output, multiplied by the gain if not a voltage follower.
    It is the same with DC. With DC voltages, the common-mode will still show up on the output. It is not rejected to cause a 0V output.

    Terrible definitions, or at least very disorganized.
     
  2. thingmaker3

    Retired Moderator

    May 16, 2005
    5,072
    6
  3. Distort10n

    Thread Starter Active Member

    Dec 25, 2006
    429
    1
    A typical inverting amplifier has no common-mode voltage because of the non-inverting input being at ground potential. Unless one wants to say that ground is '0V common-mode.'

    A typical non-inverting amplifier does indeed have common-mode voltage. It has to. Any voltage at the non-inverting pin will be imposed on the inverting pin by virtue of negative feedback. Thus the same voltage is on both inputs.

    Analog Device's 'A Designer's Guide to Instrumentation Amplifiers' pg. 1-3 actually comes right out and confirms the above when comparing op-amps and in-amps.

    Jiri Dostral's book 'Operational Amplifiers' defines common-mode voltage (Vcm) as the voltage on the non-inverting input.

    It's just a pain that one has to dig this far to find these types of answers for simple specifications.
     
  4. Dave

    Retired Moderator

    Nov 17, 2003
    6,960
    143
    Although it is not something I have been particularly bothered about before, standard texts on this subject often dive into a dicsussion on common-mode rejection before they actually define the common-mode voltage - this appears to be because CMRR deals with Acm, i.e. the common-mode gain, rather than Vcm, the common-mode voltage.

    Its seems convention to approach this topic in this fashion.

    knightofsolamnus, what is you opinion of the description in the AAC e-book?

    Dave
     
  5. Distort10n

    Thread Starter Active Member

    Dec 25, 2006
    429
    1
    I will take a look at it this weekend. When I try to explain common-mode voltages and common-mode rejection to people, I find that many believe their output should be zero. A typical op-amp configuration, to harp on the non-inverting configuration again, will obviously pass its common-mode voltage to its output.
    You need a differential amplifier (subtractor) to make sure you 'strip off' the common-mode voltage. The common-mode rejection will reject the *error* induced by the common-mode voltage.
    My old college text book didn't offer anything better. I struggled learning this concept, and now that I grasp it I can understand why it is a pain to get straight.
     
  6. Dave

    Retired Moderator

    Nov 17, 2003
    6,960
    143
    Thanks, it would be interesting to see what you think in light of you comments in this thread. If you are finding that the explanation in the e-book is difficult to understand, then we can look at the possibility of modifying it to overcome the ambiguity that you feel is present in other texts on common-mode voltages and CMRR.

    Dave
     
  7. Distort10n

    Thread Starter Active Member

    Dec 25, 2006
    429
    1
    I think the e-book is a good start, but I would like to see more examples. The long-tailed pair shorted to the same common-mode voltage will 'reject' the that common-mode voltage and the output should be 0V...ideally.
    Given my example of a typical non-inverting amplifier, the common-mode voltages are still the same yet an output voltage exists.

    This begs two questions:
    1) Why is there an output if the amplifier supposedly rejects common-mode?
    2) Why is there an output if the amplifier only responds to differential voltages?

    These are part and parcel with one another.

    I would like to re-write it, and offer more examples for clarification.
     
  8. thingmaker3

    Retired Moderator

    May 16, 2005
    5,072
    6
    I don't understand. What is "common" about the voltage on the inputs of a non-inverting amplifier?

    The common mode rejection ratio is the gain of the differential amp (or comparator) in normal operation divided by the gain of a differential amplifier with i/p shorted together. Admittedly, inverting and non-inverting amps could be thought of as "differential amps with one input at ground potential." But there would only be a common mode voltage for an inverting or non-inverting amp when the signal input is also at ground potential. Common mode gain problems only become noticable as the common mode voltage rises above zero. Inverting and non-inverting amps therefore have no problem with common mode gain. Differential amps and comparators do - and it becomes annoyingly noticable when trying to measure tiny differences.
     
  9. beenthere

    Retired Moderator

    Apr 20, 2004
    15,815
    282
    I never had it right until I started to work with strain gauge bridges. When you see that both inputs are at five volts - the common-mode voltage - and responding to the few millivolts differential voltage, it helps clear things up.
     
  10. Distort10n

    Thread Starter Active Member

    Dec 25, 2006
    429
    1
    Whenever you see an example showing common-mode voltage, and common-mode rejection, it simply shows the op-amps inputs shorted to a common voltage. The output would be ideally zero in this case; however, no feedback is ever being applied in these explanations and examples. This is far from being practical...how many op-amps do you see running in open loop? Once you apply feedback, things change. The output must drive the voltage at V- to equal V+. Even though there would be no difference in voltage, and a common-mode voltage exists, an output voltage exists.
    This MUST happen in order to drive the differential (error) voltage to zero. This is the way I always understood it.

    Freaks out your mind, doesn't it? :eek:

    Most text books will say common-mode is the average voltage on the pins: (Vp + Vn)/2. Then this follows that an inverting, or non-inverting amplifier have common-mode voltage. Since feedback makes the voltage on the pins equal then it can also follow that common-mode voltage is the same voltage applied to V+.


    Unless I am reading this wrong, CMRR is the ratio of differential voltage gain divided by common-mode gain in dB.

    I disagree with this statement. I can see that nitpicking can happen especially in the case of a typical non-inverting configuration.

    Open up any op-amp datasheet and you will find 'input voltage range', 'common-mode input voltage range' and many derivatives of names.

    National Semiconductor defines 'input voltage range' as the acceptable voltages on individual pins. They define common-mode input voltage range as the range of common-mode voltages acceptable for the op-amp. Common-mode voltages being voltages applied to both pins simulatenously.

    These two definitions are so intimately related; although, I could see a difference. Please see attached...I work better with pictures.

    Notice in the last example, the common-mode shows up on the output. You need a subtractor to strip this off, and this is what the last op-amp in an instrumentation amplifier does. It ensures that the output is 0V.

    An inverting op-amp will not have problems related to common-mode because the voltage at the pins is 0V. A non-inverting op-amp will. The voltage at the pins is not 0V, and this shows up as additional offset. You have intial offset + offset from PSRR + offset from CMRR, etc.
     
Loading...