Fully differential op-amp expression

Discussion in 'Homework Help' started by kdillinger, Jul 26, 2009.

  1. kdillinger

    Thread Starter Active Member

    Jul 26, 2009
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    I need some algebra help with the fully differential amplifier circuit attached. I have attached two PDF's both using superposition but using different variables. I am trying to get the solution in the form of Vin*A=Vout where Vin and Vout are differential.
    I am confident that my math is correct up until I can no longer simplify the expression to Vin*A=Vout.
    If the input resistors were equal (same variable, R1) and the feedback resistors were equal (same variable, R2) I can easily derive the transfer function. But the goal is to derive the transfer function when the resistors are not equal (R1, R2, R3, R4) and this is proving to be very cumbersome in the math!
    Can anyone help me with this?
     
  2. steveb

    Senior Member

    Jul 3, 2008
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    Before getting into the math, let's make sure we're all on the same page as far as the setup of the problem. The circuit you show appears to use an opamp with a differential output stage which is not as common as a single output OPAMP. This is confirmed by you labeling two of the voltages as output. Is this what you intended? If so, do you have the basic OPAMP model you are using to get your answer? There are a number of opamp models one can use, and the use of a differential output stage just adds to the uncertainty.

    If you are using a very ideal model, we would probably assume that the outputs are anti-symmetrical. If so, R3 and R4 will make a negative feedback path and R1 and R2 will also make a negative feedback path. Does this sound like the problem you are trying to solve? It looks unusual to me, which is why I'm asking. I just want to make sure I understand the question before putting time into checking the math.
     
    Last edited: Jul 26, 2009
  3. kdillinger

    Thread Starter Active Member

    Jul 26, 2009
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    The circuit is a fully differential amplifier. The input is a differential signal and the output is indeed a differential output stage. I do not know what you mean when you ask if I have the basic op amp model. I am assuming an ideal amplifier (infinite input impedance, infinite gain, etc.).
    I am starting with superposition to find the voltage expression of the V+ (non-inverting) input node and also assume that the inverting input node must be the same.

    Correct.
     
  4. steveb

    Senior Member

    Jul 3, 2008
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    Yes, I understand. This makes sense and I was able to pick up this information tracing through your derivation. I'm still checking thru the second page, but the first page looks correct.
     
  5. steveb

    Senior Member

    Jul 3, 2008
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    Looking through your derivation, I'd have to say that you did it correctly. I don't see any obvious mistakes.

    Just to be sure, I will derive it using my own approach and double check that the expression is correct.
     
  6. steveb

    Senior Member

    Jul 3, 2008
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    I did an independent derivation and got the same answer. Please see attached PDF. I basically agree with you.

    My scanner just died, so I took a picture. Sorry it's not very clear.
     
  7. kdillinger

    Thread Starter Active Member

    Jul 26, 2009
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    Thanks Steve. You are a smarter man than I am using signal flow graphs! It is nice to see that there are several methods that can be used to derive equations (signal flow graphs, block diagram algebra, superposition, KVL, KCL, etc.).

    Now my question is this: Is it possible to simply the expression? For example, if both feedback paths had R1, R2 then it would be easy to factor and reduce to Vin(R2/R1)=Vout where Vin is (V2-V1) and Vout would be (Vout- - Vout+).

    I was hoping that this could be done when there are for resistor variables, but it looks like the solution we both derived is as simple as it gets.
     
  8. Tesla23

    Active Member

    May 10, 2009
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    As you have an ideal op-amp, I think that the simplest derivation simply equates the voltages at the two inputs:

    V_N=V_P

    <br />
\frac{R_1\frac{V_{out}}{2}-R_2\frac{V_{in}}{2}}{R_1+R_2}=\frac{R_4\frac{V_{in}}{2}-R_3\frac{V_{out}}{2}}{R_3+R_4}<br />

    which simplifies in two lines to your expressions
     
  9. steveb

    Senior Member

    Jul 3, 2008
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    I think this is just about as simple as you can express it. Although, there are many different variations you could come up with.

    You are asking a very intelligent question about common-mode and differential mode performance of an opamp circuit.

    The way you did your derivations is very revealing. You did it in two ways. The first way is more general and the second way forces the input signal to be purely differential. The first page shows that you could not get the form Vout=A Vin. However, on the second page you succeeded because you forced the symmetry on the input. This is an important fact to be aware of. Common mode ratio is different as the resistor mismatch changes.
     
    Last edited: Jul 27, 2009
  10. Tesla23

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    May 10, 2009
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    If my quick analysis of the circuit with all four voltages is correct, what he should be able to show is that to get common mode rejection he needs

    R_1R_4=R_2R_3

    which is pretty obvious when you look at the schematic
     
  11. steveb

    Senior Member

    Jul 3, 2008
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    Good comment!

    I agree. That is a more precise expression of the matching constraint.

    He pretty much showed this result on his first page derivation. It's clear that that constraint allows him to get the equation in the form he wants.
     
  12. The Electrician

    AAC Fanatic!

    Oct 9, 2007
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    An interesting consideration is this: The OP's second page shows the inputs as two voltage sources, one the negative of the other, and both presumably ideal voltage sources with zero output impedance.

    Imagine that the "differential" input was a floating source.

    If we had an ideal transformer with 1:1 turns ratio and with a center-tapped secondary, the OP's circuit would be what we would have if the primary were driven from an ideal voltage source, the center tap were grounded and the ends of the center-tapped winding provided Vin/2 and -Vin/2.

    Now disconnect the center tap from ground, and apply the ends of the secondary to the left ends of R1 and R2. Now we have floating drive.

    Or, you could simply lift the grounds of the two input sources in the OP's second page circuit, and connect those two previously grounded terminals of the sources together, but don't connect them to the circuit reference anymore.

    What is the gain function now?

    Suppose we applied a small voltage (from an ideal voltage source) to the center tap with respect to ground; this would constitute a common mode signal. What is the expression for the common mode gain with this drive arrangement?
     
  13. Tesla23

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    May 10, 2009
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    That's how I analysed it, I essentially created a transformer in the equations I used. I just set

    <br />
V_1=V_{cm}-\frac{V_i}{2}<br />

    and

    <br />
V_2=V_{cm}+\frac{V_i}{2}<br />

    and you find that if R_1R_4=R_2R_3 then there is no V_{cm} in the output
     
  14. The Electrician

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    What about the gain function if the center tap is disconnected?

    And what would be the effect of a resistance in series with the center tap and the small voltage source producing the common mode signal?
     
    Last edited: Jul 27, 2009
  15. Tesla23

    Active Member

    May 10, 2009
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    I don't know - but you could do the analysis to find out if you needed to.

    Typically for differential amplifier applications that I have used, you want to maximise the CMRR, ideally you want the gain from V_{cm} to the output to be zero. Then it doesn't matter if you tie them together and float the voltage up or down, there is no output. Practically there are limitations on the common mode voltage range, so if you had a transformer coupling I would have thought you'd want to set the centre tap voltage. In other words I'd have thought that having V_{cm} coming from a very high (ideally infinite) source impedance is not a good idea.
     
  16. The Electrician

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    I would have thought you would realize that my question was not a plea for desperately needed information, but that it was intended to provoke the reader to engage in some further investigation. I know that steveb and t_n_k are generally interested in this sort of thing.

    I would have thought that such a very symmetrical circuit would not behave differently with a floating source, but that's not the result I get.

    The question is, how much does setting the center tap voltage from a finite source impedance degrade the CMRR, if any.
     
  17. Tesla23

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    May 10, 2009
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    without doing any detailed analysis, I would have thought that it would generally improve it by the factor

    <br />
\frac{R_{incm}}{R_{incm} + Z_{cms}}<br />

    where R_{incm} is the common mode input resistance and Z_{cms} is the common mode source impedance.
     
  18. steveb

    Senior Member

    Jul 3, 2008
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    I corrected something in my previous post. Without thinking, I mentioned that CMRR degrades with resistor mismatch in this circuit. But, this was a concept I erroneously brought over from single ended differential amplifiers. This circuit has equal gain for both common-mode and differential-mode, if the resistor matching condition is met. Hence CMRR is unity, not infinity, and mismatch can make the ratio either go up or down. I just didn't think it though very well.

    On The-Electrician's question, I did a quick derivation of common mode gain with matched resistors R2=R4 and R1=R3, and assuming a source resistance of Rsc in the common mode voltage driving both inputs. I get the following for gain:

     A_c=-{{R_2}\over{R_1}}\;  {{R_2+R_1}\over{R_2+R_1+1.5 R_{sc}}}

    This is different than a balanced differential drive where I would expect the gain to be

     A_d={{R_2}\over{R_1+0.5 R_{sd}}}

    since the source resistance Rsd will act like it is split equally between the two inputs.
     
  19. The Electrician

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    I get a different result, steveb. I get:

     A_c=\frac{R1R4-R2R3}{R1R4+R2R3+2R1R3}

    For the case where the input is floating (Rs=∞), I get:

     A_d=\frac{R2+R4}{R1+R3}
     
  20. steveb

    Senior Member

    Jul 3, 2008
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    I have to say your answers make more sense to me and are in line with my initial intuition. I'm inclined to think that I screwed up the derivation and then over-thought it.

    I'll double check.
     
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