Tuesday, 9 May 2017

operational amplifier - Question about differential signals and feedback


Let's consider, for example, this circuit:


enter image description here


During lessons, our professor always assumed perfect differential input (two signals with the same dc value and with equal and opposite amplitudes). As a consequence node 1 will be an ac ground due to symmetry and the small signal differential gain can be easily found:


enter image description here


Now the question: when I close this circuit (or in general every circuit with a differential pair stage, which is the input block of an op-amp) with negative feedback, I will not have a perfect differential input, thus I am not allowed to use the previous differential gain (which was actually found under the assumption of differential input). Let's consider for example this basic circuit:


enter image description here


You can see that the non-inverting terminal is fixed to the analog ground, thus it can not change in a differential way with respect to the inverting terminal. In a similar question I wrote, I've been answered that actually you can always write a couple of signals as the sum of a common mode signal and a differential signal, and since a well-designed op-amp has a common mode gain wihich is much smaller than the differential gain, we can neglect the common mode gain (and thus use only the previous expression for the differential gain). Now I would like to have some hints on how to proceed with the analysis in this case. For example, considering the previous inverting configuration, I tried to decompose the input of the op-amp:


enter image description here


where vx is the voltage at the inverting terminal. Is it correct? How to proceed with the analysis?



Thank you


Edit for the comment:


For the telescopic configuration, the differential gain was found under the hypothesis of differential input signals:


enter image description here


When we close the feedback around it we get:


enter image description here



Answer



Brief Background
Suppose you have a linear network which has two input ports with input voltages \$V_1\$ and \$V_2\$ as shown in figure below:


enter image description here



Then, since \$V_1 = \frac{V_1-V_2}{2}+\frac{V_1+V_2}{2}\$ and \$V_2=\frac{V_2-V_1}{2}+\frac{V_1+V_2}{2}\$. Thus we have:


enter image description here


Then you can transform the circuit as shown below:


enter image description here


Here the common mode voltage is: \$V_{cm} = \frac{V_1+V_2}{2}\$ and the differential voltage is: \$\frac{V_{diff}}{2} = \frac{V_1-V_2}{2}\$. Since the circuit is linear, superposition is valid. So we can say that the total response will be sum of these two.


enter image description here


enter image description here


The first one is the the common-mode circuit and the second one is the differential circuit. Here you can use all the tricks for the differential half and the common-mode half which you may know.
Your Example
The complete circuit for the example you provided will be:



enter image description here


Here the two inputs are: \$V_1=V_{cm}+V_{in}\$ and \$V_2 = V_{cm}\$.
If you use superposition here with \$V_{cm}=0\$, you get the circuit which you have shown in your question. This is the differential part of the circuit.
If you instead make \$V_{in}=0\$, you get the common-mode circuit: enter image description here


I leave it to you now to analyze it.


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