# Question about differential signals and feedback

Let's consider, for example, this circuit:

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:

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:

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:

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:

When we close the feedback around it we get:

• But did you manage to computed the of a diff amp for this case? Is there any practical difference in the finally value? Also if you ground the gate of a first mosfet and apply the input signal to M2 gate. We can still write Vin = Vid = Vgs2 + Vsg1 and because M1 and M2 are identical Vgs1 = Vid/2 and Vsg2 = Vid/2 (Vsg2 = - Vgs1) So, you can be worried or not? – G36 Jul 26 '19 at 14:42

## 2 Answers

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:

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:

Then you can transform the circuit as shown below:

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.

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:

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:

I leave it to you now to analyze it.

• I read and understood the "brief background" section, but I don't understand "your example" section. What do you mean with "differential part" and who are V1 and V2 in the inverting configuration? Again thank you – Stefanino Jul 26 '19 at 10:21
• @Stefanino Please see the edits. Hope this clarifies things. If not, let me know. – sarthak Jul 26 '19 at 11:08
• I thank you for the edit, now it is much more clear. However the doubt still remains in me: when you split the two input signals (vin and ground) into common mode signal and differential signal, you still don't have at the input terminals of the op-amp a perfect differential (or perfect common) signal, because you have a voltage drop on R1. Instead in the telescopic circuit (for example) you have a differential signal ditectly applied to input terminals. Thus, in your differential part circuit, who says I can use Ad (i.e. the differential gain)? I hope I was clear. Thank you for your patience! – Stefanino Jul 26 '19 at 12:09
• I also added two pictures to my question in order to be more clear – Stefanino Jul 26 '19 at 12:43
• @Stefanino The gain is the same for a differential input and single ended output amplifier. You would have a schematic like this: google.com/…: – sarthak Jul 26 '19 at 13:13

I considered very positive your patience to restate the question you posted before, and, like others, I thought that the problem was your concept of "perfect differential input". May I suggest that you evaluate that, in fact, the differential input stage that you consider in your analysis is simply not rail-to-rail capable? Please kindly take a look at the modified picture:

If you want to connect one of these inputs to $$\0 V\$$ you should consider your input stage powered by split supplies (e.g. $$\\pm 15 V\$$).

• Yes, you're right, that node must be connected to VSS and not ground. However, even if I adjust the schematic as you correctly said, my doubt still exists: in order to use Ad, we should have those red differential signals (because Ad was found under that hypothesis); how to prove (if I am right) that they are differential? Thank you – Stefanino Jul 26 '19 at 13:26
• I see your point. Thanks for clarifying. The lack of symmetry would break the "ac ground" assumption that facilitates the differential gain analysis. – vangelo Jul 26 '19 at 13:46
• +2 I was going to comment on this but see that you point it out neatly. From the first reading and the OPs first diagram I was nervous about below rail input operation and erroneous assumptions leading from there as a start. – KalleMP Jul 26 '19 at 21:14
• I made a mistake because I copied the schematic of my book, but this is not the problem: obviously the source of M9 must be connected to a negative voltage. The doubt I have still remains: how can be proved (if I'm right) that the red signals are differential? – Stefanino Jul 27 '19 at 8:29