# What actually is the differential gain of an operational amplifier and why does its value change when we consider the common-mode gain?

Okay, this may be a stupid question, but at the beginning of my Applied Electronics course they taught us that the differential gain $$\A_d\$$ of an operational amplifier is ideally infinite; however, in real op amps, we get a limited bandwidth. Which is fine, because it's still very high. My problem raises when I was introduced to common-mode gain and differential-mode gains. In particular, I read that an amplifier subject to common-mode input signals has as a (non-ideal) output a linear combination of its differential mode input voltage mode and its common mode input voltage, i.e., $$A_dv_d + A_{cm}v_{cm}.$$ And for now, everything's fine. But let's consider the practical example in the picture below: simulate this circuit – Schematic created using CircuitLab

According to what I read, in the schematic above $$V_u=(1+\frac{R_1}{R_2})(v_+-v_-)=\frac{R_1+R_2}{R_2}V_1-\frac{R_1}{R_2}V_2.$$ And since $$\\frac{V_1+\frac{R_1}{R_1+R_2}V_2}{2}=v_{cm}\$$ and $$\v_d=v_+-v_-\$$, then $$\V_1=v_{cm}+\frac{v_d}{2}\$$ and $$\V_2=v_{cm}-\frac{v_d}{2}\$$. Moreover: $$V_u=\frac{R_1+R_2}{R_2}(v_{cm}+\frac{v_d}{2})-\frac{R_1}{R_2}(v_{cm}-\frac{v_d}{2})=v_{cm}(\frac{R_1+R_2}{R_2}-\frac{R_1}{R_2})+v_d(\frac{R_1+R_2}{2R_2}+\frac{R_1}{2R_2})=\Big(\frac{1}{2}+\frac{R_1}{R_2}\Big)v_d+v_{cm}=A_dv_d+A_{cm}v_{cm}.$$ How are we even just able to say that $$\A_d=\frac{1}{2}+\frac{R_1}{R_2}\$$? It's limited, fine, but I would've imagined bigger numbers. I guess I can live with a common gain of 1, but since they called both differential gains $$\A_d\$$, are they seriously the same?

Already from the beginning, I couldn't even understand why would we use $$V_u=(1+\frac{R_1}{R_2})v_d,$$since $$\V_u=A_dv_d\$$, with $$\A_d\$$ a big value and, instead, considering an ideal op-amp, as $$\A_d\$$ approaches infinity, $$\V_u=\frac{1}{\beta}v_+=(1+\frac{R_1}{R_2})v_+.\$$ But shouldn't $$\(1+\frac{R_1}{R_2})\$$ be the gain from the non inverting terminal to the output (i.e., closed loop gain), not of the op-amp, or am I wrong? What am I missing? Thank you in advance!

Edit: I am starting to assume that the answer is due to the principle of superposition of effect, i.e., shutting $$\V_2\$$ off and compute $$\Vu\$$ due to $$\V_1\$$, and then shut $$\V_1\$$ off and compute the overall gain due to $$\V_2\$$, and adding then the two results together. This could make sense, I suppose, but I can't still figure out why the result is equivalent to $$\v_d\frac{R_1+R_2}{R_2}\$$, since $$\v_d\$$ is supposed to be very small and it should therefore be amplified by a high gain $$\A_d\$$.

Edit 2: fixed typo that made the result wrong, but the question still holds.

Edit 3: fixed the value of $$\\beta\$$ in the formulas for this schematic. Unfortunately, I chose an unconventional position for $$\R_1\$$ and $$\R_2\$$ in the schematic.

• have you read about the "virtual ground" method of thinking about typical opamp circuits? – analogsystemsrf Nov 14 '19 at 1:54
• Open loop gain is infinite (ideally), closed loop gain is not so. With extremely large gain, op would be + or - Vsat even with small noise.. using feedback, you can control gain. – Deep Nov 14 '19 at 3:33
• $V_u=(1+\frac{R_2}{R_1})(v_+-v_-)$ - This relation is wrong because $(v_+-v_-)$ is zero here because of the -ve feedback .... – Meenie Leis Nov 14 '19 at 19:12
• You are confusing two issues. There is the differential gain of the op amp. This is a very high number, infinite in the ideal. This is the ONLY gain an op amp has. Then, there are differential gains and common mode gains for op amp circuits -- i.e., amplifiers constructed out of op amps. – Scott Seidman Nov 15 '19 at 15:30
• I have not seen in ANY books, the very first relationship you have written on Vu. You just misinterpreted it. It's simply wrong assumption you started with. – Meenie Leis Nov 16 '19 at 10:20

$$\bbox[4px,border:1px solid red]{V_u=(1+\frac{R_2}{R_1})(v_+-v_-)}$$ The above assumption where you start from is actually wrong.

$$\(v_+-v_-)\$$ is infinitesimaly small or zero ideally.

The correct relation is: $$V_u = a_d.v_d + a_c.v_c$$ $$ie.,V_u = a_d.(v_+-v_-) + a_c.\frac{(v_+ +v_-)}{2}$$ $$\bbox[8px,border:1px solid black]{V_u = G_{OL}.(v_+-v_-)}$$ Where $$\a_d\$$ or $$\G_{OL}\$$ is the open-loop differential gain or simply open-loop gain, and it is very big or infinite for ideal op amp. And open-loop common-mode gain $$\a_c = 0\$$.

Coming to closed-loop gains in your circuit:

$$V_u=(\frac{R_1+R_2}{R_1})V_1-(\frac{R_2}{R_1})V_2 \tag1$$

The circuit has a negative feedback and you can calculate closed-loop differential and common-mode gains using the relations: $$V_1 = V_c+\frac{V_d}{2}$$ $$V_2 = V_c - \frac{V_d}{2}$$ where $$\V_d\$$ and $$\V_c\$$ are differential and common-mode components of $$\V_1\$$ and$$\V_2\$$:

$$V_d =V_1-V_2$$ $$V_c = (V_1+V_2)/2$$

Equation (1) can be simplified as: $$V_u=(\frac{R_1+R_2}{R_1}).(V_c+\frac{V_d}{2})-(\frac{R_2}{R_1}).(V_c - \frac{V_d}{2})$$ $$\implies V_u = (\frac{1}{2}+\frac{R_2}{R_1}).V_d+1.V_c \tag2$$ Compare (2) with: $$V_u = A_d.V_d+A_c.V_c$$ $$\bbox[8px,border:1px solid black]{\therefore A_d = (\frac{1}{2}+\frac{R_2}{R_1}), A_c = 1 }$$ where $$\A_d\$$ and $$\A_c\$$ are closed-loop differential and common-mode gains respectively.

• Thank you! I actually made a typo for the result, 'cos I forgot to write a - for $\frac{v_d}{2}$. However, I still wonder: when you write $V_u=(\frac{R_1+R_2}{R_1})V_1-(\frac{R_2}{R_1})V_2$, since $V_1=v_+$ (isn't it? Maybe that's where I'm getting confused) and $v_-=V_2\frac{R_1}{R_1+R_2}$ (?), isn't the expression equal to $V_u=(\frac{R_1+R_2}{R_1})(v_+-v_-)$? For sure, I got the definition of $V_d$ and $V_{c}$ wrong, as I defined them for $v_+$ and $v_-$ instead of $V_1$ and $V_2$, but I still don't get why the expression I wrote is wrong. – Maurizio Carcassona Nov 16 '19 at 22:10
• V1 is equal to v+.. But your relationship between v- and V2 is simply wrong ....we know v- is equal to v+....Or in other words you are claiming that V1 is equal to V2. R1/(R1+R2)... . Which is not at all true...... – Meenie Leis Nov 16 '19 at 23:22
• Hmm you're right, it makes sense now. $v_-$ can't be that value. However, this gets me to another doubt: if $v_-=v_+=V_1$, then due to feedback $v_-=\beta V_u=\frac{R_2}{R_1+R_2}V_u=\frac{R_2}{R_1+R_2}\Big[V_1\Big(1+\frac{R_1}{R_2}\Big)-V_2\frac{R_1}{R_2}\Big]=V_1-V_2\frac{R_1}{R_1+R_2}$. But if $v_-=v_+=V_1$, then $V_1-V_2\frac{R_1}{R_1+R_2}=V_1\implies V_2\frac{R_1}{R_1+R_2}=0$, which means that either $V_2$ is 0 or $R_1=0$. But how is that possible, since it's not general? There must be something I'm missing in the computations. – Maurizio Carcassona Nov 17 '19 at 1:33