# How should the differential gain of an op-amp be interpreted?

Let us consider an open-loop op-amp circuit:

The output voltage including the common mode signal is given by: $$\mathbf{v}_{\mathbf{o}}=\mathbf{A}_{\mathbf{d}} \mathbf{v}_{\mathbf{d}}+\mathbf{A}_{\mathbf{c}} \mathbf{v}_{\mathbf{c}}$$

Here Ad and Ac refer to the differential mode gain and the common mode gain of the circuit and the Vd and the Vc refer to the differential mode voltage and the common mode voltage of the circuit.

Since the circuit is the op-amp itself, so all the parameters are for the op-amp and hence Ad becomes the differential mode gain of the op-amp which is equal to the open-loop gain of the op-amp, Vd is internal differential voltage of the op-amp, Acm and Vcm are internal common-mode signal parameters of the op-amp. Hence,

$$\mathrm{v}_{\mathrm{d}}=\mathrm{v}_{\mathrm{+}}-\mathrm{v}_{\mathrm{-}}$$ $$v_{\text {c}}= \frac{v_{\text {+}}+v_{\text {- }}}{2}$$ $$A_{\text {d}}= A_{\text {OL}}$$

Now consider the second circuit in a closed-loop configuration (negative feedback):

Output voltage including the common-mode signal is given by: $$\mathbf{v}_{\mathbf{o}}=\mathbf{A}_{\mathbf{d}} \mathbf{v}_{\mathbf{d}}+\mathbf{A}_{\mathbf{c}} \mathbf{v}_{\mathbf{c}}=\mathbf{A}_{\mathbf{OL}} \mathbf{v}_{\mathbf{id}}$$ $$\mathrm{v}_{\mathrm{id}}=\mathrm{v}_{\mathrm{+}}-\mathrm{v}_{\mathrm{-}}$$ $$\mathrm{v}_{\mathrm{d}}=\mathrm{v}_{\mathrm{I2}}-\mathrm{v}_{\mathrm{I1}}$$ $$v_{\text {c}}= \frac{v_{\text {I1}}+v_{\text {I2 }}}{2}$$ $$A_{\text {d}}= A_{\text {CL}}$$

Now in a closed-loop configuration, the differential gain of the circuit is the closed-loop gain of the circuit and is no longer equal to the differetial gain of the op-amp (also referred to as the open-loop gain of the op-amp).

The same can be said about the differential mode voltage Vd, common-mode voltage Vc and the common mode gain Ac of the circuit. The Vid is the differential voltage of the op-amp which can still be related to output voltage of the op-amp (same as th output voltage of the circuit) using the open loop gain of the op-amp.

But why is it that the internal common mode signal parameters are not included in the op-amp closed configuration circuit? I mean to say: why doesn't this equation exist?: $$\mathbf{v}_{\mathbf{o}}=\mathbf{A}_{\mathbf{d}} \mathbf{v}_{\mathbf{d}}+\mathbf{A}_{\mathbf{c}} \mathbf{v}_{\mathbf{c}}=\mathbf{A}_{\mathbf{OL}} \mathbf{v}_{\mathbf{id}}+\mathbf{A}_{\mathbf{c}}^{'} \mathbf{v}_{\mathbf{c}}^{'}$$ $$v_{\text {c}}^{'}= \frac{v_{\text {+}}+v_{\text {-}}}{2}$$

Why are internal common mode signal parameters not included in the internal gain equation of the op-amp? Is my interpretation of all these concepts correct?

Sidenote:In this way, CMRR (common mode rejection ratio) which is the ratio of Ad to Ac can also be defined in two says: CMRR of the op-amp and CMRR of the circuit.