Closed-loop gain

The closed-loop gain $G$ is the gain of an op-amp circuit with feedback connected: the gain you actually get and design for, as opposed to the bare-chip Open-loop gain $A$. In the ideal limit it is a pure ratio of external resistors, $-R_2/R_1$ for the Inverting amplifier (op-amp) and $1+R_2/R_1$ for the Non-inverting amplifier (op-amp). It does not depend on the op-amp, on temperature, or on the active device. That independence is the whole reason Negative feedback is used.

Effect of finite open-loop gain

The ideal derivations set $A=\infty$ to get $v_{+}=v_{-}$. What if $A$ is large but finite? Redo the inverting amplifier without that shortcut. The output still obeys $v_O=A(v_{+}-v_{-})$. With $v_{+}=0$ (grounded), this means $v_{-}=-v_O/A$, not exactly zero but a tiny nonzero error voltage. Write KCL at the inverting node with this corrected $v_{-}$. Current in from the source equals current out through the feedback resistor:

$\frac{v_I-v_{-}}{R_1}=\frac{v_{-}-v_O}{R_2}$

Substitute $v_{-}=-v_O/A$:

$\frac{v_I+v_O/A}{R_1}=\frac{-v_O/A-v_O}{R_2}$

Multiply through and collect $v_O$ terms. The right side is $-\frac{v_O}{R_2}\left(1+\frac{1}{A}\right)$; the left is $\frac{v_I}{R_1}+\frac{v_O}{A{R}_1}$. Gathering $v_O$:

$\frac{v_I}{R_1}=-v_O\left[\frac{1}{R_2}\left(1+\frac{1}{A}\right)+\frac{1}{A{R}_1}\right]$

Solve for $v_O/v_I$ and tidy (multiply numerator and denominator by $R_2$):

$\boxed{\frac{v_O}{v_I}=\frac{-R_2/R_1}{1+\frac{1+R_2/R_1}{A}}}$

The result is the ideal gain $-R_2/R_1$ multiplied by a correction factor $1/[1+(1+R_2/R_1)/A]$ that approaches $1$ as $A\to \infty$. The finite-$A$ error is small precisely because $A$ is enormous.

Finite-$A$ correction factor $1/[1+(1+R_2/R_1)/A]\to 1$ as $A\to \infty$.

Worked number. A 741 has $A=10^5$. Build a $\times 100$ inverter, so $R_2/R_1=100$. The correction factor is

$\frac{1}{1+(1+100)/10^5}=\frac{1}{1+0.00101}\approx 0.999$

A $0.1\%$ gain error. You almost never need the corrected formula in design; the op-amp’s huge $A$ is what makes the ideal model trustworthy.

Loop gain — the general lesson

Reframe the result. Write the closed-loop gain as the ideal gain times $1/(1+1/T)$, where $T$ is the loop gain, the total gain going once around the feedback loop (the op-amp’s $A$ times the feedback fraction). For the inverter, $1/T\approx (1+R_2/R_1)/A$, so $T\approx A/(1+R_2/R_1)$:

$G\approx G_{\text{ideal}}\cdot \frac{1}{1+1/T}$

When the loop gain $T$ is large (open-loop gain $\gg$ closed-loop gain), $1/T\to 0$ and the closed-loop gain locks onto the feedback network, completely indifferent to $A$. That is the engineering meaning of “large loop gain”: you deliberately throw away open-loop performance to buy closed-loop precision. The op-amp itself is mediocre; the loop makes it good. Finite and frequency-dependent $A$ also limits closed-loop bandwidth to roughly $f_t/|G|$, one of the Real op-amp imperfections.