Top: \(V_{\mathrm{OUT}}\) vs ±5% · Bottom: \(I_{\mathrm{LOAD}}\)
K = 0 is open loop: idle is trimmed to 3.3 V. The step drops Vout by ΔI·Rout. Nothing pulls it back.
Vout · blue in-spec · red out · green = ±5%
Iload · step up, then release
Error at the summer is \(e = V_{\mathrm{ref}} - F V_{\mathrm{out}}\). The forward chain is \(L = C P\). Then
\[ \Delta V = \frac{\Delta I\,R_{\mathrm{out}}}{1+K} \]Idle is trimmed to \(V_{\mathrm{ref}}\). Extra load current through the plant’s \(R_{\mathrm{out}}\) is a disturbance. Closed-loop it is divided by \(1+K\). Open loop (\(K=0\)) takes the full \(\Delta I R_{\mathrm{out}}\) hit. A later page puts a pole at the origin so even the leftover DC error goes to zero, and asks whether a large \(K\) is still stable.