Analog Intuition

Load step vs the window

Top: \(V_{\mathrm{OUT}}\) vs ±5% · Bottom: \(I_{\mathrm{LOAD}}\)

droop \(\Delta I R_{\mathrm{out}}/(1+K)\) → —
vs open-loop \(\Delta I R_{\mathrm{out}}\) → —

K = 0 is open loop: idle is trimmed to 3.3 V. The step drops Vout by ΔI·Rout. Nothing pulls it back.

Open loop: feedback path is open Closed loop · C, P, F Open loop · FB is open Vref Σ + − C(s) Comp P(s) Plant Vout F(s) FB Iload
Forward: \(C\) then \(P\). Return is open — \(K=0\).

Vout · blue in-spec · red out · green = ±5%

Iload · step up, then release

Vout in spec Vout out of spec Vref Iload playhead
Control loop algebra

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.