Analog Intuition

Part 4 · Control Loops

Loop gain \(T = P\,C\,F\)

Key point: break the loop at the summer and ask what comes back. That product is loop gain \(T(s)=P(s)\,C(s)\,F(s)\) — what we measure and shape. It is not the closed-loop map \(V_{\mathrm{out}}/V_{\mathrm{ref}}\).

Loop-gain Bode

\(T = K\cdot F/(1+s/\omega_p)\) · plant is one pole · 0 dB is crossover

\(f_c\) →
\(|T|\) at DC →

T(s) = K · F / (1 + s/ωp). dB add on the Bode. K lifts |T| without moving the plant pole.

Loop opened at the summer: Vtest into Comp then Plant, Feedback returns Vreturn, no connection back to Vtest
Opened at Σ. \(T = V_{\mathrm{return}}/V_{\mathrm{test}} = P\,C\,F\).

Bode of T · 10 Hz–1 MHz

|T| (dB) Phase 0 dB fc
Dig deeper — \(T\) vs \(T_{\mathrm{cl}}\)

Forward path \(L=CP\). Closed-loop from reference to output:

\[ T_{\mathrm{cl}} = \frac{V_{\mathrm{out}}}{V_{\mathrm{ref}}} = \frac{L}{1+LF} \qquad T = L F = P\,C\,F \]

We shape \(T\) because it is what a network analyzer measures when you break the loop. \(K\) in textbooks is \(T(s)=K\,P\,C\,F\) — a bag for modulator gain, sense gain, and so on. It is not a third capacitor. Sliding \(K\) does not move pole/zero frequencies; it only moves crossover. Instability is \(1+T=0\): unity gain and −180° together. That reading is the next page.