Type II Compensator — Analog Intuition

Along the way

1
Start at ITH. The triangle is \(g_m\): it turns the error into a current. Everything on the output of the \(g_m\) amp is the compensator — \(R_o\) to ground, \(C_{\mathrm{THP}}\) on the pin, and \(R_{\mathrm{TH}}\) in series with \(C_{\mathrm{TH}}\).
2
You start as an integrator. \(R_o\) and \(C_{\mathrm{TH}}\) set a low pole \(f_{p0}\). Mid-band gain is \(g_m R_o\). Above \(f_{p0}\) the blue trace falls and phase sits near −90°. Without a zero, that phase never comes back.
3
The zero is the recovery. \(R_{\mathrm{TH}}C_{\mathrm{TH}}\) places \(f_{z1}\). After the zero the slope flattens and phase climbs — that climb is the boost you came for.
4
The high pole ends the boost. \(C_{\mathrm{THP}}\) places \(f_{p2}\) above the zero. Phase peaks at the geometric mean \(f_b=\sqrt{f_{z1}f_{p2}}\), then the extra pole takes the phase away again.
5
Read the violet line. That is \(f_b\): how much extra phase the zero–pole pair actually gives, and what \(|A|\) is right there. Mid-band gain is higher; by \(f_b\) the integrator has already pulled the blue trace down.

Try it: Start on Textbook 10× and read the violet numbers. Switch to Almost no boost — \(f_{z1}\) and \(f_{p2}\) sit on top of each other and the hump disappears. Then More gm: the corners stay put; only the gain moves. Flip to Frequencies if you think in kHz — the grayed parts are what the network became.

Tune

\( f_{z1} = \dfrac{1}{2\pi R_{\mathrm{TH}} C_{\mathrm{TH}}} \)

\( f_{p0} = \dfrac{1}{2\pi R_{o} C_{\mathrm{TH}}} \)

\( f_{p2} = \dfrac{C_{\mathrm{TH}}+C_{\mathrm{THP}}}{2\pi R_{\mathrm{TH}} C_{\mathrm{TH}} C_{\mathrm{THP}}} \)

RTH (kΩ) CTHP (pF) gm (mS) Ro (kΩ) CTH (pF)

fz1 (kHz) fp2 (kHz) fp0 (kHz) Mid-band gain (dB)

Freeze traces

Compensator Bode · |A| blue · phase rose · violet is max boost

Hover the plot

max boost at — kHz phase boost —° |A| there — dB fp/fz — fz — fp0 — fp2 — gain —

A(s) = 60.0 dB · 1 + s / (2π · 15.92 kHz) (1 + s / (2π · 0.318 kHz)) (1 + s / (2π · 175.1 kHz))

\[ |A|_{\mathrm{dB}}=20\log_{10}(g_m R_o),\quad f_b=\sqrt{f_{z1}f_{p2}},\quad \phi_{\mathrm{boost}}=\tan^{-1}\!\frac{f_b}{f_{z1}}-\tan^{-1}\!\frac{f_b}{f_{p2}} \]

A zero, then a pole, on the same ITH pin