M3 · Locate the boost's right-half-plane zero

Goal

• Explain in your own words why a step of the boost’s duty ratio first moves its output the wrong way.
• Compute the right-half-plane (RHP) zero and the double pole of the boost example.
• Find how the zero moves with the load and the duty ratio, and which operating point limits a feedback loop’s crossover frequency.

Before you start

• Read the right-half-plane zero and small-signal models.
• The equations give angular frequencies in rad/s. Divide by $2\pi$ for hertz.

Steps

1. The physical picture

Read how the diode’s shorter conduction after a step of $D$ lowers the average current into the output before the inductor current has time to grow. Write it down in two sentences. Then say why the buck has no such zero: its inductor feeds the output in both intervals [Erickson & Maksimović 2020, Ch. 8].

2. The zero and the poles of the example

The boost example has $V_g$ = 12 V, $D$ = 0.5, $L$ = 100 µH, $C$ = 10 µF and $R$ = 20 Ω. Work out its zero (the boost's RHP zero) and its double pole (the boost's double pole), and enter both below.

3. How the zero moves

Open the zero in the equation explorer and sweep $R$ towards a heavier load, then set $R$ back and sweep $D$:

▶ Open in the equation explorer: Boost: right-half-plane zero
Preset from the example boost-basic

4. The buck-boost

The buck-boost has the same kind of zero (the buck-boost's RHP zero). Work it out for the buck-boost example, at the same duty ratio, and enter it below.

5. The operating point that limits the loop

Over a range of loads and input voltages, the zero is lowest at the heaviest load and the largest duty ratio, that is, at the lowest input voltage. A loop’s crossover frequency must stay well below that lowest zero [Erickson & Maksimović 2020, Ch. 8]. Say which corner of your range that is for the example, and why designing at the nominal point would overestimate the crossover you may use.

Acceptance criteria

• In two sentences, why a step of $D$ first lowers the boost's output, and why the buck has no such zero.
• The boost example's RHP zero $\omega_z$.
  Your value (rad/s)Check
  Show the example's value

  50 krad/s (from boost-basic: Boost example)

• Its double pole $\omega_0$.
  Your value (rad/s)Check
  Show the example's value

  15.81 krad/s (from boost-basic: Boost example)

• In the explorer: halving $R$ halves $\omega_z$, and a larger $D$ lowers it.
• The buck-boost example's RHP zero.
  Your value (rad/s)Check
  Show the example's value

  50 krad/s (from buckboost-basic: Buck-boost example)

• The operating point that sets the lowest zero, and why the crossover must stay well below it.

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Gotchas

• $\omega_z$ is in rad/s. Taking it for hertz puts the crossover limit more than six times too high.
• The zero moves with the operating point. The nominal point is not the worst case: the heaviest load at the lowest input voltage is.
• A compensator cannot cancel an RHP zero. That would take a pole in the right half-plane, which is unstable [Erickson & Maksimović 2020, Ch. 8].

Go deeper

• Book chapter Converter Transfer Functions (Ch. 8) — R. W. Erickson, D. Maksimović · retrieved 2026-09-23
  Bode plots of the control-to-output transfer functions, including the right-half-plane zero of the boost and buck-boost.
• Course Converter Control — University of Colorado Boulder (Coursera) · retrieved 2026-09-23
  Averaged small-signal models, converter transfer functions and feedback loop design.

Quiz

1. Right after a step increase of the boost's duty ratio, the output voltage:

   1. Rises at once
   2. Dips briefly, then rises to its new value
   3. Stays put until the inductor saturates
   4. Oscillates at the switching frequency
   Answer and explanation

   Correct answer: B The diode conducts for a shorter part of each period before the inductor current has grown, so the average current into the output first falls. That is the right-half-plane zero at work.

2. A boost at $D$ = 0.5 with $R$ = 40 ohm and $L$ = 200 µH. What is its RHP zero, $(1-D)^2 R / L$?

   1. 50 krad/s
   2. 200 krad/s
   3. 100 krad/s
   4. 12.5 krad/s
   Answer and explanation

   Correct answer: A 0.25 times 40 ohm is 10 ohm, over 200 µH gives 50 000 rad/s, about 8 kHz.

3. The load current of that boost doubles (its $R$ halves). The zero:

   1. Doubles
   2. Halves
   3. Stays where it is
   4. Moves into the left half-plane
   Answer and explanation

   Correct answer: B The zero is proportional to $R$: a heavier load pulls it down, which is why full load is the worst case.

4. Which operating point of a boost with a regulated output sets the lowest RHP zero?

   1. Light load at the highest input voltage
   2. Full load at the lowest input voltage
   3. Light load at the lowest input voltage
   4. Full load at the highest input voltage
   Answer and explanation

   Correct answer: B The zero falls with a smaller $R$ (heavier load) and with a larger $D$, and the regulated boost needs its largest $D$ at its lowest input voltage.

5. Why can a compensator not cancel the RHP zero with a pole at the same place?

   1. The pole would have to be in the right half-plane, which is unstable
   2. Poles cannot be placed above the switching frequency
   3. The zero is not a real number
   4. The pole would slow the loop down too much
   Answer and explanation

   Correct answer: A Cancelling a zero at $+\omega_z$ takes a pole at $+\omega_z$: an unstable pole inside the loop.