The right-half-plane zero

Intent

• Explain physically why the boost and the buck-boost have a right-half-plane (RHP) zero.
• Compute the RHP zero and the double-pole parameters of both converters from circuit values.
• Know how the RHP zero limits the crossover frequency of a feedback loop.

Theory

Why the output first moves the wrong way

In the boost, energy reaches the output only while the diode conducts, that is, during the off-interval. A step increase of the duty ratio lengthens the on-interval and shortens the off-interval. Eventually the inductor current rises and the output rises with it. Immediately after the step, however, the inductor current has not changed yet while the diode conducts for a shorter part of each period, so the average diode current, and the output voltage, first fall [Erickson & Maksimović 2020, Ch. 8]. The buck-boost behaves the same way.

A zero in the right half of the $s$-plane produces exactly this: its magnitude rises with frequency like that of an ordinary zero, but its phase lags like that of a pole.

Boost control-to-output

The boost’s control-to-output transfer function has a dc gain, a pair of poles and the RHP zero [Erickson & Maksimović 2020, Ch. 8]:

$$G_{d0} = \frac{V_g}{\left(1 - D\right)^{2}}$$

$G_{d0}$

dc control-to-output gain

$D$

duty ratio

$V_g$

input voltage

Boost control-to-output dc gain, CCM✓ verified by 7 tests

Assumptions, convention and source

Assumptions

• ideal (lossless) switches and passive components
• continuous conduction mode (CCM)
• small-signal (linearised) model around a CCM operating point
• resistive load R

Notes

equals $\partial V/\partial D$; with $V = V_g/(1 - D)$ this is also $V/(1 - D)$

Source

Erickson & Maksimović 2020, Ch. 8 (Converter Transfer Functions)

Derivation

step-by-step sympy derivation

Equation id

boost.ss.Gd0 · equations.yaml entry

$$\omega_0 = \frac{1 - D}{\sqrt{L C}}$$

$\omega_0$

LC corner frequency (rad/s)

$C$

capacitance (output capacitor)

$D$

duty ratio

$L$

inductance

Boost: angular frequency of the LC poles✓ verified by 7 tests

Assumptions, convention and source

Assumptions

• ideal (lossless) switches and passive components
• continuous conduction mode (CCM)
• small-signal (linearised) model around a CCM operating point
• resistive load R

Source

Erickson & Maksimović 2020, Ch. 8 (Converter Transfer Functions)

Derivation

step-by-step sympy derivation

Equation id

boost.ss.w0 · equations.yaml entry

$$Q = \left(1 - D\right) R \sqrt{\frac{C}{L}}$$

$Q$

quality factor

$C$

capacitance (output capacitor)

$D$

duty ratio

$L$

inductance

$R$

load resistance

Boost: quality factor of the LC poles✓ verified by 7 tests

Assumptions, convention and source

Assumptions

• ideal (lossless) switches and passive components
• continuous conduction mode (CCM)
• small-signal (linearised) model around a CCM operating point
• resistive load R

Source

Erickson & Maksimović 2020, Ch. 8 (Converter Transfer Functions)

Derivation

step-by-step sympy derivation

Equation id

boost.ss.Q · equations.yaml entry

$$\omega_z = \frac{\left(1 - D\right)^{2} R}{L}$$

$\omega_z$

RHP-zero frequency (rad/s)

$D$

duty ratio

$L$

inductance

$R$

load resistance

Boost: right-half-plane zero✓ verified by 7 tests

Assumptions, convention and source

Assumptions

• ideal (lossless) switches and passive components
• continuous conduction mode (CCM)
• small-signal (linearised) model around a CCM operating point
• resistive load R

Notes

moves down as the load gets heavier (smaller $R$) and as $D$ increases

Source

Erickson & Maksimović 2020, Ch. 8 (Converter Transfer Functions)

Derivation

step-by-step sympy derivation

Equation id

boost.ss.wz · equations.yaml entry

The poles belong to the inductor and the capacitor, but seen through the dc transformer of the averaged model the inductance appears divided by $(1 - D)^2$, which moves the poles down in frequency as $D$ grows.

Buck-boost control-to-output

$$G_{d0} = \frac{V_g}{\left(1 - D\right)^{2}}$$

$G_{d0}$

dc control-to-output gain

$D$

duty ratio

$V_g$

input voltage

Buck-boost control-to-output dc gain (magnitude), CCM✓ verified by 7 tests

Assumptions, convention and source

Assumptions

• ideal (lossless) switches and passive components
• continuous conduction mode (CCM)
• small-signal (linearised) model around a CCM operating point
• resistive load R

Notes

magnitude of $\partial V/\partial D$; the inverting output makes the gain itself negative

Source

Erickson & Maksimović 2020, Ch. 8 (Converter Transfer Functions)

Derivation

step-by-step sympy derivation

Equation id

buckboost.ss.Gd0 · equations.yaml entry

$$\omega_0 = \frac{1 - D}{\sqrt{L C}}$$

$\omega_0$

LC corner frequency (rad/s)

$C$

capacitance (output capacitor)

$D$

duty ratio

$L$

inductance

Buck-boost: angular frequency of the LC poles✓ verified by 7 tests

Assumptions, convention and source

Assumptions

• ideal (lossless) switches and passive components
• continuous conduction mode (CCM)
• small-signal (linearised) model around a CCM operating point
• resistive load R

Source

Erickson & Maksimović 2020, Ch. 8 (Converter Transfer Functions)

Derivation

step-by-step sympy derivation

Equation id

buckboost.ss.w0 · equations.yaml entry

$$Q = \left(1 - D\right) R \sqrt{\frac{C}{L}}$$

$Q$

quality factor

$C$

capacitance (output capacitor)

$D$

duty ratio

$L$

inductance

$R$

load resistance

Buck-boost: quality factor of the LC poles✓ verified by 7 tests

Assumptions, convention and source

Assumptions

• ideal (lossless) switches and passive components
• continuous conduction mode (CCM)
• small-signal (linearised) model around a CCM operating point
• resistive load R

Source

Erickson & Maksimović 2020, Ch. 8 (Converter Transfer Functions)

Derivation

step-by-step sympy derivation

Equation id

buckboost.ss.Q · equations.yaml entry

$$\omega_z = \frac{\left(1 - D\right)^{2} R}{D L}$$

$\omega_z$

RHP-zero frequency (rad/s)

$D$

duty ratio

$L$

inductance

$R$

load resistance

Buck-boost: right-half-plane zero✓ verified by 7 tests

Assumptions, convention and source

Assumptions

• ideal (lossless) switches and passive components
• continuous conduction mode (CCM)
• small-signal (linearised) model around a CCM operating point
• resistive load R

Source

Erickson & Maksimović 2020, Ch. 8 (Converter Transfer Functions)

Derivation

step-by-step sympy derivation

Equation id

buckboost.ss.wz · equations.yaml entry

The derivations of all eight quantities from the averaged circuit are on the derivations page.

Consequences for control

The RHP zero moves down in frequency as the load gets heavier (smaller $R$) and as $D$ increases, so its lowest frequency occurs at full load and minimum input voltage. Because it adds phase lag while the magnitude keeps rising, the loop crossover frequency has to stay well below the lowest $\omega_z$ over the operating range [Erickson & Maksimović 2020, Ch. 8]. A compensator cannot cancel the zero: that would take a pole in the right half-plane, which is unstable.

Worked example

The boost and buck-boost examples, both at the same duty ratio:

Boost example

Symbol	Value	How it is obtained
$V_g$ input voltage	12 V	given
$D$ duty ratio	0.5	given
$L$ inductance	100 µH	given
$C$ capacitance (output capacitor)	10 µF	given
$R$ load resistance	20 Ω	given
$G_{d0}$ dc control-to-output gain	48 V	Boost control-to-output dc gain, CCM
$\omega_0$ LC corner frequency (rad/s)	15.8 krad/s	Boost: angular frequency of the LC poles
$Q$ quality factor	3.16	Boost: quality factor of the LC poles
$\omega_z$ RHP-zero frequency (rad/s)	50 krad/s	Boost: right-half-plane zero

Buck-boost example

Symbol	Value	How it is obtained
$V_g$ input voltage	12 V	given
$D$ duty ratio	0.5	given
$L$ inductance	100 µH	given
$C$ capacitance (output capacitor)	10 µF	given
$R$ load resistance	10 Ω	given
$G_{d0}$ dc control-to-output gain	48 V	Buck-boost control-to-output dc gain (magnitude), CCM
$\omega_0$ LC corner frequency (rad/s)	15.8 krad/s	Buck-boost: angular frequency of the LC poles
$Q$ quality factor	1.58	Buck-boost: quality factor of the LC poles
$\omega_z$ RHP-zero frequency (rad/s)	50 krad/s	Buck-boost: right-half-plane zero

In the boost the RHP zero at 50 krad/s lies not far above the double pole at 15.8 krad/s: a loop that crossed over near the poles would already feel the zero’s phase lag.

Try it

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

Sweep $R$ downwards (heavier load): the zero falls in proportion to $R$. Then set $R$ back and sweep $D$ instead.

Bench exercise

On an open-loop boost at low power, apply a small step to the duty ratio, for example by stepping the modulator’s control input. Watch the output voltage with AC coupling, triggered on the step: it dips briefly before it rises to its new value. Repeat at a heavier load and compare the size of the dip.

Gotchas

• Design for the worst case. The lowest RHP-zero frequency occurs at the heaviest load and the highest duty ratio, that is, the lowest input voltage.
• The buck has no RHP zero. Its duty ratio acts on the switch node, which feeds the output filter directly; the boost and buck-boost deliver energy to the output only during the off-interval [Erickson & Maksimović 2020, Ch. 8].
• Phase margin estimates must include the zero. Its phase lag starts well below $\omega_z$, over the same frequencies where an ordinary zero’s phase lead would start.

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.
• Application note Understanding Boost Power Stages in Switchmode Power Supplies (SLVA061) — E. Rogers, Texas Instruments · retrieved 2026-09-23
  The boost counterpart: steady-state and small-signal analysis in CCM and DCM.

Quiz

1. Right after a step increase of the duty ratio, the output voltage of a boost converter briefly falls. Why?

   1. The inductor saturates
   2. The diode now conducts for a shorter part of each period while the inductor current has not risen yet
   3. The output capacitor discharges into the switch
   4. The control loop is unstable
   Answer and explanation

   Correct answer: B Energy reaches the output only while the diode conducts. A longer on-interval shortens the diode interval at once, but the inductor current needs time to grow, so the average diode current first drops.

2. How does a right-half-plane zero affect a Bode plot?

   1. Its magnitude falls like a pole and its phase leads like a zero
   2. Its magnitude rises like a zero but its phase lags like a pole
   3. It changes only the magnitude
   4. It changes only the phase
   Answer and explanation

   Correct answer: B The magnitude of $1 - s/\omega_z$ grows with frequency exactly like that of $1 + s/\omega_z$, but its phase goes towards $-90^\circ$ instead of $+90^\circ$.

3. A boost converter runs at $D = 0.5$ with $R = 10\,\Omega$ and $L = 100\,\mu\mathrm{H}$. Where is its RHP zero?

   1. $12.5\,\mathrm{krad/s}$
   2. $25\,\mathrm{krad/s}$
   3. $50\,\mathrm{krad/s}$
   4. $100\,\mathrm{krad/s}$
   Answer and explanation

   Correct answer: B The zero is at $(1 - D)^2 R/L$: a quarter of $10\,\Omega$ divided by $100\,\mu\mathrm{H}$.

4. Under which conditions is the RHP zero of a boost converter at its lowest frequency?

   1. Light load and low duty ratio
   2. Heavy load and high duty ratio
   3. Light load and high duty ratio
   4. It does not depend on the operating point
   Answer and explanation

   Correct answer: B The zero is proportional to $(1 - D)^2 R$: a small load resistance (heavy load) and a large duty ratio (low input voltage) both move it down.

5. Why can the RHP zero not be cancelled by the compensator?

   1. Cancelling it would need a pole in the right half-plane, which is unstable
   2. Compensators cannot contain poles
   3. The zero is above the switching frequency
   4. It can be cancelled with a left-half-plane zero
   Answer and explanation

   Correct answer: A Cancellation places a pole at the same location as the zero. For a right-half-plane zero that pole would be unstable, so the crossover frequency has to stay well below the zero instead.

6. With the same $L$ and $C$, how does the double-pole frequency of a CCM boost at $D = 0.5$ compare with that of a buck?

   1. It is the same
   2. It is twice as high
   3. It is half as high
   4. It is four times lower
   Answer and explanation

   Correct answer: C For the boost $\omega_0$ is $(1 - D)/\sqrt{LC}$, the buck value times $1 - D$, because the inductance appears as $L/(1 - D)^2$ in the averaged model.