Small-signal models

Intent

• Explain what a small-signal model is and when it is valid.
• Describe the buck converter’s control-to-output transfer function: dc gain, double pole and quality factor.
• Compute $G_{d0}$, $\omega_0$ and $Q$ from the circuit values and predict the resonant peak.

Theory

Perturb and linearize

The averaged model of a converter is nonlinear, because the duty ratio multiplies voltages and currents. To design a feedback loop we need its response to small variations about the operating point. Each quantity is written as its dc value plus a small ac variation: the duty ratio, for example, as $D$ plus $\hat{d}(t)$. Products of two ac variations are dropped. What remains is a linear circuit, the small-signal model, valid for variations much smaller than the dc values and for frequencies well below the switching frequency [Erickson & Maksimović 2020, Ch. 7].

The buck’s control-to-output transfer function

In the buck, the duty ratio acts directly on the switch-node voltage, which the L-C filter and the load then filter. The control-to-output transfer function $G_{vd}(s)$, from $\hat{d}$ to the output voltage variation, is therefore a dc gain followed by a second-order low-pass response: a pair of poles at the resonant angular frequency $\omega_0$ of the filter, with a quality factor $Q$ set by the load [Erickson & Maksimović 2020, Ch. 8]. The dc gain is the slope of the static characteristic:

$$G_{d0} = V_g$$

$G_{d0}$

dc control-to-output gain

$V_g$

input voltage

Buck 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 $\partial V/\partial D$ of the static characteristic

Source

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

Derivation

step-by-step sympy derivation

Equation id

buck.ss.Gd0 · equations.yaml entry

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

$\omega_0$

LC corner frequency (rad/s)

$C$

capacitance (output capacitor)

$L$

inductance

Buck: 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

buck.ss.w0 · equations.yaml entry

$$Q = R \sqrt{\frac{C}{L}}$$

$Q$

quality factor

$C$

capacitance (output capacitor)

$L$

inductance

$R$

load resistance

Buck: 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

buck.ss.Q · equations.yaml entry

The derivation of all three from the averaged circuit is on the derivations page.

Reading the Bode plot

Well below $\omega_0$ the magnitude is flat at $G_{d0}$ and the phase is near zero. At $\omega_0$ the magnitude is exactly $Q$ times the dc gain and the phase has fallen by 90°; above $\omega_0$ the magnitude falls with the square of frequency and the phase approaches −180° [Erickson & Maksimović 2020, Ch. 8]. A light load (large $R$) gives a high $Q$: a tall, narrow resonant peak and an abrupt phase change, which the feedback loop has to accommodate. For hertz, divide $\omega_0$ by $2\pi$.

Worked example

The buck example from the earlier pages, now read as a small-signal system:

Buck example

Symbol	Value	How it is obtained
$V_g$ input voltage	24 V	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	24 V	Buck control-to-output dc gain, CCM
$\omega_0$ LC corner frequency (rad/s)	31.6 krad/s	Buck: angular frequency of the LC poles
$Q$ quality factor	3.16	Buck: quality factor of the LC poles

The dc gain equals the input voltage, 24 V per unit of duty ratio. The poles sit at 31.6 krad/s, and with $Q$ = 3.16 the magnitude at that frequency is that many times the dc gain.

Try it

▶ Open in the equation explorer: Buck: quality factor of the LC poles
Preset from the example buck-basic

Sweep $R$ over several decades (logarithmic axis): $Q$ grows in proportion to the load resistance, so the resonant peak is tallest at light load.

Bench exercise

With a frequency response analyzer, inject a small sinusoidal variation into the control input of an open-loop buck’s PWM modulator and measure the ratio of output-voltage variation to control-input variation over frequency. Find the resonant peak and compare its frequency with $\omega_0/(2\pi)$. Repeat at a lighter load and confirm that the peak grows while its frequency stays put. Keep the injected amplitude small enough that the response does not change when you halve it.

Gotchas

• Real components damp the resonance. The ideal model has no loss other than the load. Winding resistance and capacitor series resistance lower $Q$, and the capacitor’s series resistance adds a zero: its impedance is $1/(sC)$ plus that resistance.
• The model is only valid well below the switching frequency and for small variations [Erickson & Maksimović 2020, Ch. 7]. Near $f_s$ the averaged model no longer describes the circuit.
• DCM changes the transfer function. In DCM the inductor current no longer carries a state from one period to the next, and the L-C resonance disappears from the low-frequency response [Erickson & Maksimović 2020, Ch. 15].

Go deeper

• Book chapter AC Equivalent Circuit Modeling (Ch. 7) — R. W. Erickson, D. Maksimović · retrieved 2026-09-23
  Averaging, perturbation and linearisation of converter equations, and the small-signal model of the PWM modulator.
• 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. Why must the averaged converter model be linearized before a feedback loop can be designed with transfer functions?

   1. Because the duty ratio multiplies voltages and currents, which makes the averaged model nonlinear
   2. Because the switches are ideal
   3. Because the inductor current is discontinuous
   4. Because the switching frequency is too high
   Answer and explanation

   Correct answer: A Transfer functions describe linear systems. Products such as duty ratio times input voltage are nonlinear; writing each quantity as dc value plus small variation and dropping products of variations gives a linear model around the operating point.

2. What is the control-to-output dc gain $G_{d0}$ of an ideal CCM buck converter with a $24\,\mathrm{V}$ input?

   1. $12\,\mathrm{V}$
   2. $24\,\mathrm{V}$
   3. $48\,\mathrm{V}$
   4. It depends on the load
   Answer and explanation

   Correct answer: B The static characteristic is $D V_g$, whose slope with respect to $D$ is $V_g$. The gain is in volts per unit of duty ratio and does not depend on the load in the ideal model.

3. A buck filter has $L = 100\,\mu\mathrm{H}$ and $C = 100\,\mu\mathrm{F}$. What is $\omega_0$?

   1. $1\,\mathrm{krad/s}$
   2. $10\,\mathrm{krad/s}$
   3. $100\,\mathrm{krad/s}$
   4. $1\,\mathrm{Mrad/s}$
   Answer and explanation

   Correct answer: B The product $L C$ is $10^{-8}\,\mathrm{s}^2$, whose square root is $10^{-4}\,\mathrm{s}$, so $\omega_0$ is $10^{4}\,\mathrm{rad/s}$.

4. The load resistance of a buck converter is doubled. What happens to the $Q$ of its control-to-output poles?

   1. It halves
   2. It is unchanged
   3. It doubles
   4. It quadruples
   Answer and explanation

   Correct answer: C $Q$ is $R\sqrt{C/L}$, proportional to $R$. A lighter load damps the L-C resonance less, so the peak grows.

5. At the resonant angular frequency $\omega_0$, how large is the magnitude of the second-order control-to-output response?

   1. Equal to the dc gain
   2. $Q$ times the dc gain
   3. The dc gain divided by $Q$
   4. Zero
   Answer and explanation

   Correct answer: B At $\omega_0$ the constant term and the $s^2$ term of the denominator cancel, leaving $j/Q$; the magnitude is $Q$ times the dc gain and the phase is $-90^\circ$.

6. For which perturbations is the small-signal model valid?

   1. Any size, at any frequency
   2. Variations much smaller than the dc values, at frequencies well below the switching frequency
   3. Only at the switching frequency
   4. Only in DCM
   Answer and explanation

   Correct answer: B Linearization requires the variations to be small; averaging requires the frequencies of interest to be well below the switching frequency.