Control basics

Intent

• Model a pulse-width modulator (PWM) and compute its small-signal gain.
• Compute the loop gain of a voltage-mode feedback loop at a given frequency.
• Quantify how feedback suppresses disturbances, and recognise where it stops helping.

Theory

The PWM modulator

A voltage-mode controller compares a control voltage $V_\mathrm{ctrl}$ with a sawtooth ramp of peak amplitude $V_M$. The switch is on while the ramp is below the control voltage, so the duty ratio is proportional to the control voltage:

$$D = \frac{V_\mathrm{ctrl}}{V_M}$$

$D$

duty ratio

$V_M$

PWM ramp amplitude

$V_\mathrm{ctrl}$

PWM control voltage

Duty ratio set by a PWM comparator✓ verified by 7 tests

Assumptions, convention and source

Assumptions

• trailing-edge PWM with a ramp from 0 to V_M

Notes

so the small-signal modulator gain is $1/V_M$

Source

Erickson & Maksimović 2020, Ch. 7 (AC Equivalent Circuit Modeling)

Derivation

step-by-step sympy derivation

Equation id

pwm.d · equations.yaml entry

The modulator’s small-signal gain, from control voltage to duty ratio, is therefore $1/V_M$ [Erickson & Maksimović 2020, Ch. 7].

The loop gain

In a voltage-mode loop the output voltage is measured by a sensor of gain $H$ (often a resistive divider) and compared with a reference. The difference is amplified by the compensator $G_c$ and turned into a duty ratio by the modulator, and the converter’s control-to-output transfer function $G_{vd}$ closes the loop. The loop gain is the product of the gains around the loop [Erickson & Maksimović 2020, Ch. 9]:

$$T = \frac{G_c G_{vd} H}{V_M}$$

$T$

loop gain

$G_c$

compensator gain

$G_{vd}$

control-to-output gain

$H$

sensor (divider) gain

$V_M$

PWM ramp amplitude

Loop gain of a voltage-mode feedback loop✓ verified by 7 tests

Assumptions, convention and source

Assumptions

• small-signal (linearised) model around a CCM operating point

Notes

magnitudes at one frequency; the phases of the factors add

Source

Erickson & Maksimović 2020, Ch. 9 (Controller Design)

Derivation

step-by-step sympy derivation

Equation id

loop.T · equations.yaml entry

At each frequency the magnitudes multiply and the phases add. For the buck, $G_{vd}$ is the transfer function of the small-signal page; for the boost and buck-boost it includes the right-half-plane zero.

What feedback buys

A disturbance that enters the loop, from the input voltage or from the load, reaches the output divided by $1 + T$ [Erickson & Maksimović 2020, Ch. 9]:

$$S = \frac{1}{1 + T}$$

$S$

share of a disturbance that remains

$T$

loop gain

Disturbance suppression by feedback✓ verified by 7 tests

Assumptions, convention and source

Assumptions

• small-signal (linearised) model around a CCM operating point
• loop gain treated as a real magnitude (valid well below crossover)

Notes

at frequencies where $T \gg 1$ a disturbance is divided by roughly $T$

Source

Erickson & Maksimović 2020, Ch. 9 (Controller Design)

Derivation

step-by-step sympy derivation

Equation id

loop.suppression · equations.yaml entry

Where the loop gain is large (at low frequencies), disturbances are suppressed by roughly the loop gain, and the output follows the reference divided by $H$: it depends on the sensor and the reference, not on the converter’s gain. Where the loop gain is much smaller than 1 (above the crossover frequency), feedback has no effect. Near the crossover the loop gain is complex; with a small phase margin the magnitude of $1 + T$ drops below 1, and disturbances there are amplified rather than suppressed [Erickson & Maksimović 2020, Ch. 9].

Worked example

The voltage-mode buck loop example, evaluated at dc:

Voltage-mode loop example

Symbol	Value	How it is obtained
$V_g$ input voltage	24 V	given
$V_M$ PWM ramp amplitude	2 V	given
$V_\mathrm{ctrl}$ PWM control voltage	1 V	given
$G_c$ compensator gain	10	given
$H$ sensor (divider) gain	0.2	given
$D$ duty ratio	0.5	Duty ratio set by a PWM comparator
$G_{vd}$ control-to-output gain	24 V	Buck control-to-output dc gain, CCM
$T$ loop gain	24	Loop gain of a voltage-mode feedback loop
$S$ share of a disturbance that remains	0.04	Disturbance suppression by feedback

The modulator turns the control voltage into a duty ratio of 0.5. At dc the loop gain is 24, so a slow disturbance reaches the output scaled by 0.04 relative to its effect without feedback.

Try it

▶ Open in the equation explorer: Disturbance suppression by feedback
Preset from the example control-basic

Sweep the loop gain over several decades (logarithmic axis): each tenfold increase of a large loop gain buys another tenfold suppression.

Bench exercise

Measure the loop gain of a closed-loop converter by injection. Insert a small resistor in the feedback path, between the output and the sensing divider, and inject a small sinusoidal signal across it through an isolation transformer. The ratio of the signals on the two sides of the resistor, measured with a frequency response analyzer, is the loop gain [Erickson & Maksimović 2020, Ch. 9]. Find the crossover frequency and the phase margin, then compare how the output responds to an input-voltage step with the loop’s suppression at low frequency.

Gotchas

• The modulator gain depends on the ramp amplitude. A controller whose ramp amplitude scales with the input voltage makes the buck’s loop gain independent of the input voltage, since the buck’s dc gain equals the input voltage (Buck control-to-output dc gain, CCM).
• Use the complex loop gain away from dc. The real-valued form above is exact only where $T$ is real. Elsewhere, divide by the magnitude of the complex sum $1 + T$; where the loop gain is large, the two agree closely.
• A saturated modulator opens the loop. Once the control voltage leaves the ramp’s range, the duty ratio is clamped and the loop gain collapses; the equations on this page no longer apply.

Go deeper

• Book chapter Controller Design (Ch. 9) — R. W. Erickson, D. Maksimović · retrieved 2026-09-23
  Loop gain, crossover frequency, phase margin and the design of PD/PI compensators.
• 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.
• 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. A PWM modulator uses a sawtooth of peak amplitude $2\,\mathrm{V}$. What duty ratio does a control voltage of $0.6\,\mathrm{V}$ produce?

   1. 0.12
   2. 0.3
   3. 0.6
   4. 3.3
   Answer and explanation

   Correct answer: B The switch is on while the ramp is below the control voltage, so the duty ratio is the control voltage divided by the ramp amplitude.

2. A voltage-mode loop has $G_c = 5$, $G_{vd} = 12\,\mathrm{V}$, $H = 0.25$ and $V_M = 1.5\,\mathrm{V}$ at some frequency. What is the loop-gain magnitude there?

   1. 2.5
   2. 10
   3. 40
   4. 60
   Answer and explanation

   Correct answer: B The magnitudes multiply around the loop, and the modulator contributes $1/V_M$: five times 12 times 0.25, divided by 1.5.

3. At dc a loop gain of 19 is measured. What fraction of its effect without feedback does a slow input-voltage disturbance still have at the output?

   1. 19
   2. 20
   3. 1/19
   4. 1/20
   Answer and explanation

   Correct answer: D The disturbance is divided by $1 + T$, that is by 20; it reaches the output at a twentieth of its effect without feedback.

4. How much does feedback suppress a disturbance at frequencies well above the crossover frequency?

   1. Almost not at all, because the loop gain is much smaller than 1
   2. By the dc loop gain
   3. By the phase margin
   4. Completely
   Answer and explanation

   Correct answer: A Where the loop gain magnitude is much smaller than 1, $1 + T$ is close to 1, so the disturbance passes as if there were no feedback.

5. A loop has a small phase margin. What happens to disturbances near the crossover frequency?

   1. They are suppressed more strongly
   2. They are unaffected
   3. They can be amplified, because the magnitude of $1 + T$ drops below 1
   4. They are shifted to the switching frequency
   Answer and explanation

   Correct answer: C Near crossover the loop-gain magnitude is about 1, and with a phase close to $-180^\circ$ the complex sum $1 + T$ becomes small, so dividing by it amplifies the disturbance.

6. Why does scaling the PWM ramp amplitude with the input voltage keep the loop gain of a buck constant over the input range?

   1. Because it keeps the duty ratio constant
   2. Because the buck control-to-output gain is proportional to the input voltage, and the ramp amplitude divides it
   3. Because it changes the switching frequency
   4. Because it removes the right-half-plane zero
   Answer and explanation

   Correct answer: B The buck dc gain $G_{d0}$ equals $V_g$, and the modulator contributes $1/V_M$. If $V_M$ is proportional to $V_g$, their product, and with it the loop gain, no longer depends on $V_g$.