PCB layout for switching converters

Intent

Find the loop in a converter whose current switches on and off at every edge.
Estimate the voltage that loop’s inductance develops over an edge.
Place the input capacitor, the ground layer, the switch node and the feedback trace accordingly.

Theory

The hot loop

In a buck converter the switch draws a pulse of the inductor’s current from the input capacitor while it is on; the source, behind its wiring, supplies about the average and recharges the capacitor while the switch is off. At every edge the currents of the switch, of the diode (or the second switch) and of the input capacitor step by the inductor’s current, while the inductor’s current flows on [Erickson & Maksimović 2020, Ch. 2]: the loop of those three is the one whose current switches. Every loop has inductance, and over an edge that changes the current by $\Delta I$ in $t_\mathrm{edge}$ it develops a voltage (inductor law):

V_\mathrm{spike} = \frac{L_\mathrm{loop} \Delta I}{t_\mathrm{edge}}

$V_\mathrm{spike}$: voltage across the loop inductance
$\Delta I$: current change over the edge
$L_\mathrm{loop}$: loop inductance
$t_\mathrm{edge}$: edge duration

Voltage across a loop's inductance during a switching edge✓ verified by 8 tests

Assumptions, convention and source

Assumptions

the current changes linearly over the edge

Notes

for the same edge the voltage is proportional to the loop's inductance, and a faster edge raises it in proportion: 10 nH carrying 10 A that stops in 10 ns develops 10 V

Source

Alexander & Sadiku 2017, Ch. 6 (Capacitors and Inductors); derived here

Derivation

step-by-step sympy derivation

Equation id

layout.v_spike · equations.yaml entry

By Kirchhoff’s voltage law that voltage adds to the input voltage across the switch while its current falls [Alexander & Sadiku 2017, Ch. 2 (Basic Laws)]: the overshoot at every turn-off edge. For the same edge it is proportional to the loop’s inductance, and a shorter, tighter loop has less. The input capacitor closes that loop, so it goes right at the switches: TI calls it the single most important component for the reliable operation of a step-down converter [Texas Instruments 2015].

After the turn-off edge the loop’s inductance rings with the switch node’s capacitance, and the ring’s frequency gives the inductance without a current probe in the loop:

L_\mathrm{loop} = \frac{1}{4 \pi^{2} f_\mathrm{ring}^{2} C_\mathrm{node}}

$L_\mathrm{loop}$: loop inductance
$C_\mathrm{node}$: switch-node capacitance
$f_\mathrm{ring}$: ringing frequency

Loop inductance from the ring after a switching edge✓ verified by 7 tests

Assumptions, convention and source

Assumptions

undamped L-C resonance

Notes

after the turn-off edge the loop's inductance rings with the switch node's capacitance; the ring's frequency, read on an oscilloscope, gives the inductance without a current probe in the loop. The ring counts every capacitance at the node, the probe's included: leaving one out overstates the inductance

Source

Alexander & Sadiku 2017, Ch. 14 (Frequency Response: series resonance); derived here

Derivation

step-by-step sympy derivation

Equation id

layout.L_ring · equations.yaml entry

The ground layer and the switch node

A ground layer directly next to the outside layer that carries the power stage belongs in every such board [Analog Devices (Linear Technology), AN-136]. The switch node swings between the input voltage and ground at every edge, and the stray capacitance between its copper and a neighbouring trace carries a current in proportion to that swing’s rate (capacitor law). But the node also carries the inductor’s current and draws heat from the switches, so its area cannot be too small either: AN-136 treats it as a compromise [Analog Devices (Linear Technology), AN-136].

The feedback trace

The controller regulates what the feedback trace brings it: a voltage picked up on the way is regulated into the output as if it were the output’s own. Route the trace as far from the inductor and the noisy power traces as possible [Texas Instruments 2013], and take it from the point where the output voltage matters, with its own return (a Kelvin connection, see current sensing).

Worked example

Two loops with the same edge:

Layout example

Symbol	Value	How it is obtained
$L_\mathrm{loop}$ loop inductance	10 nH	given
${L_\mathrm{loop}}_{\mathrm{tight}}$ loop inductance	2 nH	given
$\Delta I$ current change over the edge	10 A	given
$t_\mathrm{edge}$ edge duration	10 ns	given
$V_\mathrm{spike}$ voltage across the loop inductance	10 V	Voltage across a loop's inductance during a switching edge
${V_\mathrm{spike}}_{\mathrm{tight}}$ voltage across the loop inductance	2 V	Voltage across a loop's inductance during a switching edge

The same current step develops 10 V in 10 nH and only 2 V in the tighter loop of 2 nH.

Try it

▶ Open in the equation explorer: Voltage across a loop's inductance during a switching edge
Preset from the example bench-layout

Sweep the loop’s inductance: the voltage follows it in proportion. Halve $t_\mathrm{edge}$ , as a faster switch would, and the voltage doubles.

Bench exercise

On a converter board, trace the hot loop: the input capacitor, the switch and the diode or second switch. Measure the overshoot at turn-off across the low-side switch or diode, whose one end is at ground, with the shortest ground connection your probe allows (see probes); a switch whose terminals both move needs a differential probe. A current probe needs a wire loop added to the hot loop, which adds inductance of its own, so estimate the loop’s inductance from the ring that follows the edge instead, with the whole capacitance at the switch node: the switch’s and the diode’s output capacitances from their data sheets, plus the probe’s input capacitance and the board’s. The ring measured again with a known capacitor added gives all of it at once (as in snubbers and clamps). With the edge’s current step and fall time, compare the overshoot with the loop-voltage equation. Then add a small ceramic capacitor right across the switch pair, making the loop shorter, and measure again.

Gotchas

An input capacitor placed far away makes the loop that switches its current large, and the overshoot with it (loop voltage).
A switch node with too little copper runs hot, and one with too much couples its swing into its neighbours: its area is a compromise [Analog Devices (Linear Technology), AN-136].
A long probe ground lead adds its own loop and makes the overshoot look larger than it is [Analog Devices, AN-1144] (see probes).

Go deeper

Application note AN-136: PCB Layout Considerations for Non-Isolated Switching Power Supplies — Analog Devices (Linear Technology) · retrieved 2026-09-24
The hot loop, the layer stack and the switch node of a buck converter's board, step by step.
Technical article Five steps to a great PCB layout for a step-down converter (SLYT614) — Texas Instruments · retrieved 2026-09-24
Five steps to a step-down converter's board, with the input capacitor as its most important component.
Application note AN-1149 Layout Guidelines for Switching Power Supplies (SNVA021C) — Texas Instruments · retrieved 2026-09-24
Current loops, ground planes, the feedback trace and the input capacitor, in a short note.

Quiz

In a buck converter, which loop's current switches at every edge?
1. The inductor and the output capacitor
2. The output capacitor and the load
3. The input capacitor, the switch and the diode (or second switch)
4. The feedback divider
Answer and explanation
Correct answer: C At every edge the currents of the switch, the diode and the input capacitor step by the inductor's current, while the inductor's current flows on: the loop of those three is the one whose current switches.
A current of 10 A falls to zero in 10 ns in a loop of 5 nH. What voltage does the loop develop?
1. 5 V
2. 0.5 V
3. 50 V
4. 500 mV per nanosecond, whatever the loop
Answer and explanation
Correct answer: A The loop inductance times the change of current over the edge time: 5 nH times 10 A over 10 ns.
The switch becomes twice as fast. What happens to the voltage across the same loop?
1. It halves
2. It stays the same
3. It quadruples
4. It doubles
Answer and explanation
Correct answer: D The voltage is the loop inductance times the current step over the edge time; halving the edge time doubles it.
Where does the voltage across the loop inductance appear at turn-off?
1. Across the load only
2. Across the switch, added to the input voltage
3. Across the output capacitor
4. Nowhere: an inductance stores energy but develops no voltage
Answer and explanation
Correct answer: B Kirchhoff's voltage law around the loop: while the current falls, the inductance's voltage adds to the input voltage across the switch.
Which component must sit closest to the switches of a step-down converter?
1. The output capacitor
2. The feedback divider
3. The input capacitor
4. The output connector
Answer and explanation
Correct answer: C It closes the loop whose current switches; the shorter that loop, the smaller its inductance and the voltage each edge develops across it.
Why is the switch node's copper area a compromise?
1. It carries current and cools the switches, but its voltage swing couples into its neighbours
2. Copper is expensive
3. A larger area lowers the switching frequency
4. The switch node carries no current
Answer and explanation
Correct answer: A The node needs copper for current and heat, and its swing drives current through its stray capacitance to nearby traces.
How should the feedback trace be routed?
1. Next to the switch node, to keep it short
2. As far as possible from the inductor and the noisy power traces
3. Under the inductor
4. It does not matter: the controller filters it
Answer and explanation
Correct answer: B The controller regulates what the trace brings it; what the trace picks up on the way ends up in the output.

The numbers in the examples, presets and exercises on this site are synthetic: round teaching values, not data from any real design.