Probes and measurement bandwidth

Intent

• Know what a probe’s ground lead adds to a measurement of a fast edge.
• Relate a measurement system’s bandwidth to the fastest rise it can show.
• Measure a converter’s output ripple with a short ground connection.

Theory

The ground lead

A probe measures between its tip and its ground connection, and the leads that reach them form a loop. That loop introduces parasitic inductance, which inaccurately increases the amplitude of switching transients [Analog Devices, AN-1144]. With the probe’s input capacitance it forms a series L-C that every fast edge excites, ringing at its resonant frequency: a ringing of the probe, not of the circuit.

$$f_\mathrm{probe} = \frac{1}{2 \pi \sqrt{L_\mathrm{gnd} C_\mathrm{probe}}}$$

$f_\mathrm{probe}$

probe ringing frequency

$C_\mathrm{probe}$

probe input capacitance

$L_\mathrm{gnd}$

probe ground-loop inductance

Ringing frequency of a probe's ground loop✓ verified by 7 tests

Assumptions, convention and source

Assumptions

• undamped L-C resonance

Notes

the loop's inductance and the probe's input capacitance form a series L-C that every fast edge excites: a ringing of the probe, not of the circuit. A shorter ground makes the inductance smaller and the ring faster

Source

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

Derivation

step-by-step sympy derivation

Equation id

probe.f_ring · equations.yaml entry

A shorter ground makes the loop smaller: the tip-and-barrel method removes the long ground lead and connects the probe’s barrel to the ground reference directly [Analog Devices, AN-1144].

Bandwidth and rise time

A probe and an oscilloscope together respond to a step no faster than their bandwidth allows. For a single-pole response, the 10-90 % rise time follows from the bandwidth:

$$t_\mathrm{rise} = \frac{\ln{\left(9 \right)}}{2 \pi f_\mathrm{BW}}$$

$t_\mathrm{rise}$

10-90 % rise time

$f_\mathrm{BW}$

-3 dB bandwidth

Rise time of a single-pole system✓ verified by 7 tests

Assumptions, convention and source

Assumptions

• single-pole (first-order) response to a step

Notes

ln 9/(2π) = 0.350, so a 100 MHz single-pole system shows an ideal step rising in 3.5 ns; a response of another shape has another constant

Source

Alexander & Sadiku 2017, Ch. 7 (First-Order Circuits) and Ch. 14 (Frequency Response); derived here

Derivation

step-by-step sympy derivation

Equation id

probe.t_rise · equations.yaml entry

Historically, oscilloscopes’ responses tended to follow this rule approximately [Tektronix, How is bandwidth related to rise time for oscilloscopes?]; the constant depends on the shape of the response, and a flat-response oscilloscope measures rise times more accurately than a Gaussian-response one of the same bandwidth [Keysight Technologies (Agilent Technologies), Understanding Oscilloscope Frequency Response and Its Effect on Rise-Time Accuracy]. An ideal step, infinitely fast, is shown rising in the system’s own rise time: to see an edge’s own rise time, the system’s must be well below it.

Output ripple

A converter’s output ripple is small next to the output voltage, and the switching transients that ride on it are fast: the loop of a long ground lead exaggerates them [Analog Devices, AN-1144], and they look like ripple. Measure it with the tip-and-barrel method at the output capacitor [Analog Devices, AN-1144]. A bandwidth limit on the oscilloscope’s channel attenuates what lies above it (first-order low-pass): set it well above the ripple’s own frequencies, or it attenuates the ripple as well.

Worked example

A measurement system and a probe with a long ground lead:

Probe example

Symbol	Value	How it is obtained
$f_\mathrm{BW}$ -3 dB bandwidth	100 MHz	given
$L_\mathrm{gnd}$ probe ground-loop inductance	100 nH	given
$C_\mathrm{probe}$ probe input capacitance	10 pF	given
$t_\mathrm{rise}$ 10-90 % rise time	3.5 ns	Rise time of a single-pole system
$f_\mathrm{probe}$ probe ringing frequency	159 MHz	Ringing frequency of a probe's ground loop

The system shows an ideal step rising in 3.5 ns, and the ground loop’s 100 nH with the probe’s 10 pF rings at 159 MHz.

Try it

▶ Open in the equation explorer: Rise time of a single-pole system
Preset from the example bench-probe

Sweep the bandwidth: the rise time falls in inverse proportion.

▶ Open in the equation explorer: Ringing frequency of a probe's ground loop
Preset from the example bench-probe

Sweep the ground loop’s inductance: a shorter ground rings faster, and further above the edges you measure.

Bench exercise

Measure the same switching edge twice: with the probe’s long ground lead, and with a ground spring or the tip-and-barrel method. Compare the overshoot and the ringing’s frequency, and estimate the lead’s inductance from that frequency and the probe’s input capacitance on its data sheet. Then measure the output ripple both ways, with the channel’s bandwidth limit on and off.

Gotchas

• A ringing that moves with the ground lead is the probe’s, not the circuit’s: shorten the ground and it changes [Analog Devices, AN-1144].
• An edge much faster than the system is shown with about the system’s rise time (rise time).
• A bandwidth limit near the ripple’s frequencies attenuates the ripple itself (first-order low-pass).

Go deeper

• Application note AN-1144: Measuring Output Ripple and Switching Transients in Switching Regulators — Analog Devices · retrieved 2026-09-24
  How a probe's ground lead exaggerates switching transients, and the tip-and-barrel method for output ripple.
• Application note Understanding Oscilloscope Frequency Response and Its Effect on Rise-Time Accuracy (5988-8008) — Keysight Technologies · retrieved 2026-09-25
  Why the rise time an oscilloscope shows depends on the shape of its frequency response, not on its bandwidth alone.
• Technical article How is bandwidth related to rise time for oscilloscopes? (Tektronix FAQ) — Tektronix · retrieved 2026-09-24
  The 0.35 rule between an oscilloscope's bandwidth and its rise time, and where it holds.

Quiz

1. What does the loop of a probe's long ground lead do to a measured switching edge?

   1. It adds inductance that makes the transients look larger than they are
   2. It filters the edge, so the transients look smaller
   3. Nothing: the ground lead carries no current
   4. It shifts the edge in time only
   Answer and explanation

   Correct answer: A The leads form a loop whose inductance exaggerates the transients and, with the probe's capacitance, rings at every edge.

2. A ground lead of 50 nH rings with a probe capacitance of 20 pF. Where does the ringing move if the lead is made four times shorter (a quarter of the inductance)?

   1. Up by a factor of four
   2. Up by a factor of two
   3. Down by a factor of two
   4. Nowhere: the capacitance sets it
   Answer and explanation

   Correct answer: B The resonant frequency goes with one over the square root of the inductance: a quarter of the inductance doubles it.

3. A single-pole measurement system has a bandwidth of 350 MHz. How fast does it show an ideal step rising?

   1. About 0.35 ns
   2. About 3.5 ns
   3. Instantly
   4. About 1 ns
   Answer and explanation

   Correct answer: D The rise time is about 0.35 over the bandwidth: 0.35 over 350 MHz.

4. Why can two oscilloscopes of the same bandwidth show different rise times for the same edge?

   1. The constant between bandwidth and rise time depends on the shape of each one's frequency response
   2. Their probes have different colours
   3. Rise time does not depend on the oscilloscope
   4. One of them must be broken
   Answer and explanation

   Correct answer: A A flat response and a Gaussian one of equal bandwidth measure rise times differently.

5. How should an output ripple be measured?

   1. With the longest ground lead available, to reach far
   2. Across the inductor
   3. At the output capacitor, with the tip-and-barrel method or a very short ground
   4. With the probe set to measure current
   Answer and explanation

   Correct answer: C A short ground keeps the probe's loop, and the transients it exaggerates, small.

6. A channel's bandwidth limit is set close to the ripple's frequency. What happens?

   1. The ripple reads higher
   2. It attenuates the ripple itself, which reads low
   3. Nothing: the limit only removes noise
   4. The oscilloscope stops triggering
   Answer and explanation

   Correct answer: B A low-pass filter attenuates everything near and above its corner, the ripple included.