Oscilloscope Bandwidth and Rise Time Calculator
Oscilloscope Bandwidth and Rise Time Calculator
Rise time and bandwidth through the 0.35 constant — derived here from the single-pole step response as ln 9 ÷ 2π, with the 0.40–0.45 figures that flatter instruments use — plus the system rise time from scope, probe and signal, the bandwidth a given edge needs, the amplitude error at a frequency, and what a 10× probe’s tip capacitance does to the edge you came to measure.
Bandwidth, rise time and what you will actually see
a 200 MHz scope with a 500 MHz probe, looking at an edge whose true rise time is 2 ns, with 10 pF of probe tip capacitance on a 50 Ω node
Where 0.35 comes from, and what it assumes
so tr · BW = ln 9 ÷ 2π = 0.34970
tmeasured = √(tsignal² + tscope² + tprobe²) (an approximation)
Amplitude shown at f: 1 ÷ √(1 + (f/BW)²) (70.7% at BW itself)
Probe loading: tr = ln 9 · Rsource · Ctip
- K
- the rise-time constant. 0.3497 for one pole, 0.34 for a true Gaussian, 0.40–0.45 for a flat-response instrument
- ln 9
- 2.1972 — because 10% to 90% of an exponential is a factor of nine in what is left to go
- t probe
- the probe’s own rise time. Often the slower of the two, and then it is what limits the measurement
- C tip
- the probe’s input capacitance. It loads the circuit, which is a different problem from bandwidth and is not fixed by a faster scope
Worked example
a 200 MHz scope with a 500 MHz probe, looking at an edge whose true rise time is 2 ns, with 10 pF of probe tip capacitance on a 50 Ω node
The scope's own rise time is 0.3497 ÷ 200 MHz = 1.748 ns; the probe's is 0.3497 ÷ 500 MHz = 699.4 ps
Together they are √(ts² + tp²) = 1.883 ns, which is a system bandwidth of 185.7 MHz — less than either instrument on its own
On a 2 ns edge you would therefore measure √(2² + 1.7485² + 0.6994²) = 2.747 ns — 37.4% more than the real edge
To keep the error under a few per cent you want the scope's own rise time well under the signal's: the signal implies 174.8 MHz, so the 3× rule asks for 524.5 MHz and the 5× rule for 874.2 MHz — leaving 5.4% and 2.0% of error respectively
At 100 MHz this scope shows 89.4% of the real amplitude, which is -0.97 dB
And before any of that: 10 pF on a 50 Ω node is a time constant of 500 ps, so the probe alone slows the edge at the node to 2.282 ns — 14.1% slower than it was before you touched it
The rise-time constant is a property of the instrument
| Response | t_r × BW | Where it comes from | Typical instrument |
|---|---|---|---|
| Single pole (RC) | 0.3497 | ln 9 ÷ 2π, exactly | most scopes up to about 1 GHz |
| True Gaussian | 0.3396 | the 10–90% width of an erf step | an idealisation, rarely exact |
| Flatter than Gaussian | 0.40 | measured, not derived | transitional designs |
| Maximally flat / brick wall | 0.45 | measured, not derived | many scopes above 1 GHz |
How much bandwidth an edge really needs
| Scope BW ÷ signal BW | Error in the displayed rise time | Amplitude error at the signal’s knee |
|---|---|---|
| 1× | 41.4% | large |
| 2× | 11.8% | noticeable |
| 3× | 5.4% | small |
| 5× | 2.0% | negligible |
| 10× | 0.5% | negligible |
One pole, one constant, and three things that add in quadrature
Drive a single-pole low-pass filter with a step and the output is 1 − e−2πf₃t. The 10% point is at ln(1/0.9)/ω and the 90% point at ln(1/0.1)/ω, so the 10–90% rise time is ln(9)/(2πf₃) — and ln 9 ÷ 2π is 0.34970. That is the whole of the famous 0.35. It is not measured, it is not a rule of thumb, and it is not, strictly, Gaussian: a genuinely Gaussian response gives 0.3396, computed on this page from the error-function step rather than quoted. What matters in practice is that slow scopes roll off gently and behave like one pole, so 0.35 fits them, while fast ones use sharper filtering that trades a flatter passband for overshoot and run 0.40 to 0.45. Their data sheets say which, and dividing the quoted rise time into the quoted bandwidth gives you the instrument’s own constant.
Everything in the path adds. The signal has a rise time, the probe has one, the scope has one, and what you see is roughly the root sum of the squares. Roughly: the rule is exact only for Gaussian channels, and this page’s check against the exact step response of cascaded poles finds it up to 7.5% low when two of the contributions are similar. It is still the right tool, because the useful direction is backwards — measure the displayed edge, subtract the instrument in quadrature, and what is left is the signal. The same sum explains why a fast scope with a slow probe is a slow measurement: bandwidths combine as 1/BW² and the slower part dominates.
Three times or five times? Both are choices about how much error you will accept, and neither is a law. At 3× the scope adds about 5% to the displayed edge; at 5×, about 2%; at 1× it adds 41% and you are photographing the instrument. If you are looking for a glitch, 3× is plenty. If you are reporting a rise time in a compliance document, 5× and subtract the instrument anyway.
The probe changes the circuit. Bandwidth is about what the instrument can show; loading is about what the instrument does to the thing being shown, and no amount of scope fixes it. A 10× passive probe puts its tip capacitance — 9 to 13 pF for an ordinary one, and remember the ground lead adds inductance on top — straight across the node. Into a source impedance R that is a single pole, so it imposes a rise time of ln(9)·R·C on the node itself. Ten picofarads on a 50 Ω source is 1.099 ns, which is invisible on a slow edge and ruinous on a fast one; the same probe on a 5 kΩ node imposes 109.9 ns and there is nothing left to measure. This is why good 10× probes advertise 3–4 pF and why active probes, which get under 1 pF, exist at all — and why a 1× probe, at 40 to 110 pF, should never go near a fast edge.
What this page leaves out: overshoot and ringing, which a flat-response scope has and a single-pole model cannot represent; the sample rate, which has to be several times the bandwidth before the bandwidth means anything; and the ground lead, whose inductance with the tip capacitance rings at a few hundred megahertz and produces the classic decaying wobble on a fast edge. For the same rise-time thinking applied to a PCB trace — how long a track can be before its delay matters against an edge — see the propagation delay calculator; for sampling see the ADC resolution calculator and the anti-aliasing filter calculator.
Frequently asked questions
Why is the oscilloscope rise time constant 0.35?
Because a single-pole low-pass filter’s step response reaches 10% and 90% ln(9)/(2πf₃) apart, and ln 9 ÷ 2π = 0.34970. It is an exact consequence of that response shape, not an empirical constant. Scopes with a flatter, more brick-wall response do not have that shape and use 0.40 to 0.45 instead.
How much oscilloscope bandwidth do I need?
It depends on the error you will accept. Three times the signal’s own bandwidth (K ÷ its rise time) leaves about 5% on the displayed rise time; five times leaves about 2%. The 5× rule of thumb is that choice, not a physical limit — and the probe has to be counted in the same sum.
How do I find the signal’s true rise time from what the scope shows?
Subtract in quadrature: t_signal = √(t_measured² − t_scope² − t_probe²). It breaks down when the instrument is the larger term — if the measured and system rise times are close, the answer is a small difference between two large numbers and means very little.
Does the probe’s bandwidth matter as much as the scope’s?
More often than not it matters more, because probes are frequently the slower of the two. The pair combines as 1/BW_system² = 1/BW_scope² + 1/BW_probe², so a 200 MHz probe on a 1 GHz scope gives a 196 MHz system and the scope’s specification stops mattering.
What does a 10× probe’s capacitance do to a fast edge?
It loads the node. The tip capacitance across the source impedance is a single pole, so it imposes a rise time of ln(9)·R·C on the signal itself — 10 pF on a 50 Ω source is about 1.1 ns, and on a 5 kΩ node about 110 ns. That is a change to the circuit, not a limitation of the display, and a faster scope does not help.
What amplitude does a scope show at its own bandwidth?
70.7% of the real thing — that is what −3 dB means. For 1% accuracy on a sine you need the frequency to be about a seventh of the bandwidth; for 10% accuracy, about half of it. This page prints both frequencies for your scope.
Related calculators
References
- Tektronix, Understanding Oscilloscope Bandwidth, Rise Time and Signal Fidelity, technical brief 55W-18024. Gives the 0.35 factor as a “simple one-pole model for 10–90% rise time” based on an RC low-pass, states that a maximum-flat-envelope-delay response can approach 0.45, and gives the 5× bandwidth rule of thumb this page presents as a choice.
- Tektronix, How Oscilloscope Probes Affect Your Measurement, application note 51W-30013. 10 MΩ input resistance for a standard 10× passive probe, tip capacitance from 9.5 pF on an older design down to 3.9 pF on a newer one, and the measured comparison showing the higher-capacitance probe producing “significantly slower” rise times and a visibly degraded edge.
- Johnson H, Graham M. High-Speed Digital Design: A Handbook of Black Magic. Prentice Hall, 1993. The knee frequency F_knee = 0.5/T_r used here as the frequency above which a digital edge has no significant energy, and the treatment of probe loading as a circuit change rather than an instrument limitation.
- Smith S. Relating wideband DSO rise time to bandwidth: Lose the 0.35! EDN, 2013. The case for not applying the single-pole constant to instruments whose response is not single-pole — the reason this page makes the constant a choice rather than a fixed 0.35.