Signal

Rise Time to Bandwidth Calculator

Convert digital rise time to signal bandwidth. Get the knee frequency, 3 dB bandwidth, IPC-2251 spectral bandwidth, and the highest significant harmonic of your clock.

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Formula

f_{knee} = \frac{0.5}{t_r},\qquad f_{3dB} = \frac{\ln 9}{2\pi t_r} \approx \frac{0.35}{t_r}

Reference: H. Johnson & M. Graham, "High-Speed Digital Design: A Handbook of Black Magic", Prentice Hall 1993, §1.3; IPC-2251

t_r— Rise time, 10–90% (s)

f_knee— Knee frequency (Hz)

f_3dB— 3 dB bandwidth (Hz)

f_clk— Clock frequency (Hz)

How It Works

Digital designers talk in nanoseconds; interconnect designers talk in gigahertz. The bridge between them is the rise time, and there are two different constants for crossing it — used for two different purposes, and routinely confused.

The 0.35 constant is exact

For a single-pole response, the step output crosses 10% at $t = -\tau\ln(0.9)$ and 90% at $t = -\tau\ln(0.1)$ . Subtracting gives $t_r = \tau\ln 9$ . With the 3 dB bandwidth at $f_{3dB} = 1/(2\pi\tau)$ , the product falls out:

f_{3dB}\, t_r = \frac{\ln 9}{2\pi} = 0.34965\ldots

So 0.35 is not a rule of thumb — it is $\ln 9 / 2\pi$ rounded to two figures. It answers the question what analogue bandwidth does my amplifier or scope need to reproduce this edge?

The knee frequency is a design bound

Johnson and Graham's knee frequency answers a different question: how far up in frequency does my interconnect have to keep behaving?

f_{knee} = \frac{0.5}{t_r}

Above the knee, a digital edge carries so little energy that channel behaviour there stops mattering. It is deliberately conservative, and it is the number you should use when deciding how far to trust a via model, a connector S-parameter file, or a trace loss estimate.

The two differ by roughly 43%, which is why quoting the wrong one leads to either an over-specified amplifier or an under-modelled channel.

Rise time conventions

Datasheets quote either 10–90% or 20–80% transitions, and the two are not interchangeable. For a single pole, the ratio is $\ln 9/\ln 4 = 1.585$ — a 20–80% figure must be multiplied by 1.585 before feeding it into either bandwidth relation. Modern SerDes datasheets increasingly use 20–80% because it avoids the noisy tails of the waveform, so check which convention you are reading.

What sets the rise time

It is the driver, not the clock. A 25 MHz oscillator with a modern CMOS output stage can have a 500 ps edge, which puts its knee at 1 GHz — the interconnect must behave to 1 GHz even though the fundamental is 25 MHz. This is the single most common misjudgement in board design: slow clock, fast edges, and a channel modelled only to the fundamental.

Worked Example

Problem: A 100 MHz clock is driven by a buffer specified at 500 ps (20-80%). Determine the channel bandwidth requirement and how many harmonics matter.

Step 1 - Convert the rise time convention. The datasheet gives 20-80%, so scale to the 10-90% equivalent: t_r(10-90) = 500 ps ln(9)/ln(4) = 500 1.5849 = 792 ps

Step 2 - Knee frequency: f_knee = 0.5 / 792 ps = 631 MHz The interconnect must behave to 631 MHz.

Step 3 - 3 dB bandwidth: f_3dB = 0.34965 / 792 ps = 441 MHz A receiver or scope front end below 441 MHz will visibly slow the edge.

Step 4 - Harmonic content relative to the clock: n = 631 MHz / 100 MHz = 6.3 Harmonics through the 5th (500 MHz) carry meaningful energy; the 7th (700 MHz) is already above the knee.

Step 5 - Design consequences. Model the via, connector, and trace loss to at least 631 MHz, not to 100 MHz. Choose a scope with at least 3x the 3 dB bandwidth, so roughly 1.5 GHz, if you want to measure the edge without the instrument dominating the result.

Step 6 - Sanity check against the bit period. The clock period is 10 ns and the edge is 0.79 ns, under 8% of the period, so there is no intersymbol interference concern here. Had the edge exceeded about 40% of the period, ISI would become the limiting factor.

Practical Tips

✓Always design to the fastest edge the driver can produce, not the typical. Fast-corner silicon at light load sets the emissions.
✓Use series termination or a deliberately slew-limited driver where speed is not needed — slowing an edge is the cheapest EMC fix available.
✓Pick a scope with at least three times the signal's 3 dB bandwidth so the instrument contributes under about 5% to the measured rise time.
✓When a datasheet omits the convention, assume 10-90% for older logic families and check carefully for SerDes and DDR parts, which often use 20-80%.
✓Remember rise times add in quadrature through a cascade: total is the square root of the sum of squares of each stage.
✓Translate the knee frequency into a channel requirement early — it tells you whether an S-parameter model or a connector spec covers your band before you commit to the part.

Common Mistakes

✗Designing the channel to the clock frequency instead of the knee frequency. A slow clock with fast edges still demands a wideband interconnect — this is the classic reason a 25 MHz design fails EMC.
✗Mixing the 0.35 and 0.5 constants. Use 0.35 divided by rise time for amplifier and instrument bandwidth; use 0.5 divided by rise time for how far the interconnect model must remain valid.
✗Feeding a 20-80% datasheet number into a 10-90% formula. That under-predicts rise time by 37% and over-predicts bandwidth by the same factor.
✗Forgetting that a measurement instrument adds its own rise time in quadrature. A 500 ps edge measured on a 500 ps scope reads about 707 ps.
✗Assuming the specified typical rise time is what you will get. Lightly loaded fast-corner silicon can be two or three times quicker than the datasheet typical, and that is the corner that sets your EMC result.

Frequently Asked Questions

Where does the 0.35 constant come from?

It is exactly ln(9) divided by 2 pi, which is 0.34965, derived from the single-pole step response. The 10% and 90% crossings occur at fixed multiples of the time constant, giving a rise time of tau times ln(9), and with the 3 dB bandwidth at one over 2 pi tau the product is fixed. It is not empirical.

Should I use 0.35 or 0.5 divided by rise time?

Both, for different jobs. The 0.35 form gives the 3 dB bandwidth — what an amplifier, receiver, or oscilloscope needs to reproduce the edge. The 0.5 form gives the knee frequency — how far up the interconnect model has to stay valid. Design the channel to the knee and the electronics to the 3 dB figure.

How do I convert a 20-80% rise time?

Multiply by ln(9)/ln(4) = 1.585 to get the 10-90% equivalent, then apply the bandwidth relations. Going the other way, multiply by 0.631. Note that 1.585 is the single-pole value and the largest factor in circulation — a Gaussian edge gives about 1.52 and a linear ramp 1.33 — so for a real driver it slightly over-states the rise time and therefore under-states the knee frequency. Design to the shorter estimate where that margin matters.

Why does my 25 MHz clock cause EMC problems?

Because emissions follow the edge rate, not the clock rate. A 25 MHz clock with a 500 ps edge has a 1 GHz knee, so it radiates meaningfully up to 1 GHz. Slowing the edge with series resistance is usually the fastest fix.

How much scope bandwidth do I need?

At least three times the signal's 3 dB bandwidth to keep the instrument's contribution under about 5%. Rise times add in quadrature, so a scope matching the signal's rise time inflates the reading by 41%.

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