Rise Time: What is it? (Equation & How To Calculate It)

Contents

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Key learnings:

Rise Time Definition: Rise time is defined as the duration it takes for a signal to increase from 10% to 90% of its steady value, indicating how quickly a signal changes.
Rise Time Formula: The rise time formula varies based on the system type, with a common calculation for a first-order system being 𝑡 𝑟 = 2.2 𝑇 t r =2.2T.
Calculation Method: To calculate rise time, use the transfer function to determine time constants and calculate the times when the signal reaches 10% and 90% of its final value.
Practical Measurement: Rise time is typically measured using an oscilloscope, which helps analyze the responsiveness of electronic systems.
Significance in Electronics: Understanding and calculating rise time is crucial for optimizing the performance of electronic and control systems, ensuring devices operate efficiently and responsively.

What is Rise Time?

Rise time is the time a waveform takes to go from a stated low level to a stated high level. In analog and digital work those levels are usually 10% and 90% of the final (steady) value. Some logic families quote 20% to 80% instead.

In analog circuits it is the edge duration of a step. In digital circuits it is the time spent between invalid logic levels, which sets how much timing margin you have.

Control texts pick different X% and Y% by damping. The pair is not one universal pair.

A common textbook split for second-order systems is 0% to 100% if underdamped (first crossing of the final value), 5% to 95% if critically damped, and 10% to 90% if overdamped. Instrument rise time stays 10% to 90% unless a standard says otherwise.

Rise Time Equation

The next derivations use a first-order lag and a second-order plant.

The formula depends on the model. First-order and second-order cases differ.

Rise Time of a First Order System

Take this closed-loop first-order system transfer function:

$G(s) = \frac{1}{Ts+1} = \frac{b}{s+a}$

T is the time constant. For G(s)=1/(Ts+1) the unit-step response is 1−exp(−t/T). (The stored latex later drops some minus signs; the inverse Laplace is still 1−exp(−t/T).)

Drive it with a unit step. The Laplace transform of that input is

$R(s) = \frac{1}{s}$

The output is

$C(s) = G(s) R(s)$

$C(s) = \frac{1}{Ts+1} \times \frac{1}{s}$

$C(s) = \frac{\frac{1}{T}}{s+\frac{1}{T}} \times \frac{1}{s}$

$C(s) = \frac{\frac{1}{T}}{s(s+\frac{1}{T})}$

Partial fractions:

$C(s) = \frac{A_1}{s} + \frac{A_2}{s+\frac{1}{T}}$

Find A₁ and A₂:

$\frac{A_1}{s} + \frac{A_2}{s+\frac{1}{T}} = \frac{\frac{1}{T}}{s(s+\frac{1}{T})}$

$A_1 (s+\frac{1}{T}) + A_2 s = \frac{1}{T}$

At s = 0:

$A_1(0+\frac{1}{T}) + A_2 (0) = \frac{1}{T}$

$A_1 \frac{1}{T} = \frac{1}{T}$

$A_1 = 1$

At s = −1/T:

$A_1(\frac{-1}{T} + \frac{1}{T}) + A_2 (\frac{-1}{T}) = \frac{1}{T}$

$A_1 (0) - A_2 \frac{1}{T} = \frac{1}{T}$

$A_2 = -1$

$C(s) = \frac{1}{s} + \frac{-1}{s+\frac{1}{T}}$

Inverse Laplace:

$C(t) = L^{-1} \left[ \frac{1}{s} -\frac{1}{s+\frac{1}{T}} \right]$

$C(t) = 1-e^{\frac{-t}{T}}$

10% to 90% of the final value (here 1):

$C(t_{10}) = 0.10 \quad and \quad C(t_{90}) = 0.90$

$0.10 = 1 - e^{\frac{t_{10}}{T}}$

$e^{\frac{t_{10}}{T}} = 1-0.10$

$e^{\frac{t_{10}}{T}} = 0.9$

$\frac{-t_{10}}{T} = ln(0.9)$

$t_{10} = -T ln(0.9)$

$t_{10} = -T (-0.1053)$

$t_{10} = 0.1053T$

At 90%:

$0.90 = 1 - e^{\frac{t_{90}}{T}}$

$e^{\frac{t_{90}}{T}} = 1 - 0.9$

$e^{\frac{t_{90}}{T}} = 0.1$

$\frac{-t_{90}}{T} = ln(0.1)$

$t_{90} = -T (-2.3025)$

$t_{90} = 2.3025T$

Then t_r is

$t_r = t_{90} - t_{10}$

$t_r = 2.3025T - 0.1053T$

$t_r = 2.197 T$

$t_r \approx 2.2T = \frac{2.2}{a}$

Rise Time of a Second-Order System

For a second-order step, the 0–100% formula below is for the underdamped case (first time the output hits 1). Overdamped 10–90% and critically damped 5–95% need a different calculation. Do not use (π−φ)/ωd if ζ ≥ 1.

Underdamped unit-step form (ζ < 1):

$C(t) = 1-\frac{e^{-\zeta \omega_n t}}{\sqrt{1-\zeta^2}} sin(\omega_d t_r + \phi)$

Set t = t_r at the first crossing of the final value.

$C(t) = C(t_r) = 1$

$1 = 1 - \frac{e^{-\zeta \omega_n t}}{\sqrt{1-\zeta^2}} sin(\omega_d t_r + \phi)$

$\frac{e^{-\zeta \omega_n t}}{\sqrt{1-\zeta^2}} sin(\omega_d t_r + \phi) = 0$

$sin(\omega_d t_r + \phi) = 0$

$sin(\omega_d t_r + \phi) = sin(\pi)$

$(\omega_d t_r + \phi) = (\pi)$

$\omega_d t_r = \pi - \phi$

$t_r = \frac{\pi - \phi}{\omega_d}$

Where:

$\omega_d = \omega_n \sqrt{1-\zeta^2}$

$\phi = tan^{-1} (\frac{\sqrt{1-\zeta^2})}{\zeta}$

Underdamped 0–100% rise time:

$t_r = \frac{\pi - tan^{-1} (\frac{\sqrt{1-\zeta^2})}{\zeta}}{\omega_n \sqrt{1-\zeta^2} }$

How to Calculate Rise Time?

First-Order System

Example: first-order G(s) = 5/(s+2). The 10–90% rise time still uses 2.2/a, because DC gain only scales the height, not the 10% and 90% times.

$G(s) = \frac{5}{s+2}$

Match

$G(s) = \frac{b}{s+a}$

So a = 2 and b = 5. T = 1/a = 0.5 s. Final value is b/a = 2.5, not 1.

10–90% rise time:

$t_r = \frac{2.2}{a}$

$t_r = \frac{2.2}{2}$

$t_r = 1.1 sec$

Second-Order System

Second example: underdamped plant, natural frequency 5 rad/s, damping 0.6. Use 0–100% first-crossing time.

$\omega_n = 5 rad/sec$

$\zeta = 0.6$

Use

$t_r = \frac{\pi - \phi}{\omega_d}$

Need φ and ω_d.

$\phi = tan^{-1} \left( \frac{\sqrt{1-\zeta^2}}{\zeta} \right)$

$\phi = tan^{-1} \left( \frac{\sqrt{1-0.6 ^2}}{0.6} \right)$

$\phi = tan^{-1} \left( \frac{\sqrt{1-0.36}}{0.6} \right)$

$\phi = tan^{-1} \left( \frac{0.8}{0.6} \right)$

$\phi = tan^{-1} (1.33)$

$\phi = 0.9272 rad$

Damped natural frequency ω_d:

$\omega_d = \omega_n \sqrt{1-\zeta^2}$

$\omega_d = 5 \times 0.8$

$\omega_d = 4 rad/sec$

Then

$t_r = \frac{3.14-0.9272}{4}$

$t_r = \frac{2.2128}{4}$

$t_r = 0.5532 sec$

Why is Rise Time 10% to 90%?

You may use other percentages if a standard names them.

Electronics and scope work still default to 10% to 90%.

The start and the last few percent of a step often depend on the source and on settling. The middle of the edge is more repeatable.

Example edge:

The trace sits near zero, then the edge starts.

Timing from the long zero wait is not the edge. The interval marked T_r is the transition after the trigger.

90% is used because many edges crawl into the last percent (noise, RC tails). They do reach 100% in an ideal first-order model; real traces often make 100% a poor mark.

An exponential 1−exp(−t/T) only meets 100% as t goes to infinity, so 90% is a finite, useful mark.

Start and finish of an edge vary by device.

The 10% to 90% slice of that transition is the usual comparison between parts.

So 10% to 90% is the default unless the control definition (0–100% underdamped) applies.

Rise Time vs Fall Time

Fall time is the time from a stated high level to a stated low level.

Usual marks are 90% down to 10% of the step. The figure below shows that interval.

rise time vs fall time — Rise Time vs Fall Time

The percentage pair is the reverse of rise time. The seconds are not automatically equal.

A CMOS pin, a BJT or a diode-clamped line can fall faster or slower than it rises.

Only a symmetric wave (for example a sine) has matching rise and fall. A square-ish logic edge usually does not.

There is no general formula that converts one into the other. Both still matter for timing and EMI.

Rise Time and Bandwidth

You measure an edge with an oscilloscope. A 10–90% rise time also estimates the signal’s harmonic content, which sets the scope bandwidth you need.

Pick an oscilloscope whose bandwidth is at least that of the edge, or the displayed rise time will look slower than the real edge.

Displayed tr is roughly √(tr_signal² + tr_scope²) when both are 10–90% Gaussian-like edges.

A common 10–90% rule for a simple one-pole response is BW ≈ 0.35 / t_r:

$BW \approx \frac{0.35}{t_r}$

That 0.35 applies to 10–90% on a first-order (or Gaussian) edge. 20–80% uses a smaller constant.

Keep units consistent: tr in seconds gives BW in hertz. 1 ns and MHz are a usual pair.

A near-Gaussian or single-pole front end matches 0.35 well.

Flatter, brick-wall-like scopes use about 0.40 to 0.45 (Tektronix and others quote this). Use the figure in the scope manual.

Example: a displayed 10–90% edge of 1 ns. If that edge is scope-limited, the scope bandwidth is about 0.35 / 1 ns = 350 MHz. The stored latex writes 3.5 / 10^-9 and 3.5×10^-9; that algebra is wrong. 0.35 / 1e-9 s = 3.5e8 Hz = 350 MHz.

Using 0.35 / 1 ns:

$BW = \frac{3.5}{10^{-9}} = 3.5 \times 10^{-9} = 350MHz$