Definitions

# Rise time

In electronics, when describing a voltage or current step function, rise time (also risetime) refers to the time required for a signal to change from a specified low value to a specified high value. Typically, these values are 10% and 90% of the step height. The output signal of a system is characterized also by fall time: both parameters depend on rise and fall times of input signal and on the characteristics of the system.

## Overview

Rise time is an analog parameter of fundamental importance in high speed electronics, since it is a measure of the ability of a circuit to respond to fast input signals. Many efforts over the years have been made to reduce the rise times of generators, analog and digital circuits, measuring and data transmission equipment, focused on the research of faster electron devices and on techniques of reduction of stray circuit parameters (mainly capacitances and inductances). For applications outside the realm of high speed electronics, long (compared to the attainable state of the art) rise times are sometimes desirable: examples are the dimming of a light, where a longer rise-time results, amongst other things, in a longer life for the bulb, or digital signals apt to the control of analog ones, where a longer rise time means lower capacitive feedthough, and thus lower coupling noise.

## Simple examples of calculation of rise time

The aim of this section is the calculation of rise time of step response for some simple systems: all notations and assumptions required for the following analysis are listed here.

• $t_r,$ is the rise time of the analyzed system, measured in seconds.
• $f_L,$ is the low frequency cutoff (-3 dB point) of the analyzed system, measured in hertz.
• $f_H,$ is high frequency cutoff (-3 dB point) of the analyzed system, measured in hertz.
• $h\left(t\right),$ is the impulse response of the analyzed system in the time domain.
• $H\left(omega\right),$ is the frequency response of the analyzed system in the frequency domain.
• The bandwidth is defined as

$BW = f_\left\{H\right\} - f_\left\{L\right\},$

and since the low frequency cutoff $f_L$ is usually several decades lower than the high frequency cutoff $f_H$,

$BWcong f_H,$

• All systems analyzed here have a frequency response which extends to 0 (low-pass systems), thus

$f_L=0,Leftrightarrow,f_H=BW$ exactly.

### Gaussian response system

A system is said to have a Gaussian response if it is characterized by the following frequency response

$H\left(omega\right)=e^\left\{-frac\left\{omega^2\right\}\left\{sigma^2\right\}\right\}$

where $sigma>0$ is a constant, related to the high frequency cutoff by the following relation:

$f_H = frac\left\{sigma\right\}\left\{2pi\right\} sqrt\left\{frac\left\{3\right\}\left\{20log e\right\}\right\} cong 0.0935 sigma$

The corresponding impulse response can be calculated using the inverse Fourier transform of the shown frequency response

$mathcal\left\{F\right\}^\left\{-1\right\}\left\{H\right\}\left(t\right)=h\left(t\right)=frac\left\{1\right\}\left\{2pi\right\}intlimits_\left\{-infty\right\}^\left\{+infty\right\} \left\{e^\left\{-frac\left\{omega^2\right\}\left\{sigma^2\right\}\right\}e^\left\{iomega t\right\}\right\} domega=frac\left\{sigma\right\}\left\{2sqrt\left\{pi\right\}\right\}e^\left\{-frac\left\{1\right\}\left\{4\right\}sigma^2t^2\right\}$

Applying directly the definition of step response

$V\left(t\right) = V_0\left\{H*h\right\}\left(t\right) = frac\left\{V_0\right\}\left\{sqrt\left\{pi\right\}\right\}intlimits_\left\{-infty\right\}^\left\{frac\left\{sigma t\right\}\left\{2\right\}\right\}e^\left\{-tau^2\right\}dtau = frac\left\{V_0\right\}\left\{2\right\}left\left[1+mathrm\left\{erf\right\}left\left(frac\left\{sigma t\right\}\left\{2\right\}right\right)right\right]Leftrightarrowfrac\left\{V\left(t\right)\right\}\left\{V_0\right\}=frac\left\{1\right\}\left\{2\right\}left\left[1+mathrm\left\{erf\right\}left\left(frac\left\{sigma t\right\}\left\{2\right\}right\right)right\right]$

Solving for t's the two following equations by using known properties of the error function

$0.1=frac\left\{1\right\}\left\{2\right\}left\left[1+mathrm\left\{erf\right\}left\left(frac\left\{sigma t_1\right\}\left\{2\right\}right\right)right\right]$

the value $t=-t_1=t_2$ is then known and since $t_r=t_2-t_1=2t$

$t_r=frac\left\{4\right\}\left\{sigma\right\}\left\{mathrm\left\{erf\right\}^\left\{-1\right\}\left(0.8\right)\right\}congfrac\left\{0.3394\right\}\left\{f_H\right\}$

and then

$t_rcongfrac\left\{0.34\right\}\left\{BW\right\}quadLongleftrightarrowquad BWcdot t_rcong 0.34$

### One stage low pass RC network

For a simple one stage low pass $RC$ network, rise time is proportional to the network time constant $tau=RC$:

$t_rcong 2.197tau,$

The proportionality constant can be derived by using the output response of the network to a step function input Signal (electrical engineering) of $V_0$ amplitude, aka its step response:

$V\left(t\right) = V_0\left(1-e^\left\{-frac\left\{t\right\}\left\{tau\right\}\right\}\right)quadiffquadfrac\left\{V\left(t\right)\right\}\left\{V_0\right\}=\left(1-e^\left\{-frac\left\{t\right\}\left\{tau\right\}\right\}\right)$

Solving for t's the two equations

$0.1 = \left(1-e^\left\{-frac\left\{t_1\right\}\left\{tau\right\}\right\}\right)qquad 0.9 = \left(1-e^\left\{-frac\left\{t_2\right\}\left\{tau\right\}\right\}\right),$

the times $t_1$ and $t_2$ to 10% and 90% of steady-state value thus known

$t_1=tau\left(\left\{ln 10\right\}-\left\{ln 9\right\}\right)qquad t_2=tauln\left\{10\right\},$

Subtracting $t_1$ from $t_2$

$t_2-t_1=taucdot\left\{ln 9\right\}$

which is the rise time. Therefore rise time is proportional to the time constant:

$t_r =taucdotln 9congtaucdot 2.197$

Now, noting that

$tau = RC = frac\left\{1\right\}\left\{2pi f_H\right\}$

then

$t_rcongfrac\left\{2.197\right\}\left\{2pi f_H\right\}congfrac\left\{0.349\right\}\left\{f_H\right\}$

and since the high frequency cutoff is equal to the bandwidth

$t_rcongfrac\left\{0.35\right\}\left\{BW\right\}quadLongleftrightarrowquad BWcdot t_rcong 0.35$

This formula implies that if the bandwidth of an oscilloscope is 350 MHz, its 10% to 90% risetime is 1 nanosecond.

## Rise time of cascaded blocks

Consider a system composed by $n$ cascaded non interacting blocks, each having a rise time $t_\left\{r_i\right\}$ and no overshoot, whose input signal has a rise time $t_\left\{r_S\right\}$: then its output signal has a rise time equal to

$t_\left\{r_O\right\}=sqrt\left\{t_\left\{r_S\right\}^2+t_\left\{r_1\right\}^2+dots+t_\left\{r_n\right\}^2\right\}$

This result is a consequence of the central limit theorem, as reported in and proved by Henry Wallman in .

## Factors affecting rise time

Rise time values in a resistive circuit are primarily due to stray capacitance and inductance in the circuit. Because every circuit has not only resistance, but also capacitance and inductance, a delay in voltage and/or current at the load is apparent until the steady state is reached. In a pure RC circuit, the output risetime (10% to 90%), as shown above, is approximately equal to $2.2 RC$.

## Rise time in control applications

In control theory, it is often defined as the 10% to 90% time from a former setpoint to new setpoint. The quadratic approximation for normalized rise time for a 2nd-order system, step response, no zeros is:
$t_r cdotomega_0= 2.230zeta^2-0.078zeta+1.12,$
where ζ is the damping ratio and ω0 is the natural frequency of the network.

However, the proper calculation for rise time of a system of this type is:

$t_r cdotomega_n= frac\left\{1\right\}\left\{sqrt\left\{1-zeta^2\right\}\right\}tan^\left\{-1\right\}left \left(\left\{frac\left\{sqrt\left\{1-zeta^2\right\}\right\}\left\{zeta\right\}\right\} right \right)$
where ζ is the damping ratio and ωn is the natural frequency of the network.