# RE: [SI-LIST] : Frequency based on rise time for drivers

From: Daniel, Erik S. (Daniel.Erik@mayo.edu)
Date: Mon May 28 2001 - 07:19:39 PDT

Richard-

This formula is NOT derived assuming a sinusoid. It relates the 3 dB
bandwidth of a system limited by a single pole roll-off (e.g. a simple R-C
filter) to the 10%-90% rise time of the same system with a perfect step
input. For systems with more complicated roll-off characteristics, it is
only an approximation.

A quick derivation is below.

- Erik

==================================================================
Erik Daniel, Ph.D. Voice: (507) 538-5461
Mayo Foundation Fax: (507) 284-9171
200 First Street SW E-mail: daniel.erik@mayo.edu
Rochester, MN 55905 Web: www.mayo.edu/sppdg/
==================================================================

Assume the following circuit:

R
Vin o----\/\/\/------+----o Vout
|
|
--- C
---
|
|
V (GND)

FREQUENCY DOMAIN:
-----------------
1 1
Vout = Vin * ----------- ; |Vout|^2 = |Vin|^2 * ---------------
1 + j w C R 1 + w^2 R^2 C^2

3 dB bandwidth => Pout = 1/2 * Pin => |Vout|^2 = 1/2 * |Vin|^2 => w_3dB
= 1/(R C)
=> f_3dB
= 1/(2 Pi R C)

TIME DOMAIN:
------------

Assume Vin is a step input from 0 volts at t<0 to 1 volt at t=>0. Then

Vout = 1 - exp[-t/(R C)]

Find times for which Vout = 10%, 90% * 1 Volt:

0.1 = 1 - exp[-t_10% /(R C)] => t_10% = -(R C) * log(0.9)

0.9 = 1 - exp[-t_90% / (R C)] => t_90% = -(R C) * log(0.1)

=> t_rise = t_90% - t_10% = (R C) * (ln 0.9 - ln 0.1) = (R C) * ln 9

=> ln 9 0.35
t_rise = ------ = ----
2 Pi f f

==================================================================
Erik Daniel, Ph.D. Voice: (507) 538-5461
Mayo Foundation Fax: (507) 284-9171
200 First Street SW E-mail: daniel.erik@mayo.edu
Rochester, MN 55905 Web: www.mayo.edu/sppdg/
==================================================================

> -----Original Message-----
> From: richard hill [mailto:richard2636@excite.com]
> Sent: Thursday, May 24, 2001 12:32 PM
> To: si-list@silab.eng.sun.com
> Subject: [SI-LIST] : Frequency based on rise time for drivers
>
>
> I have often seen people estimating the frequency of a
> driver based on the rise time for the driver using the
> formula Freq = .35/Tr (This seems to be true for a
> perfect sinusoid). There can be two different
> drivers outputing signal at the same frequency but
> different edge rates, won't this assumption be
> invalid. Some times there is a big difference in the
> edge rates of drivers running at the same frequency.
> If we use the above formula and estimate the edge rates
> using the frequency for the two drivers we will get
> similar no.s.
>
>
> Rich
>
>
>
>
>
> _______________________________________________________
>
>
>
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