why apply 0.35 in formula: rise time=0.35* bandwidth

could someone please help to describe why number 0.35 is applied in rise time?

thank you in advance.

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musnag said:

could someone please help to describe why number 0.35 is applied in rise time?

thank you in advance.

Click to expand...

As you know already tr is the rise time measured between 10% and 90% of steady state output voltage.

Using a simple RC circuit equation....

Vout= Vin(1-exp[-t/(RC)]-------------------------------1 (RC= time constant)

For instance,assuming that Vin=1 V and the 10% voltage is obtained at time t1..the eqn 1 becomes

0.1= 1-exp(-t1/(RC))------------------------------------2

And assuming 90% voltage is at time t2

0.9=1-exp(-t2/(RC))-------------------------------------3

Rise time tr= t2-t1


i.e tr=eqn3 - eqn2

If you solve this,you will get tr=2.197 RC----------------4

Also we know that cut off frequency is defined as


fc= 1/[2.pi.(RC)]

from 4 fc=BW= 0.35/tr-----------------------------------5

Clear now..????


Ciao,
Shiva

thank you both for answer and clarification.