Why Rise Time and Bandwidth's Product is 0.35?

0.35/tr

tr × BW = 0.35
I want know reason.
Thanks.

rise time bandwidth

This is how it works: the circuit is assumed or approximated to behave like a simple RC integrator.

The time constant will then be: τ=RC
The bandwidth is then BW=1/(2*π*RC)=1/(2*π*τ)

On the other hand, the rise time for this simple RC circuit is calculated starting from the equation of the voltage across the cap when a step impulse is applied at the input:
u(t)=U*(1-exp(-t/τ))

For u(t)=0.1U you get 0.1U=U*(1-exp(-t10/τ)), where t10 is the time when the voltage reaches 10% of the final value.

After simplification
0.9=exp(-t10/τ)

Applying the natural log to the equation you get:
ln0.9=-t10/τ or t10=-τ*ln0.9

Similarly, for u(t)=0.9U, we have t90=-τ*ln0.1

The rise time is then:
tr=t90-t10=τ*(ln0.9-ln0.1)≈2.2*τ or τ=tr/2.2

Substitute this value of τ into the expression for bandwidth:
BW=1/(2*π*tr/(2.2))=2.2/(2*π*tr)≈0.35/tr

or

BW*tr≈0.35

rise time 0.35

Good and clear derivation of the formula.
This is only to repeat the restriction that this formula applies only for a lowpass system of first order.

tr * bw = 0.35

LvW said:

Good and clear derivation of the formula.
This is only to repeat the restriction that this formula applies only for a lowpass system of first order.

Click to expand...

One additional comment:
The mentioned formula can be used with good accuracy also for higher order systems when there is a dominant pole which is far below the other poles.
This, for example, is true for universal-compensated opamps.

rise time bandwidth product

Yes, generallly for two pole system this factor can be in a range around 0.35 (for example 0.3-0.45), which depends on n a Q factor.

tr * bw = 0.35

thanks VVV,LvW and pixel.