I'm in total agreement with you Papabravo - that's why I questioned the OP as to what was their understanding of the term "propagation delay" - because it seemed to me this was not applicable in the circuit under consideration. The notion of rise time made more sense to me and I thought that might be what was really required......?I don't disagree with the rise time analysis, but it is not the same thing as propagation delay. Propagation delay is how long it takes for the output to "begin to respond" to the input and may not be calculable from the circuit parameters of a passive circuit.
Normally, it involves active devices and is determined by the complex mechanics of operating a semiconductor between saturation and cutoff. There is also an asymmetry in time between going from saturation to cutoff and cutoff to saturation.
Now in an ideal 1st order RC circuit there is no propagation delay because the capacitor begins charging immediately. I think cascading two single pole RC filters does not alter the fact that it is a linear circuit without the active components usually responsible for propagation delay.
That is only an opinion -- I could be wrong.
Well you can find the time domain response to a unit step for this circuit asHi everyone,
I was just trying to ask, what would be the delay added to the circuit because of charging of capacitor. The two capacitor takes some time for charging. I wanted to know the time taken by the capacitor to charge to 99% of the input voltage..
May be the step rise time is the answer for my query..
Nothing like being precise.Well you can find the time domain response to a unit step for this circuit as
\(V_o(t)=1-1.2e^{-0.4774*{10}^4t}+0.2e^{-{2.8774*}{10}^4t}\)
You then solve for the 99% value by setting Vo(t)=0.99 and solving for the unknown t.
Nothing like being precise.
The second term contributes 5.9e-14V.
by Aaron Carman
by Aaron Carman
by Jake Hertz
by Jake Hertz