Hi,

  I have a typical series RLC circuit. A Voltage Source followed by a
ressistor, than an inductor and finally a capacitor upon I wish to
measure the output voltage. The source voltage is a digital signal
rising (or falling) from V-Low to V-High. I have the R, L & C values.
I am not sure whether the Current is neccessary, but I can find it if
neccessary for the solution. Most of the solutions I found include a
2nd order differential equation and I am almost sure there is a
simplier solution.

How can I calculate the rising and falling time for such a circuit? I
need a straight forward answer in which I can simply insert the values
and recieve a numerical value for the required times.

Thanks for the help.

---

Request for Question Clarification by hedgie-ga on 02 Jun 2004 10:49 PDT

Hi

You have a circuit shown here:
http://en.wikipedia.org/wiki/RLC_circuit
 and want the fall time 
http://www.fact-index.com/f/fa/fall_time.html

and similarly the  rise time,

You can get formula calculating these times from R L and C
and use those expressions without dealing with calculus;
you do not need current - your circuit is linear.

However, formulas are derived by solving the 2nd order diff. equation.
There is no way to avoid it.
It is possible to derive the waveforms by integral transforms - by which
one can represent the circuit as a filter, but it still is solving the
differential equation...

So,
 do you want formulas - without derivation - as an answer?

hedgie

---

Clarification of Question by raamee-ga on 02 Jun 2004 23:44 PDT

Good morning,

   After reading your answer I understand that bypassing the 2nd order
diff. equation is impossible. However, I would appriciate if you could
elaborate the explanation on them for my case.

My case as I wrote is a series RLC circuit with a Step signal as the
Input (voltage varies from 0v to 3.33v).

Thanks.

Ok, raamee-ga 

  Let's start in the  frequency domain,
  (implicitely using the integral transform)
  since it will give us some estimate of time constant quickly.
  
   The impedance of the circuit with components in series is
   (same as in DC case) simply sum of impedances:
   
   Z(s)= L *s + R + C/s
   
   s is complex frequency (2 * pi * f) and of course, impedance is a 
   cpmplex number now.

   Impedance is reduced to the ordinary resistance as s--> 0.
   (in our case R--> oo = infinity, as capacitor becames an 'open')
   
   Reference: (requires mathML fonts - highly recommended anyway)
   http://cnx.rice.edu/content/m0024/latest/
   
   Here is a nice applet (java is required) which allows you to play
with RLC circuit:
   http://users.erols.com/renau/impedance.html
   
   HAnd here are two pdf texts with  examples how these complex
impedances can be used solving more complex circuits and connection to
 the differential equations:
   Combined Impedances
 Resonance and the Transfer
Up: Sinusoidal Sources and Complex Previous: Inductive Impedance. ...
www.phys.ualberta.ca/~gingrich/ phys395/notes/node33.html 

[PDF] Notes Linear Di erential Equations Complex Impedance Circuits and ...
File Format: PDF/Adobe Acrobat 
Notes Linear Di erential Equations Complex Impedance RLC Circuits and Resonance
The purpose of these notes is to inform you about a very simple and elegant ...
pupgg.princeton.edu/~phys104/ lguides/pdf/lg06_notes.pdf 

  To conclude with a simple and brief summary:
  
  Once you have your impedance Z(s) you can set it to zero and solve
that quadratic equation for s.
 The roots of Z(s)=0 are the two resonant frequancies
 (this the same equation you would have to solve to get two particular
solutions of the differential eq.)
  
  Once you have those two roots, s1, and s2, you know almost everything you want:
  
  The solution (current wavefront) I(t) is a linear combination of 
  expt(t * s1)   and (t * s2)  and so the real part of s1, s2
determines the time constant (directly related to the fall time and
rise time) and
 imaginary part determines the duration of overshoots and undershoots.

 Feel free to ask for clarification if needed,
If all is clear, please do rate the answer.

  
  hedgie

The formula Z(s)= L *s + R + C/s should be corrected to :
Z(s) = L*s + R + 1/(C*s).
the part of 1/(C*s) is the impedence of the capacitor. 
when using DC for the input voltage, the frequency of DC is zero, thus
, the impedence of the 1/(C*s) when f->0+ is infinity.
And this is why the total impedence of the circuit Z(s) is infinity.