I drew this low pass filter circuit to have a output voltage of 5v across capacitor at a frequency of 100Hz sine wave of amplitude 10v. But I am only getting 4.5 volt as output in steady state. I am doing simulation in LTspice on Ubuntu. What mistake I am making?

R = sqrt(3)/(2*pi*fC) - This is how I calculated R for a given C to have a half voltage drop across C

enter image description here

  • \$\begingroup\$ The \$\sqrt 3\$ term is not part of the calculation for this type of circuit. \$\endgroup\$ May 22 '16 at 14:03
  • \$\begingroup\$ How did you read this 4.5V ?? I get 4.9956V in my LTspice. \$\endgroup\$
    – G36
    May 22 '16 at 14:05
  • \$\begingroup\$ @G36 - My LTSpice gives somewhere around 4.60 volt. Don't know why? I did a transient analysis for 8 second. \$\endgroup\$
    – InQusitive
    May 22 '16 at 15:12
  • \$\begingroup\$ @G36:- When I put exactly 200ms as you did in transient, I am getting the correct value. I think I am making something related to transient timing? Can't I use a large value for stop time in transient analysis, like 10 seconds? \$\endgroup\$
    – InQusitive
    May 22 '16 at 15:16
  • \$\begingroup\$ @InQusitive try set Maximum Timestep at around 10us in transient analysis window. \$\endgroup\$
    – G36
    May 22 '16 at 15:34

To get 5V peak from a 10V peak supply across \$1 \mu F\$ capacitor at 100Hz we need current equal to I = 5V/Xc = 5V/1.592kOhm = 3.1407mA and the voltage drop across resistor is $$V_R = \sqrt{(10V^2 - 5V^2)} = 8.66V peak $$
therefore $$R = \frac{8.66V}{3.1407mA} =2.757k\Omega $$ So yes, your calculations are correct.
And the simulation result look like this enter image description here

As you can see the peak voltage is 5V but RMS value is around 3.5V


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.