0
\$\begingroup\$

What does it mean to have a large time constant of a RC network than the time period of an input AC signal?

Can someone tell me how to understand this and what sense does it make?

\$\endgroup\$
2
  • \$\begingroup\$ Trying to decompose this pretty broad question into smaller steps: so, do you understand what the importance of time constant for RC filters is? \$\endgroup\$ Dec 7, 2021 at 16:53
  • \$\begingroup\$ yes. I was reading about clamper circuits and found this term which I didn't understand. I understand that the time constant of an RC circuit is the time the RC network takes to get charged to 63.2% value of the input voltage. \$\endgroup\$
    – Newbie
    Dec 7, 2021 at 16:59

1 Answer 1

3
\$\begingroup\$

What does it mean to have a large time constant of a RC network than the time period of an input AC signal?

If you apply a 1 kHz (0 volts to 5 volts) square waveform to three low pass filters having time constants of 1RC, 10RC and 100RC you get this (a meaning): -

enter image description here

1RC at the top and 100RC at the bottom (R = 100 Ω, C = 100 nF). Schematic: -

enter image description here

\$\endgroup\$
6
  • \$\begingroup\$ So, when we use a considerably higher time constant RC network, the output will also be a flat line, is it? We might not get the AC waveform as much as the input? \$\endgroup\$
    – Newbie
    Dec 7, 2021 at 17:00
  • 1
    \$\begingroup\$ @Newbie the input should be unaffected but, if the input source has a significant output resistance then, that resistance will increase the effective RC time and apply more filtering to the signal AND the input signal will appear to be partially filtered. The higher the RC constant, the flatter the output will be. \$\endgroup\$
    – Andy aka
    Dec 7, 2021 at 17:02
  • \$\begingroup\$ Oh, thank you very much. So, in that case, with higher time constant, the output voltage can never follow the input voltage and also reach the value of the input voltage. Is my understanding correct? \$\endgroup\$
    – Newbie
    Dec 7, 2021 at 17:06
  • 1
    \$\begingroup\$ Correct @Newbie except if the frequency is much much less than the cut-off frequency = \$\dfrac{1}{2\pi RC}\$ \$\endgroup\$
    – Andy aka
    Dec 7, 2021 at 17:07
  • 1
    \$\begingroup\$ If it's a sinewave then the amplitude progressively reduces proportional to frequency increasing. Below the cut-off we generally say (although an approximation) that the input and output levels are equal. \$\endgroup\$
    – Andy aka
    Dec 7, 2021 at 17:18

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.