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?
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\$\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\$– Marcus MüllerDec 7, 2021 at 16:53
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\$\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\$– NewbieDec 7, 2021 at 16:59
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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): -
1RC at the top and 100RC at the bottom (R = 100 Ω, C = 100 nF). Schematic: -
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\$\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\$– NewbieDec 7, 2021 at 17:00
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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 akaDec 7, 2021 at 17:02
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\$\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\$– NewbieDec 7, 2021 at 17:06
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1\$\begingroup\$ Correct @Newbie except if the frequency is much much less than the cut-off frequency = \$\dfrac{1}{2\pi RC}\$ \$\endgroup\$– Andy akaDec 7, 2021 at 17:07
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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 akaDec 7, 2021 at 17:18