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I am trying to implement a delayed ON/OFF by using low pass filter implemented by a capacitor and a resistor. I can find the R,C, and cut off frequency by using calculators I found in the internet.

The problem is, I dont fully understand how the calculation works.

To be more precise, if I need a delay, say, of $$\tau = 1~\textrm{msec}$$, I can easily find a pair of R and C sizes that will give me the needed delay.

So my question is:

Why do I need to deal with the frequency cuttoff point?

Obviously it is important, because everyone calculate it. But, I dont understand why?

(Does it have something to do with the step function of the "ON" signal?)

Thanks!

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  • \$\begingroup\$ The cutoff and the delay are intertwined. You can calculate the RC for a given time constant, and the cutoff will be \$\frac{1}{2\pi RC}\$, or you can calculate a 1st order lowpass, and the delay for a step will be given by \$V_{ON}(1-exp(\frac{t}{RC}))\$. Flip the coin. \$\endgroup\$ Apr 29, 2018 at 13:42
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    \$\begingroup\$ You have to consider the threshold voltage too ! \$\endgroup\$
    – Long Pham
    Apr 29, 2018 at 13:46
  • \$\begingroup\$ @LongPham I simplified the formula, since OP said he/she understands the delay, but not the cutoff. Notice that I only said "step", that usually means 0 to 1. It's just for exemplification. \$\endgroup\$ Apr 29, 2018 at 14:06
  • \$\begingroup\$ You do not get a "delayed ON/OFF from a low pass filter. Think again what your question might be or what you need. \$\endgroup\$
    – Andy aka
    Apr 29, 2018 at 14:08
  • \$\begingroup\$ @Andyaka I was guessing it was something similar to this circuit, which does get you a delayed digital signal. \$\endgroup\$
    – Hearth
    Apr 29, 2018 at 20:03

2 Answers 2

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Your circuit is essentially a low-pass filter, filtering out the rapidly-changing high-frequency components of the step function. In the context of filters, cutoff frequency is more often used, but in the context of timing circuitry like this, it's probably better to think in terms of the time constant, $$τ=RC=\frac{1}{2πf_c}$$

The reason these are related is simple: The cutoff frequency is the frequency at which the circuit is no longer able to change rapidly enough to keep up with the changing input, and the time constant tells you how rapidly the circuit can change.

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  • \$\begingroup\$ Or simpler, T = RC. \$\endgroup\$
    – Long Pham
    Apr 29, 2018 at 13:45
  • \$\begingroup\$ @LongPham Good point. I meant to emphasize the connection to the cutoff frequency, but it's probably better to include the RC form as well. Edited. \$\endgroup\$
    – Hearth
    Apr 29, 2018 at 13:46
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This is a case where you want to analyze in the time domain, not the frequency domain. In other words, you don't care directly about the rolloff frequency, but rather the time constant.

The time constant of a RC filter is simply R*C. When R is in Ohms and C in Farads, the the result is in seconds.

If a step is put into such a RC filter, it will exponentially decay towards the new input value. For example, if both input and output are at 1 and the input goes to 0 at t = 0, then the output is:

    OUT = e-t/RC

This means that every RC seconds, the output gets another factor of e closer to the input. From this you can compute the time it takes to get to any particular output level.

For example, let's say you have a low pass filter made from 4.7 kΩ in series followed by 2 µF to ground. A Schmitt trigger with 20% and 80% thresholds looks at this signal and produces a digital output in response. What is the delay to a change in the input when that input is a digital signal that has been steady for a long time?

The time constant is (4.7 kΩ)(2 µF) = 9.4 ms. In the equation above, OUT needs to go to 20% for the circuit to trigger. We therefore know that -t/RC = ln(0.2) = -1.61. That means it takes 1.61 time constants to get to 20% change remaining (80% settled). 1.61(9.4 ms) = 15.1 ms, which is how long the circuit in this example will delay digital edges after long steady periods.

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