RC time constant

[Kevin Weddle]

Perhaps the most overlooked by most enthusiasts is the RC time constant. The best way to determine if your circuit is good is to compare the time of the RC with the time of the signal. If the time of the signal is short, such a ramp of a square wave, the RC time looks very long to it. If the time of the signal is long, such as the DC of a square wave, the RC time looks very short to it.

So, is it good to have a long RC or a short RC compared to the signal. Well a long RC of course. A long RC means that it takes a long time to charge or discharge.

So what happens if the RC looks short compared to the time of the signal. Well a short RC means the charge and discharge time is short. If the 5RC is 1 second, it means the capacitor will charge completely to the applied voltage in 1 second. If the 5RC is 5 seconds, it will take 5 seconds to completely charge to the applied voltage. That being established, it is not difficult to see that when your signal goes DC, the signal time is long, the RC looks short, you get quick charge and discharge, and the voltage is indicated by the quickly charging and discharging capacitor. If the capacitor were not allowed to charge and discharge, the voltage would remain pretty much the same.

[audioguru]

Kevin,
Are you talking about an RC lowpass filter? It attenuates high frequencies gradually. If you select a long RC time constant for better attenuation, then it might take too much time to charge and a multiple-stage filter would be best.

You contradict yourself when you talk about the capacitor charging and discharging:

So, is it good to have a long RC or a short RC compared to the signal. Well a long RC of course. A long RC means that it takes a long time to charge or discharge.

Yes, of course. But the charge and discharge times might be too long.

Well a short RC means the charge and discharge time is short.

Yes, of course. But then it won't be a filter.

If the 5RC is 5 seconds, it will take 5 seconds to completely charge to the applied voltage. That being established, it is not difficult to see that when your signal goes DC, the signal time is long, the RC looks short, you get quick charge and discharge, and the voltage is indicated by the quickly charging and discharging capacitor.

Certainly not! The capacitor won't charge or discharge completely in less than 5 seconds.

[prateeksikka]

HI audioguru!
i feel that RC time constant chosen depends on the application you are putting the filter into and the type of filter.recently i saw a timer circuit in which RC time constant was as high as 5s .i have seen circuits with time constant as low as 2-3 micro seconds.One of the book on analog electronics says the famous quote:
  "IT IS THE TIME CONSTANT WHICH DIFFENTIATES A HIGH PASS AND LOW PASS CIRCUITS AND NOT EXACTLY THE POSITION OF R&C"
BY CHOSING R&C OF EXTREME MAGNITUDES YOU CAN CONVERT HIGH PASS TO LOW PASS AND VICE VERSA TOO!! :o
so i feel time constant depends on application whats ur view? ::)
prateek

[Guest Alun]

audioguru,
In theory the capacitor will never full charge or discharge, but it is often considered to be fully charged or discharged after 5RC time constants.

[Miles Prower]

Perhaps the most overlooked by most enthusiasts is the RC time constant. The best way to determine if your circuit is good is to compare the time of the RC with the time of the signal. If the time of the signal is short, such a ramp of a square wave, the RC time looks very long to it. If the time of the signal is long, such as the DC of a square wave, the RC time looks very short to it.

That's not the way to look at it. The circuit shown here is a basic RC highpass filter. The maximum gain for this circuit is simply: G(s)= R_o/(R_i + R_o) which would occur at a theoretical frequency of infinity. At the critical frequency, the gain will be -3.0db of the max (theoretical) value. That's what determines the low frequency cutoff.

In this case, what you want is Fourier (frequency domain) analysis, not LaPlace (time domain) analysis.

[prateeksikka]

hi there!
even in this ckt, a high value of RC time constant makes its cutoff freq lower i.e high pass while very low RC makes cutoff freq high i.e low pass filter.
So its just a game of RC!
while most people think that positioning of R&C makes difference although the easiest method is to exchange their places.
:)
prateek

[audioguru]

No, with a very low RC the circuit is not a lowpass filter, it is still a highpass but its cutoff frequency is even higher.

[prateeksikka]

yes audioguru!
through my post i wished to state that broadly,frequency control the most important part for a filter is mainly governed by RC time constant and not the physical arrangement of the components.Basically our aim to pass a signal through a filter can be attained by relative high or low cutoff freq of the filter wrt signal .Any way,what frequencies are considered as low ,for design of filters?any idea? :)
prateek

[audioguru]

The cutoff frequency of an RC filter is calculated as: 1 divided by 2 pi RC.
I make it simpler by rounding it off a bit: 0.16 divided by RC.
R is in ohms and C is in Farads.

You must physically swap the parts to change from a lowpass to a highpass filter and vice-versa. In a lowpass filter, the capacitor grounds high frequencies. In a highpass filter the capacitor blocks low frequencies.

[prateeksikka]

yes audioguru!
and that too low pass and high pass filters used in respect of signals are nothing but differentiators and integrators.
:)
prateek

[prateeksikka]

hey!
what is the constant of integration for an integrator?just like we have Sin(x)+C
as integral of Cos(x)?
prateek

[Miles Prower]

hey!
what is the constant of integration for an integrator?just like we have Sin(x)+C
as integral of Cos(x)?

In the mathematical sense, C is completely arbitrary. Use it to turn the general solution of a differential equation (of which there are an infinite number) into a particular solution that depends upon initial conditions.

In the electrical sense, since you're dealing with time varying signals, C is a signal that does not vary with time. IOWs: a superimposed DC voltage.

[Kevin Weddle]

Here's the link. It's a bit long, but look where it talks about the time of the signal and the RC time constant.

http://www.kpsec.freeuk.com/capacit.htm

I you want to pass a sine wave through a capacitor, you have to know how a capacitor charges. There is a graph of voltage and time. So, you have a sine wave and you have the v(t) of the capacitor. If the RC is short, the instaneous voltage level will DC charge the capacitor fully and you get the charging curve. Close examination will reveal the sine wave is now only a vestige of the original and the resulting waveform is filled with that voltage charging capacitor. It is best to only use part of that curve, this would be the linear part.

[audioguru]

Kevin,
Your 2nd link doesn't work.
The linearity of an AC signal isn't affected by an RC time constant. A highpass RC time constant reduces the level of low frequencies and a lowpass RC time constant reduces the level of high frequencies. The frequencies reduced in level remain linear.