Calculating frequency rolloff in RC networks?

[mateusborges]

Heyas,

Brain freeze question... What if I remove this resistor?



Answer would be ~34kHz right?

Cheers!

[marcelomd]

Heyas,

Brain freeze question... What if I remove this resistor?



Answer would be ~34kHz right?

Cheers!

Try to picture the circuit with the resistor removed.
- If LPF: Open circuit. No signal.
- If HPF: No resistor to ground, filter will depend on the next block.

1 Ohm would give you the 34kHz, but a short/wire is waaaay less than that.
If you mean a short. Well, that's a lot less than 1 Ohm, which would give 34kHz.

[Rob Strand]

In theory if you pull the resistor there is no filtering effect, you can consider the cut-off as being at infinite frequency.   What does happen is current still flows through the capacitor.

In practice the thing feeding the RC filter has an output impedance, R0.    It is often assumed the R for the filter is much larger than R0 so we ignore R0 in the calculations a get  f3 = 1/(2*pi*R*C) but really it should be f3 = 1/(2*pi(R + R0)).  So now when we remove R we are left with R0 and the cut off is  f3 = 1/(2*pi*R0*C) and we don't have to think hard about  R not being there.

If for example the stage driving the filter is a transistor gain stage the R0 is roughly equal to value of the collector resistor, Rc.    Quite often Rc is not *way* bigger then the R for filter so by calculating f3 = 1/(2*pi*(R + Rc)*C) you get a more accurate result, however as a ball-park estimate  f3 = 1/(2*pi * R*C)  is fine.

If the stage driving the filter is an opamp or buffer, R0 is quite low and can be ignored.You can still calculate f3 = 1/(2*pi*R0*C) as the cut-off which will be some high frequency.      Normally you try to avoid putting a capacitor directly across the terminals in these cases as bad things can happen (like oscillations).    The point is you shouldn't try to make a filter  that way and for basic calculations it's more of academic value.  For detail engineering calculations you have a lot more to worry about.

A common but more complicated case is connecting a cap across a guitar pickup.  The R0 is actually the guitar pickup impedance (and pots).   The impedance isn't just resistive it's quite complicated.   You can calculate the cut-off but it's not done using the simple formula.

[mateusborges]

If you mean a short. Well, that's a lot less than 1 Ohm, which would give 34kHz.

Sorry, yup, I meant a short.

The deal is, I can hear what it's doing to the circuit but I'm trying to understand what and why, cause it's a lot more than just a filter.

The circuit in this case is the Proco Rat with a LM308 opamp.

Cheers!

[mateusborges]

Thanks a lot @Rob!

I think I can see what you pointing and why it won't ever be a fixed value, and thus will give me different results depending on what I plug in.

I have put this hi pass resistor/short in a switch for trying something different.

[Rob Strand]

I have put this hi pass resistor/short in a switch for trying something different.

For some circuit shorting the switch will work.     

If the thing driving the filter is an opamp what you can do is instead of switching to a short you switch to 1k (or more).   A value small enough that the filter cut off is so high it does nothing but large enough so it doesn't cause oscillation problems.