One of the things that's always confused me about moving from general AC / DC analysis to actual functional circuits, like RC Filters is the seeming abandonment of series-parallel dynamics.

For instance, this simple voltage divider as a low pass filter. It was explained to me that higher frequencies will encounter lower impedance in the capacitor while lower frequencies will encounter a higher impedance - causing high frequencies to pass to ground easier, thereby directing lower frequences to Vout.

But, I have a hard time even getting out of the gate, because (if I'm looking at this right) the capacitor is in parallel with Vout, both of which are in series with Vin. In real life Vout is more downstream circuitry that will have an overall impedance felt in parallel with that cap. And in turn effect the overall resistance when added in series with R1. And all of that will entirely effect how this low pass filter works.

This is a problem for me because this kind of schematic, with dangling Vins and Vouts in various configurations with all kinds of electronic components, is used over and over again in subtending study. It's made it hard for me to understand more advanced material now that I've *just* made it out of DC/AC Analysis.

What is it I'm not *getting* here? No one else seems to have this problem, from what I can tell...

 

Do you understand the principle of a voltage divider? - http://en.wikipedia.org/wiki/Voltage_divider

If you have that operation in mind, then look at how the Xc changes with increasing frequency and how that affects the voltage between the fixed resistor and the capacitor.

 

Do you understand the principle of a voltage divider? - http://en.wikipedia.org/wiki/Voltage_divider

If you have that operation in mind, then look at how the Xc changes with increasing frequency and how that affects the voltage between the fixed resistor and the capacitor.

Yes, I understand the principle of a voltage divider. That wasn't my question. In fact, you just linked the same page that I got the RC filter link from.

I'm asking about why Vout being in parallel with C1 is not being considered - it seems like we just abandoned the behavior of parallel circuits. If Vout is a resistor, that changes everything doesn't it?

In other words, why does it matter how this filter behaves without a load at Vout if in real life there *will* be a load at Vout?

 

Vout is not in parallel with C1 - it is the voltage across C1 that is the result of Irms and the Xc impedance of the capacitor at the selected frequency.

Vout can hardly be a resistor, but a resistor that exactly matches Xc can be substituted for C1 and give the same Vout.

In other words, why does it matter how this filter behaves without a load at Vout if in real life there *will* be a load at Vout?

Designers are aware of loading effects and provide a buffer for the divider output such that the divider's Vout is not loaded.

 

Vout is not in parallel with C1 - it is the voltage across C1 that is the result of Irms and the Xc impedance of the capacitor at the selected frequency.

Vout can hardly be a resistor, but a resistor that exactly matches Xc can be substituted for C1 and give the same Vout.

Hmm, that's definitely the source of my confusion. How can Vout *not* be in parallel with C? Maybe it's how I'm drawing this out in my mind...is this drawing the equivalent? (sorry for my sloppy paint skills...)

If this is the equivalent circuit, then Vout is clearly a parallel measurement across C. If this is a filter, then Vout is the output of this filter, right? And if that's true, then that means another circuit will be connected to Vout - it's *that* circuit that is now in parallel with C. And that's what is bugging the hell out of me.

When that circuit connects across C, it will have some kind of associated impedance, right? Being in parallel with C will directly effect the impedance there and that will change everything observed from when Vout was just open.

Am I freaking crazy? It's intirely possible...

 

Hmm, that's definitely the source of my confusion. How can Vout *not* be in parallel with C? Maybe it's how I'm drawing this out in my mind...is this drawing the equivalent? (sorry for my sloppy paint skills...)

If this is the equivalent circuit, then Vout is clearly a parallel measurement across C. If this is a filter, then Vout is the output of this filter, right? And if that's true, then that means another circuit will be connected to Vout - it's *that* circuit that is now in parallel with C. And that's what is bugging the hell out of me.

When that circuit connects across C, it will have some kind of associated impedance, right? Being in parallel with C will directly effect the impedance there and that will change everything observed from when Vout was just open.

Am I freaking crazy? It's intirely possible...

Vout is the voltage across C, and it is often permissible to ignore downstream loading, such as in the case that beenthere mentioned, where a buffer amplifier is added whose input impedance is high enough over the frequency of interest that the loading is negligible ("negligible" is determined by the designer).
If the loading is significant, then the designer has to include the load in his/her analysis and design.

 

Vout is the voltage across C, and it is often permissible to ignore downstream loading, such as in the case that beenthere mentioned, where a buffer amplifier is added whose input impedance is high enough over the frequency of interest that the loading is negligible ("negligible" is determined by the designer).
If the loading is significant, then the designer has to include the load in his/her analysis and design.

Well much thanks, both of you.

I'm going to look at some more of these and try to consider what you're saying. I understand it now, at least in the context of this RC filter. I remember seeing some op-amp circuits that looked similarly confusing...only a bit worse - ha!