What's the deal here?

[earthtonesaudio]

I have seen all of these at one time or another.  I know people have discussed this, but it still seems unclear. 

Questions:
1. Doesn't version "a" make the input impedance 500k?
2. I know version "c" is a single-pole lowpass filter, but what is version "b"?

Thanks!

[R.G.]

1. Doesn't version "a" make the input impedance 500k?

Yes, at some high enough frequency. At frequencies high enough that the capacitive impedance is small compared to 1M, the impedance approaches 1M || 1M. Where is that?

F = 1/(2*pi*1M*0.1uF) is where it starts. At that point, both C and the second 1M are 1M and the net impedance is 1M parallel with 1M -jw1M. At ten times that frequency, you're pretty much at 500K resistive.

2. I know version "c" is a single-pole lowpass filter, but what is version "b"?

"b" is a nothing. If the signal into "b" is low enough impedance that it's not loaded down by 1M, it sees only the capacitor - and nothing else. The capacitor in this one is a monode. You can't say anything about what b does until you say what load the cap drives.

[alanlan]

2. I know version "c" is a single-pole lowpass filter...

I'm sure it was a typo. but "c" is a high pass filter.

[earthtonesaudio]

Thanks for the explanation, and the homework, RG.  I just see version "a" all the time.  Often it seems the case is that somebody has "modernized" an old circuit by placing pull-down resistors on the input, but input impedance is not considered, or only as an afterthought.

As for the frequency where this would take effect, by solving for f, I get the result that the impedance is 1M at 1.59Hz, and so according to your post, at 15.9Hz the input impedance would be 500k.  Does that mean that at f=159Hz, input Z=250k?

...I don't know what "-jw1M" means.  I'm getting there, I think, but still a little new at this.

I'm sure it was a typo. but "c" is a high pass filter.

Yep, I thought about it after I posted it, but had already turned off the computer.   

[R.G.]

As for the frequency where this would take effect, by solving for f, I get the result that the impedance is 1M at 1.59Hz, and so according to your post, at 15.9Hz the input impedance would be 500k.  Does that mean that at f=159Hz, input Z=250k?

The cap itself has an impedance of 1M at 1.59Hz; at 15.9Hz, it's 100K, at 159Hz, it's 10K. Just looking at the 0.1uF and the right hand 1M for the moment, the impedance of the two of them gets ever closer to 1M as frequency rises and the cap looks more like a short circuit to a 1M resistor. In fact, at 15.9Hz, it's pretty much a short compared to the 1M.

At DC, the cap is an open circuit, so the impedance of the whole mess is just that first 1M. At an unspecified high frequency, the cap's impedance is so low that you might as well call it a wire, and the impedance is two 1M's in parallel, 500K. In between, the size of the impedance declines as the cap's reactance declines. The combined impedance never gets lower than 500K (ignoring parasitic inductances and capacitances for the moment). No, input Z never gets below 500K.

...I don't know what "-jw1M" means.  I'm getting there, I think, but still a little new at this.

Sorry - I'm always walking a line between keeping it simple and keeping it scrupulously correct.  The "-jw1M" is part of the secret EE's language.  In this context, it means a magnitude of 1M at a phase angle of 90 degrees lagging compared to the input signal. "jw" is the way one writes " J-omega" on a computer with only characters. "J" is the imaginary number which stands for the square root of negative one, which has no real existence. However, it is omnipresent in the advanced math that describes engineering, physics and other sciences. "Omega" is the shorthand notation for the frequency, measured in radians; this is 2*pi*frequency.

That "-jw1M" just kind of slipped out. Ignore it for now. We don't need complex plane math here for the moment.

[earthtonesaudio]

"J" is the imaginary number which stands for the square root of negative one, which has no real existence.

Oh crap.  Well I'd rather not get into Euler's formula at the moment.  The main reason I included circuit "b" is because my boss at work was designing a lowpass noise filter for a high-frequency pulse, and it just so happened that "b" worked, and "c" had the effect of annihilating the signal completely.  Because it was a pulse signal, not a sine type signal, we're pretty sure there was some imaginary number-related explanation, but ended up scrapping the filter and just dealing with the noise.

...Getting back to the DIYstompboxes world, I guess my main problem is that caps and resistors appear so simple and straightforward on the surface, but when you really get to the nitty-gritty, they are crazy-complex. 

So let me rephrase my original post in the form of a question: which circuit is preferred for effect inputs (in general)?  Can we get away with using "c," or should we use "a" for pop protection?  Maybe it depends on the size of the input cap, or the gain of the circuit, or...?


Thanks for your help!

[wampcat1]

I don't see "a" much at all... however, the second 1m resistor is for bias on an opamp. Instead of ground, it connects to 4.5v or half the full voltage.
bw