original Studio Projects C1 HPF

[deb611]

I've been goofing around with an old 797-era SP C1 and checking out the schematic online, just learning more - and I cannot figure out for the life of me how the C1 high-pass filter works.

According to SP, the original C1 had a 150Hz HPF @dB/octave...but I can't see how I'd change the rolloff corner frequency even if I wanted to.

Anyone want to enlighten me?

schematic attached. From my trace of the circuit, the resistor value of 1Gohm (or 1KM, as printed on the resistor) is correct. 

It seems they also put out a newer C1 v2 with a separate switch for the rolloff and pad, but this one makes no sense to me.

 

[PRR]

Ground the right side of "R". What happens? Bass rolloff determined by capsule capacitance and R.

This may be more clever, taking R to a point with gain and inversion. Miller effect. This will typically give an effect like, say 10Meg, but with the RTI hiss of a 1Gig. (I amp pulling numbers out of my butt; that image is too large for my monitor, and larger than it needs to be.)

 

[deb611]

Ground the right side of "R". What happens? Bass rolloff determined by capsule capacitance and R.

Isn't the HPF point already determined by the 1Gohm R18?

This may be more clever, taking R to a point with gain and inversion. Miller effect. This will typically give an effect like, say 10Meg, but with the RTI hiss of a 1Gig. (I amp pulling numbers out of my butt; that image is too large for my monitor, and larger than it needs to be.)

I tried changing R to a lower value and it didn't do a darn thing. I had a 150M and that did nothing. Even tried 330K. So I'm wondering if the corner frequency is set by a different component? Or am I missing something insanely and blindingly obvious?

Image link is here for a smaller version, sorry!

http://recordinghacks.com/images/mic_extras/studioprojects/C1-797-schematic.png

 

[abbey road d enfer]

I've been goofing around with an old 797-era SP C1 and checking out the schematic online, just learning more - and I cannot figure out for the life of me how the C1 high-pass filter works.

According to SP, the original C1 had a 150Hz HPF @dB/octave...but I can't see how I'd change the rolloff corner frequency even if I wanted to.

Anyone want to enlighten me?

schematic attached. From my trace of the circuit, the resistor value of 1Gohm (or 1KM, as printed on the resistor) is correct. 

It seems they also put out a newer C1 v2 with a separate switch for the rolloff and pad, but this one makes no sense to me.

The HPF works by feeding back an inverted signal to the capsule.
The capsule sees an additional load of about 1/10th of R (because the gain from the capsule to the collector of Q3 is about 10.
If R is actually 1G, that would mean that the cut-off frequency goes from about 4Hz to 40 Hz; no wonder you don't hear much difference on vocal stuff.
But when you say that reducing R to 150Meg, even much less than that result in no change, I believe there is something wrong in your connections.

 

[deb611]

The HPF works by feeding back an inverted signal to the capsule.
The capsule sees an additional load of about 1/10th of R (because the gain from the capsule to the collector of Q3 is about 10.
If R is actually 1G, that would mean that the cut-off frequency goes from about 4Hz to 40 Hz; no wonder you don't hear much difference on vocal stuff.
But when you say that reducing R to 150Meg, even much less than that result in no change, I believe there is something wrong in your connections.

Thanks! That makes a lot of sense. So its pulling the inverted signal off the source of the JFET and feeding it back.

For the benefit of my understanding - so this means the inverted signal is LPF'd, and therefore the inverted LPF'd signal is cancelling the low frequencies of the original signal? where does the LPF occur? I've been scanning the schematic and I can't seem to find a place where  a simple RC network LPF would occur between Q3 and the switch.

 

[abbey road d enfer]

For the benefit of my understanding - so this means the inverted signal is LPF'd, and therefore the inverted LPF'd signal is cancelling the low frequencies of the original signal?

No. The signal is somewhat band-limited but not really LPF'd in the sense you're making. The High-Pas effect comes from the capsule's capacitance meeting a resistance. By applying gain to the fedback signal, resistor R appears like it's 10 times smaller. The advantage of this technique is that in terms of noise and distortion, it still appears like a 1G.
As I wrote earlier, I have doubts about the value of resistor R. If 1G, the resulting HPF corner frequency would be about 25Hz, which would probably diminish impact and handling noises, but not plosives, which are the main incentive for a HPF.

BTW, what you describe (feeding an LPF'd version of the signal back to the capsule) is the technique used by Neumann in the U67 and 87.

 

[deb611]

No. The signal is somewhat band-limited but not really LPF'd in the sense you're making. The High-Pas effect comes from the capsule's capacitance meeting a resistance. By applying gain to the fedback signal, resistor R appears like it's 10 times smaller. The advantage of this technique is that in terms of noise and distortion, it still appears like a 1G.
As I wrote earlier, I have doubts about the value of resistor R. If 1G, the resulting HPF corner frequency would be about 25Hz, which would probably diminish impact and handling noises, but not plosives, which are the main incentive for a HPF.

BTW, what you describe (feeding an LPF'd version of the signal back to the capsule) is the technique used by Neumann in the U67 and 87.

That makes a lot of sense. So technically changing the value of "R" sets an RC filter corner point, where C is the capsule capacitance and R is roughly 1/10th of the resistor "R"? Or do the other 2 1G resistors, R18 and R17, also have an effect on the R value in the RC filter?

 

[abbey road d enfer]

Or do the other 2 1G resistors, R18 and R17, also have an effect on the R value in the RC filter?

Yes they do, but the virtual resistance is without doubt dominant.