Question about buffer interaction with other circuits

[joelap]

OK, question.  At risk of sounding like a complete noob.



All resistors are 1Meg, except for the source resistor, say 4k7.

OK.  Obviously the JFet Source follower is a buffer.  Consider R5 and C3 to be the input stage to a fuzz face.  Wouldn't C2 interact with C3, causing the equivalent series capacitance (input cap to the fuzz face) to be (C2*C3)/(C2+C3)?  I ask because I had an idea in which I could get PCB mounted jacks, but have a buffer permanently wired on the board the in jack is on, so whether the circuit is on or off, it is still buffered and won't suffer from cable capacitance.  Now, if the equivalent series capacitance would in fact be (C2*C3)/(C2+C3), then essentially to get a 2.2uf input cap for the fuzz face, I could eliminate C2 (and subsequently R5) and everythings dandy.  However, if I want to wire a 3PDT switch to turn the effect on and off, while leaving the buffer always in the signal path, I'm thinking I'd need to leave C2 in there.  Otherwise, by removing C2 and going to a 3PDT, essentially I'd have a DC voltage on the switch equivalent to the source voltage, which would cause a loud pop when switching on and off, and when in bypass would be sending a DC across the patch cable to the next effect too.... is my logic making sense?  Its just something I've been thinking about... is there a simple way to wiring up an always-on buffer, without screwing with the input capacitance of the actual effect itself?

Come to think of it, if I had an external buffer at the input of my pedal board, and another effect immediately following, wouldn't it also screw with the input capcitor of the next effect?

Actually, while writing this, I'm thinking twice about doing this as now if power is removed, even the bypassed signal wont be audible, whereas with the buffer out of the circuit, it's not a problem.  But still curious to see... am I thinking right, am I crazy?

[R.G.]

All resistors are 1Meg.

OK. So your input resistance is 333K - three 1M's in parallel, R1, 2, 3.

OK.  Obviously the JFet Source follower is a buffer.  Consider R5 and C3 to be the input stage to a fuzz face.  Wouldn't C2 interact with C3, causing the equivalent series capacitance (input cap to the fuzz face) to be (C2*C3)/(C2+C3)?

To a first approximation, yes. The 1M pulldown affects it a little, not much.

I ask because I had an idea in which I could get PCB mounted jacks, but have a buffer permanently wired on the board the in jack is on, so whether the circuit is on or off, it is still buffered and won't suffer from cable capacitance.  Now, if the equivalent series capacitance would in fact be (C2*C3)/(C2+C3), then essentially to get a 2.2uf input cap for the fuzz face, I could eliminate C2 (and subsequently R5) and everythings dandy.  However, if I want to wire a 3PDT switch to turn the effect on and off, while leaving the buffer always in the signal path, I'm thinking I'd need to leave C2 in there.  Otherwise, by removing C2 and going to a 3PDT, essentially I'd have a DC voltage on the switch equivalent to the source, which would cause a loud pop when switching on and off... is my logic making sense?

That's correct.

Its just something I've been thinking about... is there a simple way to wiring up an always-on buffer, without screwing with the input capacitance of the actual effect itself?

Define "screwing up". Make C2 be 22uF - or 100uF. It then has essentially no effect on C3, and both costs about the same and is about the same size.

Come to think of it, if I had an external buffer at the input of my pedal board, and another effect immediately following, wouldn't it also screw with the input capcitor of the next effect?

Once again, define "screwing up". Make one cap big enough so it doesn't matter.

Actually, while writing this, I'm thinking twice about doing this as now if power is removed, even the bypassed signal wont be audible, whereas with the buffer out of the circuit, it's not a problem.  But still curious to see... am I thinking right, am I crazy?

Nope, you're right. But it's not much of a real problem.

[joelap]

RG, thanks for the response.

So basically, by making C2 very large in comparison to C3, its effect would be negligible on the circuit.  Gotcha.

Also, input pulldown resistors have that much of an effect on the input impedance of a circuit?  I didn't realize they'd have that much of an effect.

[PerroGrande]

Also, input pulldown resistors have that much of an effect on the input impedance of a circuit?  I didn't realize they'd have that much of an effect.

Sure they do.  They represent alternate paths to ground for the signal - thus introducing a parallel resistance into the mix.  How *much* they effect the input impedance depends both on their size and the nature of the circuit that follows.   If your first active stage is a JFET op-amp with rated input impedance into the non-inverting input of, say, 10^12 Ohms, a 3.3 Meg pulldown resistor will dominate the voltage divider.   On the other hand, if your active stage is a bipolar transistor circuit with an input of 15k, a 3.3 pulldown isn't going to have as much relative impact on the impedance of the circuit.