Series capacitors, but separated by a resistor? How is this calculated?

[aion]

The original Marshall Shredmaster uses a switching system where there's a 68n cap after the first op-amp stage, then a 4k1 resistor, then another 4k1 resistor followed by a 100n capacitor. The junction point between the resistors is sent to ground when the circuit is bypassed.

I understand why they do this, but what is the capacitance between the two op-amp stages when the circuit is engaged? Do normal series capacitor calculations apply when they are separated by only a resistor, or is there a different way of calculating it? How would it be simplified into 1 resistor and 1 capacitor in a true-bypass arrangement?

[R.G.]

Things in series all have identical currents passing through them. They have to - it's the same current.

So things in series will have the same voltage at the ends of the string no matter what order they're in. The voltages at the internal points may be different, but they will always sum to the same thing at the ends of the series string. Always.

So - as long as you don't look at the internal voltages, THE ORDER OF THINGS IN SERIES DOES NOT MATTER. You may rearrange them in any order and it won't affect the voltages at the end of the string.

In the case you're proposing, the voltage at the ends of the C-R-R-C string will be the same if you rearrange it to C-C-R-R and add the Cs and Rs.

But while we're already doing some thinking, let's think about why the original designer went to the trouble and expense of putting in two Cs and two Rs when one of each would have done fine. There had to be a reason to go to that work and expense, right? Is it possible that the point between the two resistors needs to be grounded to keep the thing quiet when it's bypassed?

Just sayin'.   

[Perrow]

But the "Cs" in series shouldn't be added, should they? I seem to remember that they should be divided (at least when they have the same value), or does my memory work as intermittently as the circuits I design?

[aion]

Things in series all have identical currents passing through them. They have to - it's the same current.

So things in series will have the same voltage at the ends of the string no matter what order they're in. The voltages at the internal points may be different, but they will always sum to the same thing at the ends of the series string. Always.

So - as long as you don't look at the internal voltages, THE ORDER OF THINGS IN SERIES DOES NOT MATTER. You may rearrange them in any order and it won't affect the voltages at the end of the string.

In the case you're proposing, the voltage at the ends of the C-R-R-C string will be the same if you rearrange it to C-C-R-R and add the Cs and Rs.

Awesome, that makes sense. So it would simplify down to an 8k2 resistor + 40n capacitor in this case.

But while we're already doing some thinking, let's think about why the original designer went to the trouble and expense of putting in two Cs and two Rs when one of each would have done fine. There had to be a reason to go to that work and expense, right? Is it possible that the point between the two resistors needs to be grounded to keep the thing quiet when it's bypassed?

A 3PDT true bypass arrangement with circuit-input grounding would make this a non-issue, right? That was maybe for the specific case of DPDT bypass since 3PDT switches weren't really a thing in '92.

I'm treading lightly, though, because I know you've got more than a little experience with this particular circuit and the J&H maintains the same grounding setup.

[slacker]

That was maybe for the specific case of DPDT bypass since 3PDT switches weren't really a thing in '92.

The originals use non true bypass switching where the input is always connected, like shown at General Guitar Gadgets in the "Non True Bypass with LED, circuit input unbypassed" example. The spare throw on the switch is used like R.G said to ground the point between the first two opamp sections. Input grounding with a 3PDT would probably achieve the same thing, or you can use a 3PDT to get true bypass and still ground the same place as the originals.

[R.G.]

But the "Cs" in series shouldn't be added, should they? I seem to remember that they should be divided (at least when they have the same value), or does my memory work as intermittently as the circuits I design?

C's in series add the way C's in series add - as conductances. That is, the equivalent capacitance is Ceq = 1/C1 + 1/C2 = (C1*C2)/(C1+C2).

A conductance is amps per volt, an impedance is volts per amp.  The conductance of a resistor is 1/R.

Being able to think in both conductance and impedance at the same time helps in calculation. Figuring parallel resistances, for instance. If you have two resistors, it's easy to remember R1*R2/(R1+R2), but what if you have three resistors in parallel? You can go do the math and figure it out, but if you use your calculator, which does 1/X easily, then you can do 1/R1 + 1/R2 +1/R3, then take 1/X and have the answer. Works for any number of resistors - just like adding capacitors in parallel is simply adding the capacitances.

To add capacitors in series, do 1/C1 + 1/C2 + ... then 1/X. This works because capacitance is a number that's a conductance (amps per volt) and resistances are numbers that are volts per amp.