Input loading question involving two buffers

[midwayfair]

I'm working on a bypass scheme for a stereo in/stereo out effect that switches an H11F1 on when nothing is plugged into the right channel jack, so that it can either be two independent channels or a stereo splitter.

This is how I have it drawn up right now:

Left input > Q1 Buffer ------> ---> Q2 left side effect > Left output
                                          |
                                           \
                                          Sw (H11F1)
                                          |
Right input > Q3 buffer ----> --> Q4 right side effect > right output
        |
         \
      Sw jack grounds the gate when nothing is plugged into the Right jack -- not sure if this is needed

Description:
In "mono in" mode, the signal is split after the LEFT buffer and connects directly to the right channel "effect" section. My concern is that this could do something goofy with Q3 being unconnected, or perhaps that the output impedance of Q3 might affect the output of Q1 (can that happen)?

It would be a little more convenient to do this (simplifies the question of what Q1 is doing in "mono in"):

Left input > Q1 Buffer ------> Q2 left side trem > Left output
|
\
H11F1 (impedance = 300M when something is plugged into the Right jack, 200R otherwise)
|
Right input > Q3 buffer ----> Q4 right side trem > right output

Description:
In "mono in" mode, the signal is presented to two buffers, Q1 and Q3, and split directly from the input. Input impedance is ... the pulldown resistors in parallel (~1M in my case)? Or is it worse than that?

Essentially, my question is: Do I have to worry about loading if presenting an unbuffered signal simultaneously to two buffers with 2.2M input impedance?

[dwmorrin]

I think a more detailed schematic is needed to really answer your questions.
Are the Qs transistors?  What kind?  Opamps?
Depending on the details of your first plan, it should work.  You should be able to mix the two outputs together passively... I would think you've got some kind of output resistors on Q1, Q3.  But I'm assuming a lot.
Regarding the second plan, a 2.2M input impedance is a light load to a guitar pickup... but again the details of what exactly Q1, Q3 are are needed for a definitive answer.
And when you say, "grounds the gate" - what gate?

[midwayfair]

I think a more detailed schematic is needed to really answer your questions.
Are the Qs transistors?  What kind?  Opamps?
Depending on the details of your first plan, it should work.  You should be able to mix the two outputs together passively... I would think you've got some kind of output resistors on Q1, Q3.  But I'm assuming a lot.
Regarding the second plan, a 2.2M input impedance is a light load to a guitar pickup... but again the details of what exactly Q1, Q3 are are needed for a definitive answer.
And when you say, "grounds the gate" - what gate?

All FETs. Q1 and Q2 would be MPF102 or similar. Q3 and Q4 would either be 2N5457, 2SK170, or J201 or something else with similar gain.

There's a 33K between the output of Q1/Q3 (after the decoupling cap) and the gates of Q2/Q4. LIke this:

Left channel looks like this in the first method:
       +9V
         |                           bias
         D    10uF--33K--*     |
------G    |                       D
|      S ---10uF--33K-------G
|      |                 |           S
2.2    10K          2.2M         |
G      G                G           bias

Right channel looks like this:
       +9V
         |                        bias
         D                *__     |
------G                     |    D
|      S --10uF--33K--^---G
|      |                 |         S
2.2    10K          2.2M       |
G      G                G         bias

*is the optical switch and they are only connected when the right input jack is not used.

In the second example:
Left channel looks like this in the first method:
       +9V
         |                           bias
         D                            |
*-----G                            D
|      S ---10uF--33K-------G
|      |                 |          S
2.2    10K          2.2M        |
G      G                G          bias

Right channel looks like this:
       +9V
         |                      bias
         D                       |
*----G                        D
|      S --10uF--33K----G
|      |                 |      S
2.2    10K          2.2M    |
G      G                G      bias

*is the optical switch and they are only connected when the right input jack is not used.

"Grounds the gate" > Gate of Q3, i.e., grounded input to avoid noise, like in most true bypass schemes. I had an extra pole on the switched jack and figured I could use it.

[dwmorrin]

I like the switch after the buffers, personally.
I don't see how Q3 is ever "unconnected."  And the impedance looking "back" at Q3 from the gate of Q4 does not look like it will be a problem.
In version 1, you have an unused transistor in "mono" mode.
In version 2, you have redundant buffering.
Take your pick.

[screamersusa]

When switching things around like that, I usually short the unused input to ground.
Less noise and or rf pickup and preventing any externally injected noise into
the power source.
I would add a 10k resistor between q1 n q2, another between q3 n q4.
I would use a dpdt to switch q3 out of circuit just before the 10k connecting it to
the output of q1 before the 10k. The other side of the switch would short the input of
q3 to ground. This way it would also work in high gain settings such as a high gain guitar amp or
a 50kw pa system.
I'm a hack and I used to mod big mixers a lot.  

[PRR]

> switches an H11F1 on when nothing is plugged into the right

Why??

*Everybody* does it with a switched jack:



If you just got one plug, it goes to Mono In and comes out both outs.

If you got two plugs, each gets its own path.

If you got no plugs, the Mono/Left jack is shorted thus the Right path is shorted, thus no surprise when some fool uses the output(s).

The buffer-butlers take each signal *gently*. Two 2Meg inputs (1Meg mono) ought to be plenty gentle for any signal on a wire. (Three feet of cable is 100pFd which is under 200K at the top of the guitar band.

Butler-buffer hands-off to the Effect which does whatever it does.

[midwayfair]

That may be the most awesome graphic ever posted in a tech help thread ....

[dwmorrin]

Amazing schematic PRR.  I am going to have the buffer/butler image burned into my head now.  Brilliant!

[bluebunny]

That may be the most awesome graphic ever posted in a tech help thread ....

+1.  Needs to go in the Pictures thread.  Talk about memorable...   

[kingswayguitar]

That may be the most awesome graphic ever posted in a tech help thread ....

+1.  Needs to go in the Pictures thread.  Talk about memorable...   

Wow !!