Do I need to bias a CD4066 when switching.

[alparent]

Would  I need to buffer all 4 inputs?
What would be the best buffer? Using simple jfet or using a quad op-amp?

[R.G.]

OK now you've done it!  
Now you have to tell me why and how you would do this?

I'd need something with no knobs or switches and something that will not tint the signal being switched.

I would use a JFET input opamp like the cheap, easily available TL07x series. Just power it from 9V and ground, bias the + input to Vref with a 1M resistor, use a 0.1uF film input cap and a 2.2M pulldown resistor on the outside of the cap. Connect output pin to inverting input pin. No knobs, no switches, and flat from below audio to mostly beyond audio, at least for small signals. No tint. Flat response. No guitar loading. It's a remarkably simple, non-distorting, non-coloring solution. It'll happily drive a 10K load resistor. In fact, depending on several imponderables, it ...may... be able to connect the output pin to the switch chip pin directly, since the opamp output is already sitting at Vref for DC purposes. There are some minor gotchas there, but you might be able to do it and have it be quiet and easy.

Would  I need to buffer all 4 inputs?

Wherever there is a guitar driving an input, a buffer is a good idea, or at least one to use unless you have specific technical reasons for why NOT to do it. So yes, if they're guitar inputs, I would.

What would be the best buffer? Using simple jfet or using a quad op-amp?

"Best" is a very emotionally loaded word. I prefer to say that a modern JFET opamp is very hard to beat for ease of use, quiet, and smooth, uncolored response. There are situations where something else works better, but "better" needs to be tightly defined to be able to show the other solution is better. And for reasons unrelated to this whole string of reasoning, I would use two duals, not a single quad. The layout is easier on PCBs.

[alparent]

I was looking at this as a buffer http://www.muzique.com/lab/superbuff.htm
Looks a lot like what you are describing (if you forget the 3 extra stages).
What about the comment about the 2 output caps he is using?

[R.G.]

I was looking at this as a buffer http://www.muzique.com/lab/superbuff.htm
Looks a lot like what you are describing (if you forget the 3 extra stages).
What about the comment about the 2 output caps he is using?

Sometimes I despair from the sheer weight of overblown garble on the internet. There's nothing particularly "super" about that. If one wants high current for cable drive, there are other ways to get it. And very, very few people will drive cables long enough to need anything like that.

I guess I should have tagged more things with "hyper-" "ultra-" "meta-", and other esoteric add-ons in the title.

Yes, use the simplest opamp buffer, it's more than enough for what you're doing. The comment about the two output caps is specious for your purposes.

[alparent]

I guess I should have tagged more things with "hyper-" "ultra-" "meta-", and other esoteric add-ons in the title.

Your a funny guy!


I have a tendency to over complicate things, I want my builds to be as good as they can be and I sometimes end-up adding to much!
As Mark tells me...........keep it simple...........your adding to much!
(I guess the more I learned why things are used I will get a better feeling on what is needed and what is not.......I hope!)

OK thanks for all the help R.G.

One last thing......can you point me to ..... or put on a lecture tape on why adding a buffer is a good thing?

[alparent]

Ok! I added the buffering stage as you suggested.
So I'm going to use 10k resistors to pull to Vr.
And the output of the op-amp connects to the 2.2uF DC blocking caps.....right?

[R.G.]

One last thing......can you point me to ..... or put on a lecture tape on why adding a buffer is a good thing?

It's pretty simple. A guitar pickup closely resembles a voltage source for the signal, but in series with about 2K-10K of resistance in the wire, and 1/2 to 4 **Henries** of inductance. The impedance of that inductor goes up linearly with frequency. So at low frequencies, like 82Hz, low E, the impedance of a middle-of-the-road 2H pickup would be
Xl = 2*pi*82Hz*2H = 1030 ohms. That's in series with, say, 4K resistance.

However, at 7kHz, upper harmonics of strings, that becomes
Xl = 2*pi*7,000Hz*2H = 87920 ohms, with the same 4K resistance. It's not quite this simple, but if we think of that 87920 ohms of inductor as a resistor, and we load the output with an 82K resistor, then the total signal voltage will be divided across the inductor impedance and the load resistance. Ohm's law tells us that the voltage we'll see at the output resistor will be about half of what the pickup generates. We've lost half the signal.

But only at 7kHz. At lower frequencies, we lose less. What this comes out as is a loss of treble, since we lose more treble frequencies than bass. And so far this is the argument for using high impedance amp inputs. If we loaded it with a 1M resistor instead of an 82K, the treble losses would be almost unnoticeable.

If you put a buffer right at the guitar, with an input impedance of 1M or more, the buffer will have a quite low output impedance. It can drive the capacitance of a long cable with little or no treble loss. Capacitance is even worse than resistance loading, because the impedance of a capacitor goes *down* with frequency, meaning it makes even more treble go away as frequency goes higher. The important point is that a buffer can drive a capacitive load and not be bogged down and lose treble.

So buffers are good because (a) they have a big input impedance to not load whatever drives their input, and (b) they have a low output impedance so they can drive long cables without treble loss. They make the other loading issues not matter, or not matter so much.

And the output of the op-amp connects to the 2.2uF DC blocking caps.....right?

Yep.

[egasimus]

Another question closely related to the topic: If I use one of a 4053's switches as a SPST switch, do I still need to bias the unused output so that I don't get pops?

[R.G.]

Whenever you're using **CMOS switches of any kind** to switch any signal you will be listening to, you need to worry about the DC levels on the switch pins. If you want lowest clicking, always bias the pins to half the power supply voltage on the CMOS switch.

Do you need to bias the unused output? Only if you don't care that switching to the unused output may cause a different voltage that the unused output may have drifted to. *You* may not need to, depending on your circuit layout and whatever else on that board and on your tolerance for clicking.

But *I* would, for anything I did for myself or a customer. Up to you.

[alparent]

OK this is coming along great!

Little question about the DC blocking caps?

My setup is basically this:

              ____                                                                                                                ____
in a  * --  |  4  | --?  all connected to "to effects"     then     all connected to "from effects" ?--  |  4  | -- * out a
in b  * --  |  0  | --?                                                                                                  ?--  |  0  | -- * out b
in c  * --  |  6  | --?                                                                                                  ?--  |  6  | -- * out c
in d  * --  |  6  | --?                                                                                                  ?--  |  6  | -- * out d
              -----                                                                                                              ------

My question is do I put the 2.2uf DC blocking caps only at the inputs and output (identified by a *) or do I also need to add caps where the (?) are?

I hope my question is clear?