Chaining CD4066 for more switching poles?

[sfr]

Say I wanted to be able to switch a whole lot of bits of audio signal - the sort of thing that might require a 5PDT or 6PDT switch or something, maybe more. 

I'm thinking right now about being able to have two effects in the same box or same slot in a unit, using the the same the bypass switch, in/out jacks and pot, with a small toggle to switch which effect is using those external components.  (Plus a couple of LEDs to know which effect is in use because I'm a guitarist and shiny things make me happy.)  Like being able to switch between two different phasers or something, since I don't see myself needing to use both at once.

Now, I suppose I could use a rotary switch; but say I want to use a micromini switch, or I want to use a footswitch so I can do this on the fly - seems like electronic switching is the way to go.  (I do actually have some mini 4PDT, but the uses I'm thinking of may need one more pole)

If I'm using CD4066 to do my switching, can I connect the control points of two of these chips to the same inverter gates (like in a CD40106) or flip flop, to get more switching poles?   I'm probably planning on ordering a bunch of CD4066 and CD40106 anyway, so being able to use what I'll soon have in my parts bins would be handy. 

[R.G.]

Using several CMOS switches driven by the same signal is entirely feasible. You mention the 4066, but I like the CD4053, which is three SPDT switches each independently controlled by one of three control pins on the package. The control pins are side by side, so just connecting all of the control pins makes the 4053 into a 3PDT. You can then use the same control signal to switch a second 4053, or a third. All you have to worry about is whether the part doing the switching can drive so many inputs. In this usage, I think the answer will always be yes, as you don't really care whether the parts switch in 100nS or not. So it's easy to make a 9PDT with three 4053's.

[sfr]

Ugh, that's annoying to look at.  I accidently used brackets instead of parentheses when writing "pot(s)" so half of my post came up in strikethrough!  Is there a way I can change that?

Thanks R.G. - sounds like it should work.  I haven't decided whether I'm going 4053 or 4066.  4066 I can get locally, so I've been experimenting with one of those.  (I supposed I should have just driven out and gotten a second one rather than ask my question here.)   I think I'll probably order a bunch of 4053 at some point here.  The versatility seems useful.

So here's another couple of questions for anyone that wants to chime in - say I'm running a transistor into a 4013 to control my 4053, or using the 4049 to control the 4053.  (Like the circuits posted over at geofex :  http://geofex.com/Article_Folders/cd4053/cd4053.htm )  The whole thing driven by a momentary switch. 

Say I also wanted to implement momentary switching as an option for my effect - ( a flanger for example, I often use on just maybe the last bar of a phrase, it'd be convenient to have it only on when I'm standing on it.)  I should be able to simply use a toggle switch and cut out the 4013 or 4049 based circuitry that does the "latching" part, and just drive the control lines of the 4053 solely with the momentary switch, right?  Do I need to do anything with the latching circuitry when it's disconnected?  Ground both ends of it or anything? 

This opens up some other possibilites with having one footswitch capable of doing other things (particularly with that extra SPDT in the 4053) but I'm not certain what happens when I disconnect the footswitch from the latching circuitry; I don't want it changing states on me after I toggle the switch over to something else, particularly if I leave it connected to switches in the 4053 that are controlling bypass. 

Some of the things I'm dreaming up, I'm afraid I'm going to end up with a recursive nightmare of electronic switches controlling electronics switches controlling electronics switches . . .

[R.G.]

I mentioned the CD4053 because you will be making SPDTs from the CD4066's, two of them per IC package. And each single switch needs the control and the inverted control run to the two switches that make it up. The 4053 only needs one control per switch and you don't have to generate the negative of that signal to run the switch.

Said another way, a 4066 is not a complete SPDT. With a 4066 and two inverters you can wire up two SPDTs. But the inverters are a necessary part of the setup. A single 4053 gives you three SPDTs each driven by a single control signal, no inverted control signal needed.

CMOS logic lets you do what pieces of switching you like, not what the mechanism likes. For instance:
(a) you can make an electronic SPDT from 1/3 of a 4053 or from half a 4066 and 1/6 of a 4049.
(b) Once you have that SPDT wired up, it's an SPDT and you can quit worrying about what happens inside it. All you need to know is that when the control line changes state, so does the switch.
(c) The switch doesn't care what happens outside it. It only looks at the control signal and does as it's told. So if you want to drive it from a latch, half a 4013 is a good way. Or a mechanically latched switch can provide the control signal. Or a momentary switch where your foot is the mechanical latch can provide the control signal. The SPDT switch doesn't care.

Here's an interesting thought: a section of a 4053 can switch audio signals, but it can also route digital signals. You could use one section of the 4053 as a selector switch to select whether a momentary switch provides the control signal to the two other sections, or a latch driven by that same momentary switch. So let's look at that.
We have:
Momentary footswitch -> CD4013 latch -> throw of one section of the 4053.
The same momentary footswitch  -> other throw of the same section of the 4053.
pole of the section of the 4053 -> both control lines of the two remaining sections of the 4053.

There is one more switch, let's say it's a slide switch. The slide switch controls the state of the section of the 4053 that routes the control signals. When this slide switch is in one position, the momentary footswitch is routed to the two control lines of the "remaining" sections of the 4053 which do the switching that you want. The latch is disconnected from the analog-conducting two sections. The latch still changes state each time the footswitch is pressed, but it doesn't care that it doesn't control anything. It just counts its own toes all by itself.

When the slide switch is thrown the other way, the latch output is connected to the audio switching sections, and the momentary footswitch is disconnected from the two audio sections. All it controls is the latch. Now the footswitch acts like a standard latching footswitch. So with the same two chips, you have a momentary AND a latched DPDT analog switch.

[sfr]

this just keeps getting cooler and cooler.   I'm starting to get my head wrapped around this, thanks! 

Now to get my hands on some chips to play with.

Looking at Mouser, I can get the STMicroelectronics HCF4053 for less than half the price of a TI or Fairchild CD4053 (at least in a DIP package - maybe it's time to brave surface mount; but not yet) It appears to basically be the same chip; a few minor things seem different, but I'm imagine nothing that would effect me in this application.  Still, anyone know what exactly I should be looking at to make sure?  I'm assuming the resistance of the "on" state is all I'd want to worry about, and it seems the same.