Using voltage reg. for Vr on multiple CD4053 based switches?

[sfr]

I've been doing a PCB layout for the CD4053 based switching over at Geofex.  (located here)  I'm looking to use these for my bypass switching on a multiple-effects-in-one-enclosure type deal.  Since all these pedals are in one enclosure, I'm looking at using a voltage regulator to get my 4.5V, and running that out to each switching board, rather than putting a voltage divider on each switching board. 

I assume as long as I don't run into the limits of current draw for the voltage regulator I'm using (probably an LM317 since that's what I have handy; which has up to a 1.5A output current, if I recall)  I'll probably be okay. 

My only questions - how would I calculate the current requirements for this application?  I'm assuming it's just Ohm's law - the 4.5V Vr voltage divided by the 1M resistor used on each input or output, (which gives me about .0045 mA) times the number of resistors used (5), so a current draw on the Vr line of about 0.0225mA  for each electronic switch.  Is this right? 

Am I apt to run into any other problems moving the Vr off of the switching boards and onto one unit for the mess?  Is there anything else I could move off the switching boards to reduce redundancy?  ( I don't see how there could be, but I'm not the brightest.)

[R.G.]

I would use an LM386 for this one. Just ground the inputs and take 1/2 of the power supply off the output. This has the advantage that it can pull down as well as up.

I would put a 10 ohm resistor in series with the output  to keep the amp from oscillating and then put a 10uF cap on each local board where it's used. That keeps any signal funnies from Vr impedance or power/ground impedance local to the board that makes the funnies.

A possible move to reduce complexity. Since you're already making a power supply regulator, if you made a minus voltage regulator, you could use the Vee pin of the 4053 connected to -5V and do all your switching at ground.

The 4053 has the internal ability to use +/- power supplies. Putting the Vee pin at -5V and the +V pin at +5V lets you pull all of your signals to the reference voltage of ... ground, which your pulldown resistors already do.

There are some special cases to observe. The CD4053 can run from as high as 15V, so the total power supply voltage on it has to be less than +/-7.5V. I mentioned 5V because it's easy. So you could take your power supply, generate a negative voltage and a positive voltage, then use a 7809, a 7805 and a 7905 to get the voltages. Now your 4053 works with biasing to ground. Be sure that if you do this, the positive signal to the A, B, C inputs is no higher than the 5V supply, or if you feed them 9V/ground signals, you use a series resistor and a diode clamp to keep the input from being pulled above the chip's positive supply.

[sfr]

Hrm, +/- power hadn't occured to me.  If I go that route, I can obviously get rid off the 1M resistors between Vr and the inputs/outputs, but the 2.2uF caps on the input lines and 1M pulldown resistors should stay, correct?

If I tie Vee to -5V, I'm thinking Vss still goes to ground?  And what exactly do I do with "Inh" (pin six) ?  I'm not really understanding what that pin is, although I'm still slogging my way through the data sheet, hopefully I'll figure it out.

[R.G.]

Hrm, +/- power hadn't occured to me.  If I go that route, I can obviously get rid off the 1M resistors between Vr and the inputs/outputs,

Well, actually, you can't. The analog switch pins need to be tied to the reference voltage with high value resistors. In this case, you've just made the reference be ground. However, if there are pulldown resistors that already do that for you, you're all set. But pop performance will be unpredictable if there is no DC reference for an input/output.

but the 2.2uF caps on the input lines and 1M pulldown resistors should stay, correct?

It's more like "any signals you put on the CMOS switches need to have a 0V DC average value". Guitar signals do this, as do the inputs and outputs of effects which are capacitor coupled with pulldowns. But there needs to be a DC path to reference (ground) on each signal on the switches.

If I tie Vee to -5V, I'm thinking Vss still goes to ground?

That's correct. The logic part of the chip still works from Vss (ground) to Vdd (+V).

And what exactly do I do with "Inh" (pin six) ?  I'm not really understanding what that pin is,

"Inh" is a logic input, so it needs to be always 0V or +V. When Inh is at +V, all of the switches are open/off, all directions. Nothing conducts at all. When Inh is at Vss, then the switches work according to the A, B, C inputs. It's an "everything off" logic input pin.

[sfr]

Okay.  But if I've switched over to a +/- 5V supply, I'm no longer using the 4.5Vr that originally tied to the inputs (X, Y, Z) and the outputs (X1, X0, Y1 ...  etc.) through those 1M resistors - so I tie those 1M resistors to  . . . ground?  Since that's my "middle" now?  Since I've already go 1M resistors to ground on the other side of those 2.2uF caps before X, Y and Z, do I need one on both sides of that cap?

Reading R.G.'s reply, and making my way through things here in my head trying to understand whats going on, it's obvious I don't fully comprehend the idea of the DC reference - I'm used to trying to block DC coming out of an amplifier stage in a pedal, not this referencing stuff.  Anyone here  want to point me in the right direction of somewhere I can read up and understand this more, so I don't continue to ask silly questions? 

"Inh" is a logic input, so it needs to be always 0V or +V. When Inh is at +V, all of the switches are open/off, all directions. Nothing conducts at all. When Inh is at Vss, then the switches work according to the A, B, C inputs. It's an "everything off" logic input pin.

Oh, okay.  Thank you. From the datasheet I was gathering it was something like that but I just wasn't making the connection.  Good to know.  Might even turn out handy at some point.