Using CMOS switching instead of a 3PDT?

[spargo]

I'm still in the process of creating a momentary XLR A/B box.  I got some momentary 3PDT switches from Mammoth but am not satisfied with the quality of the switches, and the force it takes to depress them as well as the click of the switch (even though it's momentary).  I very much so like the soft touch SPST momentary switches I've gotten in the past - they are perfect.

Is there a way, using some sort of CMOS switching, that I could utilize these SPST switches to accomplish a 3PDT task?  If it makes a difference, the switch will accommodate condenser mics as well so it must allow 48v phantom power in the circuit.  I'm completely unfamiliar with CMOS switching - or maybe someone can point me to some good resources.

[Hides-His-Eyes]

Nnnng...

This is one instance where I'd prefer to use relays. You can't take a CMOS switch above the power supply voltage, and I doubt they like running on 48V.

[km-r]

have you tried venturing with relay-switching? you can use flipflops to trigger relays [with a driver of course]. and a lower voltage high current supply.

flipflops dont take much current so you can use a standard IC regulator and a voltage drop resistor to power it. you may also use LDRs?

or this
http://www.geofex.com/article_folders/cd4053/cd4053.htm

[spargo]

I'm not familiar with relay switching.  Do you have any links?  I was beginning to look at that geofx article...

I guess what I really need is a 3PST.  I won't be using the second poles at all if that is helpful.  Is there such thing for a high quality momentary switch?

[egasimus]

Nnnng...

This is one instance where I'd prefer to use relays. You can't take a CMOS switch above the power supply voltage, and I doubt they like running on 48V.

Hmm. 48V is the DC bias, not the signal's peak or peak-to-peak voltage. You could remove it and re-add it at the output. It could be done in an elegant way, probably.

[km-r]

heres a link from the geofex website. the diagram indicates the use of a SPST switch or a latching SPST pushbutton switch. a 3PST switch may not be used as an A/B box, you may be referring to a SPDT switch?

http://www.geofex.com/Article_Folders/relays/relays_for_switching_audio_signa.htm

[stringsthings]

I'm completely unfamiliar with CMOS switching - or maybe someone can point me to some good resources.

http://www.paia.com/epfm.asp

project #15, p.127

[spargo]

Well I made up "3PST".  I basically need three SPST switches, so I don't need a full 3PDT but that's what I have because I didn't know of another way.  I just can't find any quality momentary 3PDT switches so I was looking to see if another route would be better.

I may try the CD4053 route, but I find the geofx article a bit confusing as I'm not setting it up for a stompbox.  The other things is for this application I'm very limited on board space, and if the CD4053 requires me to add another 10 resistors and another 10 capacitors, it probably won't be possible.

[R.G.]

Well I made up "3PST".  I basically need three SPST switches, so I don't need a full 3PDT but that's what I have because I didn't know of another way.  I just can't find any quality momentary 3PDT switches so I was looking to see if another route would be better.

Every switching job requires that you **know** ahead of time what signals you'll be switching.

If you're switching mike levels, CMOS will do it. Using CMOS requires only a signal voltage that does not go outside the power supply to the chip. For a 4053, according to the datasheet, that's 15V max. If you use a 4053, and you *have to* have 48V tolerance, you're going to need to split out the AC signal, then recombine with the phantom 48V.

... which turns out not to be too hard, for the 4053's part. Quiet switching with a 4053 requires you to force the DC levels to be what the 4053 wants, not what's on the outside anyway. So you'd use the same capacitor input/output to break the DC voltage from the audio line, then re-add the DC after signal switching. The problem ois that you need to let the DC through from the phantom supply line and pass it to the other side of the switch for recombining with the switched signal. That needs not a capacitor, but an inductor. The inductor has to be big, and probably will be expensive.

The other things is for this application I'm very limited on board space, and if the CD4053 requires me to add another 10 resistors and another 10 capacitors, it probably won't be possible.

That's is a real problem. A 4053 and resistors/capacitors are not very large compared to the other solutions which are possible.

I suggest you read up on relays, and count on using some of the low power relays. However, you'll probably need to use two DPDT relays to do the switching. That is going to be larger than a 4053 plus support parts. You *may* just need a larger box. I can't do a better job of guessing without knowing more ab out the application.

Another possible solution was to use the H11F1/F2/F3 LED-photoFET devices for three SPST switches, but again that's three six-pin DIPs and is a fair amount of PCB space.

[Hides-His-Eyes]

For more space, what's your vertical clearance like? You could use header pins for a "two storey" section.

[spargo]

This will be in a standard 125B enclosure - another size is not an option at this point.  The XLR jacks will be mounted on one of the sides straight into the PCB with horizontal pins.  So, the PCB is vertical and size is limited to the dimensions of the side of a 125B (around 4.5x1.25in).

I have been able to fit the CD4053 and extra resistors and capacitors on the board layout, but I'm not sure it's all I need.  I basically included the parts in the first diagram of the geofx article - "Using the CD4053 for analog signal switching", and then treated its inputs and outputs as a basic 3PDT, applying voltage to A, B, and C using a SPST momentary switch.  I think I'm missing something though.

The XLR signal (pins 2 and 3) will be going through the CD4053 pins, and must be able to accept phantom power.  If I need to remove and then re-add the voltage, how do I do that?  I suppose I could always just give it a go and see if the chip explodes.

[spargo]

So I've wired up the CD4053 layout to my switch box setup.  I won't be able to test it until later tonight (left a power cable somewhere), but here's a question.  I didn't fully follow what R.G. was saying earlier about removing and re-adding the voltage.  If you look at this document: http://www.geofex.com/article_folders/cd4053/cd4053.htm

How I have it setup is I have the mic signals running through the SPDT switches (pins 1, 2, 12, 13, 14, 15).  These signals may contain up to 48v of phantom power.  However, the voltage being applied to pins 10/11 for the switching to take place will be a standard 9v.  R.G. mentioned the CD4053 could only support 15v.  Does that only apply to the voltage being supplied to the chip for its power and switching, or does that included what's running through the switches as well?

Later tonight I plan to test it and see if it works, pops, or explodes.

[spargo]

So I've got everything wired up and went to test the A/B switching just with a dynamic mic first.  I was surprised to find an incredible amount of hum/buzzing coming through the channel (enough to clip the track on the sound board if I had the gain up to a reasonable level).  I just have this laid out on a breadboard right now.

After doing some digging, even with ALL power connections on the board severed to the CD4053 chip and simply the 9VDC power running through the breadboard, the noise is there in the signal.  It persists even if I only plug in either the + or - of the 9V power connector.  I'm very confused - just putting 9V into the rails of the breadboard and connecting nothing to it makes all this noise (and some faint radio stations as well).

I've added a power filtering section but there has been no change.  Again, I'm starting simple just with a dynamic mic that is not phantom powered.  Any ideas?

[spargo]

I've fixed the vicious hum by grounding the XLR grounds (pin 1) to the circuit ground as well.  Now on to the next issue.

I'm basically making a more complex version of this (schematic at bottom): http://www.procosound.com/download/datasheets/Old%20Version%20of%20Cough%20Drop.pdf

When run through the CD4053 as a switch, the XLR channel does not fully mute, it just reduces the volume.  Anybody know why that might be the case?

[spargo]

I think so far I've determined that within the CD4053 it's not a straight path when switched...there is some resistance which ultimately makes it so the mic signals are not full cut.  Anybody have an idea of how to get at this?  Maybe somehow instead of actually running the mic signals through the CD4053 to be switched, the chip would simply trigger something else to do the switch.  Is that possible?  I'm not able to use relays.

[Processaurus]

The XLR signal (pins 2 and 3) will be going through the CD4053 pins, and must be able to accept phantom power.  If I need to remove and then re-add the voltage, how do I do that?  I suppose I could always just give it a go and see if the chip explodes.

Don't worry, it will.  In any case, you'd want the phantom power to go to both mics at once, because there is no guarantee they will turn on instantly and silently.  Same reason you don't turn the power off and on to stompboxes with the bypass switch to save battery.  POP!  

Has anyone tried a circuit for splitting phantom power?  I can't think of an immediate way to do it without adding an unacceptable amount of series resistance and possibly sagging the voltage the mic sees; phantom power spec I believe has, in the source device, two 5k resistors from the +48 power supply to mic pins 2 and 3, so the only way I could imagine it is in spargo's splitter box, 5K resistors from the box's output pins 2 and 3 to a BFC (big freakin capacitor) to ground, and from there, 5K resistors out to the pins 2 and 3 of both mic inputs.  You would need to AC couple the signal lines, and add zener diodes or reverse biased diodes to the power rails, on the cmos switch side of the cap to protect the CMOS IC from seeing a big 48v transient sneak through the cap and hit the IC when the phantom power gets turned on.

How bad do you need phantom power?  

[spargo]

I need phantom power...but maybe I should clarify this.  I have the circuit 100% working without any pops or clicks on both dynamic and condenser mics.  The issue I ran into is this: my design needs a DPDT momentary switch, but I'd also like an LED indicator.  This pushes me to a 3PDT momentary switch.  I got some of these from Mammoth, but am very disappointed in the quality.  They require too much force and I could see them breaking at some point.  On the other hand, I LOVE the soft touch SPST momentary switches I've gotten from Smallbear, so I'm using the CD4053 as a means to explore using the SPST switch to perform a 3PDT task.

I've gotten the CD4053 to perform clickless switching, but it doesn't fully mute the channels as with the original design, which I suspect is from the resistance encountered within the CD4053 switching mechanism, and I'm trying to think of a way to solve that.

[R.G.]

... it doesn't fully mute the channels as with the original design, which I suspect is from the resistance encountered within the CD4053 switching mechanism, and I'm trying to think of a way to solve that.

The traditional fix is to not use a single switch, but to use a switch that opens the direct signal line, and then resistance shorts the output line.

If you're redoing the noted schematic, then try attaching the output line to the pole of the 4053 switch, and the incoming signal to one of the throws. The other throw goes only to a capacitor-to-ground-shorted reference voltage. What that does is put the **open** resistance of the 4053 in series with the incoming signal and the closed resistance of the 4053 to reference. The open resistance of the 4053 is in the hundreds or thousands of megohms, so even a few hundred ohms series resistance on the closed side gives you near-infinite muting.

That is, you're copying too closely for the actual hardware you're using.

[spargo]

... it doesn't fully mute the channels as with the original design, which I suspect is from the resistance encountered within the CD4053 switching mechanism, and I'm trying to think of a way to solve that.

The traditional fix is to not use a single switch, but to use a switch that opens the direct signal line, and then resistance shorts the output line.

If you're redoing the noted schematic, then try attaching the output line to the pole of the 4053 switch, and the incoming signal to one of the throws. The other throw goes only to a capacitor-to-ground-shorted reference voltage. What that does is put the **open** resistance of the 4053 in series with the incoming signal and the closed resistance of the 4053 to reference. The open resistance of the 4053 is in the hundreds or thousands of megohms, so even a few hundred ohms series resistance on the closed side gives you near-infinite muting.

That is, you're copying too closely for the actual hardware you're using.

Could you elaborate on that some?  The switch box doesn't have an input/output per se running through the switch.  The switch is being used as a means to short a signal which effectively mutes the mic.  Very similar to this diagram: http://www.procosound.com/download/datasheets/Old%20Version%20of%20Cough%20Drop.pdf

Actually, the best luck I've had so far (that created no pops) was to remove ALL of the resistors and capacitors in that CD4053 switching diagram and to simply plug things straight through the chip.  However, I still had the volume issue.

[R.G.]

Actually, the best luck I've had so far (that created no pops) was to remove ALL of the resistors and capacitors in that CD4053 switching diagram and to simply plug things straight through the chip.  However, I still had the volume issue.

OK, I'm confused now. Can you post a conceptual schemo of what you're trying to do? Use switches instead of CMOS, and ignore anti-pop stuff.