Input jack power switching for +/- 9V dual-battery supply?

[poojalooba_cow]

I'm building an R.G. Keen Neutron ECF filter and I've got everything up and running but can't think of how to switch the power with the jack when there's a two-battery supply.  The in/out jacks are supposed to go to the ground - not +9V or -9V, and R.G's layouts aren't helpful either because both of them feature his optional single-battery supply with a 7660 power transformer, where the power is switched by attaching the base of a transistor to the jack, effectively blocking the circuit until a jack is connected.

I have the parts for the 7660 power transformer option but I'd prefer to use two 9V batteries so I don't have to replace batteries quite as often, but with two batteries I can't figure out how to get the circuit to turn on/off when a jack is inserted.  I could just use like a SPST switch to do it (between battery leads, but this doesn't work with the jack because of grounding), but is there another way? I did Google this before I posted here, and I didn't turn up anything except one un-verified post where somebody mentioned that you could use two stereo jacks for input and output, and use one to switch the +9V on and the other to switch the -9V on, but won't that cause grounding issues with the jacks?  And what do you do about the jacks supposed to go to the 0V ground then?



This is a rough drawing of what was suggested in the thread I found when I Googled it.  This wouldn't work, right?  This is the page where it's referenced:
http://www.diystompboxes.com/smfforum/index.php?topic=54534.msg417052#msg417052

[italianguy63]

Might I suggest??

I just built one of these....  love it BTW!

I would use the 7660 charge pump circuit.  It looks to me like this was designed some time back when the 7660 was in it's infancy.  Instead of the 7660 use a TC1044S or equivilent charge pump IC.  If you use a "modern" one, you will have to add a "boost" trace on the PCB, and omit the chokes/inductors.

Then, just use the single battery configuration.  If you want more "life," then just put (2) 9V batteries in parrallel.

I'd be glad to share pics of my build if you need them-- just PM me and I can get back to you.

EDIT:  I also did not like the way the signal/jack power was with the DPDT.  It did not shut off the circuitry when the effect was "off/bypassed."  The envelope light would still light.  I modified it to a 3PDT switch with a seperate power LED too...

Mark

[KazooMan]

I think that the usual method of switching the ground connection should work just fine.  From the information at RG's site I see that he has the dual battery supply connected to the board at -9V, +9V, and "0" points on the board.  The signal ground (ground lug on the input jack) is also shown as connected to the "0".  This is the ground point for the entire pcb.  You should be able to run the "0" output of the dual battery supply to the sleeve connection on the input jack and have the ground from that jack go to the board.  The ground from the battery would be open without a plug, turning the power off.  Inserting the plug completes the circuit as usual.

Oooops........ Forget that!  I took a closer look and I see the real problem now.      RG mentions the problem of trying to switch the -9V side and a potential solution to prevent the whining that results.  I'll have to think some more about just what would hapen if you wired it as I suggested.  Probably a big cloud of nasty smelling smoke. 

I think the "best" solution if you want to use batteries might be an honest-to-goodness power toggle switch.  A double pole switch could be wired to switch both the pos and neg leads.  Incorporating a power on indicator LED would be easy.

I agree with the post that appeared while I was typing my initial response that the charge pump with a power input jack would be ideal.

[ashcat_lt]

In order for this to work, you have to break the power supply circuit.   The "ground" 0V point usually is not really part of the power supply circuit.  DC doesn't actually flow through those caps that "create" the 0V AC reference point.  I'm thinking that if you put the bottom of the top battery on either R or S and the top of the bottom battery on the other, it should work just fine.

[Seljer]

I can confirm that this works just fine: http://www.geofex.com/FX_images/PNP_power_switching.pdf

Mirror the circuit with an NPN transistor on a negative rail and you're set,.

[italianguy63]

R.G. already has the switching circuit built into the PCB on the charge pump section...  if he chooses to use it.

[R.G.]

This is probably a good place to repeat my advice on input jack power switching.

That is: don't do this, at least the way it's most commonly done.

That being to connect the battery/power supply negative to the circuit ground by using a stereo jack and letting the input cord sleeve short the power source to ground.

That forces 100% of all the current the power supply provides to flow through the possibly dirty or not very solid connection on the input jack. That being the case, every possible bad effect of poor grounding is forced to happen because any voltage drop between the input grounding and the circuit is forced to be in series with the input voltage. It's an amazing example of how very much we get away with because pedals are such small systems that it works for any high gain pedals at all.

It's OK to use the stereo jack trick, but it's a lot better to use it only as a signal that activates the power flow through a separate switch.

But of course, it always works  - until it doesn't.

[Fender3D]

Have you room in your enclosure for this?
http://www.banzaimusic.com/Taiwan-jack-DPDT-closed-Stereo.html

[R.G.]

Bad idea, version II.

Then there is the whole issue of interrupting power by pulling out an input (or output!) plug. Back in the day when a guitar player had one or perhaps two pedals, this was a GREAT idea. It saved a whole power switch, and pedals were always unplugged when playing was over.

Fast forward to today. Guitar players have one or in some cases TWO pedalBOARDS. Is it really a clever idea to stop the batteries from draining by inplugging all the signal cords on a pedalboard? Doh!

Ah, you don't use batteries, you use a power adapter because batteries are not really practical in 25-per-case lots.

In that case, why is there a power switch on the pedal at all? Why doesn't it simply get power from the power jack all the time, and the power switch becomes plugging the power adapter in?

Like true bypass, input jack power switching is a solution to a problem we no longer have very much.

[Fender3D]

lol 

I agree with you about batteries and in special way about true bypassing, but if someone is willing to devolve his whole wage to Duracell....

[ashcat_lt]

That forces 100% of all the current the power supply provides to flow through the possibly dirty or not very solid connection on the input jack. That being the case, every possible bad effect of poor grounding is forced to happen because any voltage drop between the input grounding and the circuit is forced to be in series with the input voltage.

Yeah, but here it's halfway between the ends of the power supply, so it's twice as bad!    

All that said, would it "work" the way I indicated above?

[amptramp]

Of course, you could always switch the positive battery from the input jack and the negative battery from the output jack (or vice versa).  But batteries are a colossal pain at the best of times, so I would go with external power supplies only.  Normal 9VDC supplies on two jacks that are insulated from the chassis so one can be outer shell to ground and the other can be inner pin to ground.  That allows you to use two identical wall warts.  If you don't have internal batteries, you don't have to reserve volume inside the enclosure for internal batteries - jacks are smaller.