Power switch for Op-amp w/dual P/S

[guitician]

I'm working out a pedal circuit that uses +/- 18v from four 9v batteries. How would I add a power switch to it? Can I just lift one leg, leaving the other connected? Or would that drain current off the other supply? Can I just lift the Common ground leg from the op-amp leaving the +/- connections on the op-amp? The usual input jack grounding the power supply was what I was thinking instead of using a DPDT switch. The chip is a TL071CP.

[eschertron]

Off the top of my head, I can't see how to use one ground-lift switch to disable two essentially independent power rails. At least not mechanically. Use something to gate two FETs, interrupting the power at the +/- rails, with only the 'monitor' circuit active at all times? Just my WAG.

[The Dude]

You could use a jack with 2 switches built in and interrupt both supplies.

[guitician]

I think leaving one supplies connections hooked up will work because it needs the other supply to complete the full +/- supply loop. Because when used on a single supply, you ground the negative supply pin. I'll hook it up and see if any current is drawn with only one supply connected. Maybe I should just use the voltage divider/Ref., and run a single supply voltage.

Just tested it...no current with one supply connected. I guess I'll do it that way, just switch one set on/off.

Dude, I may use that idea...a stereo jack...grounding the positive/negative from both sets.

[g1]

Did you try with signal when checking current?
I guess it doesn't matter much as long as it's not draining the battery when turned off.

[tedmich]

Jeeze 4 batteries? Its not like you jump around with a pedal, why not a nice linear PS?
If you are set on batts, Keystone makes a handy series dual snap connector
229 Keystone Electronics | Mouser
I'll have onboard on my new EMG equipped project, combined with vertical clips
http://www.mouser.com/Search/Product...tualkey534-080
to take minimum space
Also gonna try a linear Acopian supply regulated down the 18VDC, since its gonna be corded anyway...

[guitician]

I did the 4 batts for simplicity more than anything else. I thought about borrowing some juice from a wall-wart that my other effect uses, and making a through-jack with a plug going into the original effect. But then, I would have to do that split bias thing. I bought clips from mouser and mounted them right to the box where the batts are externally mounted, insulated of course.

[Alan0354]

Can you run pedal circuit from +36 to ground, ground reference at half voltage. I do all my pedal circuit with 9V or 18V to ground and AC couple signal so I can use half battery voltage as reference ground. Then you have only one supply to switch. I designed a few pedals like this.

BTW, why do you need 36V?

[tedmich]

I'm guessing its the twin goals of voltage swing and headroom! (or maybe owns EverReady or Duracell stock )

[R.G.]

There is a circuit to do exactly that on my web site, geofex.com. You use PNP transistor for the + supply, an NPN for the - supply, and one transistor to turn both NPN and PNP on when it's base is pulled to ground by a jack being inserted. Transistors can be switches, too, not just amplifiers.

[Alan0354]

If you start to consider extra transistor, use MOSFET, plenty of small MOSFETs with low Rds on that you don't even have to worry about the 0.3V drop of a BJT as switch. But I always bias the ground reference at half voltage and just use single supply.

[guitician]

It seems that if I just connect the -18v, it will draw .00003A just waiting for the +18 to be connected, which works for me. I can then just connect the neg. for the +18 with the stereo jack. Simple is the name of the game with this build.

Yeah, I thought about really having a lot of headroom, but then again the long life of the cells with so much extra to work with is a plus.

[R.G.]

It seems that if I just connect the -18v, it will draw .00003A just waiting for the +18 to be connected, which works for me. I can then just connect the neg. for the +18 with the stereo jack. Simple is the name of the game with this build.

What happens inside the circuit when you disconnect just the V- or V+ supply is heavily dependent on the insides of the IC. Some chips it'll be fine on, others it will discharge the batteries, and others it will violate the common mode input voltage range and could cause SCR latchup that would then fry the chip when the other power supply is turned on. No way to tell without trying it. That uncertainty is what makes me like switching both V+ and V-.

There is a problem with using the input ground lead to connect power to your circuit when you use a single supply. This forces 100% of all the power current used by the chip to flow in the same wire that carries the delicate ground side of the input signal. Often you get away with it, as witness the many pedals that don't have a problem. But there are circumstances, even as simple as a dirty, oily, or corroded jack or plug, where this will cause unwanted feedback from the power current through the resistance of the connection.

MOSFETs can give lower on voltages than bipolars, and vice versa, depending on the conditions and especially how hard the base is driven. It's pretty easy to get under 100mV with the currents drawn by a single opamp chip.

[R.G.]

Ah. Found it. I've attached it locally.



Notice that the base current for the middle transistor comes from the ring connection on the socket, and that its current is returned through the two NPNs to the V- supply. Any voltage this causes is no longer in series with the signal ground, so it cannot cause feedback or notch/peak response issues. The base current is low, and fixed, and there is beta-squared gain from the turn-on signal to the power supplies and this is not voltage limited, so there is good saturation for low Vce on the actual switch transistors.

[loudthud]

With one rail floating, current will want to flow out the output pin or the inputs. If the input transistors are JFETs, the current could be substantial but it depends on which rail is disconnected. Many chips have parasitic diodes to the substrate and those can turn on and leak current out the inputs. Any cap in the feedback network might see the full rail voltage so polarity is important if it's an electrolytic.

Ever wonder how those automatic toilet flushers work? There is an opamp with an enable pin. The opamp gets turned on for short periods of time to read the IR sensor. The battery will last 10 years. Look for an opamp with an enable function or programable bias current. You can usually program them to be essentially off.