Op-amp output floating?

[JRay]

Hi All

Is it OK to leave an op-amp output floating? I have a buffer stage that uses one input and 2 op-amps, one non-inverting and the other inverting.  I intend to switch the output volume pot (10K log) between each output using a SPDT switch and then out to the mixer/amplifier etc and so have a polarity switch.  Obviously this would always leave one of the op-amps outputs floating with respect to the signal ground. Is this OK or do I need to add a resistor from output of each to ground, which will appear in parallel to the Volume pot when switched in?

Cheers
Ray

[petemoore]

Is this OK or do I need to add a resistor from output of each to ground, which will appear in parallel to the Volume pot when switched in?
  The parallel resistor would reduce the R value of the potwafer R [from output to ground.
  "Typical'' wiring of volume pot puts a resistive wafer [inside the pot] between signal path to ground. Reducing the value of the pot [by replacing with smaller pot, or a parallel resistor fixed across the outside pot lugs]. Smaller values here increase the loading of the output.
   Try opamp data sheets to get a better ideas about the ways OA's can and can't be operational.

[ashcat_lt]

Won't the output of the opamp be held to audio ground via the feedback path?

Anyway, somewhere around here not too long ago I saw a scheme that used just one opamp stage switchable between inverting and non-inverting.  That would moot the point.

[merlinb]

Obviously this would always leave one of the op-amps outputs floating with respect to the signal ground.

They won't be floating because the feedback resistors will still be attached, no matter which opamp is selected, so you don't need to worry about it.

[Fender56]

Some useful info:

http://www.maxim-ic.com/app-notes/index.mvp/id/1957

[JRay]

Ah of course I completely ignored the feedback resistors!! You guys are great!!

OK I will continue with my plan!

AshCat, you are correct, I did see a scheme that offered inverting and non-inverting outputs using only one op-amp and I thought my prayers were answered. Unfortunately the configuration changed the input impedance considerably depending on which polarity was switched in and hence was no good for my needs as I need a constant input impedance.

If anyone here knows of a scheme that does offer both polarities with only one op-amp AND keeps the input impedance the same, there is a beer in it for you as it would save me about 4mA of battery drain!

Cheers
Ray

[JRay]

Interesting link there Fender56... thanks!

Ray

[slacker]

Roughly what input impedance do you need? Why does the impedance have to be the same in either configuration, couldn't you just make it high enough not to matter and use the single opamp circuit.

[Gurner]

I'd be inclined to go with a low power dual jfet opamp (you should be able to find one that draws *way* under 4mA per channel)...then it'd be just a simple high impedance voltage follower ....followed on by the one opamp polarity selector....

http://www.beis.de/Sonstiges/Inv-NonInv.GIF  (all resistors being of equal value)

[R.G.]

Opamp outputs are never floating, feedback resistor or not. Their outputs are driven to match what the inputs tell them to do, if that's possible, or driven to one or the other power supply. In no case are they floating. So it is always OK to leave an opamp output unconnected. The feedback resistor is driven by the output; the output is not held in place by the feedback resistor.

If you want to have some say about what voltage it sits at, you may want to set the inputs to where you want them. In the case you mention, one inverting and one noninverting opamp, simply disconnecting the unused one is OK for the opamp.

The only amplifiers I can think of that are not OK in general being left floating are (1) tube audio amps (2) RF amps which may overheat if not properly loaded (3) some current-source output devices.

It is never OK to leave CMOS logic *inputs* floating. But outputs are fine.

[JRay]

Thanks for the replies!

I am after 1 meg or more input impedance. The pre is for acoustic guitar and I am using the OPA2134 as I started off with a TL072 but it sounded a little crashy, The OPA2134 sounds much. much smoother with great depth and detail and gives a smidge more headroom from a single 9V supply as you can run closer to the rails. OPA2134 runs about 4mA per amp whereas the TL072 runs at 1.4mA or so. I am willing to sacrifice battery time to the superior sound I get from the OPA chip however. But If I could find a scheme that offers polarity reversal with one chip (OPA134) then obviously it would be an advantage. I could run the input with a higher input impedance and use 1meg as the minimum, as Slacker suggested, but I wish to avoid a situation where I get a live sound eq'd in and then find that if I need to switch polarity, the sound alters enough to need a re-tweak. I know the difference would only be very slight but it would aggravate me!

R.G thanks for the informative reply, very reassuring too!

Cheers
Ray

[R.G.]

Have you looked at the "switch hitter" circuit? It lets you flip the phase of an opamp with a single switch. It's probably better after a buffer opamp than trying to make it be high impedance.

A switch hitter circuit is a development of the single-opamp differential amplifier. A differential amplifier has a resistor from incoming signal to the (-) input, a resistor from output to the (-) input, a resistor from the input signal to the (+) input, and another resistor to signal ground from the (+) input. If all resistors are equal, the output is equal to the voltage difference between the two input resistors. Tied to a single input, both inputs have the same voltage so the output is the difference - which is zero.

If instead of two resistors to the (+) input, you use a potentiometer from input signal to ground, and connect the (+) input pin to the wiper, then the opamp output is equal to the input signal when the wiper is closest to the input signal, decreases to zero output signal when the pot wiper is in the middle of the pot, and increases again but in reversed phase as the wiper travels to audio ground.

You can get the switch-hitter function by using not a pot but one resistor into the (+) input pin of the opamp, and using a switch to short across the (+) input to ground. When the switch is open, the gain is +1. When it's closed, the gain is -1.

You need a buffer in front of this because the input impedance changes when you throw the switch.

[slacker]

If you use switch hitter like this one, same as R.G. described, but replace R1 with a resistor and then switch the + input either to ground or to the junction of R1/R2, then isn't the input impedance the same in either case?
To my simple way of looking at it, the input impedance in inverting mode is the parallel value of R1 and R2, and in non inverting mode it's the same but in parallel with the impedance of the + input of the opamp which is so massive it makes no odds.
By the way all the resistors should be the same value, not like it says on the schemo.

[R.G.]

Yep. That works, and does not change the input impedance. But it does leave you with a low input impedance. As a practical matter, the input resistance is half the value of the resistors. That's awfully low for bare guitar signals unless you go to 2M or more for the resistance value, and that gets you firmly into secondary effects from stray capacitance and perhaps thermal noise issues if there is much gain after it.

If you know what comes before it is always an active device or circuit, 10K is probably fine. If it will be guitar ever, using a buffer in front will help a lot.

[slacker]

Sorry, what I meant was using that but with bigger resistors to get what ever input impedance Ray wants, forgot to mention that. Like you said though then you're into the old problem of trying to make a high impedance inverting stage.

[JRay]

R.G, Slacker

Thanks guys, that is a sneaky little circuit and it would be ideal if I didn't need the 1meg or more input Z. As R.G said putting a buffer in front would do the trick but then I would need another op-amp for the circuit and so I may as well stick with the scheme I have of one non-inverting op-amp providing high input Z and required gain, with the inverting op-amp handling the polarity swap on the output. To save battery drain my original idea was to use a couple of FETs to provide the input buffer and then feed the 500R or so buffer output Z into the op-amps.  I could use this in front of a single device, OPA134, configured as the switch hitter, however I only have a board space of 53 x 40mm and I ran out of room so I went with the OPA2134 providing both features as it equate to less components.

Really appreciating your input here!

Cheers
Ray

[Fender3D]

If it's just a buffer with no gain involved, might this be of any use?

[PRR]

> isn't the input impedance the same in either case? ... in non inverting mode it's the same but...

No, in non-inv mode the signal voltage across R2 is zero, "bootstrapped" by opamp action, so that leg looks like infinity. If you have R1, then Zin == R1, as opposed to in inverting mode Zin = R1||R2.

I'm not clear on ETA's values, Surely gain runs from -2 to +1?

> might this be of any use?

Maybe. The cathodyne or split-load should normally be biased with cathode or emitter near 1/4 of supply voltage, to get equal swings on both sides. But that drawing shows bias supply at 1/2 of supply voltage. Maximum emitter swing is to 4.5V. so max output swing is 4.5V-3.4V= 1.1V. Incidently the unnecessarily low 10K values at R6 R7 limit the other swing to just about 1.05V. OK, may work fine for straight guitar. Would have more headroom if R6 R7 and loads could be higher than 10K and if R4 were more like 47K or 56K.