Quick question… if you supplied power to a dual op-amp IC (say a TL072) and only used one of the op-amps on board, would the other one that’s not being used still continue to draw power from the supply? 
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According to Texas Instruments you want to properly wire unused op-amps to prevent power consumption and noise.
Like this:
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Thanks for the fast response @CTorp! That’s great news. I was worried it wouldn’t be doable.
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Op-amp question #2 
I was trying to simulate a noninverting buffer I was thinking of adding to the Arduino sequencer. On paper it looked like the output (yellow signal) should be the same as the input(blue) but with a gain alittle greater than 1. Instead I got this nonsense:
Any clues as to what might be keeping the output from falling back to zero? I thought maybe it was a slew rate issue but the circuit is only operating at 500 Hz which is super slow compared to what the TL072 can handle.
When none of the diode conducts, the non-inv input of the op-amp is floating…
Add a very high R from the non-inv input to gnd (like 10M Ω).
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What are you trying to achieve with the diodes? And what is the bottom right circuit good for (it does not seem to be connected to a power source)?
If you leave out the diodes, you can leave out R8.
For what I understood a month ago, they “simulated” two of the output stages of the simple arduino sequencer.
Presumably the lower right circuit is for another input channel, which at that point they hadn’t yet added to the simulation. Each of the two sections on the right, including the diodes, is more or less a simplified version of two channels of the LMNC 8 step sequencer, which they mentioned.
If the diodes are omitted (and there’s a voltage applied on both channels), then a problem occurs when both pots are at maximum; then two different voltage sources both are connected directly to the same point (the sequencer output, or in the above schematic the buffer input) with bad consequences.
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if there is a power source, indeed.
Like @analogoutput and @eric mentioned, the bottom right circuit was a second channel. Since only one channel is supposed to be on at a time I figured I didn’t need to add another source to test the idea out. I went back and simulated it again with a second source just to double check and everything looks tip top
Yellow is channel 1
Blue is channel 2
Pink is the output
Figured as much. Note, though, the difference between not on and not connected. In the earlier diagram the voltage at the top of R2 is floating; in this one it’s 0 V when it’s not 5 V. I find that an easy thing to forget, but an op amp with a floating input behaves very differently from an op amp with a 0 V input. (In this particular instance it doesn’t matter, because of the diode and the pulldown resistor, but in general it’s something to watch out for.)
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Good point good point. I keep overlooking that.
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I sort of guessed that but found the setup of the simulation odd because leaving out a power supply is not the same as ‘powering something with 0 Volts’.
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Yea for some reason I assumed the simulation would treat it as being the same. I guess it’s true what they say about making assumptions 
Floating pins are the cause of a lot of problems around here.
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I’m a bit confused about the decoupling caps used for the mixer (and a lot of other modules I’ve ran into).
It’s hard for me to tell from the above schematic, but is this how I should be connecting the capacitors? With the ceramics close to the chip?
Also do you need an additional pair (one per rail) of ceramics and/or electrolytics for each additional op-amp IC?
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These are properly called bypass caps (decoupling caps are series capacitors intended to remove DC component of a signal). The large electrolytics go on the power rails close to the power header. The small ceramics go close to the chips from each power rail to ground. Best practice is to have one 100 nF cap for each power connection on each chip, so two per op amp (with split supply) but just one per logic chip. But just one pair of electrolytics.
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