EE newbie here.

I am trying to put an additional op-amp circuit onto a crowded half-sized "perma-proto" breadboard; I have no room remaining on the centerline to mount a DIP package.

What I do have is a dual op-amp in 8-pin DIP; I only need one of the amps, so on one side of the package only the +V pin is in use; I can straddle the DIP across the gap between the horizontals and the power rail, and the power pin will land in the right place.

Obviously in the interest of sanity I will clip the unused leads and put an insulator between the stubs and the power rail, and I know this is not an elegant solution or something I should be doing routinely.

But if I were to take a common 8-pin dual op-amp, and put all four pins on the right side (+V, inverting input, non inverting input, and output) onto a common +V rail, what would happen to the op-amp and the power rail?

I'm not asking for solutions to my layout problem; it's solved. I'm curious what the behavior of the part would be if connected in that way.


I would not be shorting the output to the rail.

The inputs will be fine tied to the rail. Do not leave them floating or the thing may sit there howling with noise consuming large currents and radiating EMI. That noise would be coupled to the rail.

However, both inputs tied to one rail means there is a 50-50 chance the output will try to pull to the opposite rail. As such it will be straining and consuming current, though the latter will likely be limited internally. It will get a little warm. The op-amp will likely not significantly affect the rail voltage unless it is a really high current op-amp or the rail is higher source resistance.

For your hack solution, just lift, or break off, the output pin.

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  • \$\begingroup\$ Did you not read my third paragraph? I have no intention of doing that but I am interested in knowing what would happen if I did. \$\endgroup\$ – Russell Borogove Dec 27 '17 at 19:55
  • \$\begingroup\$ @RussellBorogove yup I expanded it a bit anyway \$\endgroup\$ – Trevor_G Dec 27 '17 at 19:57
  • \$\begingroup\$ @RussellBorogove and expanded some more. \$\endgroup\$ – Trevor_G Dec 27 '17 at 20:06

The standard way to deal with an unused op-amp is to short the (-) pin to the output pin and connect the (+) pin to ground.

If you are ambitious, place dummy components on your layout to allow those connections and then route shorting traces across the pads of those components. This gives you the opportunity to make use of the op-amp in the future by cutting the shorting traces.

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  • \$\begingroup\$ I Agree that's a better way to park an unused op-amp, but for a quick fix my answer is sufficient. \$\endgroup\$ – Trevor_G Dec 27 '17 at 19:52
  • \$\begingroup\$ That's a good suggestion for my eventual formal layout, thank you. However, it doesn't address my actual question. \$\endgroup\$ – Russell Borogove Dec 27 '17 at 19:57
  • \$\begingroup\$ I'm using single-supply (i.e. opamp V- is ground), so from what I'm reading, I want to tie ground on the unused OA to the split-rail point, right? \$\endgroup\$ – Russell Borogove Dec 28 '17 at 0:30
  • \$\begingroup\$ It normally does not matter where you tie the (+) input into so long as it is within the allowable Common Mode voltage range. Tie it to whatever is convenient. It can even be the output pin of another op-amp. \$\endgroup\$ – Dwayne Reid Dec 28 '17 at 0:44
  • \$\begingroup\$ (Sorry, meant to say "tie + on the unused" instead of "tie ground on the unused", obviously) \$\endgroup\$ – Russell Borogove Dec 28 '17 at 0:51

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