Engineer It - How to terminate an unused op amp

Find out how to properly terminate an unused op amp, the benefits of doing so, commonly used circuit limitations, and how to select a proper configuration from senior applications field engineer, Todd Toporski.

[@antoleyo3857]

I am an oldschool designer, I mostly use LM358, TL072, CA3240. For unused LM358 and CA3240 I connect -in to output and +in to negative rail. For unused TL072 I connect -in to output and +in to positive rail. Both run perfect. Thank you TI, for your huge amount of priceless application notes. Cheers from Indonesia

[@stephenweidner1187]

Thanks Todd This video on how to terminate unused op amp is the best I have seen so far And not just single supply but split supply. Thank you

[@autechrak]

so how many you have seen so far?

[@jtb8631]

Super helpful, something I had read had implied that the process of neutralizing unused opamps was a pain, but you made it pretty easy.

[@BrandonClearwater]

I have been using the suggested layout in my circuits. But while watching this video I thought about it more in depth. In split supply configurations you may not have a guarantee that both supplies come up at the same time, or even unfortunate situations when one supply isn't up at all. I think it would be a safer default circuit to also include a current limiting resistor between the positive input and ground.

[@w6wdh]

Thanks, Todd. Interesting and useful. Note that if you insert a resistor (say 10K) between the op amp’s output and (-) input, and you have a resistor or resistor divider on the (+) input, you can later unsolder the resistors to use the spare op amp to implement a Field Change Order. It’s nice not to have to cut traces to use the op amp. Bonus points for including test points on the op amp terminals to solder wires or leaded components! (I am, of course, speaking from experience: oooh, if only I could hook up that op amp to buffer a signal...)

[@roberto660260]

Thank you very much TOD, for this information , I never thought that the unused part was subjected to this stress.I think I will review several projects. See you TOD

[@youtuuba]

I don't think I ever did a design where I had any left over ap-amps, but I always thought that if I DID, I would tie (-) input to output, and tie (+) input to the output of the nearest op-amp that WAS being used.

[@iblesbosuok]

You know something,sir? TI had given me hugh amount of knowledge about mixed signal. Great thank you. Cheers from Indonesia

[@paultopace1234]

Great video, really well done, it was relaxing to watch and learn and useful. I wish all education was like this. I have noticed unused op amp halves in my Mackie powered speakers. They have tied the inputs to ground. cheap op amps though, I wish they would budget for better op amps from Texas Instruments.

[@michaelmounts1269]

great video--love the circuit examples!!

[@emorag]

Excellent video. Very helpful.

[@JohnClulow]

Great explanation of this topic !!

[@fpgamachine]

Excellent video, thanks!

[@braselectron]

Thanks excellent best practice tip.

[@netdudeuk]

Great video thanks

[@MrMamanDon]

This is a great video. Unrelated to this, I've seen many tutorial videos where electronic component manufacturer's employees describe something about a product or a concept and a large number of times those employees were shaking like 6:13, it almost seems like they're forced to do this with a gun on their heads. Like wtf?

[@Regimantas_Baublys]

Supper Thanks :) werry informatible video :)

[@gwav1a]

Am working with a quad LM324 and have seen a drop in current as I remove the terminations- it's actually opposite to what I expected. Am also observing a drop in current as I transition from linear to saturated output region. Am using 1 opamp as a non inverting amplifier for 50Hz signal of 20mV rms.

[@TexasInstruments]

Hi gwav1a, thanks for your interest! You can search our TI E2E Support Forums for an answer to your question here: e2e.ti.com

[@power-max]

If your op amp was connected to a 12v single supply and you have a 5v or 3v3 digial logic voltage somewhere you could use that instead of extra resistors.

[@w6wdh]

Not a good idea, due to power supply sequencing. What happens if only one of those power supplies is present, even for a moment? Many op amps have clamp diodes that prevent their inputs from exceeding the op amp’s power supply voltages. So if +3.3 or +5V power is present but the op amp’s +12V power is not, and you tied the (+) input to +3.3 or +5V, then a clamp diode inside the op amp will try to feed +3.3 or +5V power into the +12V supply. Highly not recommended. The clamp diodes are often rated for no more than 10mA.

[@shantk7378]

I recommend watching the entire presentation, though the answer is at 12:06.

[@mx.olydian2111]

Hmm does the unity gain follower become a problem if your opamp isn't unity stable? I'm suspicious that I'd have to configure such opamps with gain which means board space is committed to the two resistors

[@bergweg]

If the op-amp IC has 4 or more op-amp units, could you just use one voltage divider for all of the non-inverting inputs of the unused units?

[@TexasInstruments]

Yes, you can use one voltage divider to feed the inputs of any/all of the unused units. The divider is then acting as a reference for all of them. If you have two or more units that aren’t used, then you may also want to consider a “reduced” version of that op amp device that comes in a single or dual (instead of a quad, etc.). That way, you’ll likely save board space and cost. Many TI op amps come in single, dual and quad versions. If you have any other questions, I encourage you ask them in our precision amplifier forum, which is actively monitored by our applications team: e2e.ti.com/support/amplifiers/precision_amplifiers/.

[@Bubu567]

FYI, ideally, the inputs do not pull current, they simply read the voltage. That isn't what actually happens in the real world, but it's close enough for you to treat it as such.

[@jonathang4963]

With the unused op amp, not just use it as a buffer for an op amp that is being used?

[@w6wdh]

Sure, but a buffer adds its offset voltage, noise, and delay. If those contributions don’t matter, then you’re set.

[@miguelfernandes333]

I have a circuit that has an unsed opamp which is a tl072, what resistor values should I use to terminate it, would 10k also work?

[@JoaoAntonioCardoso]

definitely yes.

[@jasondaniel3594]

Why not give unity feedback to the inverting input and ground the non-inverting input for the single supply op amp?

[@Graham_Wideman]

Because (a) many op amps can't accept inputs down to their negative rail, and (b) if you drive the output to the negative limit it will saturate and draw more current and produce more heat.

[@jasondaniel3594]

@@Graham_Wideman Thank you so much for your response Graham. That helps a lot.

[@RahulGavade]

Would i be committing a mistake if I did not connect anything to the unused op-amp (just left all unused pins floating, in other words High impedance) ?

[@autechrak]

no

[@RexxSchneider]

Yes

[@RahulGavade]

@@RexxSchneider Thanks for the reply! It's funny how I get an email notification about a reply I got for my KZbin comment that I hardly remember I made about four years ago! 😂

[@RexxSchneider]

@@RahulGavade It's quite funny, I agree. But If I can come across a video and its comments from 4 years ago, so can anybody else, and I sometimes feel compelled to correct the errors I see. For all the folks who eventually see this thread, they ought not to be fed disinformation. The fact is that it is a mistake to leave an unused opamp floating. The resulting behaviour is unpredictable and may cause the problems outlined in this video. So as the video explains, it's always sensible to tie the output to the inverting input and the non-inverting input to a fixed voltage within the opamp's common mode range.

[@andymouse]

Well done for putting all correct, the universe can relax now LOL ! :)@@RexxSchneider

[@KariArgillander]

Does this applied to comparators also?

[@RexxSchneider]

You need to check the datasheets, but comparators are designed to deal with large differential input voltages, so are less likely to be problematic. Usually they have open collector outputs and are designed to run open-loop, so there is less of a problem with the outputs and feedback is often not needed. To take an example, the LM393 has a common mode range that goes from the negative rail to Vcc - 1.5V. You could just tie both inputs to ground and leave it at that.

[@sabahoudini]

What happens if you leave input and outputs floating?

[@RexxSchneider]

It depends on the configuration of the input stage, so you need to check the datasheet for any particular opamp. Without any input bias current, a bipolar opamp will swing its output to one rail or the other, but it's hard to predict which one. An FET-input opamp is likely to be similar, but it could just oscillate because of stray capacitances. You really are better off tying the inverting input to the output and the non-inverting input to a voltage somewhere in the amp's common mode range, preferably via a few K resistance in each case.

[@RexxSchneider]

I think you're over-selling the advantages in terms of excess supply current and dissipation, at least for the OPA2227 when the inputs are tied together. I can't really agree that 8.21mA is "much lower" than 7.74mA. Of course, there may be other opamps where the difference is more, but some opamps (like the humble LM358) include the negative rail in their input common mode range, so there are some where the first scheme is fine. As you say, you need to look at the datasheet. The OPA2227 has an absolute maximum input current of 20mA (in abnormal conditions, of course), so I'd be very tempted to have enough resistance in series with any unused input to ensure that limit is not exceeded, even under transient conditions such as switch-on or switch off. Unless I was working for penny-pinching bean counters, I'd have a feedback resistor of around 1.5K between output and inverting input and the same value between non-inverting input and ground for the split-supply case. As one of your other commentators states, it would also then make it far easier to press the unused opamp into service with a minimum of bodging at a later date.

[@CyberArtem]

I don't see much difference between first and the last termination methods really. The current consumption is only 1/2mA lower but you need two more additional parts , which is an additional effort i think, unless there is more to it than just a power consumption.

[@1337ASM]

+Artiom K It's outside of spec, so operation is not warranted/characterised. Open-loop configuration of the op-amp can (easily) lead to oscillations in the output, and induce noise on the power-rails, and it can impact the neighbouring op amp. It's bad, very bad.

[@CyberArtem]

SamBell Thank you.

[@ingr6889]

+Artiom K, some op amp can not work stable with shorted feedback. Then one more pair of resistor in feedback can feel be good.

[@sebaogal1]

Because op-amps have a very high gain in open loop, which drives the output to very large voltages (until it meets the rails) even with the most subtle input difference level.

[@RexxSchneider]

@@ingr6889 Opamps which are only stable with closed loop gains greater than 10 (for example) are as rare as hen's teeth these days. It's not a consideration likely to bother modern designers. Nevertheless, adding a couple of dummy resistors on the pcb to mimic a working part can be a great aid to development when you find later that you now need to make use of that previously unused opamp.

[@aerohk]

Free OpAmp 🥰

[@RoGeorgeRoGeorge]

I'll be back. :o)

[@therandomchannel9226]

Gnd one input. Leave other floating.. it will become an antenna ;-)

[@DeeegerD]

Hmmm, half of 12v is 6v and that exceeds the +/- 2v input range. Shouldn't the voltage divider limit the input reference to the specified range in the datasheet?

[@kurdim4628]

I had the same question, Did you find out the answer?

[@DeeegerD]

Kurdim Nope

[@cue108]

You did not read the datasheet correct. The +/-2V is with respect to the op-amps supply voltage! (-V)+2 to (+V)-2 is for example (0V)+2 to (12V)-2 -> 2 to 10V common mode voltage range.

[@armandine2]

@@cue108 seems obvious but I missed the same thing on the input range - "that looks a very small input" LOL

[@ishigamiyu1991]

Whose here after blowing up their op amp. Me for sure

[@AB-yu2tj]

If I could make a suggestion for the video, instead of zooming in to the point of no help to anyone😄 have an overhead camera that would show the breadboard magnified please. Thank you. Then idiots know-nothings like me would enjoy watching too. And might even learn something. Sorry I can't help it and just enjoy, I got things to say after two lattes with four shots of espresso in each and 4 spoons of sugar added lololol

[@felixcat4346]

You would think these things could be designed so if you're not using part of the op-ap the part you're not using would do nothing. Go figure.

[@Graham_Wideman]

I realize this is an old comment, but... how would you expect the op amp to detect that it's not being used?

[@ChaseNoStraighter]

Click bait. I was expecting some smoke!

[@PhuckHue2]

I just cut the legs

[@Graham_Wideman]

Hahahaha