Voltage Follower - Operational Amplifier | Basic Circuits | Electronics Tutorials

[@Jnglfvr]

At 5:28 you have it backwards. The output should provide negative feedback into the inverting terminal and the 47K ohm resistor should be fed into the non-inverting terminal. The circuit will not work the way you describe. It did work, however, because you mis-identified the inverting and non-inverting terminals and actually connected them opposite to your narrative. Take a look at the circuit diagram at 3:28.

[@CircuitBread]

Dang. You are right, I misidentified those when describing the setup at 5:28. The circuit was setup correctly, it was just my description. I'll pin this comment so others can see if they're confused. Thanks for catching that!

[@JonDeth]

I had minor confusion about the inverting terminal last night because what I was looking at vs what I heard lol. I watched it again before bed, revisited the video this afternoon, and your comment just helped me out. ATM, my primary area of interest is designing and building musical equipment, and there will always be a need for some devices to be a low impedance output.

[@zainabashai7546]

@@CircuitBread Why can the 47k not be connected to non-inverting terminal of op amp? Why is it necessary to connect feedback to inverting terminal?

[@petera5790]

@@CircuitBread are you sure the end of the 1k load resistor not connected to pin 1 is connected to ground? Looks like it is connected to Vcc ?

[@mrduck1817]

@@zainabashai7546 lets assume that we give the feedback loop to non inverting and give the input signal at inverting if we do that the output of op amp will tend to saturate towards the rails (+vcc and -vee) as lets assume the non inverting terminal signal at one instance of time has higher voltage than inverting ,when this happens the o/p voltage is amplified and fed back again into the non inverting causing the same stuff to happen again and the o/p is amplified until it saturates to +vcc similarly in other case if inverting is grater than non inverting we go on until we reach -vee

[@brianm1916]

The WHY explanation was exactly what I needed! Please keep telling us the why in all of your videos.

[@notanymore9471]

Thank you! I’ve been struggling to understand what applications op amps in general are used for and this totally made sense.

[@CircuitBread]

Glad it helped!

[@bothwellchingombe4746]

I’m glad I found this channel. Makes understanding electronics much easier. Please keep up the good works my good Sir.

[@CircuitBread]

Thank you for the feedback!

[@crystalmacau]

Thanks for the explanation, i was wondering what high impedance meant and now i know, and what that voltage buffer is good for!

[@zetaconvex1987]

Thanks Josh. It's worth pointing out that the output draws current from the power supply rather than the input source itself. So there's no magic involved. Something that I struggle with - and something that is rarely mentioned - is what to do about powering the opamp. This is to do with how close the input voltage can get to the power rails. It's a subject that tutorials skimp over and taken as a given, but can leave nebies like me puzzled as to why I'm not getting the output I expect.

[@CircuitBread]

Good feedback, thank you! You are totally right - all the power from the output comes from the power supply. I did finish shooting all of the op-amps videos we have planned and while mentioning the rail-to-rail output, I didn't mention the input voltage versus power rails in them either...

[@edgeeffect]

I've just this minute, finally got my head around what high/low impedance really means... Great stuff... thanks. Be interesting to see what it'd look like to build a voltage follower using BJTs or FETs.

[@CircuitBread]

I definitely remember when that clicked for me as well. A memorable moment!

[@colyjiang412]

Thanks for these videos, they really have improved my knowledge of circuits. You also consistently answer questions in the comments, making sure your viewers understand the concepts, which is something I haven't seen other youtubers do. Keep up with the great work.

[@CircuitBread]

Thanks for the feedback! We do try and answer as much as we can in the comments but I know I miss a lot. I think our Discord channel is starting to pick up more as well, with some good answers and insights there as well.

[@Mischiefcity2013]

Best explanation I've found by far! Thanks, seriously appreciate it.

[@ovi_4]

Wow....what a great resolve. I was trying to understand someone's circuit. He made a little circuit using a dual op-amp (LM358) to amplify a small signal needed for an analog movement in an old radio. By analog movement I mean the signal strength indicator (the meter. I believe it must be a baby size micro-amp-meter. Possibly 50-to-150uAmps). So he used the 2 op-amps in a series configuration but didn't know that the very first op-am was configured as a voltage follower (so the existing signal could be amplified without unbalancing the signal itself. Wow..this is what I needed to snow. I have subscribed right away!. Thank you. PS: I'm just beginning to understand the world of OP-amps. They are so versatile is unreal. I hope to master it one day (fingers crossed) :)))

[@lad7534]

I am binge watching this video's, they are awesome, better than my classes

[@CircuitBread]

Awesome! I'm so glad they've been helpful!

[@_yourNameHere__]

Besides the mistake in the pinned comment, I think that RL is connected to Vcc rather than to ground. Shouldn't that be connected to Ground? Thank you for the this series. It's pretty helpful.

[@flaviodanca2057]

Few words but a lot of meaning. Nice job.

[@JonDeth]

I'm assuming that op amp has sentimental value considering when it was purchased. My first serious dual op amp circuit turned out to become a high gain distortion pedal with soft clipping germanium diodes, and soon I will retire it to showcase more like a trophy. When I "designed" it, I really had no scientific understanding and relied on tutorial basics of setting gain, RC filters and diodes on the feedback path to produce soft clipping.

[@wysskey1]

I use this circuit often to feed remote sensors . If you stick a blocking diode into the supply voltage of the op amp you get non destructive short circuit protection as well as reverse battery protection without damage to the op-amp. ?¿

[@sunrayseducation]

Thank You Sir, for this content.

[@maovichet1683]

Great i like electronic so much,thank for your sharing.

[@henrysavioezeorah5852]

I love your tutorials! ❤ I wish to know if this configuration can be applied in any way to build an optical transmitter. Pls be of help 🤝

[@DasIllu]

I always had an issue with the labeling and the subsequent misdirection by calling the inputs positive and negative. I made it an effort to get that out of my head as soon as i started to understand OpAmps a bit better. Inverting- and Non-Inverting input is a better description, imho. It might seem "even more techno babble" to the uninitiated, but i think it helps in the long run. Once you start seeing the inputs like this and not "so this is where positive voltage goes and this is where negative voltage goes", you can start using the opamps to manipulate signals to your own liking. That is why i see the labeling as a pitfall for people just starting to learn electronics.

[@CircuitBread]

Yeah, while doing these I was debating on that terminology. All of the opamp videos have been shot now and I use the terms inverting and non-inverting more and more as they progress for exactly those reasons. As you get to know them better, it makes more sense to say inverting and non-inverting. Hopefully the transition eases people into it rather than causing a confusing stumbling block.

[@andymouse]

Good advice...cheers.

[@jimedgar6789]

SUPER helpful. My "current" project is using a LiPo to power an ATTiny as an LED flasher in a remote area, and I need to know how often I need to swap the LiPo out. mAh math aside, cause who really knows in real life. So I am going to set up an MCU to read a pulse from the Tiny every so often, but I don't want that current to participate so I THINK with a the near infinite impedance of the opamp input, I could get a timeline of the battery life. Meaning the MCU and the opamp are their own circuit. Hmmm?

[@andymouse]

Great video....cheers.

[@adrianzayas4456]

Hi, thank you for your clarifications. I have a question about the impedance of the voltage follower. If the Vi=Vo in this type of op-amp, would you say that the impedance of the op-amp is 0 or 1? If there is no Rf would you consider (Rf/Ri)= 0 or 1.

[@CircuitBread]

Hi Adrian, for an ideal op-amp, you would consider the output impedance zero and the input impedance (to the non-inverting input) to be infinity. I'm not sure if I would look at it in the context of Rf/Ri, that may be unnecessarily complicating things, as your input isn't connected to the same leg as your output (as it is in an inverting or non-inverting amplifier configuration).

[@andylaurie326]

a buffer also isolates that tiny input from the nasty load

[@zackwatson755]

Helpful, thank you

[@aryanpandey188]

best explanation

[@CircuitBread]

Thanks!

[@aungkyawmoe8023]

why are you doing positive feedback circuit when you explain it is the negative feedback in first half of the video?

[@clusterbx]

Thanks for the video. Helps refreshed my memory on the use of OP-AMP learnt during my schools days. May i ask, if i want to transmit 6.4Vdc across a 100m 18AWG cable, how many OP-AMP do i have to insert along the 100m length? Will every 10m suffice?

[@CircuitBread]

That's both an interesting question in that I'm curious what the application is and that I can't answer it with that information. The key missing info here is the current you anticipate pushing through the wire, as that will affect your voltage drop. If there is no current, then it's theoretically possible a single op-amp will work, as you won't get any voltage drop and op-amps can output a non-negligible amount of current. My understanding (a quick Google search) is that non-PoE ethernet is only around 1-2 volts and that can go about 100 meters but it doesn't move any significant current.

[@clusterbx]

@@CircuitBread Hi. I am trying to transmit 6Vdc over 200m to operate a small servo. The 200m cable is 12ohms and the servo has a rated stall current of 2A. I actually did that and the servo was not really moving. It's like there's not enough power.

[@CircuitBread]

@@clusterbx Sorry about this extremely delayed response, KZbin doesn't notify me about follow-up questions. I think a little Ohm's Law here will clear things up. You have 12 ohms of cable and are running upwards of 2A. With V=IR, and I=2 and R = 12, you'll need 24V to "push" that much current just through the wire, let alone to run the servo. If your op-amps are running off of that same power supply, they're going to run into the same problem. They don't magically create the current, they need a power supply as well. If they're hooked into another source of power, it should be doable, because you won't actually be running much power through that 200m wire. But I imagine that there isn't anything else (or else you wouldn't be running a 200m cable) so I don't think this will work with that large of a load.

[@Electronzap]

Good info.

[@KA9DSL]

Simple, a voltage to current convertor.

[@jogreeen]

Thanks

[@zainabashai7546]

Can you provide a link to the op amp used? I want to buy it. Is it any op amp?

[@CircuitBread]

There are many differences with op-amps even though, with these examples, we're not pushing them hard enough to make a big difference in performance but there will still be changes in how they're powered and their pin-outs. We used the LM2903P from Texas Instruments for these examples if you want to ensure that everything works as it does in the examples: www.digikey.com/en/products/detail/texas-instruments/LM2903P/372786

[@Dc_tech386]

I use the lm393comparator on my power supply I design for 1000watt the issues I have is the voltage drop still and I don’t understand way is that the voltage not holding even though the duty cycle is adjusted from the comparator am only getting current not voltage stabilized the voltage is 55 input 20amp and 14.80 volt out put when I add a low the voltage drop to 14 volt not 14 .80 and I see the tl494 only stabilized the volt in negative drive for p channel the voltage don’t drop but in positive pulse the n channel design the comparator have to use but voltage drops on the positive side can use explain why is this and should I not use n channel for

[@Cm0nd00d]

woah! you changed your shirt!

[@CircuitBread]

😂 Fashion is not my thing...