Very useful, thanks. And it's always good when a psychic instructor addresses things that are bothering me, like the issue with the output's wandering DC level
Many many thanks, just a quick question if I may, when we design a passive integrator , RC should be a lot larger than T, the period of the Square wave, so is this also the case with the active integrator in this video? Thank you.
RC gives us the information on how long the capacitor will take to charge to its maximum. After 3 RC your capacitor is somewhat "full" In this case the opAmp saturated. To avoid that you could make a bigger RC, which will make for a less steep curve, or you could reduce the period of the square wave
What is the purpose of this circuit from a practical usage perspective? like what problems does it solve? what do I use it for? And can you make an integrator without the use of an op amp?
Integrators are used in digital multimeters to measure voltage (or the voltage drop across a current shunt resistor). They do this by "run-up" and "run down" phases. During the run-up phase, the switch selects the measured voltage as the input to the integrator. The integrator is allowed to ramp for a fixed period of time to allow a charge to build on the integrator capacitor. During the run-down phase, the switch selects the reference voltage as the input to the integrator. The time that it takes for the integrator's output to return to zero is measured during this phase. For more info, see the Wikipedia article titled "Integrating ADC". To answer your second question, any op-amp can be replicated by discrete components. Sometimes, the best choice for a particular circuit is a judicious choice between op amps and discrete componets that may do the same thing as an op amp.
The circuit turns into a differentiator where the output is related to the derivative of the input voltage. Something I've never really thought about before is why do we typically create a differentiator circuit with a capacitor on the input rather than an inductor on the feedback loop? Hmmmm... my first thoughts would be cost, size, and performance at DC but now I'm curious if those are valid reasons.
3:13 hm... i'm no expert but is'nt that a inverting opamp circuit? (you said a non-inverting, non-inverting has the signal coming at the + input of the opamp)
Yep, you're right! Sorry, I really struggled with this series saying inverting and non-inverting when I meant the other. I thought I'd caught or corrected all the mistakes but it seems like I didn't. Thanks for pointing this out!
is the shrinking of output when we boost the frequency related to how the capacitor becomes more and more like a conducting line as frequency is increased? and i wonder what would happen if we take an inductor into the mix, knowing that Vʳ = R I , Vˡ = L dI/dt , and Vᶜ = 1/C ∫ I dt
It's probably a combination of both the capacitor starting to act more like a short as frequency goes up as well as the fact that the op-amp output will not be fast enough / able to source enough current to fully make the swings. I think that would be an interesting experiment to try and in practical applications there are many other components involved to make these circuits more linear and less error-prone.
We're using TI's LF412 and, yeah, pin one is the output for those chips. It is a bit irritating but I do understand why different companies would want to change the pinouts on otherwise similar (in performance) devices.
i never saw that when we integrate triangular wave, we get a sinusoid. why is it like that? i know that the triangle wave looks like a 'low resolution' jagged sine wave, but there's gotta be more into it than that
This would be difficult to explain via text but just look at the area under the curve of the triangular wave. While a triangular wave is linear, the area under that curve is not, and when it changes direction abruptly at its peak, the area under the curve does not change abruptly, it just starts to get bigger at a slower rate. It isn't until it crosses the 0 point that the total area under the curve will start to get smaller (which will be the inflection point at the top of the sinusoid). I just recommend trying, without any math, to visually integrate a triangular wave, I think that will make it click.
@@CircuitBread hmm... the integral of a linear function in time is a parabola ∝ t², which makes the integral of a triangular wave look like a parabola and upside down parabola side by side, which *kind of* looks like a sine wave although not exactly ∩ ∩ ∩ ∩ ∩ U U U U U this also tells me that a full wavelength of the triangular wave only makes one parabolic section, making the frequency of the output wave 1/2 input wave frequency
Hey there, Op amps can't directly affect their inputs, they only affect their outputs. In fact they are just amplifiers that blindly output a voltage that is proportionally higher (huge gain) than the difference in their inputs. The only way for the op amp to somehow indirectly affect one of their inputs is through the electrical connection of the op amp's output with that input (feedback loop). The idea that an op amp "wants" or "forces" its inputs to be the same comes from the form of balance that is achieved between the output and the input through the electrical connection of the feedback loop.
@@7GIGEO7 So after all of that you could have just said no then ? So if you are correct then the input with no feedback won't try to be the same as the input with feedback, that's not how an op amp works my friend. Due to the nature of the op amp it outputs the difference between the two inputs right, and golden rule 1 no current shall flow into, thus forcing the inverting an non inverting to be near dam the same, but your saying in open loop mode you magically turn this property off, strange something that needs to be built and or simulated I'm thinking.
@@SlyerFox666 Hey well, for most explanations i usually go by the quote "tell me and i'll forget, teach me and i may remember, involve me and i will learn" which is true in every learning procedure. The point of every technical question is to learn and not to be told the short answer. As for the rest of your post, the op amp's inputs are indeed very high impedance, they offer some kind of isolation and no significant current is flowing through them. Also in open loop configurations the op amp is just a huge amplifier of the input voltage differential, in fact a very small input difference (in open loop config) will most likely saturate the output. So the inputs won't try to do anything by themselves because they can't, they are just high impedance sampling points. As you said trying out a few configurations will be a good idea.
@@7GIGEO7 Simulating it I can actually get it to do both thing in both configurations so so much for that idea 🤣 il set it up on the bench when I get some free time at work. Everyone has a different learning style as we are all individuals, that's the whole purpose of skilled teachers as they can see when the penny has dropped in each person they teach, not preach their amazing knowledge and just walk away which is really the difference between teaching and lecturing. If an individual is interested the short answer is usually more than enough to put them on the journey to self discovery because as you rightly say you learn from doing. 👍
Darn! At 14:20, it is 1000 Hertz - I misspoke and managed to miss it when I was reviewing it as well. 1 kilohertz or 1000 hertz. It's showing 1kHz on the scope.
not a criticism, just something i noticed the differential dt inside the integrals is missing nice job btw, i've learned about integrator before but not this in-depth
Thanks! And me making a math mistake is not unheard of... 😬 but could you give me a time stamp to refer to? Just trying to reduce the amount of mistakes I make in the future.
Yep, dt had to go to the left side. Meaning, we should also have multiplied both sides by dt, then everything will go right. Also, there is always a constant after integration, and it depends on the initial state. If we start the process with input equal to 0, then the constant will also be zero.
Hello, I appreciate these videos but I have a question about an AO setup that I couldn't find online with a resistor and a capacitor in series together in the loop. I hope You understood my problem and I am waiting for an answer
I'm sorry but unfortunately, I did not understand the problem. I can give some general advice to use a simulator to help figure out whatever problem you're having. LTSpice, PartSim, Flux.ai, something along those lines should be very helpful.
@@CircuitBread thank you for reply I'll search at those sites and hopefully will find a solution, wish U the best and I hope You keep making these educational videos. So helpful
Very useful, thanks. And it's always good when a psychic instructor addresses things that are bothering me, like the issue with the output's wandering DC level
I'm doing Electronics 1, and I forgot everything about intergrals and derivatives, this helped refresh my memory some! Thank you!
Thank you! Great video!
Great Job for the explanation!
Great video you made a fabulous effort🅰
Great video, thanks a lot.
do you have a tutorial for cutoff-frequencies on op-amps?
Many many thanks, just a quick question if I may, when we design a passive integrator , RC should be a lot larger than T, the period of the Square wave, so is this also the case with the active integrator in this video? Thank you.
RC gives us the information on how long the capacitor will take to charge to its maximum. After 3 RC your capacitor is somewhat "full"
In this case the opAmp saturated. To avoid that you could make a bigger RC, which will make for a less steep curve, or you could reduce the period of the square wave
What is the purpose of this circuit from a practical usage perspective? like what problems does it solve? what do I use it for? And can you make an integrator without the use of an op amp?
Integrators are used in digital multimeters to measure voltage (or the voltage drop across a current shunt resistor). They do this by "run-up" and "run down" phases. During the run-up phase, the switch selects the measured voltage as the input to the integrator. The integrator is allowed to ramp for a fixed period of time to allow a charge to build on the integrator capacitor. During the run-down phase, the switch selects the reference voltage as the input to the integrator. The time that it takes for the integrator's output to return to zero is measured during this phase. For more info, see the Wikipedia article titled "Integrating ADC". To answer your second question, any op-amp can be replicated by discrete components. Sometimes, the best choice for a particular circuit is a judicious choice between op amps and discrete componets that may do the same thing as an op amp.
appreciate it.
So it's like a Real-time calculator that records your voltage every time then adds it to what it already has?
And that means an AC signal will provide 0v Output while a DC will provide the saturation output
As long as there's no DC bias, that's correct!
Cool thanks @@CircuitBread
What happens when you replace the feedback capacitor with a feedback inductor?
The circuit turns into a differentiator where the output is related to the derivative of the input voltage. Something I've never really thought about before is why do we typically create a differentiator circuit with a capacitor on the input rather than an inductor on the feedback loop? Hmmmm... my first thoughts would be cost, size, and performance at DC but now I'm curious if those are valid reasons.
So what is the integral of Vin?
3:13 hm... i'm no expert but is'nt that a inverting opamp circuit? (you said a non-inverting, non-inverting has the signal coming at the + input of the opamp)
Yep, you're right! Sorry, I really struggled with this series saying inverting and non-inverting when I meant the other. I thought I'd caught or corrected all the mistakes but it seems like I didn't. Thanks for pointing this out!
script...... almost as bad as the learnelectronics dude hahaha jk. no one is that dumb
is the shrinking of output when we boost the frequency related to how the capacitor becomes more and more like a conducting line as frequency is increased?
and i wonder what would happen if we take an inductor into the mix, knowing that Vʳ = R I , Vˡ = L dI/dt , and Vᶜ = 1/C ∫ I dt
It's probably a combination of both the capacitor starting to act more like a short as frequency goes up as well as the fact that the op-amp output will not be fast enough / able to source enough current to fully make the swings.
I think that would be an interesting experiment to try and in practical applications there are many other components involved to make these circuits more linear and less error-prone.
What op amp chip are you using in the example? I’m used to pin six being the output but in your case it looked like pin One was the output.
We're using TI's LF412 and, yeah, pin one is the output for those chips. It is a bit irritating but I do understand why different companies would want to change the pinouts on otherwise similar (in performance) devices.
i never saw that when we integrate triangular wave, we get a sinusoid. why is it like that? i know that the triangle wave looks like a 'low resolution' jagged sine wave, but there's gotta be more into it than that
This would be difficult to explain via text but just look at the area under the curve of the triangular wave. While a triangular wave is linear, the area under that curve is not, and when it changes direction abruptly at its peak, the area under the curve does not change abruptly, it just starts to get bigger at a slower rate. It isn't until it crosses the 0 point that the total area under the curve will start to get smaller (which will be the inflection point at the top of the sinusoid). I just recommend trying, without any math, to visually integrate a triangular wave, I think that will make it click.
@@CircuitBread hmm... the integral of a linear function in time is a parabola ∝ t², which makes the integral of a triangular wave look like a parabola and upside down parabola side by side, which *kind of* looks like a sine wave although not exactly
∩ ∩ ∩ ∩ ∩
U U U U U
this also tells me that a full wavelength of the triangular wave only makes one parabolic section, making the frequency of the output wave 1/2 input wave frequency
Does an op amp always force it's inputs to be nearly the same in all configurations or is it only when using feed back ? 👍
Hey there, Op amps can't directly affect their inputs, they only affect their outputs. In fact they are just amplifiers that blindly output a voltage that is proportionally higher (huge gain) than the difference in their inputs.
The only way for the op amp to somehow indirectly affect one of their inputs is through the electrical connection of the op amp's output with that input (feedback loop). The idea that an op amp "wants" or "forces" its inputs to be the same comes from the form of balance that is achieved between the output and the input through the electrical connection of the feedback loop.
@@7GIGEO7 So after all of that you could have just said no then ? So if you are correct then the input with no feedback won't try to be the same as the input with feedback, that's not how an op amp works my friend. Due to the nature of the op amp it outputs the difference between the two inputs right, and golden rule 1 no current shall flow into, thus forcing the inverting an non inverting to be near dam the same, but your saying in open loop mode you magically turn this property off, strange something that needs to be built and or simulated I'm thinking.
@@SlyerFox666 Hey well, for most explanations i usually go by the quote "tell me and i'll forget, teach me and i may remember, involve me and i will learn" which is true in every learning procedure. The point of every technical question is to learn and not to be told the short answer. As for the rest of your post, the op amp's inputs are indeed very high impedance, they offer some kind of isolation and no significant current is flowing through them. Also in open loop configurations the op amp is just a huge amplifier of the input voltage differential, in fact a very small input difference (in open loop config) will most likely saturate the output. So the inputs won't try to do anything by themselves because they can't, they are just high impedance sampling points. As you said trying out a few configurations will be a good idea.
@@7GIGEO7 Simulating it I can actually get it to do both thing in both configurations so so much for that idea 🤣 il set it up on the bench when I get some free time at work. Everyone has a different learning style as we are all individuals, that's the whole purpose of skilled teachers as they can see when the penny has dropped in each person they teach, not preach their amazing knowledge and just walk away which is really the difference between teaching and lecturing. If an individual is interested the short answer is usually more than enough to put them on the journey to self discovery because as you rightly say you learn from doing. 👍
Im just not getting this for some reason....
Great videos
Thanks Jack!
what is the C value ? @15:20 you say that freq is one thousand kHz ! yet crobshows approx 1000 HZ. What is the opamp you are using?
Darn! At 14:20, it is 1000 Hertz - I misspoke and managed to miss it when I was reviewing it as well. 1 kilohertz or 1000 hertz. It's showing 1kHz on the scope.
what is C value and opamp used?
not a criticism, just something i noticed
the differential dt inside the integrals is missing
nice job btw, i've learned about integrator before but not this in-depth
Thanks! And me making a math mistake is not unheard of... 😬 but could you give me a time stamp to refer to? Just trying to reduce the amount of mistakes I make in the future.
the math bit starting from 7:20
Yep, dt had to go to the left side. Meaning, we should also have multiplied both sides by dt, then everything will go right. Also, there is always a constant after integration, and it depends on the initial state. If we start the process with input equal to 0, then the constant will also be zero.
Hello, I appreciate these videos but I have a question about an AO setup that I couldn't find online with a resistor and a capacitor in series together in the loop. I hope You understood my problem and I am waiting for an answer
I'm sorry but unfortunately, I did not understand the problem. I can give some general advice to use a simulator to help figure out whatever problem you're having. LTSpice, PartSim, Flux.ai, something along those lines should be very helpful.
@@CircuitBread thank you for reply I'll search at those sites and hopefully will find a solution, wish U the best and I hope You keep making these educational videos. So helpful