Hi guys..much after my previous thread " Exact role of opamp in integration circuit"..took lots of effort ..especially thanks to Mike,Jony and crutschow for their valuble inputs..now I got stuck into another problem(though now I understood the previous one ..

The text says that "opamp integrator circuit suffer from two distortions first is nonlinearity due to parallel feedback resistor across the capacitor and other is "corner rounding".Corner rounding occurs on the output ramp waveform when the opamp does not have a fast slew rate .To avoid this type of distortion the opamp slew rate should be 100 times faster than the ramp rate of change (the ramp rise time)".

How SR is related with waveform rounding?
Please simplify the statement so that I could gain some insight....

 

Basically it is a limitation in bandwidth. The corners which require high bandwidth (quick response) get rounded.

 

How SR is related with waveform rounding?

An opamp's slew rate limit and its gain-bandwidth product both limit an opamp's ability to respond quickly to rapidly changing inputs.

No opamp can respond infinitely fast to a change in input. Not only does it take a finite amount of time for the opamp's output to jump from one value to another in the case of an input step, but it also takes a finite amount of time for the output's first derivative (that is, its dV/dT) to reverse sign when its inputs suddenly change direction.

Thus, all sharp corners on waveforms get rounded off.

If you look at an opamp data sheet, you'll usually find somewhere a pair of graphs showing waveforms; one is called the Large Signal Response, and the other is called the Small Signal Response. The Large Signal Response shows the action of slew rate limiting in the opamp, while the Small Signal Response shows the effects of the opamp's limited gain-bandwidth product (that is, its reduced gain at high frequencies).

 

Thus, all sharp corners on waveforms get rounded off.

...If slew rate limitation is there then there may be some phase difference between input and output due to difference response time of opamp...but why rounding off...of the waveform..? Not exactly getting...

 

...If slew rate limitation is there then there may be some phase difference between input and output due to difference response time of opamp...

Usually when we speak of "phase difference" this implies that the system (whether an opamp circuit or whatever) is operating in a linear fashion; but slew rate limiting is nonlinear so strictly speaking, it is not correct to describe it in terms of a "phase difference."

but why rounding off...of the waveform..? Not exactly getting...

You need to try harder, then.

Look at it this way: suppose you're driving along the highway and decide to turn left. No matter how fast or hard you yank on the steering wheel, you can't make the car instantaneously change direction; rather, the car describes an arc joining its path before you decided to turn, and the path you end up on after the turn. Just as your turn gets "rounded off" by the car's momentum and the fact that it takes you time to turn the steering wheel, so does the opamp's slew rate limit "round off" the peaks in a triangle waveform.

I hope I've explained it simply enough for you; I can't explain it any more simply. Go think about it for a while before asking any more questions, OK?

 

...If slew rate limitation is there then there may be some phase difference between input and output due to difference response time of opamp...but why rounding off...of the waveform..? Not exactly getting...

When you say "phase difference" you mean delay between output and input but that is not what happens because the amplifier has no problem following the parts that change slowly. But it cannot follow the parts that change rapidly and, as has been said, it is impossible for an electrical magnitude to change instantly because inductance and capacitance are everywhere and you cannot get rid of them. You cannot instantly stop an electron any more than you can instantly stop the Titanic.

Any wave, no matter its shape, through Fourier analysis, can be analyzed by decomposing it into sinusoidal waves. If you take a perfectly triangular wave it decomposes into a fundamental sine of the same frequency and harmonics that go into infinity. Any real world device is going to block the higher harmonics all the way to infinity.

The same happens with mechanical movement. No magnitude can change instantly.

 

it is impossible for an electrical magnitude to change instantly because inductance and capacitance

I got the point ...is it only just because of parasitic inductance and capacitance or any other phenomenon that prevents an electrical magnitude to change instantly....??

or can it be said that newton laws of motion are applied to electron as well.....

 

I got the point ...is it only just because of parasitic inductance and capacitance or any other phenomenon that prevents an electrical magnitude to change instantly....??

or can it be said that newton laws of motion are applied to electron as well.....

For a voltage to change instantly you would need infinite current. For a current to change instantly you would need infinite voltage. For a body to change motion instantly you would need infinite acceleration....

Electronic devices have their own limitations which manufacturers try to improve but the basic limitations are there and you cannot totally get rid of them. Capacitance and inductance are an intrinsic aspect of any moving electrical particle. Just like inertia is intrinsic to any mass.

 

thanks guys..

 

I got the point ...is it only just because of parasitic inductance and capacitance or any other phenomenon that prevents an electrical magnitude to change instantly....??

No. Inductance has absolutely nothing whatsoever to do with an operational amplifier's slew rate limiting behavior. It is primarily caused by the way in which frequency compensation is implemented in internally-compensated opamps.

There's a nice, very straightforward explanation here: http://www.edn.com/electronics-blogs/the-signal/4415482/Slew-Rate-the-op-amp-speed-limit

 

Hello there,

The simplest explanation is that there is a small capacitor inside the op amp that must be charged or discharged using a limited current and limited frequency response current source. This capacitance can be looked at like a transistors base emitter charge storage, only larger, and in some cases may actually be the base emitter base area charge storage. So we can look at it as a common emitter transistor with a capacitor from base to emitter and the base (and thus the capacitor also) is being driven by a current source that has limited frequency response and a limited peak value.

The limited peak value of the current source sets the slew rate. The limited frequency response of the current source means that the current can not change too abruptly. The two act together to form a source who's response is the convolution of a frequency response limited current source that is also limited in peak value and an impedance that is mostly capacitive.
So it takes some finite time t1 for the current source to respond to the input in the first place, and while that current source is responding the slew rate is possibly slowing down the voltage change across the capacitor as well (once the change from the limited frequency response becomes greater than the slew rate, if it ever does).
This can be looked at roughly as the convolution of at least a step and a ramp, which in the time domain comes out to the shape of a parabola, which as you know, has a rounded mid section in time in the vicinity of t=0.