A question about slewing in an opamp

[naalald]

To my knowledge, an opamp slews if its differential input is large enough to turn off one of the two input transistor pairs. In this case the output current is limited and the output capacitor is charged with a constant current so that the output voltage will be
vo(t) = SR×t
So if the differential input of an opamp is so small that none of the two input transistors turns off, can we say that the opamp does not slew? How can we recognize if an opamp slews?

 

[A.Anand Srinivasan]

an opamp doesnt slew only when the diff input is large enough to turn off one of the two input transistor pairs.... an opamp slews when there is a change in the input which needs the opamp to change its output faster than it can.... whenever the opamp slews the ouptut will be a straight line with a constant slope equal to the slew rate....

 

[saro_k_82]

an opamp doesnt slew only when the diff input is large enough to turn off one of the two input transistor pairs.... an opamp slews when there is a change in the input which needs the opamp to change its output faster than it can.... whenever the opamp slews the ouptut will be a straight line with a constant slope equal to the slew rate....

Mmm...I have doubts about this. If the signal is not large, it is a high frequency small signal and first stage is fully operational. Won't the response be dictated by the linear exponential charateristic ? If both the input transistors are on, there is constant feedback about what is happening at the output which will make the response exponential. AFAIK slewing is a large signal phenomenon and it requires the feedback path to be shut off.
Or is my beleif wrong?

 

[A.Anand Srinivasan]

Mmm...I have doubts about this. If the signal is not large, it is a high frequency small signal and first stage is fully operational. Won't the response be dictated by the linear exponential charateristic ? If both the input transistors are on, there is constant feedback about what is happening at the output which will make the response exponential. AFAIK slewing is a large signal phenomenon and it requires the feedback path to be shut off.
Or is my beleif wrong?

when i said change of input at a fast rate i meant high frequency only... i didnt talk about large signal... afaik i know feedback has nothing to do with slew rate... feedback doesnt have a curative effect on slew rate... say your input stage is staying on but there is a finite gap between the input and the feedback from the output... give a square wave (high frequency of small amplitude) input to your opamp which has considerably high gain you'll be able to see the slew in the output....

the higher the gain the larger will be the chances of slew... because a small input change means a large output change... correct me if there is a fault in what i'm sayin...

 

[naalald]

an opamp doesnt slew only when the diff input is large enough to turn off one of the two input transistor pairs.... an opamp slews when there is a change in the input which needs the opamp to change its output faster than it can.... whenever the opamp slews the ouptut will be a straight line with a constant slope equal to the slew rate....

Thanks for your reply. That's right, but the change in the output rate is also dependent on the level of the input, in addition to it's rate. Is this right?

 

[LvW]

when i said change of input at a fast rate i meant high frequency only... i didnt talk about large signal... afaik i know feedback has nothing to do with slew rate... feedback doesnt have a curative effect on slew rate...

The slew rate of an opamp is specified with feedback ! (Normally, with 100% fedback). That means: feedback is the key element for the occurence of slewing.
The effect is as follows:
The input signal brings the first stage into saturation with the consequence that the compensation cap is loaded with a constant current. This effect is reponsible for the slew rate with a quasi-linear rise of the output voltage. After a certain (small) delay time the feedback path works and brings the first stage back to normal/linear operation.

 

[saro_k_82]

when i said change of input at a fast rate i meant high frequency only... i didnt talk about large signal... afaik i know feedback has nothing to do with slew rate... feedback doesnt have a curative effect on slew rate... say your input stage is staying on but there is a finite gap between the input and the feedback from the output... give a square wave (high frequency of small amplitude) input to your opamp which has considerably high gain you'll be able to see the slew in the output....

the higher the gain the larger will be the chances of slew... because a small input change means a large output change... correct me if there is a fault in what i'm sayin...

Feedback (or the lack of it) is responsible for slewing in opamps. For exponential characteristic, the rate at which the final value is reached depends on the instantaneous difference between the desired output and the instantaneous output. As long as there is a feedback link, the rise/fall can only be exponential. A major disadvantage of Slewing is that it can introduce a lot of input differential voltage which might be catastrophic (the protection diodes add lot of noise) and this is at unity gain feedback.

Have a look at this very popular paper by J.E.Solomon
**broken link removed**

 

[LvW]

Feedback (or the lack of it) is responsible for slewing in opamps. For exponential characteristic, the rate at which the final value is reached depends on the instantaneous difference between the desired output and the instantaneous output. As long as there is a feedback link, the rise/fall can only be exponential. A major disadvantage of Slewing is that it can introduce a lot of input differential voltage which might be catastrophic (the protection diodes add lot of noise) and this is at unity gain feedback.

I don´t know which "exponential characteristic" you are referring to. Moreover, what means "difference between desired output and instantaneous output" ???

Again: The slew properties of the opamp are responsible for its large signal or power bandwidth. Therefore, the slew rate is specified WITH feedback.
Case 1: If the input signal is large enough to bring the first stage into saturation and if the rise time of the input signal is smaller than the time necessary for the signal to travel through the circuit and the feedback path, then slewing will occur (quasi-linear rise of the output). This "travelling" time, of course, is determined by the bandwidth resp. the transit frequency of the opamp. The slew rate itself is dependent on the compensation capacitor.
Case 2: If the input rise time is larger than the travelling time through the whole path back to the inverting terminal, the first stage will NOT go into saturation because the feedback is effective and holds the first stage within its linear region.
And as a consequence: There will be no slewing of the signal.

 

[saro_k_82]

When you apply a step, and if the amp is linear everywhere, you expect exponential chracteristic.
I just tried to explain the basic nature of the exponential response (dy/dt=ky)

 

[LvW]

When you apply a step, and if the amp is linear everywhere, you expect exponential chracteristic.
I just tried to explain the basic nature of the exponential response (dy/dt=ky)

Yes, that´s the small signal response of an universal compensated opamp (which behaves like a first order system). But the question was related to a large signal response and the corresponding slew rate effect. OK ?

 

[saro_k_82]

Yes, that´s the small signal response of an universal compensated opamp (which behaves like a first order system). But the question was related to a large signal response and the corresponding slew rate effect. OK ?

This is exactly what I was trying to point out to Anand. Unless the loop is broken, the response cannot be anything other than exponential. He was claiming of small-signal slew response., and I tried to point this out. Cool

 

[LvW]

This is exactly what I was trying to point out to Anand. Unless the loop is broken, the response cannot be anything other than exponential. He was claiming of small-signal slew response., and I tried to point this out. Cool

I think a reason for the slight misunderstanding between us is connected with the meaning of "slewing". For my understanding, there is no "small-signal slew response" since the term "slewing" is connected with a kind of non-linearity within the circuit caused by a large signal excitation (with saturation of the input stage for a short period).
Regards
LvW