dependence of response on configuration of opamp

[srimatresri]

an opamp is configured in 2 ways, non inverting and inverting. which configuration is fast assuming single pole transfer function for opamp

 

[LvW]

an opamp is configured in 2 ways, non inverting and inverting. which configuration is fast assuming single pole transfer function for opamp

The bandwidth (response time) depends on the open-loop characteristics as well as on the feedback factor "beta".
For both configurations this factor "beta" is nearly the same (because for constant "beta" both gain values differ only by "1")

 

[srimatresri]

what if beta is 1?

 

[srimatresri]

i deduced the transfer function for both the configurations with single pole. what i observed is, the non inverting configuration has higher gain compared to inverting configuration and the pole locations are same in the closed loop.


the problem is now simpler,
we have two frequency responses one with higher gain , both having same pole location. now can u tell which is faster

 

[poorchava]

The speed of opamp (if you mean bandwidth and slew rate by this) doesn not depend on feedback configuration unless feedback network includes some reactive components (capacitor, inductor). Only on the parameters of the opamp itself.

 

[srimatresri]

The speed of opamp (if you mean bandwidth and slew rate by this) doesn not depend on feedback configuration unless feedback network includes some reactive components (capacitor, inductor). Only on the parameters of the opamp itself.

so u mean unless otherwise other non idealities of opamp are considered, both are equally faster

 

[LvW]

The speed of opamp (if you mean bandwidth and slew rate by this) doesn not depend on feedback configuration unless feedback network includes some reactive components (capacitor, inductor). Only on the parameters of the opamp itself.

Poorchava, have you really never learned that feedback widens the bandwidth and, thus, can reduce the reaction time?

---------- Post added at 16:55 ---------- Previous post was at 16:51 ----------

the problem is now simpler,
we have two frequency responses one with higher gain , both having same pole location. now can u tell which is faster

It is not the closed-loop gain that determines speed but the gain of the open loop and its bandwidth (simply "loop gain"). And the loop gain is determined by the feedback path only.

 

[srimatresri]

i asked one of my teachers and suggested to approach like this:

 

[LvW]

Some comments are necessary:
* The shown equations are somewhat confusing. What is Hs?
The non-inv. gain is twice the inv. gain for 50% feedback only - that means: Gain (non-inv.)=2 and gain (inv)=-1.

* According to the drawing the same feedback factor applies to both circuits. Hence, the same loop gain bandwidth (as indicated in the drawing).
Hence, the same small-signal response.

 

[srimatresri]

for the second image i attached,
i assumed 50% feedback only.
it means gain of non inverting =2 and that of inverting is 1(phase is 180 deg)

as u said same feedback factor for both circuits.

i took the closed loop response of both configurations as (1/(s+1)) and (2/(s+1))
i found the output response for step input.
in that the output raises fast in non inverting configuration as it has higher gain(a multiplication factor of 2).

 

[LvW]

i took the closed loop response of both configurations as (1/(s+1)) and (2/(s+1))
i found the output response for step input.
in that the output raises fast in non inverting configuration as it has higher gain(a multiplication factor of 2). so non inverting configuration is faster

To be specific: For both gain configurations the output (small-signals only!!!) is reached within the same time period. How do you define "fast"?

 

[srimatresri]

i got this question in a quiz
i am attaching the exact image.
the fourth option missing is "both are equally fast".
View attachment Fullscreen capture 10142011 25421 PM.bmp

what i thought fast means "output is raising quickly". i still have the confusion whether fast means this or "how quickly it reaches steady state"

 

[LvW]

It`s just a matter of definition.
As mentioned before: Both configurations reach their small-signal steady-state output within the same time period (settling time constant) - determined by the available bandwidth (which is equal).
However, that means that the circuit with higher gain has a lower rise time.

So - you have the choice which of the both parameters you are using to define the property "fast".

 

[srimatresri]

it was a question in quiz. i have to give only one option in that image.
he did not mentioned what fast means in the question.

 

[LvW]

Sorry, I have made an error: In my last posting I mentioned the term "rise time". That`s not correct. It should be "slope" of the rising signal.
To make it clear, take the following example:
Excite a simple passive RC lowpass with a 1V-step and a 2V-step.
In both cases, the settling time is equal (of course) but the slope is different. But I think, it is not appropriate to say "one configuration is faster than the other one".