afonsom
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Hi,
Why are you AC coupling the inputs? I believe that this your problem, since you have only positive supply you need to have some DC offsets on the inputs.
At first, it is correct and necessary to use capactive input coupling. Only in this case. the DC non-inverting gain is unity (good for single supply operation).
Secondly - how did you simulate the loop gain? It must start at Aol (open-loop gain of the opamp) due to unity gain dc feedback.
Remember: Loop gain is the gain around the feedback loop which - for measurement or simulation purposes - has to be opened at a suitable point.
In AC analysis , iprobe is ignored(shorted). Hence probing ac response on the output node will give you closed loop response and not the open loop response.
In Stb analysis, first dc operating point is evaluated (i.e. any ac signal is set to 0V), then small signal transfer response from "iprobe's" one terminal(+ve node) to the other terminal (-ve node) is reported. This basically is open loop response of the loop containing the "iprobe".
Hi! Everything you said makes sense and I undestand how it works now. But do you have any idea why the loop gain (A*Beta) that I've obtained is negative (in dB)? This mean the loop gain has crossed 0dbs close to DC so the system is unstable?
Thanks for everything!
According to your own words - something with the loop gain response seems to be wrong.
Question: Why you don`t clearly show or describe how do you have simulated the loop gain (instead of speaking only about a "probe" and a "location").
The loop gain must start (at low frequencies) with a very large value and the phase shift must be -180deg.
In order to simulate the loop gain I simply placed the iprobe in a location that opens up the loop (as shown in the schematic)
For simulating the loop gain of a system involving voltage transfer functions you must not use a current probe but a voltage probe, of course.
You need simply an ac voltage source - not available in analoglib?
For loop gain simulation, perform the following steps:
* Set all input signal sources to zero;
* Place an ac voltage source (1V) between opamp output and feedback path;
* Display vs. frequency the RATIO V(signal,out)-to-signal,in. In your case: V(out)/V(8).
Did you follow this procedure (including step 1)?
Only now I have recognized that you are working with single supply.
But I cannot verify a correct input biasing with Vdd/2.
Did you check the DC operating point?
What is the DC value of the output voltage with feedback?
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