[SOLVED] simulate open loop gain

[JanniS]

I have a problem with an open loop simulation.

I have a resistive load of 10 Kohm and a capacitive of 1pF. Usually I just insert a DC offset voltage source at one of the input terminals, run a DC sweep and chose the offset value to be where the output voltage is in the middle.

This time I can not find this point. i think it is because of the resistive load, because I usually only have capacitive load. It seems like the simulator can not find a point in the middle even with 10 significant digits of the offset source.

Another method I tried is to simulate with a feedback network consisting of a 100 Gohm resistor and a 10F capacitor to create an output voltage in the middle, but when I do that the phase response starts at -180 degrees.

So... how do you usually simulate open loop? My gain is around 80-90 dB..

Thank you!

 

[LvW]

Jannis,

* the method with a dc voltage at the input that leads to a "good" operating point in the middle of the linear region sounds good and should work - independent on load conditions.
* the other described method is an alternative (with some restrictions due to loading errors) - however what is the purpose of a 100Gohm resistor? Who told you this? Use instead a 100H inductor!

 

[JanniS]

I thought that maybe the gain was too high to find a point in the middle . The sweep response looks as I expected but it tilts no matter how close i get to the point in the middle
I also tried with the inductor but the phase response is the same and starts at -180 degrees as seen on the figure. I have never seen anything like that. This only happens when the load is at same potential as the common mode at 2.5 (rail is 0V-5V).

 

[LvW]

I thought that maybe the gain was too high to find a point in the middle . The sweep response looks as I expected but it tilts no matter how close i get to the point in the middle
I also tried with the inductor but the phase response is the same and starts at -180 degrees as seen on the figure. I have never seen anything like that. This only happens when the load is at same potential as the common mode at 2.5 (rail is 0V-5V).

JanniS, your dc response looks as if you are using single supply.
Why don`t you use normal opamp operation with double supply? This will certainly work.
And - select a good resolution for dc sweep (steps of 10µvolts for example).

 

[JanniS]

I am not sure what you mean by single supply.

My test bench looks like this:

 

[LvW]

JanniS, sorry but I don`t completely understand your "test bench".
For example, what is the purpose of the current source ?
Why do you have three inputs ?
At which node is the L-C circuit connected?

Regarding single supply: Are you not aware that in most operations the opamp is used with TWO supply voltages (+-Vsupply) ?
This has the advantage that the input does not need any bias voltages (except a small offset correction in case of missing feedback).
According to your diagram, you operate with one single supply voltage only.
Here are my recommendations:

** Open-loop measurements (normal operation with symmetrical supply voltages):
1.) Inv. input grounded, small dc voltage at the non-inv. input swept over +-5mV (10µV resolution)
2.) Fix the dc voltage at a level which gives approx. zero dc output.
3.) Add an ac source in series to the dc source.
4.) AC analysis gives you gain and phase.

** Alternative (without dc offset correction)
1.) Place a large inductor (100H) between output and the inv. input. Thus, you have full dc feedback (no dc offset problems).
2.) AC analysis with an ac source at the non-inv. input.
3.) For single supply operation of the opamp put the ac source in series to a DC source of 0.5*Vsupply.

Supplement/Correction to point 1.) Place a large capacitor (100F) between inv. input and ground.