OPA189: Open loop gain PSpice for TI

Yup that's what I was missing!  Thank you!

No problem.  Good luck!

Hello again Marek--I am currently trying to determine the stability limits for my amplifier design--For the most part, it seems to make sense--but when I increase my load capacitance to 0.8uF and look at the transient response, there is quite a bit of overshoot.  My open loop response shows I have a phase margin of ~90 degrees.  I can't imagine that's actually true.  I have attached the open loop circuit, open loop AC sweep plot, closed loop circuit and the closed loop AC sweep plot, Step response circuit and step response.  Am I doing something wrong?  FYI--I am using PSpice for TI to simulate.  My simulation also doesn't seem to agree with overshoot vs phase margin or gain peaking vs phase margin graphs.

Hi Katrina,

The phase margin is in fact close to 90 degrees but you do not correctly simulate the small-signal overshoot.  Since the input signal gets filtered out by feedback RC, in order to see the true overshoot you must apply a square current signal (+/-100uA) directly to the output - see below. Doing so shows 7% overshoot - a very stable system. 

Ok I think I understand this--one (maybe stupid) question--where does the 13.1mV value come from for percent overshoot?  Also--am I doing something wrong with the gain peaking simulation?  It looks like ~10dB of peaking, which relates to about 18 degrees of phase margin--obviously this can't be right?

It should be initial step 6.6 mV (good catch) resulting in 14% overshoot - still a very stable circuit. 

Great--thanks for clarifying.  In regards to the gain peaking--is there something I'm doing wrong there?

Hi,

where is the 170pF cap coming from?

Kai

Hi Kai--there is actually a ADG1404 mux in parallel with 10K feedback resistor.  The 170pF is the Con of the mux.  Although you bring up a point--in the case where the feedback is only the 10K (mux is off), I need to take into account the Coff value for that capacitance in my simulations, not Con.  Something more like the picture below:

Actual circuit (values are not final values):

Hi,

you may eventually improve the situation by swapping the switch contacts: Connect pin "D" to the output of OPAmp (eventually behind an isolation resistor) and connect the resistors directly to the -input of OPAmp.

Kai

How does that change improve the situation?

The 170pF Con capacitance moves from the -input of OPAmp to the output where it has less impact on the stability:

kb_opa189.TSC

Kai

And the whole circuit:

kb_opa189_1.TSC

Please note that I had to increase C7 from 10pF to 47pF.

Kai

Ah yes--I did not include the mux channel off characteristics in my circuit model. If I'm reading this correctly, the first circuit is with the 10K channel on and the second is with all channels off.  I noticed you did not include the isolation resistor?  I also noticed that you broke both the inverting input and non inverting input at AC--is this necessary?  Is the model you used for the amp input a generic model that takes into account the parasitic capacitances?  And since the loop is broken before the input to the amp, don't you have to add the input capacitance of the amplifier to the feedback?

Hi,

KB said:

If I'm reading this correctly, the first circuit is with the 10K channel on and the second is with all channels off.

Yes, I started with what I assumed to be the worst case.

KB said:

I noticed you did not include the isolation resistor?

I only wanted to demonstrate the difference when swapping the pins of analog switch. But there's still enough room for further fine tuning. So, adding an isolation resistor may be a good idea. I leave it up to you.

KB said:

I also noticed that you broke both the inverting input and non inverting input at AC--is this necessary?

No, it is not necessary when the +input is directly connected to signal ground. But if not, you may need this true differential approach. To be honest, I always use this differential approach, if necessary or not not, because I take an already fabricated simulation file in which I only change the OPAmp circuit and the component values. This is much simpler and faster than drawing the whole circuit always from scratch.

KB said:

Is the model you used for the amp input a generic model that takes into account the parasitic capacitances?  And since the loop is broken before the input to the amp, don't you have to add the input capacitance of the amplifier to the feedback?

Exactly. C4, C5 and C6 play this role.

The method I use to carry out the phase stability analysis and to simulate the phase margin is similar to what is shown in the TI's training video series on stability:

https://training.ti.com/node/1138805

But I introduce the stimulus directly to the input pins of OPAmp and look what comes back via the feedback loop. From the frequency response and phase response I can directly determine the phase margin, without needing to use the post-processor of TINA-TI.

Because the feedback loop is opened at the input pins, the feedback loop no longer sees the input capacitances of OPAmp ("C4", "C5" and "C6" in the simulation). Because of that the input capacitances have to be "mounted" externally.

"L2" and "L3" close the feedback loop for DC and allow the OPAmp inputs to be properly DC biased.

"C3" provides an AC coupling of stimulus without ruining the DC biasing of OPAmp inputs. And because the AC coupling shall be invisible even for the lowest frequencies "L2", "L3" and "C3" are chosen to be "infinitely" high.

The advantage of this a bit more opulent method is that the complex output impedance of OPAmp is also taken into calculation.

Keep in mind that only the phase shift at unity gain in the phase stability analysis is called "phase margin" and not the phase shift at any other frequency, because only the phase shift at unity gain can destabilize the circuit. I have discussed this here:

https://e2e.ti.com/support/tools/simulation-hardware-system-design-tools-group/sim-hw-system-design/f/simulation-hardware-system-design-tools-forum/1087610/tida-00489-pir-amplifier-stability-analysis

Kai

Katrina,

For your specific condition where I had noticed you have changed CL to 150nF, you would need to re-compensate the circuit by increasing feedback cap. Using C7 of 800pF would result in the phase margin above 93 degrees - see below.

Running transient analysis confirms circuit stable operation with a small-signal overshoot of around 9% - see below.

Kai--thank you for your very detailed answer!

Marek--thanks for looking at that. I actually have to design this circuit to work with a range of loads--part of what has been difficult with this design.  Also why the simulation is so important--I need to know the limits of the final circuit.  Thanks so much for all of your help!

No problem.  Also, thank you, Kai. Please let us know if you need further assistance.  Good luck.