Phade (bode) anaylsis of this opamp configuration?

**broken link removed**

Since opamps using the inverting configuration have a built in 180° lag, I would think that at low frequencies, the opamp would just drive the output to the open loop gain lagged 180° since the gain at low frequencies would just be -(Z2/R1), in which Z2 goes from infinity to a decreasingly large number from DC on.

However, the book I am reading is saying that this configuration STARTS with a lag of 270° due to the pole at the orgin. It says the pole at the origin causes an additional 90° of lag at low frequencies. Does anybody mind explaining to me why this is and why I don't see this when I simulate this circuit? Am I doing something wrong or just not understanding?

Bode plot I came up with from simulation is attached. R1 = 1k, R2 = 14.7k, C1 = 2nF, C2 = 220pF

The 90 degree pole is in the frequency domain while the inverting 180 degree shfit you speak off is more in the time domain...(not really, but I am not sure how to say it). The op-amp reacts differently to different frequencies and this is called the frequency response.

You are going to need to know about imaginary/complex numbers for this, specifically the graping of complex numbers on the complex plane and the angles.) Mathematically, a pole is of like the zero or root of a demominator in an equation. When it occurs it causes a divide by zero.

For the transfer function (the equation that represents the response of the system for a particular input frequency component), if a pole appears at zero, this causes the phase of the denominator of the transfer function to be 90 degrees at DC (because a purely imaginary number has a phase of 90 degrees). It causes the transfer function to have a phase-shift of at least 90 degrees right off the start since all frequencies must be zero or greater.

A transfer function with one pole might look something like this:
G/(1-jw),

j is the complex variable a.k.a i or sqrt(-1), w is the frequency in radians per second for the input frequency component you are interested in.

Now, you can see that the denominator 1-jw=0 to occur for a certain w and this will cause a pole at w. A pole at zero will be something like 0-jw. As you can see, this is a purely imaginary number and therefore has a phase of 90 degrees.

Also, the phase shifts introduced by all the the poles at frequencies lower than the frequency of interest accumulate, and this would cause 90+180 = 270 degrees. You see, if there is more than one pole in the system (the transfer function equation) it looks something like this for an equation with 3 poles:

G/(0-jw)(1000-jw)(10000-jw)

For the pole A-jw, when w >> A, the real component is very small relative to the imaginary component and on the complex plane it has an angle approaching 90 degrees. Similarily. when A >>w the real component is very large relative to the imaginary component and this makes an angle very close to the X-axis on the complex plane making it 0 degrees phase- at frequencyes much lower than the pole frequency of a particular pole, it contributes pretty much 0 degrees. So this means that at frequencies much greater than a particular pole frequency, that pole is contributing pretty much 90 degrees of phase shift, in addition to all the other poles lower than your current frequency. They all add up and accumulate. As w get's higher, each pole contributes more and more phase shift...the phase shift from the lower poles doesn't dissapear, you can see it is still there.



Something like that...I'm kinda rusty since summer started.

BTW, to see this your software must be able to simulate the phase response Bode Plot. You won't see it in the magnitude response graph.

As a practical matter, you could not build this circuit and measure it because it is DC open loop. Perhaps your simulator is having the same problem, it is not a measurable circuit. The circuit is an integrator which has 90 degree phase shift but due to R2, the phase goes to 180 degrees at high frequency.

The lag is 270 degrees at zero frequency only if the op amp's gain is infinity. It looks like you are using a very low open-loop-gain amplifier. Try rerunning your sim with the op amp's gain set to 1e12. It's all relative to open loop gain, frequency, and the time constant R1*C2.

Yes I know about transfer functions and how they work (I'm alittle rusty like you, summer break is making me dumb). The transfer function is the frequency domain, but the opamp's 180° lag is not part of this domain, but you still add the two together which gives you a total lag of 270° at DC? Thats alittle hard to follow.

The software I'm using is Orcad Pspice. Yes the plot you see if the phase response of the opamp configuration above. Youre saying that this plot is not going to show the additional 90° of lag? Doesnt this mean that the plot wouldnt show the lagging and leading due to the zero and the pole? Do you mind explaining why? Is there a way I have to set up the configuration to account for the pole at orgin along with the other zero and the pole?

Thanks again for your help so far.

FusionITR said:

Yes I know about transfer functions and how they work (I'm alittle rusty like you, summer break is making me dumb). The transfer function is the frequency domain, but the opamp's 180° lag is not part of this domain, but you still add the two together which gives you a total lag of 270° at DC? Thats alittle hard to follow.

The software I'm using is Orcad Pspice. Yes the plot you see if the phase response of the opamp configuration above. Youre saying that this plot is not going to show the additional 90° of lag? Doesnt this mean that the plot wouldnt show the lagging and leading due to the zero and the pole? Do you mind explaining why? Is there a way I have to set up the configuration to account for the pole at orgin along with the other zero and the pole?

Thanks again for your help so far.

Click to expand...

I'm rusty so read my stuff with a very critical and suspicous mind. The plot you provided is a magnitude plot (the Y axis is dB, not degrees). YOu need a phase plot. I don't know how to get OrCAD to graph a phse reponse...I have a feeling I'm at the same point you are in all this stuff...I learned this stuff in analog filter and amplifier design so I still don't know how to apply it all.

When you say pole at the origin, you mean the circuit and not the op-amp right (in my class we kind of treat them as a black box when designing a circuit with them...designing the op-amp itself is a different story though)? I'm pretty sure this is because of the capacitor feedback just sitting up there all by itself which introduces a 1/jwC into your circuit which would cause a pole at zero. You have to know what you want your circuit to do to compensate for it...but you already got your circuit implying you know what you want it to do...you know what I mean?

dknguyen said:

I'm rusty so read my stuff with a very critical and suspicous mind. The plot you provided is a magnitude plot (the Y axis is dB, not degrees).

Click to expand...

No the Y axis is degrees, not db.

Whenever you plot phase, the Y axis defaults to degrees (notice the "d" after each number on the Y axis).

Oh I see now that the graph is a phase plot. I don't know then...but just to throw it out there... are you sure the feedback loop shouldn't be on the inverting terminal rather than the non-inverting one? (I'm actually asking because I've always had the feedback loop on the inverting input...wouldn't doing it the other way cause positive feedback instead of negative feedback?)

Could it be that there is a zero that occurs at an ultra low frequency that counteracts the phase shift from the pole at zero very early on that you can't accurately see on the graph?

Have you worked out the transfer function for this circuit?

Read my previous post. Below are schematic and Bode plots wih Avol set to 100 and to 1e12.
Keep in mind that -270 degrees=90 degrees.

Damnit, yes you are right I forgot to flip the op amp around in the schematic but the simulation data I posted is from an opamp with the feedback going to the inverting terminal.

Ron_H, I am not sure what I am looking at in that bode plot are the dotted lines phase and the solid lines gain? And i'm assuming the blue line is the open loop gain set to 1e12 and the green line isn the open loop gain set to 100? Then yes it would seem your simulation shows the phase of -270 whent he open look gain is set high.

And yes when I turn the open loop gain up, I get a plot that looks like yours, except the frequency flat lines at 90 degrees again instead of going below 90.

Bode plot here: **broken link removed**

FusionITR said:

Damnit, yes you are right I forgot to flip the op amp around in the schematic but the simulation data I posted is from an opamp with the feedback going to the inverting terminal.

Ron_H, I am not sure what I am looking at in that bode plot are the dotted lines phase and the solid lines gain? And i'm assuming the blue line is the open loop gain set to 1e12 and the green line isn the open loop gain set to 100? Then yes it would seem your simulation shows the phase of -270 whent he open look gain is set high.

And yes when I turn the open loop gain up, I get a plot that looks like yours, except the frequency flat lines at 90 degrees again instead of going below 90.

Bode plot here: **broken link removed**

Click to expand...

Fusion, you are interpreting the plot correctly.
Your phase settles at 90 degrees because you are using an infinite (or at least very high) bandwidth amplifier. My amplifier model has a single internal pole with a gain-bandwidth product of 10MHz. The additional high-frequency phase shift is due to that pole.

A final (?) note of explanation: If you work out the transfer function of an operational integrator, the pole is actually at w=1/(k*R*C) (w in radians/sec), where k is the open loop gain of the op amp. You can see that the pole is at zero frequency only if k = ∞. As k goes down, the pole frequency goes up. You can see that on the Bode plots I posted.