What is the gain error for this opamp driven by a voltage

[Boltzman Oscillation]

Homework Statement

The op amp has a near ideal level 1 model with G = 5000V/V, ri = inf, ro = 0
How would I obtain the feedback function?

Homework Equations

I know I have to find the feedback function which is:

f = - (ΔV/Vin)

The Attempt at a Solution

I will first drive using the voltage source. My professor said to convert the op amp into a voltage source and short the nodes 1 and 2. He then said to find the feedback function by first finding ΔV = Vp - Vn which he gives (in the solutions) to be -(R1 + R3)/(R1 +R2 +R3 +R4). How did he find this value for ΔV? I can try using the standard way of solving ideal op amps and resulting in two equations:

1: (Vin-Vp)/R3 - Vp/R4 = Ip
2: (Vin-Vn)/R1 + (Vo - Vn)/R2 = In

and then I can solve for Vp and Vn but I do not get the same as he does. I get a value with In, Ip, Vo which he does not have. He only has resistance values in his ΔV. Any help would be appreciated. P.S. Since this is not ideal then I cannot use Vp = Vn and Ip = In = 0 right? thanks

 

[LvW]

Are you familiar with the DEFINITION of the feedback function (feedback factor)?
Definition: The feedback factor is the input differential voltage (directly across the opamps input terminals) divided by the driving output voltage of the opamp only - that means: With Vin=0 (short between node 1 and 2). Hence, it is a voltage divider expression (Vdiff/Vo) - and Vin must not appear in this function.
It is correct (as your prof said) that for this calculation the opamps output is regarded as an ideal voltage source.

More than that, because the input resistance of the opamp is infinite, you can set Ip=In=0.
But note that for calculating the feedback function you MUST assume, of course, that Vp is not equal to Vn.

Question: What is the complete task descrription? Just to find the feedback function?

 

[Boltzman Oscillation]

Are you familiar with the DEFINITION of the feedback function (feedback factor)?
Definition: The feedback factor is the input differential voltage (directly across the opamps input terminals) divided by the driving output voltage of the opamp only - that means: With Vin=0 (short between node 1 and 2). Hence, it is a voltage divider expression (Vdiff/Vo) - and Vin must not appear in this function.
It is correct (as your prof said) that for this calculation the opamps output is regarded as an ideal voltage source.

More than that, because the input resistance of the opamp is infinite, you can set Ip=In=0.
But note that for calculating the feedback function you MUST assume, of course, that Vp is not equal to Vn.

Question: What is the complete task descrription? Just to find the feedback function?

Well he actually does have Vin in the function. This is his solution below. I am almost certain he got his Vp - Vn using voltage divider but I can't figure out how. Care to explain?

 

[LvW]

In the hand-written diagram I an read Vin=0 !
And - yes: The feedback factor is F=7/10-4/10=3/10.
(Two simple voltage divider).

For calculating the closed-loop gain Acl you can apply the following formula (Aol=open-loop gain):

Acl=Aol*K/(1+Aol*F)

K is the "forward damping factor and can be calculated similar to F.
The forward factor K is the input differential voltage (directly across the opamps input terminals) divided by the driving output voltage Vin only - that means: With Vo=0

 

[Boltzman Oscillation]

In the hand-written diagram I an read Vin=0 !
And - yes: The feedback factor is F=7/10-4/10=3/10.
(Two simple voltage divider).

For calculating the closed-loop gain Acl you can apply the following formula (Aol=open-loop gain):

Acl=Aol*K/(1+Aol*F)

K is the "forward damping factor and can be calculated similar to F.

I guess I am not understanding the voltage divider here. Could you explain it?

 

[LvW]

For the feedback factor:

Vp/Vo=R4/(R1+R2+R3+R4)=4/10
Vn/Vo=(R1+R3+R4)/(R1+R2+R3+R4)=7/10
F=(Vp-Vn)/Vo=-3/10

 

[gneill]

I guess I am not understanding the voltage divider here. Could you explain it?

Look at the current path:

Just a bunch of resistors in series. If you can find the potential across $R_1 + R_3$ then you have $\Delta V$.

 

[Boltzman Oscillation]

Look at the current path:
View attachment 235603
Just a bunch of resistors in series. If you can find the potential across $R_1 + R_3$ then you have $\Delta V$.

thanks a lot!

 

[DaveE]

Interesting that the schematic shows a current source or a voltage source as the input, at least that's my interpretation. The circuit response is different for the two cases. Everyone's comments, including your professor's, imply that it's a voltage source, so all that is ok.
However, I would describe the process for determining the feedback factor slightly differently. You don't short circuit the input, you set the input to zero. Of course for a voltage source, this is the same as a short circuit. But for the current source case, it would be the same as an open circuit since current sources have infinite impedance. In that case you would do the same voltage divider analysis for the feedback but including the Rs=7K resistor.
In fact, you don't have to set the input to zero; you just find how much change in the output voltage appears as a correlated change in the amp input. This assumes that the circuit is linear and obeys superposition. You could set it to any value that's independent of the output, in practice everyone sets the input to zero because it's easier to think about it that way.
I guess I dislike the implication that you change the circuit by adding a short circuit to the input. You don't, the short circuit is built into the voltage source (which has zero source impedance). Sorry if I'm being too pedantic here, but I think there is a deeper level of understanding than just following a recipe.

 

[LvW]

In fact, you don't have to set the input to zero; you just find how much change in the output voltage appears as a correlated change in the amp input.

Dave - what do you mean: For which purpose?
For finding the closed-loop gain or finding the feedback factor?

 

[DaveE]

Dave - what do you mean: For which purpose?
For finding the closed-loop gain or finding the feedback factor?

Feed back factor. The closed loop gain is best done mathematically, IMO, after you know the feedback and amp response. This analysis assumes that you have control of the output voltage, which is equivalent to removing the amplifier from the circuit. i.e. you don't proceed to analyse what the amplifier output does in response to the input signal that is fed back from the output, ad nauseaum...

To analyse (or measure) the closed loop feedback, I would insert a signal source somewhere within the feedback path and look at how the signal is split between the forward and reverse direction (from R.D. Middlebrook). I'm not sure this is a great description since I'm being vague about what sort of signal (voltage, current, etc) and where it's injected. There is a better explanation here:
http://ecee.colorado.edu/~ecen5807/course_material/Lecture14.pdf

Then for the closed loop response you can just drive the input and look at the output.

Anyway, this is far beyond the OPs question. Sorry if I've hijacked the thread.

 

[DaveE]

Dave - what do you mean: For which purpose?
For finding the closed-loop gain or finding the feedback factor?

What I meant by finding the change in voltages is this, for example:
You could set the input voltage to 1.7V and the output voltage to Vo1=2.0V, then find the amplifier input Va1=Vp1-Vn1 (without any amplifier action, you have control of the output, not the amplifier). Then change the output to Vo2=2.1V and find the new amplifier input Va2. The feedback factor would be δVa/δVo = (Va2-Va1)/(Vo2-Vo1). You are free to choose the input and output voltages, there's nothing magic about zero. Of course anyone that is actually doing this will choose zero instead of 1.7V and 2.0V because it's easy to subtract zero.
As I said, it is a bit pedantic, more of a thought experiment than practical. My overall point is that you aren't really changing the circuit, you are just analyzing how the feedback network responds to changes in the output.

 

[LvW]

DaveE - thank you.
OK, I think your method is primarily for measurements and not for calcualtons resp. circuit simulation, right?

 

[DaveE]

DaveE - thank you.
OK, I think your method is primarily for measurements and not for calcualtons resp. circuit simulation, right?

No, all of those things. It's about understanding the circuit; analysis, measurement, design, function. Of course once you understand it, you will make assumptions to to simplify the actual calculations. I'm preaching understanding over blindly following someone else's instructions.
This, for example: https://authors.library.caltech.edu/63241/1/00405579.pdf

 

[DaveE]

DaveE - thank you.
OK, I think your method is primarily for measurements and not for calcualtons resp. circuit simulation, right?

Circuit simulators are a valuable tool, but are a curse for education and understanding. Just having the answers provided for you provides little in the way of insight about how the circuit really works.

 

[LvW]

DaveE - just one comment.
We all know how the feedback function/factor for an amplifier with feedback (negative, positive, or both) is defined.
Don`t you think that calculations/measurements/simulations should follow this definition?
For my opinion, THIS method would help to understand what`s really happening - in particular, because this method opens the way to find the loop gain of the circuit which is a key parameter for stability.

 

[DaveE]

Don`t you think that calculations/measurements/simulations should follow this definition?

Don't they already? I guess I don't understand.

 

[LvW]

Don't they already? I guess I don't understand.

OK - let me explain.
Let`s take a simple example: An ideal opamp with resistive feedback: R1=1k and R2 (feedback path)=2k.
Of course, we can immediately give the feedback factor with k=1/3..
That`s what I had in mind (see my post 16) writing that the calculation follows the definition.
Do you propose an alternative calculation of k ?

.

 

[DaveE]

OK - let me explain.
Let`s take a simple example: An ideal opamp with resistive feedback: R1=1k and R2 (feedback path)=2k.
Of course, we can immediately give the feedback factor with k=1/3..
That`s what I had in mind (see my post 16) writing that the calculation follows the definition.
Do you propose an alternative calculation of k ?

.

No.