Time Constant of an Opamp Circuit (1 Viewer)

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1. The problem statement, all variables and given/known data:
Find the time constant of an inverting op amp with C = 50uF, Ri = 200 Ohms, Rf = 20 Ohms.

This is a problem I have found in a prep book for the Fundamentals of Engineering Exam (Electrical component). The book gives the answer as time constant = Rf * C. I am not exactly sure what kind of op amp you would classify this as. My first thought is that it is a practical differentiator. But you got that resistor and capacitor on the input side. Surely, Ri on the input side should have some say in the time constant. I tried to go about this question by considering that the current on the input side is equal to the current on the output side. My goal was to try to prove the book's answer by finding a gain equation in terms of just Rf and C but have been unsuccessful. Looking to see if I can get some help on this. I just might be approaching this differently. These questions are supposed to be simple.

Thanks! Dan.

2. Relevant equations

3. The attempt at a solution
 

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berkeman

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Does the problem specify "time constant for a step input waveform" or something like that? Can you Upload a PDF or JPEG of the full question? Thanks.
 
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berkeman, That's what I was thinking. The circuit you provided looks like the problem with the exception being that the problem has the capacitor come first and resistor second on the input side, but I don't think that makes a difference. I will provide a screenshot of the book cover and problem tomorrow. This particular book doesn't give good explanations or tips on how to get to the answer.
 

LvW

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dvscrobe - the inverting opamp circuit is given with 3 parts; I don`t know how you could arrive at another circuit with 4 parts (2 capacitors). Such a circuit would be described with two time constants.
Therefore, the only practical circuit which makes sense (and has one single describing time constant) is the active first-order lowpass with a parallel combination RF||C as feedback element.
The transfer function is
H(s)=-[RF||(1/sC)]/Ri=...=-(RF/Ri)/(1+sRF*C).
This first order lowpass has a time constat T=RF*C and a cut-off frequency wc=1/T
 
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LvW, Hmmm, interesting. As a student in an undergrad EET program, transfer functions is something I have not gotten in to yet. I did learn about Laplace Transforms in my last calculus course, so I am able to tell that you took the equation for Voltage out / Voltage in = - Rf / Ri and converted it to Laplace since you are dealing with complex numbers. Based on the resulting transfer function -(RF/Ri)/(1+sRF*C) how are you able to tell that this is a first order low pass and how do you only come up with Rf and not including Ri? Thanks. I also attached the book cover and problem.
 

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LvW

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dvscrobe - do you realize that the shown circuit diagram contains four passive parts - in contrast to your question in which you are mentioning only three parts?
As you can see, asking for some circuit properties without showing the circuit is always problematic.
(I suppose you have edited your question showing the circuit diagram some days after my answer).
Maybe that my former answer is not a solution for your problem - but that is not my fault.
 
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Darn, my editing was me trying to fix my typo’s. I thought I drew the circuit correctly in my attached homework problem but realize my wording in the step 1 was poor. I thought someone could find something on the circuit I drew. Newbies for you. Okay, thanks for responding.
 

LvW

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dvscobe - the circuit as shown in the diagram has the following transfer function (bandpass with Q<0.5):

H(s)=sRF*C/[1+s(T1+T2)+s²(T1T2)]

with T1=Ri*C and T2=RF*C

As you can see, there are two time constants which also appear in the step response (continuation of a tangent to the falling resp. rising slope of the step response).
(Question: Are you sure about the given resistor values ? Rather uncommon !).
 
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(Question: Are you sure about the given resistor values ? Rather uncommon !).[/QUOTE]

LvW, I attached a screenshot of the problem here. I guess I should have done this on the original post. This is how I drew it on my first post in my homework though. It does look like a very odd circuit. Notice how there are only three variables, C, Ri, and Rf, but there are 2 of each one. I guess I will need to use a virtual instrument software or something to run this thing and see what it does. I am wondering if the book made a mistake. Is that possible? Dan.

upload_2019-1-15_10-26-22.png
 

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berkeman

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There is a very good reason that they have drawn copies of Ri and Rf connecting the + opamp input to ground. See if you can find the reason in some of the opamp application notes. (They could have just put a single resistor there with a value equal to the parallel combination of Ri and Rf, but it's more illustrative for teaching purposes to show it like that.)

Hint -- add to your search terms: Input Bias Current of Opamp

:smile:

EDIT / ADD -- Actually, since the input is AC coupled with that capacitor, the Ri component in the + input leg should not be needed. Can you say why? :smile:
 

LvW

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LvW, I attached a screenshot of the problem here. I guess I should have done this on the original post. This is how I drew it on my first post in my homework though. It does look like a very odd circuit. Notice how there are only three variables, C, Ri, and Rf, but there are 2 of each one. I guess I will need to use a virtual instrument software or something to run this thing and see what it does. I am wondering if the book made a mistake. Is that possible? Dan.
No, it must not be a mistake. As you can see - the given answer in the book is one of the two time constants I have mentioned. Perhaps they consider this time constant as "primary". By the way - this has, of course, something to do with the DEFINITION of a systems time constant.
Before answering such a question, It is quite normal (even necessary) to verify the definition of the quantities which are wanted - in this case the quantity called "time constant". Therefore my recommendation: At first, search for the definition of such a term.
By the way: Do you know how the step response looks like? This could help and is of course part of the answer to be given.
 
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berkeman, Thanks for the tip on Input Bias Current of Op Amp. I also did some reading on coupling methods for Op Amps. Yes, I can see how Ri would not make much of a difference because the Op Amp does not load down the source, by having a very high input impedance. So, AC would pass thru at high frequencies since the capacitive reactance would go down as frequency goes up but my original thinking was that the current in the input circuit equals the current in the feedback circuit. This is why I thought Ri played a role in this current.

LvW, From your response, I realize that this circuit can have two time constants but it is up the tester to try to understand what the problem is asking for or what would be considered as a "primary," as compared to a "secondary" answer. Your use of transfer functions is unfortunately something I am not able to follow along with. My current BSEET degree program at Excelsior College has not covered this yet. I am in the Transmission and Generation component of the degree and I think transfer functions is taught in the Power Electronics component of the degree. So, I realize I will likely miss out on what I know is very important knowledge to know if preparing for an engineer exam.

Dan
 

LvW

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dvscrobe - at first, I am afraid you have misinterpreted Berkemans contribution. Certainly he only spoke about the "copy of Ri and Rf" at the non-inverting input.
Of course, the resistor Ri in series with the capacitor plays an important role for the gain factor.
As far as the time constant(s) is (are) concerned - they are (as indicated by the name) defined in the TIME DOMAIN, in particular with the help of the step repüonse of the system.
However, due to the interrelation between time and frequency domain the time constants can also be identified in the frequency domain (based on the transfer function and/or the corresponding BODE diagram).
Do you know how the step response looks like? This is necessary to see how they are defined and to be able to answer the question
 
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LvW, Okay, I know I am killing this, but to see this thru, and I thank everyone's patience here. Your mention about step response made me think of this problem in a different way. I am aware of step response. It is like having a switch that immediately goes close and you have an input voltage that goes from zero to some constant voltage. What happens during that moment of voltage increase? So, if I just consider the input circuit without any op amp or feedback network, then the time constant is clearly Ri * C (200 Ohms * 50 uF). This lets me know that the initial current is just the voltage divided by Ri because the capacitor would be acting as a short at time = zero. So, the time constant tells me how much time it will take for the current to drop by 63%. Now, if I add in the op amp and just a Rf resistor without the parallel capacitor, V out has to change in order to keep the inverting input at zero potential. Also, V out would drop 63% by the same time constant. Now, if I add in that parallel capacitor on the feedback resistor, I can clearly see how V out would drop even faster because parallel branches gives more paths for the current to flow (I don't have a mathematical explanation for that at the moment). And, yes, I did misinterpret berkeman. I originally considered the op amp an ideal one and ignored those resistors on the non-inverting input.
 

LvW

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dvscrobe - no, you have made an error. The quantitty 63% plays a role in a first-order system only where we have an exponential change of current or voltage. But your circuit resembles a second-order system with two poles (bandpass). And the output voltage - as a result of an input step - will NOT rest at a certain voltage...it will go back to zero (because of the series capacitor).
Within the next minutes I will try to show you a step response diagram for your circuit.

Now, I have added a graphic - at t=0 it starts at 1E-18 (which is zero) and it goes back to zero. The step response goes negative because of the inverting nature of the circuit. As you can see, the slops (falling and rising) can help to define two time constants.

EDIT: dvscrobe, some time ago I have asked you about the "uncommon" values as given in your first post. However, I got no answer - hence, for my simulation I have used the values as given in your first post. Only now I have seen that you, suddenly, have mentioned other values (post 10). The step response with these new values will have, in principle, the same form - however, with another time scale.
 

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LvW

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Here is the revised step response with values as given in post 10.
Both time constants (T1=0.4ms and T2=4ms) can be identified in the drawing using the slope of the falling resp. rising part of the curve.
 

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LvW, Oh my! Those curves really helped. Not sure how you plotted that but what this tells me is that your first time constant T1=.4ms of shorter time means that the voltage out will drop rapidly shortly after time = zero, but the second time constant T2=4ms of longer time means the voltage out will go back to zero gradually. Very interesting!!! I see that in my post #15, I was describing my problem wrong. I will have to get better at learning step response and this post has helped me out with dealing with multiple time constants.
 

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berkeman

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berkeman, This was my first thread on the forum. Now that I am happy, for the sake of future readers of this thread, might be good idea to edit the original post. Letting you know since you are moderator. You will see that I messed up Ri and Rf values from the published problem. I can’t edit that anymore.
 

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