OpAmp Circuit Analysis: Finding the Differential Equation for Vo

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Homework Statement


upload_2016-11-27_23-26-9.png

Find the differential equation for Vo

Homework Equations


KCL

The Attempt at a Solution


At node v1:
##\frac{V_1-V_{in}}{R_1}+\frac{V_1-V_p}{R_2}+C_2(\dot{V_1}-\dot{V_2})=0##
At node vp:
##C_1\dot{V_P}+\frac{V_P-V_1}{R_2}=0##
At node vn:
##\frac{V_N}{R_3}+\frac{V_N-V_o}{R_4}=0##
At node v2:
##\frac{V_2}{R_6}+\frac{V_2-V_o}{R_5}+C_2(\dot{V_2}-\dot{V_1})=0##
At node Vo:
##\frac{V_o-V_N}{R_4}+\frac{V_o-V_2}{R_5}=0##This is my attempt at solving for Vo:
Take equation of node VP and solve for V1
##C_1\dot{V_P}+\frac{V_P-V_1}{R_2}=0##
##V_1=R_2C_1\dot{V_P}+V_P##
derivative
##\dot{V_1}=R_2C_1\ddot{V_P}+\dot{V_P}##

Take equation of node Vo and solve for V2
##\frac{V_o-V_N}{R_4}+\frac{V_o-V_2}{R_5}=0##
##V_2 = \frac{R_5}{R_4}(V_o-V_{N})+V_o##
derivative
##\dot{V_2} = \frac{R_5}{R_4}(\dot{V_o}-\dot{V_{N}})+\dot{V_o}##

Plug these four (V1, dV1, V2, dV2) equations into node V1 equation
##\frac{V_1-V_{in}}{R_1}+\frac{V_1-V_p}{R_2}+C_2(\dot{V_1}-\dot{V_2})=0##
##\frac{R_2C_1\dot{V_P}+V_P-V_{in}}{R_1}+\frac{R_2C_1\dot{V_P}+V_P-V_P}{R_2}+C_2(R_2C_1\ddot{V_P}+\dot{V_P}-\frac{R_5}{R_4}(\dot{V_o}-\dot{V_{N}})+\dot{V_o})=0##
Simplify
##\frac{R_2C_1\dot{V_P}+V_P-V_{in}}{R_1}+C_1\dot{V_P}+C_2(R_2C_1\ddot{V_P}+\dot{V_P}-\frac{R_5}{R_4}(\dot{V_o}-\dot{V_{N}})+\dot{V_o}=0##
By property of opamps: VN = VP
##\frac{R_2C_1\dot{V_P}+V_P-V_{in}}{R_1}+C_1\dot{V_P}+C_2(R_2C_1\ddot{V_P}+\dot{V_P}-\frac{R_5}{R_4}(\dot{V_o}-\dot{V_{P}})+\dot{V_o}=0##

This is where I am stuck... Are my original equations correct? Can someone give me a hint as to how to get rid of Vp?
 
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NascentOxygen said:
This "equation" doesn't belong. Omit it, solve the others.
Why does this equation not belong?
 
NascentOxygen said:
What rule did you use to form it?
Nodal analysis
These points all have a voltage of Vo, don't they?
upload_2016-11-28_0-16-37.png
 
NascentOxygen said:
Show precisly how you applied nodal analysis here.
upload_2016-11-28_0-19-42.png


(Vo-Vn)/R4 + (Vo-V2)/R5 = 0
 
NascentOxygen said:
That can't be right. There are 4 paths for current to/from that node, including the unknown current from the op-amp's output.

You already accommodated that node to the extent possible, there's nothing more to be done here.
So I just ignore it? I'm not really understanding why...
 
You are forming an equation based on ∑ currents into a node = 0. That node has 4 currents, yet you are ignoring two of them and writing an "equation" based on just the other two. That's not valid.

Yes, pay no further attention to this node. You already have it covered.
 
Last edited:
NascentOxygen said:
You are forming an equation based on ∑ currents into a node = 0. That node has 4 currents, yet you are ignoring two of them and writing an "equation" based on just the other two.

Yes, pay no further attention to this node. You already have it covered.
At node v1:
##\frac{V_1-V_{in}}{R_1}+\frac{V_1-V_p}{R_2}+C_2(\dot{V_1}-\dot{V_2})=0##
At node vp:
##C_1\dot{V_P}+\frac{V_P-V_1}{R_2}=0##
At node vn:
##\frac{V_N}{R_3}+\frac{V_N-V_o}{R_4}=0##
At node v2:
##\frac{V_2}{R_6}+\frac{V_2-V_o}{R_5}+C_2(\dot{V_2}-\dot{V_1})=0##

Any tips for solving these equations?
 
Last edited:
4 equations, 5 unknowns, so it's easy to solve to obtain the ratio ##\dfrac {V_o}{V_{in}}##.

The instructions say "Find the differential equation for Vo" and all I can think that means is keep the Laplace operator, s, in your final result. Is that your understanding?
 
Last edited:
NascentOxygen said:
4 equations, 5 unknowns, so it's easy to solve to obtain the ratio ##\dfrac {V_o}{V_{in}}##.

The instructions say "Find the differential equation for Vo" and all I can think that means is keep the Laplace operator, s, in your final result. Is that your understanding?
We are not supposed to use Laplace. Can you give any hints as to how I'd start solving it? I can't see any way to get rid of everything but Vo and Vin
 
It looks like a second-order system, so the answer will involve first and second derivatives, too.

If you were to use Laplace you would end up with both s and s2 terms. These can be substituted by the first and second derivatives, represented as either ##D^2V_o## or ##\ddot V_o## form, to give the general transfer function involving possibly all of the first- and second-order derivatives of both ##V_o## and ##V_{in} ##

Do you have a worked example to follow?
 
NascentOxygen said:
It looks like a second-order system, so there will be first and second derivatives, too. If you were to use Laplace you would end up with both s and s2 terms. These can be substituted by the first and second derivatives, represented as either ##D^2V_o## or ##\ddot V_o## form, to give the general transfer function involving possibly all of the first- and second-order derivatives of both ##V_o## and ##V_{in} ##

Do you have a worked example to follow?
No worked example to follow. We have to solve this using algebra. No laplace allowed.
 
Are you allowed to use mathematical software, such as Wolfram Alpha, to help with the work, or must you do it all by hand?
 
The Electrician said:
Are you allowed to use mathematical software, such as Wolfram Alpha, to help with the work, or must you do it all by hand?
Everything by hand, sadly.
 
Have you studied Gaussian elimination, or any of the related methods of systematized solution of simultaneous equations?
 
The Electrician said:
Have you studied Gaussian elimination, or any of the related methods of systematized solution of simultaneous equations?
Yes, but I'm a bit rusty. How would guassian work with differentials?
 
You would probably have to use the differential operator D technique; look up "Operational Calculus". You don't have to use Gaussian elimination; doing so just helps avoid errors.

You can just do what you were doing in post #1. You will have to find the right substitutions to get rid of Vp in your last equation.

I will tell you that your equations are correct, and solving using the Laplace variable S gives the correct answer. Even though you're not supposed to do that, you don''t have to tell your instructor that you verified your answer that way.

I'm afraid you will just have to slog it out for what you hand in.
 
The Electrician said:
You would probably have to use the differential operator D technique; look up "Operational Calculus". You don't have to use Gaussian elimination; doing so just helps avoid errors.

You can just do what you were doing in post #1. You will have to find the right substitutions to get rid of Vp in your last equation.

I will tell you that your equations are correct, and solving using the Laplace variable S gives the correct answer. Even though you're not supposed to do that, you don''t have to tell your instructor that you verified your answer that way.

I'm afraid you will just have to slog it out for what you hand in.
This will most likely be an exam question (he takes exam questions from homework) so I can't use Laplace no matter what.
NodeV1
##\frac{V_1-V_{in}}{R_1}+\frac{V_1-V_p}{R_2}+C_2(\dot{V_1}-\dot{V_2})=0##

NodeVP
##C_1\dot{V_P}+\frac{V_P-V_1}{R_2}=0##

NodeVN
##\frac{V_N}{R_3}+\frac{V_N-V_o}{R_4}=0##

NodeV2
##\frac{V_2}{R_6}+\frac{V_2-V_o}{R_5}+C_2(\dot{V_2}-\dot{V_1})=0##

I want to solve nodeV1 for VP, and then derive to get ##\dot{V_P}##

Then I can plug VP and ##\dot{V_P}## into NodeVP and solve for V2?
Am I on the right track?
 
NascentOxygen said:
Try it and see. If it works it's valid. :smile:

You'll need a way to eliminate V1 and its derivative.
I can get rid of ##V_1## and ##\dot{V_1}## by taking equation 2 and solving for ##V_1## and then deriving. However, I do not see any way to get rid of V2.
 
NascentOxygen said:
Don't forget that Vp= VN. So your third equation gives you Vp being directly proportional to Vo.
##\frac{V_N}{R_3}+\frac{V_N-V_O}{R_4}=0##
##V_N=V_P##
##\frac{V_P}{R_3}+\frac{V_P-V_O}{R_4}=0##
##V_P(1+\frac{R_4}{R_3})-V_O=0##

Solve eq1 for Vp
##V_P = \frac{R_2}{R_1}(V_1-V_{in})+V_1+C_2R_2(\dot{V_1}-\dot{V_2})##
Plug into earlier equation
##V_P(1+\frac{R_4}{R_3})-V_O=0##
##(\frac{R_2}{R_1}(V_1-V_{in})+V_1+C_2R_2(\dot{V_1}-\dot{V_2}))(1+\frac{R_4}{R_3})-V_O=0##
Expand
##\frac{R_2}{R_1}(V_1-V_{in})+V_1+C_2R_2(\dot{V_1}-\dot{V_2})+\frac{R_4R_2}{R_3R_1}(V_1-V_{in})+\frac{R_4V_1}{R_3}+\frac{R_4C_2R_2(\dot{V_1}-\dot{V_2})}{R_3}-V_O=0##

solve eq2 for V_1
##C_1\dot{V_P}+\frac{V_P-V_1}{R_2}=0##
##V_1=V_P+R_2C_1\dot{V_P}##
derive
##\dot{V_1}=\dot{V_P}+R_2C_1\ddot{V_P}##
Am I on the right track? I still don't know how to get rid of ##V_2## and ##\dot{V_2}#
 
eehelp150 said:
##\frac{V_N}{R_3}+\frac{V_N-V_O}{R_4}=0##
##V_N=V_P##
##\frac{V_P}{R_3}+\frac{V_P-V_O}{R_4}=0##
##V_P(1+\frac{R_4}{R_3})-V_O=0##

Solve eq1 for Vp
##V_P = \frac{R_2}{R_1}(V_1-V_{in})+V_1+C_2R_2(\dot{V_1}-\dot{V_2})##
Plug into earlier equation
##V_P(1+\frac{R_4}{R_3})-V_O=0##
##(\frac{R_2}{R_1}(V_1-V_{in})+V_1+C_2R_2(\dot{V_1}-\dot{V_2}))(1+\frac{R_4}{R_3})-V_O=0##
Expand
##\frac{R_2}{R_1}(V_1-V_{in})+V_1+C_2R_2(\dot{V_1}-\dot{V_2})+\frac{R_4R_2}{R_3R_1}(V_1-V_{in})+\frac{R_4V_1}{R_3}+\frac{R_4C_2R_2(\dot{V_1}-\dot{V_2})}{R_3}-V_O=0##

solve eq2 for V_1
##C_1\dot{V_P}+\frac{V_P-V_1}{R_2}=0##
##V_1=V_P+R_2C_1\dot{V_P}##
derive
##\dot{V_1}=\dot{V_P}+R_2C_1\ddot{V_P}##
Am I on the right track? I still don't know how to get rid of ##V_2## and ##\dot{V_2}#
How about trying this:
Solve this: ##\frac{V_P}{R_3}+\frac{V_P-V_O}{R_4}=0##
for Vp, then substitute that in: ##V_1=V_P+R_2C_1\dot{V_P}##
 
The Electrician said:
How about trying this:
Solve this: ##\frac{V_P}{R_3}+\frac{V_P-V_O}{R_4}=0##
for Vp, then substitute that in: ##V_1=V_P+R_2C_1\dot{V_P}##
##V_P=\frac{R_3V_O}{R_3+R_4}##
##\dot{V_P}=\frac{R_3\dot{V_O}}{R_3+R_4}##
##V_1=\frac{R_3V_O}{R_3+R_4}+R_2C_1(\frac{R_3\dot{V_O}}{R_3+R_4})##
 
OK. Now you have V1 in terms of Vo. Substitute that in some of the earlier equations. Get Vin in terms of V1. Slog away.
 
The Electrician said:
OK. Now you have V1 in terms of Vo. Substitute that in some of the earlier equations. Get Vin in terms of V1. Slog away.
I need Vin, so Equation 1 has to be used right? I still can't see any way to get rid of V2