OpAmp Circuit Analysis: Finding the Differential Equation for Vo

In summary, the differential equation for Vo can be found by using nodal analysis at v1, vp, vn, and v2 and simplifying the equations to eliminate any terms that do not involve Vo or Vin. The resulting equation will involve first and second derivatives of both Vo and Vin, and can be solved by hand without the use of Laplace transforms or mathematical software.
  • #1
eehelp150
237
0

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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  • #2
eehelp150 said:
At node Vo:
This "equation" doesn't belong. Omit it, solve the others.
 
  • #3
NascentOxygen said:
This "equation" doesn't belong. Omit it, solve the others.
Why does this equation not belong?
 
  • #4
What rule did you use to form it?
 
  • #5
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
 
  • #6
Show precisly how you applied nodal analysis here.
 
  • #7
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
 
  • #8
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.
 
  • #9
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...
 
  • #10
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.
 
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  • #11
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:
  • #12
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:
  • #13
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
 
  • #14
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?
 
  • #15
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.
 
  • #16
Are you allowed to use mathematical software, such as Wolfram Alpha, to help with the work, or must you do it all by hand?
 
  • #17
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.
 
  • #18
Have you studied Gaussian elimination, or any of the related methods of systematized solution of simultaneous equations?
 
  • #19
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?
 
  • #20
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.
 
  • #21
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?
 
  • #22
Try it and see. If it works it's valid. :smile:

You'll need a way to eliminate V1 and its derivative.
 
  • #23
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.
 
  • #24
Don't forget that Vp= VN. So your third equation gives you Vp being directly proportional to Vo.
 
  • #25
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}#
 
  • #26
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}##
 
  • #27
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})##
 
  • #28
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.
 
  • #29
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
 
  • #30
If you combine your first and fourth equations, the derivatives can disappear and you will be left with V2 in terms of things you know.
 
  • #31
NascentOxygen said:
If you combine your first and fourth equations, the derivatives can disappear and you will be left with V2 in terms of things you know.
##\frac{V_1-V_{in}}{R_1}+\frac{V_1-V_p}{R_2}+\frac{V_2}{R_6}+\frac{V_2-V_o}{R_5}=0##

becomes

##\frac{\frac{R_3V_O}{R_3+R_4}+R_2C_1(\frac{R_3\dot{V_O}}{R_3+R_4})-V_{in}}{R_1}+\frac{R_2C_1(\frac{R_3\dot{V_O}}{R_3+R_4})}{R_2}+\frac{V_2}{R_6}+\frac{V_2-V_o}{R_5}=0##
How would I get rid of V2?
 
  • #32
Isolate V2. Determine ##\dot V_2##.

Now you can substitute for all necessary terms so as to leave you with an equation relating Vo to Vin.
 
  • #33
NascentOxygen said:
Isolate V2. Determine ##\dot V_2##.

Now you can substitute for all necessary terms so as to leave you with an equation relating Vo to Vin.
I'm not really following. There is no other equation with V2.
 
  • #35
The Electrician said:
If you use the D operator technique, I think you will improve your ability to solve the system:

http://www.codecogs.com/library/maths/calculus/differential/linear-simultaneous-equations.php

http://www.solitaryroad.com/c658.html
Even with d operators, how would I get rid of V2? There is only one equation with V2
If I combine NodeV2 and NodeV1, I get rid of the derivatives.
If I derive the combined equation and then solve NodeV1 for V_2' and plug into get rid of V_2', would that work?
 
<h2>1. What is an OpAmp circuit?</h2><p>An OpAmp circuit is a type of electronic circuit that uses operational amplifiers (OpAmps) to amplify and manipulate electrical signals. OpAmps are high-gain, differential amplifiers that have two inputs and one output. They are commonly used in a variety of electronic devices, such as audio amplifiers, filters, and sensors.</p><h2>2. What is the purpose of finding the differential equation for Vo in an OpAmp circuit?</h2><p>The differential equation for Vo in an OpAmp circuit is used to analyze the behavior and performance of the circuit. It helps to determine the relationship between the input and output signals, and can be used to design and optimize the circuit for specific applications.</p><h2>3. How do you find the differential equation for Vo in an OpAmp circuit?</h2><p>To find the differential equation for Vo, you need to first determine the equivalent circuit model for the OpAmp circuit. This involves identifying the input and output nodes, and applying Kirchhoff's laws to write the equations for the circuit. Then, using the properties of OpAmps, you can simplify the equations and solve for the differential equation for Vo.</p><h2>4. What factors affect the differential equation for Vo in an OpAmp circuit?</h2><p>The differential equation for Vo in an OpAmp circuit can be affected by various factors, such as the values of the resistors and capacitors in the circuit, the input voltage, and the type of OpAmp being used. Changes in these factors can alter the behavior of the circuit and therefore, the resulting differential equation for Vo.</p><h2>5. How is the differential equation for Vo used in practical applications?</h2><p>The differential equation for Vo is used in practical applications to analyze and design OpAmp circuits for specific purposes. It can help in selecting the appropriate components and values for a circuit, and can also be used to simulate and test the circuit's performance. Additionally, the differential equation can be used to troubleshoot and identify any issues with the circuit. </p>

1. What is an OpAmp circuit?

An OpAmp circuit is a type of electronic circuit that uses operational amplifiers (OpAmps) to amplify and manipulate electrical signals. OpAmps are high-gain, differential amplifiers that have two inputs and one output. They are commonly used in a variety of electronic devices, such as audio amplifiers, filters, and sensors.

2. What is the purpose of finding the differential equation for Vo in an OpAmp circuit?

The differential equation for Vo in an OpAmp circuit is used to analyze the behavior and performance of the circuit. It helps to determine the relationship between the input and output signals, and can be used to design and optimize the circuit for specific applications.

3. How do you find the differential equation for Vo in an OpAmp circuit?

To find the differential equation for Vo, you need to first determine the equivalent circuit model for the OpAmp circuit. This involves identifying the input and output nodes, and applying Kirchhoff's laws to write the equations for the circuit. Then, using the properties of OpAmps, you can simplify the equations and solve for the differential equation for Vo.

4. What factors affect the differential equation for Vo in an OpAmp circuit?

The differential equation for Vo in an OpAmp circuit can be affected by various factors, such as the values of the resistors and capacitors in the circuit, the input voltage, and the type of OpAmp being used. Changes in these factors can alter the behavior of the circuit and therefore, the resulting differential equation for Vo.

5. How is the differential equation for Vo used in practical applications?

The differential equation for Vo is used in practical applications to analyze and design OpAmp circuits for specific purposes. It can help in selecting the appropriate components and values for a circuit, and can also be used to simulate and test the circuit's performance. Additionally, the differential equation can be used to troubleshoot and identify any issues with the circuit.

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