Transfer Function Derivation 1

Thread Starter

Lambo Av

Joined Apr 24, 2019
44
Greetings to the world, I am trying to prove the transfer function but i failed to do so, any help?

upload_2019-7-5_9-49-54.png

Below my method of working it out (which is wrong). Can I get some helping hand to reach the answer? Many thanks.
upload_2019-7-5_11-21-13.png
 

MrAl

Joined Jun 17, 2014
6,500
Hello,

The most general way to work this out is to consider that the op amp is a voltage controlled voltage source.
The inputs are the inverting input and non inverting input and the gain goes to infinity. That is the most general.

An even simpler way is to calculate the output knowing the gain for the inverting input and the gain for the non inverting input.
Then subtract the output due to the non inverting input from the output due to the inverting input and that's your solution for the total output voltage.
This is less general because often the two input circuits interact or there is positive feedback (not the case here though).

You should know what the different gains are for inverting and non inverting configurations, do you?

Using either of these two methods will result in the correct answer.
 

WBahn

Joined Mar 31, 2012
24,700
How are you coming up with your starting equation? Seems like something pulled out of thin air.

Analyze the circuit before you, don't just throw the nearest handy equation at it.
 

Thread Starter

Lambo Av

Joined Apr 24, 2019
44
Hello,

The most general way to work this out is to consider that the op amp is a voltage controlled voltage source.
The inputs are the inverting input and non inverting input and the gain goes to infinity. That is the most general.

An even simpler way is to calculate the output knowing the gain for the inverting input and the gain for the non inverting input.
Then subtract the output due to the non inverting input from the output due to the inverting input and that's your solution for the total output voltage.
This is less general because often the two input circuits interact or there is positive feedback (not the case here though).

You should know what the different gains are for inverting and non inverting configurations, do you?

Using either of these two methods will result in the correct answer.
I know how to calculate the gain for the inverting input, but as of non-inverting input, I am not sure. May I know the formula for this case?
 

Thread Starter

Lambo Av

Joined Apr 24, 2019
44
How are you coming up with your starting equation? Seems like something pulled out of thin air.

Analyze the circuit before you, don't just throw the nearest handy equation at it.
I applied parallel circuit calculations for V1(s) and V2(s)
 

WBahn

Joined Mar 31, 2012
24,700
I applied parallel circuit calculations for V1(s) and V2(s)
Why? What does that even mean? What are "parallel circuit calculations" applied to two the voltages at two points?

Forget about the opamp for the moment.

Can you develop an s-domain equation for V1 as a function of Vi and Vo?

Can you develop an s-domain equation for V2 as a function of Vi and Vo?

NOW take the opamp into account -- what relationship will it force on V1 and V2?
 

MrAl

Joined Jun 17, 2014
6,500
I know how to calculate the gain for the inverting input, but as of non-inverting input, I am not sure. May I know the formula for this case?
Hello again,

You know the gain due to the inverting input, so you know the gain of the non inverting input because it is the same plus 1. So see if you can solve it now.
This is still a special case though because there is no interaction between the two input circuits and no positive feedback.
 

RBR1317

Joined Nov 13, 2010
493
The foolproof method to apply here is nodal analysis. Write the node equations for V1(s) & V2(s) and solve for V1(s) & V2(s). Presumably the op-amp is working in its linear range, so V1(s)=V2(s). This will yield an equation relating Vi(s) & Vo(s). Solve for Vo(s), then divide by Vi(s). Vo(s)/Vi(s) is the transfer function. Works every time.
 

Thread Starter

Lambo Av

Joined Apr 24, 2019
44
Why? What does that even mean? What are "parallel circuit calculations" applied to two the voltages at two points?

Forget about the opamp for the moment.

Can you develop an s-domain equation for V1 as a function of Vi and Vo?

Can you develop an s-domain equation for V2 as a function of Vi and Vo?

NOW take the opamp into account -- what relationship will it force on V1 and V2?
s-domain equation for V1 = -1
s-domain equation for V2 = [I(s)R3]/(1+0.1s)
As of this, "what relationship will it force on V1 and V2", I am unsure, mind telling me? Is it V1(s) = V2(s)?
 

Thread Starter

Lambo Av

Joined Apr 24, 2019
44
Hello again,

You know the gain due to the inverting input, so you know the gain of the non inverting input because it is the same plus 1. So see if you can solve it now.
This is still a special case though because there is no interaction between the two input circuits and no positive feedback.
Correct me if I am wrong, its V+ = V-, so my V- is -1, as a result, V+ will also be -1, am I correct?
 

MrAl

Joined Jun 17, 2014
6,500
Correct me if I am wrong, its V+ = V-, so my V- is -1, as a result, V+ will also be -1, am I correct?
Hi,

Well no, when i said "the non inverting gain is the same as the inverting gain plus 1" i assumed you would realize that there is a positive gain for the non inverting input and negative gain for the inverting input.

So you know how to calculate the gain for the inverting input? Let's start with that. You know it is the ratio of two of the resistors right? Which resistors? Try that next. After that we can work on the non inverting gain.
 

MrAl

Joined Jun 17, 2014
6,500
The foolproof method to apply here is nodal analysis. Write the node equations for V1(s) & V2(s) and solve for V1(s) & V2(s). Presumably the op-amp is working in its linear range, so V1(s)=V2(s). This will yield an equation relating Vi(s) & Vo(s). Solve for Vo(s), then divide by Vi(s). Vo(s)/Vi(s) is the transfer function. Works every time.
Hi,

Yes that is a third method i forgot to mention, thanks for bringing that up.
The first method i always think about is to replace the op amp with a voltage controlled voltage source, and the second is to replace it with a voltage controlled current source with low output impedance. These two are very general and can handle finite internal gains as well as the theoretical infinite gain models.
 

Thread Starter

Lambo Av

Joined Apr 24, 2019
44
Hi,

Well no, when i said "the non inverting gain is the same as the inverting gain plus 1" i assumed you would realize that there is a positive gain for the non inverting input and negative gain for the inverting input.

So you know how to calculate the gain for the inverting input? Let's start with that. You know it is the ratio of two of the resistors right? Which resistors? Try that next. After that we can work on the non inverting gain.
"the non inverting gain is the same as the inverting gain plus 1"
It will be (inverting gain)+1 = -1 + 1 = 0? Mind explaining on positive gain?

"Which resistors?" It will be the ratio of R1 and R2
 

Thread Starter

Lambo Av

Joined Apr 24, 2019
44
Hi,

Yes that is a third method i forgot to mention, thanks for bringing that up.
The first method i always think about is to replace the op amp with a voltage controlled voltage source, and the second is to replace it with a voltage controlled current source with low output impedance. These two are very general and can handle finite internal gains as well as the theoretical infinite gain models.
Do you mind explaining "voltage controlled voltage source" and "voltage controlled current source with low output impedance"? I am sorry about that
 

WBahn

Joined Mar 31, 2012
24,700
s-domain equation for V1 = -1
This is why we always ask people to show their work -- it let's us get at the heart of the problem you are having so much quicker.

You are claiming that voltage at V1 is completely independent of either the voltage at Vi or the voltage at Vo. Does that make sense?

You have two voltages, Vi and Vo, connected by two resistors, R1 and R2. What is the voltage at the junction of the resistors?
s-domain equation for V2 = [I(s)R3]/(1+0.1s)
I(s) is the current in the upper branch.

Even if it was the voltage in the lower branch, I(s)R3 would be the voltage drop across R3.

As of this, "what relationship will it force on V1 and V2", I am unsure, mind telling me? Is it V1(s) = V2(s)?
That is correct. If possible, the opamp outputs whatever voltage is necessary to make the voltage differential at its inputs equal to zero.
 

WBahn

Joined Mar 31, 2012
24,700
"the non inverting gain is the same as the inverting gain plus 1"
It will be (inverting gain)+1 = -1 + 1 = 0? Mind explaining on positive gain?

"Which resistors?" It will be the ratio of R1 and R2
I would recommend forgetting about this approach (at least for now). Most people new to opamp analysis get quite confused by it and it doesn't always work, which means you have to understand it well enough to be able to determine whether it will work or not. Focus on the fundamentals, which is the approach I recommended in Post #7 (and which RBR1317 recommended in Post #11).
 

WBahn

Joined Mar 31, 2012
24,700
Do you mind explaining "voltage controlled voltage source" and "voltage controlled current source with low output impedance"? I am sorry about that
You really should have had these in your course well before you got to opamps. You might go back through your material and look for them. The quick explanation is a voltage-controlled voltage source is simply a variable voltage source whose output voltage is determined by the voltage differential somewhere else in the circuit. An opamp's behavior, in the linear region, is dominated by the output voltage being equal to the voltage differential between its inputs multiplied by a very large gain factor.

Vout = Ao·(V+ - V-)

Where Ao is somewhere between about 100,000 and 10,000,000, depending on the opamp in question. Even with the low end (which usually means very old-style opamps like the 741), a 10 V output means that the difference between the inputs is only about 0.1 mV, which is why we can analyze the circuit very well by assuming that their is no voltage difference across the inputs.
 
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