"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

Hi,

No it is the inverting gain taken as positive, so it is 1+1=2.

But really if you did not know that there's no way you should be working on this problem yet.
You need to work on op amp problems that have ONLY inverting gain, then another problem that has ONLY non inverting gain.
If you can do those two and only if you can do those two will you be able to do this problem that way.
There are other ways too as you should know now. You can try those two. You should be able to do either methods mentioned in this thread.
I'll list them here for reference:
1. Op amp as voltage controlled voltage source. (recommended)
2. Op amp as voltage controlled current source with low output impedance. (not as easy to deal with)
3. Add the inverting gain to the non inverting gain. (easy if you had both of these separately)
4. Equate the two input voltages to zero. (not too hard to do either)

 

Hi,

No it is the inverting gain taken as positive, so it is 1+1=2.

But really if you did not know that there's no way you should be working on this problem yet.
You need to work on op amp problems that have ONLY inverting gain, then another problem that has ONLY non inverting gain.
If you can do those two and only if you can do those two will you be able to do this problem that way.
There are other ways too as you should know now. You can try those two. You should be able to do either methods mentioned in this thread.
I'll list them here for reference:
1. Op amp as voltage controlled voltage source. (recommended)
2. Op amp as voltage controlled current source with low output impedance. (not as easy to deal with)
3. Add the inverting gain to the non inverting gain. (easy if you had both of these separately)
4. Equate the two input voltages to zero. (not too hard to do either)

That last method is not equating the two input voltage to zero, but simply equating the two input voltages (or equating the difference of the two input voltages to zero).

This is a common mistake made by students. They are told that the input is a "virtual ground" early on (because the non-inverting input is often tied to the common in early circuits) and after that they automatically assume that the inputs are always at ground potential. The far better description is that there is a virtual short between the two inputs -- the voltages are the same but no current can flow through this type of short.

 

That last method is not equating the two input voltage to zero, but simply equating the two input voltages (or equating the difference of the two input voltages to zero).

This is a common mistake made by students. They are told that the input is a "virtual ground" early on (because the non-inverting input is often tied to the common in early circuits) and after that they automatically assume that the inputs are always at ground potential. The far better description is that there is a virtual short between the two inputs -- the voltages are the same but no current can flow through this type of short.

So i worded it wrong.

 

So i worded it wrong.

I know what you meant and most people comfortable with opamps would not what you meant. This just happens to be one of those cases where an easy "wording it wrong" that people with some experience will look right past and draw the right meaning from happens to coincide with a reasonable, but wrong, interpretation that those not experienced enough often make.

 

That last method is not equating the two input voltage to zero, but simply equating the two input voltages (or equating the difference of the two input voltages to zero).

This is a common mistake made by students. They are told that the input is a "virtual ground" early on (because the non-inverting input is often tied to the common in early circuits) and after that they automatically assume that the inputs are always at ground potential. The far better description is that there is a virtual short between the two inputs -- the voltages are the same but no current can flow through this type of short.

Are Ei(s) and E(o)s the virtual shorts between the two inputs that you meant? based on the diagram below.

 

Hi,

No it is the inverting gain taken as positive, so it is 1+1=2.

But really if you did not know that there's no way you should be working on this problem yet.
You need to work on op amp problems that have ONLY inverting gain, then another problem that has ONLY non inverting gain.
If you can do those two and only if you can do those two will you be able to do this problem that way.
There are other ways too as you should know now. You can try those two. You should be able to do either methods mentioned in this thread.
I'll list them here for reference:
1. Op amp as voltage controlled voltage source. (recommended)
2. Op amp as voltage controlled current source with low output impedance. (not as easy to deal with)
3. Add the inverting gain to the non inverting gain. (easy if you had both of these separately)
4. Equate the two input voltages to zero. (not too hard to do either)

Thanks for the recommendations, i appreciate it a lot

 

@WBahn @MrAl @Xavier Pacheco Paulino @RBR1317

I have solved it with some help from this thread and also online sources. Allow me to explain (and correct me if I am wrong) as I need to ensure that I understand the basics correctly.

Here's the diagram which i added arrows and points

Firstly, set 2 points at A and B. Are these called junctions as stated in Post #18?

At point A, assuming the R1 isnt there, the voltage across R2 will be Vi – Vo as Vo is in a negative direction.
Since R1 = R2 and point A is in between both R1 and R2, therefore, ½ is used.
The voltage across R2 will then be ½ (Vi – Vo).
The inverting input is assumed to have infinite impedance, making 0 voltage to pass thru and also making R1 and R2 series.
The voltage across R1 will then be Vo
VA = ½ (Vi – Vo) + Vo
= ½ [Vi + Vo]
Taking Laplace Transform, VA = VA(s)
= ½ [Vi(s) + Vo(s)]

At point B, VB(s)


VA(s) = VB(s)

 

Hi,

I did not go over every step but that looks good. The only exception might be that the output you show should be multiplied by Vin.

The other way we talked about is to first find the unsigned inverting gain:
GA=R1/R2

then the non inverting gain is:
GB=GA+1

The input is Vi and the inverting input makes the gain negative so the output due to GA is:
VoA=-GA*Vi

and the output due to GB is:
VoB=GB*Hs*Vi

where Hs is the transfer function of the junction of R3 and C1 which you computed correctly.
The total output is then Vout=VoA+VoB

So you can use that method to check, or use the VCVS method to check.

 

@WBahn @MrAl @Xavier Pacheco Paulino @RBR1317

I have solved it with some help from this thread and also online sources. Allow me to explain (and correct me if I am wrong) as I need to ensure that I understand the basics correctly.

Here's the diagram which i added arrows and points
View attachment 181151
Firstly, set 2 points at A and B. Are these called junctions as stated in Post #18?

At point A, assuming the R1 isnt there, the voltage across R2 will be Vi – Vo as Vo is in a negative direction.
Since R1 = R2 and point A is in between both R1 and R2, therefore, ½ is used.
The voltage across R2 will then be ½ (Vi – Vo).
The inverting input is assumed to have infinite impedance, making 0 voltage to pass thru and also making R1 and R2 series.
The voltage across R1 will then be Vo
VA = ½ (Vi – Vo) + Vo
= ½ [Vi + Vo]
Taking Laplace Transform, VA = VA(s)
= ½ [Vi(s) + Vo(s)]

At point B, VB(s)
View attachment 181152

VA(s) = VB(s)
View attachment 181153

You went a little sideways at the beginning. What if R1 and R2 hadn't been the same?

Just analyze it as given.

Since we generally assume infinite input impedance for an opamp, R1 and R2 are in series (as you note). But it is because no CURRENT flows into the input, saying that no voltage passes through the input is meaningless. Voltage appears across things, current flows through things.

So

I(s) = (Vi(s) - Vo(s)) / (R1 + R2)

The voltage at A is simply the output voltage plus the voltage drop across R2 due to I(s)

Va(s) = Vo(s) + I(s)·R2
Va(s) = Vo(s) + [(Vi(s) - Vo(s)) / (R1 + R2)]·R2
Va(s) = Vo(s) · {1 - [R2 / (R1 + R2)]} + Vi(s) · [R2 / (R1 + R2)]
Va(s) = Vo(s) · [R1 / (R1 + R2)] + Vi(s) · [R2 / (R1 + R2)]

Vb(s) is a simple voltage divider, as you saw.

Then set Va(s) = Vb(s) and solve for Vo(s)/Vi(s)

You need to start tracking your units throughout your work. If you had, you would realize that the units on the final equation are

\( \frac{V_o(s)}{V_i(s)} \; = \; - \; \frac{10 {sec}^{-1} \; - \; s}{10 {sec}^{-1} \; + \; s}\)

It's very, very unfortunate that the convention for Laplace transforms is to use s as the independent variable given that, for time-domain functions, it ends up having units of inverse seconds, thus conflicting with the SI base unit for time. So I generally use 'sec' for seconds when working with s-domain equations.

 

...I need to ensure that I understand the basics correctly.

A tried and true method to increase understanding of the basics is to work more problems. To that end, I offer the following somewhat more difficult example to practice on:

 

GB=GA+1

Thanks for your explanation. As of the formula I quoted, thanks to you, I learnt something new as I have yet to come across it.

 

You went a little sideways at the beginning. What if R1 and R2 hadn't been the same?

Just analyze it as given.

Since we generally assume infinite input impedance for an opamp, R1 and R2 are in series (as you note). But it is because no CURRENT flows into the input, saying that no voltage passes through the input is meaningless. Voltage appears across things, current flows through things.

So

I(s) = (Vi(s) - Vo(s)) / (R1 + R2)

The voltage at A is simply the output voltage plus the voltage drop across R2 due to I(s)

Va(s) = Vo(s) + I(s)·R2
Va(s) = Vo(s) + [(Vi(s) - Vo(s)) / (R1 + R2)]·R2
Va(s) = Vo(s) · {1 - [R2 / (R1 + R2)]} + Vi(s) · [R2 / (R1 + R2)]
Va(s) = Vo(s) · [R1 / (R1 + R2)] + Vi(s) · [R2 / (R1 + R2)]

Vb(s) is a simple voltage divider, as you saw.

Then set Va(s) = Vb(s) and solve for Vo(s)/Vi(s)

You need to start tracking your units throughout your work. If you had, you would realize that the units on the final equation are

\( \frac{V_o(s)}{V_i(s)} \; = \; - \; \frac{10 {sec}^{-1} \; - \; s}{10 {sec}^{-1} \; + \; s}\)

It's very, very unfortunate that the convention for Laplace transforms is to use s as the independent variable given that, for time-domain functions, it ends up having units of inverse seconds, thus conflicting with the SI base unit for time. So I generally use 'sec' for seconds when working with s-domain equations.

Noted with thanks. Oops, I missed the unit, thanks for reminding. May I know how you type out the equations?

 

Noted with thanks. Oops, I missed the unit, thanks for reminding. May I know how you type out the equations?

For all but the last one I just used normal text.

For the last one I used the [tex] and [/tex] tags. This uses a somewhat stripped down version of the LaTeX typesetting language.

If you quote a post that contains tex code you can see the code that generated the equation.

One thing to keep in mind is that some browsers no longer properly render the tex code unless it is written in a very specific way (even though they used to).

 

I offer the following somewhat more difficult example to practice on:

The same procedure works:
1. Write the Node Equations
2. Solve for V(-) & V(+) at Op Amp input
3. Set V(-)=V(+) & solve for Vo
4. Divide Vo by Vi to solve the for transfer function

 

The same procedure works:
1. Write the Node Equations
2. Solve for V(-) & V(+) at Op Amp input
3. Set V(-)=V(+) & solve for Vo
4. Divide Vo by Vi to solve the for transfer function

View attachment 181190

I was offering the example to Lambo Av, not you; he's the one who needs practice.

 

I was offering the example to Lambo Av, not you;

The original problem in this thread came with the final answer provided; and yet you supplied a problem with no final answer given....

 

The original problem in this thread came with the final answer provided; and yet you supplied a problem with no final answer given....

You didn't just give a final answer, you show the complete working of the problem. The idea was for Lambo Av to attempt to work the problem.

 

A tried and true method to increase understanding of the basics is to work more problems. To that end, I offer the following somewhat more difficult example to practice on:

View attachment 181180

Hi,

I was going to offer two new examples myself, one with only the inverting input being used (other grounded).

But just to note, that example also has some component value constraints that have to go with it in order to make it work properly as an All Pass. Without those extra constrains i am not sure it will behave as such. I'd have to look that up though.
Still a good general example though to see how he handles it.