Checking the Gains of the Given Circuit

[NHLspl09]

Hi all,
About a month ago I had thought I completed what was the final question of my EE summer assignment... I thought wrong. My professor just posted a final problem, an easier problem involving less work. But I decided to ask for some guidance before continuing. I have also attached an example solution of the problem he's provided.

Homework Statement

(Attachment below - EE P8)

In each part of this problem, you will be given an expression for the gain of the above
circuit. Assume that you solved the circuit and found the given expression (answer).
For each given expression (answer), comment on whether the answer is either correct,
plausible, or incorrect. Please state reasoning for the choice you picked.

Homework Equations

(Attachment below - EE P8)

The Attempt at a Solution

My question is, should I begin plugging in 0 for the different resistors as he did in the example solutions? I know it's a completely different circuit than the one that I have, but I thought that might be where I start. Any thoughts and input would be greatly appreciated!

 

[gneill]

My question is, should I begin plugging in 0 for the different resistors as he did in the example solutions? I know it's a completely different circuit than the one that I have, but I thought that might be where I start. Any thoughts and input would be greatly appreciated!

You could start by solving for the gain and identifying any correct answers directly. Judging the result of setting this or that resistance to zero (or infinity) requires that you understand the circuit operation. This understanding can best be obtained by first solving the circuit!

Also, on the solution sheet, for answer (d) I think that he meant to set G₃ = 0, not R₃ = 0. If G₃ is zero then R₃ goes to infinity (equivalent to removing the resistor from the circuit) and his argument stands. If R₃ is set to zero then the gain must be zero, since the output would be shorted to ground.

 

[NHLspl09]

You could start by solving for the gain and identifying any correct answers directly. Judging the result of setting this or that resistance to zero (or infinity) requires that you understand the circuit operation. This understanding can best be obtained by first solving the circuit!

Gotcha, (I don't know if you remember helping me a month ago, I can help refresh your memory if needed) so I would solve as I did for the gains of the other circuits? Solving for $\frac{V_o}{V_i}$

 

[gneill]

Gotcha, (I don't know if you remember helping me a month ago, I can help refresh your memory if needed) so I would solve as I did for the gains of the other circuits? Solving for $\frac{V_o}{V_i}$

I remember

Yes, solve for $\frac{V_o}{V_i}$. Use what you know about the characteristics of the ideal op-amp.

 

[NHLspl09]

I remember

Yes, solve for $\frac{V_o}{V_i}$. Use what you know about the characteristics of the ideal op-amp.

Great out of curiosity, is R₃ in parallel with R₂ and R₄?

 

[gneill]

... out of curiosity, is R₃ in parallel with R₂ and R₄?

No. Although a certain property of the op-amp's input terminals will allow you to consider R₂ and R₃ to be in parallel. Not that this would be of tremendous help to you...

Consider writing a KCL equation for the node formed by the confluence of R₂, R₃, and R₄. Assign an arbitrary voltage to that node; Maybe call it V₂. You should also be able to write KCL for the "-" input terminal node.

 

[NHLspl09]

No. Although a certain property of the op-amp's input terminals will allow you to consider R₂ and R₃ to be in parallel. Not that this would be of tremendous help to you...

Consider writing a KCL equation for the node formed by the confluence of R₂, R₃, and R₄. Assign an arbitrary voltage to that node; Maybe call it V₂. You should also be able to write KCL for the "-" input terminal node.

Makes sense. I did this and these are my two equations. Now my thoughts would be to solve for $\frac{V_z}{R_2}$ on the "-" nodal equation and substitute into the V_z nodal equation..

EDIT: Apologies on the small upload, if it's difficult to read just let me know!

 

[gneill]

If you solve the first equation for V₂ then you can use it to eliminate all the V₂'s in the second equation and solve it for V_o (and then V_o/V_i).

EDIT: Note that you can repeat the operations using conductances (G₁, G₂, ...) and find the gain in terms of them, too.

 

[NHLspl09]

Ok I've done so, just asking for a check on the current work I've done. If correct I think I know where I'm going with it and will continue, just thought I'd see if I was correct so far.

 

[gneill]

From what I can make out from the Lilliputian writings , you're doing okay so far.

 

[NHLspl09]

Hahaha apologies for that, I took a different approach... to the scanning that is :tongue: How does this look so far? I know there's got to be something in my next step dealing with that R₃ I feel

 

[gneill]

Hahaha apologies for that, I took a different approach... to the scanning that is :tongue: How does this look so far? I know there's got to be something in my next step dealing with that R₃ I feel

Looks good indeed. Can you combine the terms and put them over a single denominator?

 

[NHLspl09]

Looks good indeed. Can you combine the terms and put them over s single denominator?

Great! This is what I was mentioning, my thoughts now would be to eliminate the R₃ in the denominator and bring it to the numerator

 

[NHLspl09]

Great! This is what I was mentioning, my thoughts now would be to eliminate the R₃ in the denominator and bring it to the numerator

In doing so I would multiply the right side of the equation with R₃ over 1 which would eliminate it in the denominator and leave R₁ in the denominator. Is that a legal algebraic move though?

 

[gneill]

In doing so I would multiply the right side of the equation with R₃ over 1 which would eliminate it in the denominator and leave R₁ in the denominator. Is that a legal algebraic move though?

Nope. Anything you do to one side that changes its value you must do to the other. Multiplying the numerator or denominator by a constant that is not equal to 1 will change the value of the expression.

There's nothing wrong with having a denominator that's R1*R3.
Take a look at some of the proposed gain expressions in the question statement for confirmation.

 

[NHLspl09]

Nope. Anything you do to one side that changes its value you must do to the other. Multiplying the numerator or denominator by a constant that is not equal to 1 will change the value of the expression.

There's nothing wrong with having a denominator that's R1*R3.
Take a look at some of the proposed gain expressions in the question statement for confirmation.

I see I see, I remember doing this algebraic move in one of my previous problems (only thing is all my completed homework is at school, I moved in on Saturday and thinking I wouldn't get another problem I brought all my school work/books and put them in my bedroom ). I do see a provided answer with a denominator of R1*R3 - Isn't the required move to multiply the two terms without R3 by R3/R3?

 

[gneill]

I do see a provided answer with a denominator of R1*R3 - Isn't the required move to multiply the two terms without R3 by R3/R3?

Yup. You can always multiply a term by one and leave the value unchanged. R3/R3 = 1.

 

[NHLspl09]

Got it - so with this gain, that proves that letter D is correct Now, dealing with conductance, I know a little about this and have briefly researched a little bit about it. From what I remember, and correct me if I'm wrong, it is the inverse of resistance?

 

[gneill]

Yup. G = 1/R.

Now, you could stick in 1/G_i for each R_i in the gain formula you've derived, and then clean up the algebra. Or, which might be preferable, try the gain derivation from scratch using the conductances rather than the resistances.

For conductance G = 1/R, with voltage across it V, the current is I = V/R = V*G. You'll find that your KCL equations look pretty clean when conductances are used!

 

[uart]

Got it - so with this gain, that proves that letter D is correct Now, dealing with conductance, I know a little about this and have briefly researched a little bit about it. From what I remember, and correct me if I'm wrong, it is the inverse of resistance?

Yes that's the correct answer. Also one of the other answers is correct too.

To find an expression using conductances just follow these two simple steps.

1. Replace all occurrences of resistance in the gain formula by 1/G, eg replace R1 with 1/G1 etc.

2. Multiply top and bottom by the product (G1 G2 G3 G4)

 

[NHLspl09]

Yup. G = 1/R.
Now, you could stick in 1/G_i for each R_i in the gain formula you've derived, and then clean up the algebra.

To find an expression using conductances just follow these two simple steps.
1. Replace all occurrences of resistance in the gain formula by 1/G, eg replace R1 with 1/G1 etc.
2. Multiply top and bottom by the product (G1 G2 G3 G4)

I took both of your advice and have found the other correct answer! Now, my last question with this: I can successfully explain how I found the two correct answers, but how am I to explain why the others are incorrect? I know it's entirely up to the professor, but in your guys opinion, would simply saying that 'the other answers aren't the gain equations shown in the derivations' be OK? I feel like that's fine because I've shown all the work leading up to the two formulas.

 

[gneill]

You could also point out where proposed solutions cannot be correct due to mismatched units, or having the incorrect sign.

 

[NHLspl09]

You could also point out where proposed solutions cannot be correct due to mismatched units, or having the incorrect sign.

Got it Here are my final answer sheets!

Thank you gneill and uart for the help and input, I really do appreciate your patience and help a lot!

 

[uart]

Good work. If your professor is going to ask this type of problem frequently then it's probably worth asking him/her about their proposed marking scheme for the "plausible" responses.

Personally I've got nothing against this type of question, it's an interesting approach. I am however a little uncertain about exactly how marks are going to be awarded for the "plausible" responses.

My first guess was (and this was re-enforced by looking at the solutions to the example problem) that checking either "incorrect" or "correct" as appropriate would yield the maximum marks and that "plausible" is only offered as something of a fall-back position for students who are unable to establish an exact expression.

For example, if I was marking such a test I'd probably do something like 2 marks for each correctly identified "correct" or "incorrect" solution and perhaps 1 mark for each correct solution identified as "plausible" (and no marks for an incorrect solution identified as "plausible"). Now that's just me, your professor may have other ideas.

 

[NHLspl09]

Good work. If your professor is going to ask this type of problem frequently then it's probably worth asking him/her about their proposed marking scheme for the "plausible" responses.

Personally I've got nothing against this type of question, it's an interesting approach. I am however a little uncertain about exactly how marks are going to be awarded for the "plausible" responses.

My first guess was (and this was re-enforced by looking at the solutions to the example problem) that checking either "incorrect" or "correct" as appropriate would yield the maximum marks and that "plausible" is only offered as something of a fall-back position for students who are unable to establish an exact expression.

For example, if I was marking such a test I'd probably do something like 2 marks for each correctly identified "correct" or "incorrect" solution and perhaps 1 mark for each correct solution identified as "plausible" (and no marks for an incorrect solution identified as "plausible"). Now that's just me, your professor may have other ideas.

I was wondering the same thing, but my conclusion was similar to yours in that maybe if you're not 100% about an answer you mark it as plausible and explain why you've received this answer and how.

As you have mentioned, I'll definitely ask him about it and what exactly it means if asked a similar question like this. Being my first time taking a course with him I'm not exactly certain as to what it is he means, but my guess definitely would be it's for any uncertainties.

 

[gneill]

For (a) it's not the sign of G1 that matters, it's the sign of the gain. G1's sign is positive because it's just a common conductance (unit of conductance is Siemans = 1/Ohms, usually designated S = 1/Ω). In other words, the sign is associated with the gain, not anyone component value.

For (e) the units are fine: S²/S², so the proposed gain is unitless as required. I would say that this result is plausible, if incorrect

 

[uart]

BTW. There's a super easy way to solve that opamp problem. You can represent the opamp output and the voltage divider R4,R3 with a Thevenins equivalent, which is merely an ideal voltage attenuation block of B=R3/(R3+R4) with a series Thevenins resistance of R3||R4 (that is, such that this R3||R4 is in series with R2).

Now the closed loop gain with an ideal opamp (-R2/R1 in this case) is not effected by the gain attenuation. So it follows that the gain from input to the attenuator output is -R2'/R1, where R2' is simply R2 plus the additional thevenins resistance of R3||R4.

Since our output is measured at the attenuator input then it is 1/B times the voltage given above. So basically you get a derivation in two lines that,

$$\frac{v_o}{v_i} = \frac{-R_2^'}{B \, R_1}$$

which you can easily verify is correct by substituting in the above expressions for B and R2'.

 

[uart]

For (e) the units are fine: S²/S², so the proposed gain is unitless as required. I would say that this result is plausible, if incorrect

Yeah I think that this is the difficultly with the "plausible" options. Any incorrect answer might seem "plausible" after a limited number of plausibility tests, but if it really is incorrect there must be some test that would eventually invalidate it.

For the example of (e) we could deduce that if G3=infinity (R3=0) that there would be no negative feedback and therefore that the gain magnitude should go to infinity. However, for the expression for part (e), this is not the case.

 

[NascentOxygen]

If R₃ is set to zero then the gain must be zero, since the output would be shorted to ground.

This isn't correct. With no feedback, the gain will be high--practically the open circuit gain.

"output shorted to ground"? <shudder> :yuck: :yuck:
I'm sure you mean "feedback shorted to ground".

 

[gneill]

This isn't correct. With no feedback, the gain will be high--practically the open circuit gain.

"output shorted to ground"? <shudder> :yuck: :yuck:
I'm sure you mean "feedback shorted to ground".

Nope. Go back to the original post, and the circuit in the second attachment that I was referring to with that comment

 

[NascentOxygen]

Nope. Go back to the original post, and the circuit in the second attachment that I was referring to with that comment

Oops, I was looking at the wrong diagram? Then that explains why your analysis seemed awry!

 

[NHLspl09]

For (a) it's not the sign of G1 that matters, it's the sign of the gain. G1's sign is positive because it's just a common conductance (unit of conductance is Siemans = 1/Ohms, usually designated S = 1/Ω). In other words, the sign is associated with the gain, not anyone component value.

For (e) the units are fine: S²/S², so the proposed gain is unitless as required. I would say that this result is plausible, if incorrect

I understand what you've said about (a), but not (e) - how can it be considered plausible if the terms in the denominator aren't correct?

 

[gneill]

I understand what you've said about (a), but not (e) - how can it be considered plausible if the terms in the denominator aren't correct?

It's plausible because, without doing the circuit analysis to check its details, the expression contains no errors in the form of addition of unlike units and it yields a unitless result as required; it looks like it might be a gain expression for some circuit (if not this one!).

Note that there's a vast gulf between "plausible" and "is".

 

[NHLspl09]

It's plausible because, without doing the circuit analysis to check its details, the expression contains no errors in the form of addition of unlike units and it yields a unitless result as required; it looks like it might be a gain expression for some circuit (if not this one!).

Note that there's a vast gulf between "plausible" and "is".

Ahhhh ok now I understand what you meant in saying it's plausible!