Deriving voltages for a circuit

WBahn

Joined Mar 31, 2012
30,065
Where does "not this" come from?

You have a factor, vg, that comes out of left field. What is it? Where did it come from?
 

dcbingaman

Joined Jun 30, 2021
1,065
Do you know how to solve any generic op amp circuit with negative feedback?
Do you know how to use the principle of 'Super-position' to solve for the output?

Also, there is a big assumption being made in this circuit. Vg and VL will not be what you expect if the values of R3 and R4 compared to R along with the input voltages -Vs and +Vs do not meet certain requirements. Do you understand why that is and how to check that those requirements are being met?

You need to have both of these skills to analyze the circuit in question.
 

Thread Starter

electricalnewbie123

Joined May 1, 2023
7
Do you know how to solve any generic op amp circuit with negative feedback?
Do you know how to use the principle of 'Super-position' to solve for the output?

Also, there is a big assumption being made in this circuit. Vg and VL will not be what you expect if the values of R3 and R4 compared to R along with the input voltages -Vs and +Vs do not meet certain requirements. Do you understand why that is and how to check that those requirements are being met?

You need to have both of these skills to analyze the circuit in question.
I'd like to say yes to all the above, unfortunately I can't. I understand I have some serious limitations in circuit analysis and I'm trying to work with them!
I do not understand the assumption you are talking about, neither the why and how to check if the requirements being met. Can you help?
 

Thread Starter

electricalnewbie123

Joined May 1, 2023
7
Here is my solution.
First, I assumed the circuit could be simplified to this:
1682985829745.png
And
1682985881931.png1682985918129.png1682985944751.png1682985999657.png
Considering an automatic cold junction compensation:
1682986068542.png
1682986089032.png
And then assuming that V0 must only depend on Tj:
1682986146672.png

I know that R will be something around 12kohms, R1 will be 1kohm, and that this is a type J thermocouple.

Are my assumptions correct?
 

dcbingaman

Joined Jun 30, 2021
1,065
Here is my solution.
First, I assumed the circuit could be simplified to this:
View attachment 293278
And
View attachment 293279View attachment 293280View attachment 293281View attachment 293282
Considering an automatic cold junction compensation:
View attachment 293283
View attachment 293284
And then assuming that V0 must only depend on Tj:
View attachment 293285

I know that R will be something around 12kohms, R1 will be 1kohm, and that this is a type J thermocouple.

Are my assumptions correct?
ok, for the first step, you state you are 'assuming' it can be simplified to this:
1682994363792.png
if this assumption is true or not, we can revisit later. Let's assume for the moment that it is true and proceed from there. I would like to concentrate on what you are stating as your second 'assumption' based on the preceding one (taking the preceding one as an axiom):
1682994526519.png

Is this true? That depends upon more 'basic' axioms. We might as well state it up front, we have to make some 'assumptions' call them what you will. Say axioms, which is my preferred terminology. If this formula is correct is based on a set of axioms from which you consider correct and then can prove it based on those axioms. Those axioms are basically something like the following:

1. The op amp in question is an ideal op-amp with the following properties:
a. The voltage 'gain' of the op-amp is 'infinite'. That is any difference in the op-amp input voltage on the non-inverting terminal to the inverting terminal will result in an output that is: Vout=A*((+) - (-)). Where 'A' is infinite.
b. No current flows into or out of the non-inverting input of the op-amp or the inverting input of the op-amp.
c. The output of the op-amp has zero impedance (that is it can drive any load with no voltage loss).
d. If an op-amp is 'close enough' to this stated ideal compared to the impedances, gains etc. It is safe to assume it to simplify the analysis.

2. Ohms law of V=IR always holds. Kirchhoff's law of voltage around any closed loop always sums to zero. And the current flowing into and out of a 'junction' (Kirchhoff's current law) always sums to zero.

From these assumptions you can derive this formula and 'prove' it is true assuming the other assumptions are true:

1682994526519.png

Do you know how to 'prove this'? If so please provide your proof and we can go from there. I would prefer you not 'assume' this is true but prove it via the above stated 'assumptions'.
 
Last edited:

dcbingaman

Joined Jun 30, 2021
1,065
ok, for the first step, you state you are 'assuming' it can be simplified to this:
View attachment 293289
if this assumption is true or not, we can revisit later. Let's assume for the moment that it is true and proceed from there. I would like to concentrate on what you are stating as your second 'assumption' based on the preceding one (taking the preceding one as an axiom):
View attachment 293290

Is this true? That depends upon more 'basic' axioms. We might as well state it up front, we have to make some 'assumptions' call them what you will. Say axioms, which is my preferred terminology. If this formula is correct is based on a set of axioms from which you consider correct and then can prove it based on those axioms. Those axioms are basically something like the following:

1. The op amp in question is an ideal op-amp with the following properties:
a. The voltage 'gain' of the op-amp is 'infinite'. That is any difference in the op-amp input voltage on the non-inverting terminal to the inverting terminal will result in an output that is: Vout=A*((+) - (-)). Where 'A' is infinite with the following limitations: it cannot be beyond the supply rails of +Vs and -Vs and if it is it will move towards one or the other.
b. No current flows into or out of the non-inverting input of the op-amp or the inverting input of the op-amp.
c. The output of the op-amp has zero impedance (that is it can drive any load with no voltage loss).
d. If an op-amp is 'close enough' to this stated ideal compared to the impedances, gains etc. It is safe to assume it to simplify the analysis.

2. Ohms law of V=IR always holds. Kirchhoff's law of voltage around any closed loop always sums to zero. And the current flowing into and out of a 'junction' (Kirchhoff's current law) always sums to zero.

From these assumptions you can derive this formula and 'prove' it is true assuming the other assumptions are true:

View attachment 293290

Do you know how to 'prove this'? If so please provide your proof and we can go from there. I would prefer you not 'assume' this is true but prove it via the above stated 'assumptions'.
Here is a hint to start proving the stated equation, based on the above assumptions what is Vout as a function if I1? Please show the proof for the solution from the axioms:

1682997196988.png
 

dcbingaman

Joined Jun 30, 2021
1,065
According to 1b) "No current flows into or out of the non-inverting input of the op-amp or the inverting input of the op-amp."
As such: \[ V_{out}=-I_{1}R_{1} \]
That is not a 'proof'. Though 1b is a necessary axiom for the proof it is not sufficient to lead to the solution you provided. You also need 1a, 1c and all portions of 2. Also, if you look carefully at the circuit I provided you will find that the solution you are showing is still wrong. (It is close but there is a subtle error).

With just 1b I cannot come to any conclusions, but by invoking a few more of the axioms I can disprove your solution and still be wrong because I did not invoke the other axioms about ideal op-amps. Consider:

1683045318535.png
I labeled the 'node' between the current source and R1 as Vx.

From 1b, if follows no current would flow into the inverting input of the op-amp. If no current flows in then from ohms law the impedance into the op-amp is infinite from ohms law, thus I can disconnect Vx from the inverting input and the output should be the same as the original circuit. Now the current flowing in R1 has no choice but to be flowing through the op-amp (otherwise it has no path). I am also using Kirchhoff's current law again for the node current into and out of Vx and all the current then in I1 must be going through R1 and then per ohms law the voltage across R1 is I1*R1. Then I need to invoke Kirchhoff's voltage law to find Vout. Per Kirchhoff's voltage law Vout must be the sum of the voltage across R1 and the voltage across I1 (that is Vx). Thus the output would be:

Vout=I1*R1 + Vx

This is not the correct solution, but I did not expect it to be because I never invoked the other Axioms for the 'ideal op amp' except 1b, showing that 1b only is not sufficient to get a correct solution. But the other axioms do allow me to prove that Vx has to be 0 volts which then leads to the correct solution. To get the correct answer you need to take advantage of all the axioms I presented. I am looking for a set of 'steps' taken one at a time each starting from a given axiom and leading into the next step of the proof that finally arrives as the solution.
Please attempt the proof again and we can go over it.
 
Last edited:

Thread Starter

electricalnewbie123

Joined May 1, 2023
7
That is not a 'proof'. Though 1b is a necessary axiom for the proof it is not sufficient to lead to the solution you provided. You also need 1a, 1c and all portions of 2. Also, if you look carefully at the circuit I provided you will find that the solution you are showing is still wrong. (It is close but there is a subtle error).

With just 1b I cannot come to any conclusions, but by invoking a few more of the axioms I can disprove your solution and still be wrong because I did not invoke the other axioms about ideal op-amps. Consider:

View attachment 293320
I labeled the 'node' between the current source and R1 as Vx.

From 1b, if follows no current would flow into the inverting input of the op-amp. If no current flows in then from ohms law the impedance into the op-amp is infinite from ohms law, thus I can disconnect Vx from the inverting input and the output should be the same as the original circuit. Now the current flowing in R1 has no choice but to be flowing through the op-amp (otherwise it has no path). I am also using Kirchhoff's current law again for the node current into and out of Vx and all the current then in I1 must be going through R1 and then per ohms law the voltage across R1 is I1*R1. Then I need to invoke Kirchhoff's voltage law to find Vout. Per Kirchhoff's voltage law Vout must be the sum of the voltage across R1 and the voltage across I1 (that is Vx). Thus the output would be:

Vout=I1*R1 + Vx

This is not the correct solution, but I did not expect it to be because I never invoked the other Axioms for the 'ideal op amp' except 1b, showing that 1b only is not sufficient to get a correct solution. But the other axioms do allow me to prove that Vx has to be 0 volts which then leads to the correct solution. To get the correct answer you need to take advantage of all the axioms I presented. I am looking for a set of 'steps' taken one at a time each starting from a given axiom and leading into the next step of the proof that finally arrives as the solution.
Please attempt the proof again and we can go over it.
I'm not sure I can follow your request, but I'll do my best!
Considering
And:
Axiom 1a) - V_{out} = A(V_{+}-V_{-}) = A(0-V_{-}) = -AV_{-} : considering V_{-} = -V_{out}/-A and A is infinite, V_{-} must be 0;
Axiom 1b) - If no current flows into the opamp, then all current must flow through R1. This means that V_{R1} = I_{1}*R_{1}
Axiom 1c) - V_{out} = V_{R1} = I_{1}*R_{1}
 

dcbingaman

Joined Jun 30, 2021
1,065
I'm not sure I can follow your request, but I'll do my best!
Considering

And:
Axiom 1a) - V_{out} = A(V_{+}-V_{-}) = A(0-V_{-}) = -AV_{-} : considering V_{-} = -V_{out}/-A and A is infinite, V_{-} must be 0;
Axiom 1b) - If no current flows into the opamp, then all current must flow through R1. This means that V_{R1} = I_{1}*R_{1}
Axiom 1c) - V_{out} = V_{R1} = I_{1}*R_{1}
yes, that is basically what I was looking for. You have to consider 1a.

I am building up to the one you did not understand, so this is going somewhere (trust me) :)

Now that we have that other case understood we can build on it, consider the following:

1683068903216.png
Notice that I have dashed lines going to the last current source on the left. With the last one as Ix.
In other words this is a circuit that can have any number of current sources in the configuration shown specifically 'x' current sources.

You can use your last proof as a stepping stone for this one by now making your last proof its own 'axiom'. Using just that and Kirchhoff's current law, provide the solution to this one using a similar proof to the last one.
 
Top