Engineering Why is the voltage source high potential into node 1 at -12 V in this op amp problem?

[s3a]

Homework Statement

Determine V_0 in this operational amplifier circuit.

Relevant Equations

i_- = i_+ = 0
v_- = v_+

Kirchoff's Current Law

Ohm's Law

Nodal Analysis

The last non-black frame of the video is attached as TheProblemAndItsSolution.png, as shown below.

Here is the transcript of the video in its entirety.:

"The topic of this problem is
Operational Amplifier Circuits, and the problem is to find V out
in the circuit shown below. It's a circuit with multiple
resistors in it, one voltage source, an independent 12-volt source, two op
amps, and the output is measured across the load resistance, which is 40 kilo-ohms
at the output of the second op amp.

In order to solve this problem, we have to use the properties of an ideal op amp, so we start with the symbol for an ideal op amp. It has an inverting input with a current associated with it, I-, and a volt associated with it, V-. It also has a non-inverting input with a current, I+, and a voltage, V+, associated with it. It has an output and a ground, we know that the properties of an ideal op amp are that the currents at both of the inputs are 0. We also know that the voltage is at the inverting and the non-inverting input are equal.

Those are two of the properties of an ideal op amp that we use to solve linear circuits. So, when we use along with our knowledge of nodal analysis and measure analysis in order to solve this two op amp circuit that we have in this problem.

So, let's start with finding voltages at the input. So, if we can find the voltage at this
non-inverting input of the first op amp, then we know the voltage at
the inverting input of the op amp, and that will allow us to work
further back into our circuit, ultimately getting back to our voltage
drop across our 40 kilo-ohm resistor.

So, let's sum our currents into this node, into the V+ node, into our non-inverting input node. Let's call this Node 1, so if you sum currents into Node 1, then we have first of all the current through the 40 kilo-ohm resistor, which is minus 12 minus V+. Divided by 40 kilo-ohms. We also have the current which is flowing through the 20 kiloohm resister. And that current is
going to be 0 volts because this is a ground point in our circuit, the bottom of the circuit. 0 volts- V+ Over 20k.

And we have the current which is flowing out of the non-inverting, we know that through ideal op-amp that that current is equal to 0. And so the sum of those using
Kirchhoff's current law At node one is equal to zero and so it's an equation
which just has the one variable that is the voltage at the inverting or the non
inverting input of our first op amp.

If we solve for that voltage, V plus We get minus 4 volts. So it's minus 4 volts at node 1. And it also tells us that it's minus 4 volts at the inverting input as well. So we need to work further back into our circuit. And so we rely on the first property of an op-amp to get us a little bit further into the circuit. That is, that the current into each one of those inputs of the op-amp is always equal to zero.

If that's the case, then there's no current flowing through this 10kOhm resistor, because it is the current flowing into the Inverting input of our first op-amp there's no current flow in through the 10 kilo ohm resistor therefore there's no voltage drop across the 10 kilo ohm resistor.So what that tells us is, that this voltage between the 10 and 20 kilo ohm resistor at the top of our circuit is also a minus 4 volts, at that point.

Now we can use that to work a little bit further into our circuit. Let's look at our circuit a little bit closer. If we look at V out, V out is measured across the 40 kilohm resistor. It's the voltage at this output node of our second. We also notice that there's no voltage drops between these two points along the feedback path of our second op amp. So that also tells us that V out is the voltage at this point in our circuit.

At the top of our circuit is equal to the voltage at the inverting input of our second op amp. We know that that voltage is the voltage at the non-inverting input of our second opamp as well as using our second properties of an opamp. So we have a V out which can be measured at this point in front of our second opamp. So that helps us a little bit. because what we can do is we can start summing voltages, or summing currents in this case, into nodes and come up with equations which allow us to solve for V out.

Let's see if we can do that. Let's look at the top node which is at minus four volts first and let's call that node one. Sorry if I had node one let's call this node two. So let's look at node two and
we're going to use Kirchhoff's current law at node two and we're going to
sum the current into node two. So it's a current first of all through the 10 kilohm resistor which we know is zero, it's a current through the 20 kilohm resistor Up through the 20 kilo ohms resistor.

Let's call this node 3 down below. So in this case the current up
through the 20 kilo ohms resistor is going to be V sub 3 -(- 4) volts. Dividend by. 20 kilohms that's a current up through our 20 kilohm resistor we also have the current flowing through the 20 kilohm resistor at the top of the circuit. It's going to V out minus a -4 divided by 20 kilohms for the current flowing into node two. So V out Minus a minus four volts divided by 20 kilohms. V sub 0 minus a minus four volts divided by 20k. And that's all the currents that are flowing into node 2.

Because we know that the current flowing through the 10-kilo ohm resistor is 0 when I add it in there just for completeness, and that's equal to 0. So that's an equation which has two unknowns in it. It has V3 as an unknown and also has V out as an unknown. Let's look at this node where we have VOut measured, let's call this node 4. So let's sum the currents into node 4. So all these are summing currents into the nodes.

So if we sum the current into node four, we first of all have the current up through the 10 kilohm resistor is going to be zero volts minus Vout over 40k. We have the current flowing left to right from node three To node 4 through the 20 kilohm resister, so that's going to be V3 minus V out divided by 20 k And we have the current which is flowing out of the non-inverting input of the second opamp, and we know that current = 0 through the properties of an ideal opamp. Those are all the currents flowing into node 4, and the sum = 0.

So now we have in the second equation, which is independent of our Other equation that if we used these two equations simultaneously to solve for V sub 3 and V out, we can do that. And so V out can be determined by finding these voltage levels. So if we use this two equations to solve for V out simultaneously, then we end up with a V out which
is equal to two thirds V sub f, V sub 3. And that's equal to minus sixteen-fifths of a volt."

What I don't understand is that the instructor is saying that he defines the current for node 1 as going into the node, but then gets a -12, as opposed to a +12, in (-12 - V_+) / 40k. Shouldn't it be (12 - V_+) / 40k (notice that I purposely omitted the minus in front of the 12)? My logic is that if the current is going from the voltage source to node 1, then the current would first cross the positive terminal of the voltage source.

If someone could help me understand what's going on here, I would greatly appreciate it!

 

[pbuk]

The +ve terminal of the 12V source is connected to ground, it is the -ve terminal that is connected to the 40k resistor.

 

[s3a]

The +ve terminal of the 12V source is connected to ground, it is the -ve terminal that is connected to the 40k resistor.

Oh, so, it's just a matter of recognizing that the current, when flowing into node 1 (as defined by the instructor) (as opposed to away from it) and using the passive sign convention, the terminal of the resistor where the current goes into the resistor is positive and the voltage source's positive terminal should be connected to the resistor's positive terminal since, given the aforementioned assumptions, current flows from the negative terminal of a voltage source to its positive one and then from the positive terminal of the resistor to its negative one?

 

[Tom.G]

The explanation in this paragraph is WRONG! (highlited inRed)

Homework Statement: Determine V_0 in this operational amplifier circuit.
Relevant Equations: i_- = i_+ = 0
v_- = v_+

Kirchoff's Current Law

Ohm's Law

Nodal Analysis

So, let's sum our currents into this node, into the V+ node, into our non-inverting input node. Let's call this Node 1, so if you sum currents into Node 1, then we have first of all the current through the 40 kilo-ohm resistor, which is minus 12 minus V+. Divided by 40 kilo-ohms. We also have the current which is flowing through the 20 kiloohm resister. And that current is
going to be 0 volts because this is a ground point in our circuit, the bottom of the circuit. 0 volts- V+ Over 20k.

1) Current into the V+ node is Zero Agreed

(With zero current flowing thru the 40k resistor, that voltage would be -12v. However, there is a 20k resistor also connected from that node to Ground. The author either forgot that detail or had a brain-fart.)​

2) The author did not take account of the current thru the 20k resistor.

Since the V+ op-amp input draws no current, it can and must be ignored for this case.

What is left is a 12V supply with a voltage divider across it, the values being 40k and 20k in series. That is what determines the voltage at the V+node.

What voltage do you get at the V+ node?

I have not read or solved the whole circuit; give it a try and we will (hopefully) get to the correct answer.

Cheers,
Tom

 

[s3a]

All right, so

V_out = _V_in * Z_2 / (Z_1 + Z_2)
V_out = (-12 V)*(20k ohm) / (20k ohm + 40k ohm)
V_out = (-12 V)*(1/3)
V_out = -4 V

V_+ = V_out = -4 V

This does seem to be in agreement with what he found using nodal analysis, by the way.

It seems to me, not just from the above, but just in general if I ignore the trouble I'm having with the signs and directions of the currents, that his way is also correct.

But, now, I'm more confused than ever about the signs and directions of the currents.

Why is it (-12 - V_+) / 40k + (0 - V_+) / 20k + 0 = 0 instead of (V_+ - -12) / 40k + (V_+ - 0) / 20k + 0 = 0 ⇔ (V_+ + 12) / 40k + (V_+ - 0) / 20k + 0 = 0 (since what the instructor gives is what I get from my logic but multiplied by -1, which doesn't change the final result since the other side of the equality is 0)?

I realize that those equations are equivalent, but I'd like to make sure that I don't get the right answer for the wrong reason(s).

 

[pbuk]

Oh, so, it's just a matter of recognizing that the current, when flowing into node 1 (as defined by the instructor) (as opposed to away from it) and using the passive sign convention, the terminal of the resistor where the current goes into the resistor is positive and the voltage source's positive terminal should be connected to the resistor's positive terminal since, given the aforementioned assumptions, current flows from the negative terminal of a voltage source to its positive one and then from the positive terminal of the resistor to its negative one?

I don't understand what you are trying to say here, so I will explain it in my own words.

• By Kirchoff's first law, the sum of (signed conventional) currents flowing into a node is zero.
• Current flows into Node 1 from the 40k resistor from left to right (note that this current may be, and in fact is, negative and so conventional current actually flows from right to left away from Node 1, but we don't worry about that because the sign convention will take care of everything for us).
• The left end of the 40k resistor is connected to the -ve terminal of the voltage source and is therefore at $-12\text{ V}$, the right end is connected to $V_+$ and so the current flowing into Node 1 from the 40k resistor is $\frac{-12 - V_+}{40\text{k}}\text{ A}$.
• The bottom end of the 20k resistor is connected to ground and is therefore at $0{ V}$, the top end is connected to $V_+$ and so the current flowing into Node 1 from the 20k resistor is $\frac{0 - V_+}{20\text{k}}\text{ A}$.
• Node 1 is connected to $V_+$ and because this is an ideal op amp no current flows to or from Node 1 to $V_+$.
• The sum of these three currents is equal to $0$, giving the expression in the slide.

 

[pbuk]

The explanation in this paragraph is WRONG! (highlited inRed)

1) Current into the V+ node is Zero Agreed

(With zero current flowing thru the 40k resistor, that voltage would be -12v. However, there is a 20k resistor also connected from that node to Ground. The author either forgot that detail or had a brain-fart.)​

I don't think the explanation is wrong, but it is rather confused because it is a word-for-word transcription and lacks the context given by the lecturer's phrasing. Let me rewrite it so it says what is intended (and what is written in the slide):

• So, let's sum our currents into this node, into the V+ node, into our non-inverting input node. Let's call this Node 1.
• We have first of all the current through the 40 kilo-ohm resistor, which is minus 12 volts minus V+ divided by 40 kiloohms.
• We also have the current which is flowing through the 20 kiloohm resister, which is 0 volts [because this is a ground point in our circuit] minus V+ over 20k.
• And we have the current which is flowing out of the non-inverting input, we know that through an ideal op-amp that that current is equal to 0.
• The sum of those using Kirchhoff's current law at node one is equal to zero.
• It's an equation which just has the one variable that is the voltage at the inverting or the non inverting input of our first op amp.
• If we solve for that voltage, V+ we get minus 4 volts. So it's minus 4 volts at node 1.

 

[pbuk]

Why is it (-12 - V_+) / 40k + (0 - V_+) / 20k + 0 = 0

Because we have decided to measure the conventional currents flowing into Node 1.

instead of (V_+ - -12) / 40k + (V_+ - 0) / 20k + 0 = 0

Because that measures the conventional currents flowing out from Node 1.

Each is equally valid, because by Kirchoffs current law they must each sum to zero, and so the choice is arbitrary, however in order to reduce the possibility of error it is a good idea to always choose the same one, and by convention we choose to sum the conventional currents flowing into a node.

 

[s3a]

Ah! I think I get it, but just to make sure (I suppose, by reiterating what you said, but in my own words):
* It's just about using the assumed direction of the current with the passive sign convention to know which part of the resistor is assumed to be the high potential difference and which one is the low point, right?
* Then, for the -12 V instead of 12 V, since the potential difference goes from high to low as opposed to going from low to high when the current it produces goes toward the 40k ohm resistor, we tack on a negative, right (since the voltage source is supposed to energize / incite (positive) charged particles to go to the 40k ohm resistor and then the kinetic energy of the (positive) charged particles would be lost when "consumed" by that aforementioned 40k ohm resistor)?

I hope I'm not confusing you again :P, but I think I get it. I just need to see other similar problems and see if my thoughts also work for those problems too.

 

[pbuk]

* It's just about using the assumed direction of the current with the passive sign convention to know which part of the resistor is assumed to be the high potential difference and which one is the low point, right?

No. Forget about the passive sign convention, it doesn't help you with Kirchoffs laws. And we don't need to think in advance about which way electrons, holes, or anything else is flowing or which end of the resistor is at a higher potential.

We start off by assuming that the current from the resistor into N1 i.e. from left to right, and this is the potential difference between the left hand end and the right end i.e. V_left - V_right = -12V - V_+.

If it turns out that this gives a negative value for current then that's fine, it means that the conventional current is actually flowing from right to left (although of course the only thing that is actually moving is electrons and they move in the opposite direction to conventional current).

* Then, for the -12 V instead of 12 V, since the potential difference goes from high to low as opposed to going from low to high when the current it produces goes toward the 40k ohm resistor, we tack on a negative, right (since the voltage source is supposed to energize / incite (positive) charged particles to go to the 40k ohm resistor and then the kinetic energy of the (positive) charged particles would be lost when "consumed" by that aforementioned 40k ohm resistor)?

1. We are not interested in how much power the resistor is consuming, we are only interested in the current through it.
2. Kinetic energy has nothing to do with power in electrical circuits.
3. There are no positive charged particles that flow anywhere in a conductor.

You are overthinking this, all you need for this problem is the fact that the conventional current through a resistor from A to B is given by $I_{a \to b} = \frac{V_a - V_b}{R}$.

 

[s3a]

Okay, thanks.

So,
* for both the 40k ohm resistor and the voltage source, high potential is left of N1 and low potential is right of N1 because that's what is implied when current is assumed to move toward N1, right?
* about the minus being tacked on to the 12 V, it was because the positive terminal of the voltage source coincides with the ground point and N1 is assumed to have a positive voltage value, which is a contradiction because that implies that the current goes out of N1, right?

If I'm wrong again, sorry. :(

 

[Tom.G]

The inputs labelled "+" and "-" are differential inputs, that means that the output of the OP-AMP depends on the voltage difference between the two inputs.

If the "+" (non-inverting) input is positive with respect to the "-" (inverting) input, then the op-amp output will be Positive.

If the "-" (inverting) input is positive with respect to the "+" (non-inverting) input, then the op-amp output will be Negative.

Cheers,
Tom

 

[s3a]

Thanks for the answer, but unless I'm mistaken now, I think you misunderstood what I meant to say.

Imagine $V_+$ wasn't something related to an op amp and is just the voltage at a node in a circuit with resistors and a voltage source. In that case, is what I said correct?

 

[Tom.G]

Sorry for the mis-interpretation.

I would say that your interpretation is 90% correct. The other 10% is that common usage, unless otherwise stated, is that voltages are usually referenced ground; without considering direction of electron flow.

For example in your problem statement the 12V battery has its Positive terminal connected to circuit ground.

In cases with only a single supply, a node may measure perhaps 3V relative to circuit ground without the polarity being stated.

In other cases it should be stated. Like your op-amp circuit, the polarity IS important because of multiple supplies and how op-amps work.

Yeah, very confusing! But it does get better after you get used to the conventions.

Hang in there s3a.

Hope this helps,
Tom

 

[s3a]

Okay, so, in short, if there are at least two sources or at least one op amp, then polarity matters, right?

What about if there is one voltage source and one current source? Polarity still matters then, right?

Also, the circuit of this thread only has one source, right? I ask because that's what I see, but you seem to imply it has more than one, and I want to be sure there isn't something weird going on because of the op amp or something.

 

[Tom.G]

Op-amp circuits usually have their own dedicated power supplies, and for text-book problems they frequently are not shown (less confusing [maybe]). In reality, most have two supplies, a Positive supply and a Negative one. For instance ±12V or ±15V.

This puts their voltage input range centered around 0 Volts (circuit Ground, also-known as Common).

There are some, however, that need only a single supply. A single-supply op-amp typically can not handle inputs around Ground level.

What about if there is one voltage source and one current source? Polarity still matters then, right?
Right!

Cheers,
Tom