# Op-Amp

#### nop

Joined Nov 26, 2008
5
Look up summing amplifier, the way you were doing it is not correct due to the way the op amp is set up.

As for V1, convert the current source into a voltage source then using the equation for summing amplifier, you should be able to figure out the output fairly easily.

#### ihaveaquestion

Joined May 1, 2009
314
Thanks nop, but I beg to differ...

I don't think using superposition is 'wrong'... I would think that it most definitely could be used.

I'm not familiar with techniques of turning current sources into voltages sources, so I'd rather not do that.

#### steveb

Joined Jul 3, 2008
2,436
Thanks nop, but I beg to differ...

I don't think using superposition is 'wrong'... I would think that it most definitely could be used.

I'm not familiar with techniques of turning current sources into voltages sources, so I'd rather not do that.
Personally, I think either method is OK to use. However, this is a mysterious circuit to me. The purpose of the current source is unclear. Mathematically, you will find that the value of the current source has no effect on the output voltage. The only thing I can see is that perhaps the current source is intended to support the current capability of voltage V1. In other words, if V1 is incapable of supplying much current (like a voltage reference for example), I1 can be set to provide the bulk of the current draw.

Anyway, based on the above fact. There is no need to convert the current source into an equivalent voltage source, since it has no mathematical effect on the answer.

The superposition method then becomes very simple, since the remaining circuit is a basic summing amplifier with only two inputs.

One thing you did not make clear is what OPAMP model you will use to solve the problem.

EDIT: Also, it should be clear that the resistor R1 has no effect on output voltage and can be ignored in the analysis.

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#### ihaveaquestion

Joined May 1, 2009
314
Since there is negative feedback, I plan on using the v+ = v- method...

I'm not sure what other 'methods' there are for Op-Amps...

Fast-forward to 22:00, I was just going to approach and analyze it like done in the video at that time...

Since the plus terminal is connected to ground, that means both v+ and v- are 0... so if I use voltage division to get the voltage at the top node, i.e. V2*R2 / (R2 + R3), then I'm not sure where to go from there...

#### t_n_k

Joined Mar 6, 2009
5,455
steveb is quite right.

In fact this circuit is so well known that most electronics users would write the output as

Vo = - ((R4/R2)*V1 + (R4/R3)*V2)

Still, it's worth making sure you can derive the relationship from 1st principles.

My choice would be nodal analysis.

Source I & R1 are there as a "distractor" for the unwary.

#### steveb

Joined Jul 3, 2008
2,436
Since there is negative feedback, I plan on using the v+ = v- method...
OK, that's the simplest method and is a good place to start. Basically you will be saying that v+=v- and that no current flows into the input pins of the OPAMP.

I'm not sure what other 'methods' there are for Op-Amps...
There are many other approaches, but you will learn those later. The simple model will show the basic functions of the circuit.

Fast-forward to 22:00, I was just going to approach and analyze it like done in the video at that time...

Since the plus terminal is connected to ground, that means both v+ and v- are 0... so if I use voltage division to get the voltage at the top node, i.e. V2*R2 / (R2 + R3), then I'm not sure where to go from there...
OK, you are right that that section of the video provides the tools to solve your problem. However, there is no voltage divider present. The V- terminal acts like a "virtual" ground connection. So, if you use superposition, you just use the method in the video twice. And, since both inputs are the same topology, the basic answer is the same for both voltage sources. Then, just add both answers and you will see that this is a simple inverting summing amplifier as described by t_n_k above.

#### ihaveaquestion

Joined May 1, 2009
314
Ok I think I see it now...

1) What confuses me is the voltage divider part (which I don't need to use). I would think that if I want the voltage at the node above R3, I would use voltage division as (V1*R3)/(R2+R3)... however, since V- is 0 volts.. and they are sharing the same node, this is the reason why the voltage divison relation mentioned above doesn't help?

2) What if the current source were in a different location in the circuit, for example if I had put the current source where V2 is and V2 where the current source is... would I then have to include the current source into my calculations?

3) On a side question, which has somewhat to do with confusing me on this problem, if I want the voltage at the point marked Vout in this picture, would I have to take into account anything on the left side or just use voltage division?
http://img11.imageshack.us/img11/7042/30987134.jpg

#### steveb

Joined Jul 3, 2008
2,436
1) What confuses me is the voltage divider part (which I don't need to use). I would think that if I want the voltage at the node above R3, I would use voltage division as (V1*R3)/(R2+R3)... however, since V- is 0 volts.. and they are sharing the same node, this is the reason why the voltage divison relation mentioned above doesn't help?
Yes, it sounds like you understand it.

2) What if the current source were in a different location in the circuit, for example if I had put the current source where V2 is and V2 where the current source is... would I then have to include the current source into my calculations?
Yes, any current source that injects current into the V- node is going to directly flow into the R4 resistor and add to the output voltage. A summing amplifier like this is really adding the currents. You can generate current with a voltage source and a resistor, or you can just use a current source directly.

3) On a side question, which has somewhat to do with confusing me on this problem, if I want the voltage at the point marked Vout in this picture, would I have to take into account anything on the left side or just use voltage division?
Well, in that figure you are shorting the power supply V1 which I assume is just an error you made when drawing it. However, without the short, you don't need to consider anything on the left side. The reason is that the V1 voltage source is a perfect (ideal) voltage that forces that node to be equal to V1 no matter what else (other than a short or another ideal voltage source) is in parallel with it.

#### ihaveaquestion

Joined May 1, 2009
314
1) So what is the reason that the current source in the original problem is not accounted for?

2) I actually drew the short there on purpose because I wasn't sure... so Vout in the case of my drawing would be 0?

If the short wasn't there, then would Vout just be V1 since they are parallel like you mentioned or would I have to use voltage division?

i.e. http://img515.imageshack.us/img515/8494/26837463.jpg

in that circuit, with a bunch of garbage on the left with the dot dot dot parts after the resistors and open circuit etc... would Vout simply be V (whatever the source voltage is) or (V*R2)/(R1+R2)?

#### steveb

Joined Jul 3, 2008
2,436
1) So what is the reason that the current source in the original problem is not accounted for?
Simply because it is in parallel with a voltage source.

2) I actually drew the short there on purpose because I wasn't sure... so Vout in the case of my drawing would be 0?
OK, but a short is equivalent to a voltage source with 0V, so putting both in parallel is not mathematically correct. The voltage is now undetermined.

If the short wasn't there, then would Vout just be V1 since they are parallel like you mentioned or would I have to use voltage division?
The voltage division would apply here. V1 is not in parallel with Vout.

in that circuit, with a bunch of garbage on the left with the dot dot dot parts after the resistors and open circuit etc... would Vout simply be V (whatever the source voltage is) or (V*R2)/(R1+R2)?
The latter is correct

#### ihaveaquestion

Joined May 1, 2009
314
Ok I think I understand this problem pretty well now, except to clarify on some point you made earlier:

http://img117.imageshack.us/img117/3796/26797427.jpg

Putting the current source in the same wire with R1
http://img408.imageshack.us/img408/1526/267974271.jpg
I'm thinking the current still wouldn't be taken into a account because it will want to flow in the direction of least resistance, so into the V1...

Putting the current source where V2 is
http://img408.imageshack.us/img408/1210/267974272.jpg
Vout due to the current source would simply be -IR4?

And lastly, if we made a short on the last circuit shown here
http://img408.imageshack.us/img408/1307/2679742723.jpgthis simply wouldn't make sense and wouldn't be mathematically correct?

A lot of these questions as you can see are helping me gain an overall understanding of circuitry, so thanks..

#### PRS

Joined Aug 24, 2008
989
The current source and R1 are not quantities that determine the voltage at the input to the amp. V1 is the voltage at the node joining three branches so it's a given. Tnk gave you the correct expression for the output.

#### steveb

Joined Jul 3, 2008
2,436
Ok I think I understand this problem pretty well now, except to clarify on some point you made earlier:

http://img117.imageshack.us/img117/3796/26797427.jpg

Putting the current source in the same wire with R1
http://img408.imageshack.us/img408/1526/267974271.jpg
I'm thinking the current still wouldn't be taken into a account because it will want to flow in the direction of least resistance, so into the V1...

Putting the current source where V2 is
http://img408.imageshack.us/img408/1210/267974272.jpg
Vout due to the current source would simply be -IR4?

And lastly, if we made a short on the last circuit shown here
http://img408.imageshack.us/img408/1307/2679742723.jpgthis simply wouldn't make sense and wouldn't be mathematically correct?

A lot of these questions as you can see are helping me gain an overall understanding of circuitry, so thanks..
I agree with these statements.