Differential Amplifier Confusion

[Hideki]

Could you explain why the op amp would completely block the effects V1 would have on the load the op amp (the effects that would be there if the op amp wasn't there)?

Because the effects of the opamp is like connecting a power supply directly to the load. The power supply does its best to maintain a certain voltage and basically ignores the V1 you connect to it through the resistors.

The op amp is just amplifying it's output to make the two terminals equal, it's not determining what happens to the load... just the output pin and the negative feedback line.

It completely determines what happens to the load. And in this case, since nothing else is connected, the resistors back to V1 _IS_ in fact the load.

[dannyf]

think of any (ideal) opamp as a gain block with two important attributes:

1) it has infinite gain;
2) its input terminals have infinite impedance so no current flows into them.

Take that to your circuit and you can quickly work out the math that links the output to the inputs.

Conceptually, the opamp will affect the output so that the potential on the inverting input is the same as that on the non-inverting input. aka it has isolated V1 from the output.

[JacobEdward]

The op amp is just amplifying it's output to make the two terminals equal, it's not determining what happens to the load... just the output pin and the negative feedback line.

It completely determines what happens to the load. And in this case, since nothing else is connected, the resistors back to V1 _IS_ in fact the load.

Thanks, I was using the word 'load' incorrectly, but just to confirm, does that mean the feedback resistor would always be considered the load, just on a parallel line?  Does load just mean whatever is attached to the output pin of the op amp in this case?

[w2aew]

because the negative feedback of the op amp is causing the effect of V1 to be cancelled out at the inverting input of the op amp.  In other words, the current flowing through the input resistor will be equal to the current in the feedback resistor

So if I understand this correctly, the current from the negative feedback cancels any current flowing from V1 to that negative feedback line?

The current is the same in both resistors because the op amp input draws neglible current. The op amp drives the output voltage (and whatever load is connected to it) such that the voltage at the (-) input equals the voltage at the (+) input.

[Hideki]

Thanks, I was using the word 'load' incorrectly, but just to confirm, does that mean the feedback resistor would always be considered the load, just on a parallel line?  Does load just mean whatever is attached to the output pin of the op amp in this case?

Well, yes. The value of the feedback resistor does matter in real circuits. A real opamp can't push unlimited current, so any real circuit needs to take into account not only the load it's driving but also the current going through the feedback resistor (and through whatever the feedback resistor connects to).

Ideal opamps have no such limitations of course. They are perfect beings that can output unlimited current without the output voltage moving the slightest.

[albert22]

I think that you need to review the inverting a amplifier configuration for an op amp. Be able to understand how it works and derive the equations. Also the non inverting. And then go for the differential.
Here is another explanation that seems more clear than the tutorial that you are following:
http://www.electronics-tutorials.ws/opamp/opamp_2.html

[JacobEdward]

The current is the same in both resistors because the op amp input draws neglible current. The op amp drives the output voltage (and whatever load is connected to it) such that the voltage at the (-) input equals the voltage at the (+) input.

I dont understand why just ensuring the two inputs are the same will negate the effects V1 has on the load (the effects it would have had if you removed the op amp from the circuit and it was just one series line from V1 to the load... imagine there's a 10kohm resistor on the Vout line that is connected to ground that will act as the load, that would mean without the op amp, V1 is connected to three resistors before reaching ground).

Are you saying the output of the op amp is canceling the effect of V1 through the negative feedback line (as in a current of 5A going left will combine with a current of 5A going right to make 0A or 4A going left combined with 5A going right will equal 1A going right)?

That doesn't make sense since, even though any subtraction in current through the negative feedback line will be added to the input from V1 creating the proper voltage level, there is still nothing that would prevent the output from the op amp to continue up the V1 line, thereby subtracting any current coming from V1, which makes V1 useless...

Why does V1 suddenly not have an effect on the load with an op amp there and no diodes, it's still connected to the load in parallel with the op amp?

[IanB]

Why does V1 suddenly not have an effect on the load with an op amp there and no diodes, it's still connected to the load in parallel with the op amp?

V1 does have an effect on the output. In mathematical terms we have:

    Vout = V2 - V1

Or in other words,

    Vout = f(V2, V1)

This says that the output voltage is a function of the input voltages. If the input voltages did not affect the output voltage the whole circuit would not work!

[Hideki]

V1 has no direct effect on the load because no current passes from V1 to the load. It all goes into (or out of) the opamp.

Assuming all resistors being 1k and 0 volt at the noninverting input, then V1 = 1 volt makes the opamp output -1 volt. The current from V1 into the opamp output is 1 - (-1) = 2 volt divided by 1k + 1k = 2k. 2V / 2000 ohm = 1 mA. That current goes from V1, through the resistors, into the opamp output and comes out of the negative supply pin.

This is without any additional load.

If you add a load from the opamp to ground, say a 1 kohm resistor, then, since the output is still -1 volts, the load will force the opamp to drive one more milliamp to the negative supply.
If you add a load from the opamp to ground, say a 1 ohm   resistor, then, since the output is still -1 volts, the load will force the opamp to drive one more amp to the negative supply.

[JacobEdward]

V1 does have an effect on the output. In mathematical terms we have:

    Vout = V2 - V1

I dont understand why just ensuring the two inputs are the same will negate the effects V1 has on the load (the effects it would have had if you removed the op amp from the circuit and it was just one series line from V1 to the load... imagine there's a 10kohm resistor on the Vout line that is connected to ground that will act as the load, that would mean without the op amp, V1 is connected to three resistors before reaching ground)

?? Load Resistor = (V1-voltage drop for input resistor+voltage drop for feedback resistor)+Op Amp output Voltage ??

V2 is not directly connected to the line connected to the Load Resistor (since it only goes to the Op amp and ground) so it wouldn't cause an effect.

[IanB]

I don't understand why...

Frankly, you are not going to understand why, today, no matter how much anyone tries to explain it to you.

You will need to mull over the problem for a few weeks, or months, or even years until eventually it clicks. This is how learning works. Failure of understanding is caused by knowledge gaps in other areas, leaving holes in the picture. Eventually those other holes will be filled as you learn more, and then you will wonder how you ever could have been puzzled by this. But I don't think it is going to happen today, or you would have seen the answer already.

[JacobEdward]

V1 has no direct effect on the load because no current passes from V1 to the load. It all goes into (or out of) the opamp.

Why?  I've attached a picture of the path I think current travels without diodes to prevent flow... could you explain why that is incorrect (preferably visually since we seem to be getting nowhere with just text).

[JacobEdward]

I don't understand why...

Frankly, you are not going to understand why, today, no matter how much anyone tries to explain it to you.

You will need to mull over the problem for a few weeks, or months, or even years until eventually it clicks. This is how learning works. Failure of understanding is caused by knowledge gaps in other areas, leaving holes in the picture. Eventually those other holes will be filled as you learn more, and then you will wonder how you ever could have been puzzled by this. But I don't think it is going to happen today, or you would have seen the answer already.

Is there a way to block profiles on this forum?  If not, can you simply choose to not comment next time... it's not my fault you're so bad at explaining how these concepts work...

[JacobEdward]

That current goes from V1, through the resistors, into the opamp output and comes out of the negative supply pin.

I've attached a picture of what I am visualizing when I read this... let me know if I am correct

[JohnnyBerg]

if I am correct

No

[JacobEdward]

if I am correct

No

That was just so mind boggling helpful I am so glad you decided to take the time to post that response... like seriously, what would I do without you

[katzohki]

If V- is greater than Vo, current flows from V- to Vo. If V- < Vo current flows from Vo to V- ...

[JohnnyBerg]

That was just so mind boggling helpful I am so glad you decided to take the time to post that response... like seriously, what would I do without you

You're welcome. Any time again.

[katzohki]

I'm hoping this will help you. Here is an analogy of how an opamp works:

From wikipedia (http://en.wikipedia.org/wiki/Operational_amplifier) if you read that article it might help.

Here's a look at the internal circuitry of an opamp:

From http://www.play-hookey.com/analog/op_amps/inside_741.html

The opamp is considered an active component and is actually capable of "supplying" voltage at its output, unlike inactive components (resistors, capacitors, inductors) which can't.

[Hideki]

No current flows out of the INPUT. The current flows out the negative SUPPLY connection.

So in the image right above (inside_741) it goes into pin 6 (marked OUT), through the 50 ohm resistor below, through the PNP transistor and out pin 4 (V-).

[katzohki]

OK I think I understand where your confusion is coming from. In your drawings you have drawn on the opamp the + and - input terminals and the output terminal (and that's not incorrect), but not the voltage supply terminals on the opamp. And so everyone is telling you that the opamp can supply voltage, but you must be thinking "how? Where is the voltage coming from?" Well, it comes from the power supply inputs on the opamp.

[TimFox]

Re-stating slightly what was mentioned in several replies:

For an ideal op-amp,

• connected to appropriate DC power supplies
• with input voltages at the differential input stage within the usable voltage range
• with output voltage and current within the capability of the amplifier and power supplies
• and with an appropriate DC feedback network

The amplifier will adjust its output voltage so that the two input terminals are at the same voltage.  The current required out of the output pin comes from the power supplies connected to the op amp.

If you remove the op amp from its socket, or disconnect the DC power supplies, this effect will not happen and the output voltage will be different.

[w2aew]

That current goes from V1, through the resistors, into the opamp output and comes out of the negative supply pin.

I've attached a picture of what I am visualizing when I read this... let me know if I am correct

The diagram is not correct.  It shows current flowing into the op amp inverting input, and that is wrong. Essentially no current flows in/out of the op amp inputs. This has been clearly stated (and requoted by you) in several posts.

Of course V1 affects the output, but it affects it because it causes the op amp to react in such a way to adjust the output voltage until the two op amp inputs are equal.

We all understand your question.  Several have given you perfectly accurate answers to your question. Maybe the piece of the puzzle you're missing is that the op amp output is a low impedance, thus can override any direct affect that V1 has on the output.

[mrkev]

Hi, not that I'm positive that you'll actually get it after you've dismissed everybody that tried to help, but it might help someone else. So I drew the picture of how this actually works with an ideal oamp.

The voltage between - and + input pins of OP is 0V, due to the negative feedback. Currents to both + and - pins are also 0A, they act only as "voltage sense". OA is "balancing" output voltage so that the voltage at the node between R1 and R2 at the top branch is equal to the voltage at the node between R1 and R2 at the bottom (that is literary the same thing as the first sentence).
 
In other words, your problem with current from V1 is dealt with by OA.

And btw, i drew it without any load (load would be at U_O terminals), so I1 = - I_oamp (it flows to the OA).
With load connected, the OA would provide (or sink) exactly as much current, that the voltage between + and - inp. would still be 0V.